Teaching and Assessing Junior High School Mathematics (Intermediary) (2024)

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COMPLETE COURSE CONTENT

TEACHING AND ASSESSING JUNIOR HIGH SCHOOL MATHEMATICS (INTEMEDIARY)

UNIT 1

MEASUREMENT, SHAPE AND SPACE

UNIT 2

CONSTRUCTION

UNIT 3

ANGLES AND POLYGONS

UNIT 4

DIAGNOSIS AND REMEDIATION; ASSESSMENT RESOURCES/RECORDS, AND MONITORING PROGRESS

UNIT 5

MICRO LESSONS AND USE OF TECHNOLOGY ACROSS JUNIOR HIGH SCHOOL MATHEMATICS

UNIT 6

RIGID MOTION

UNIT 7

ALGEBRA

UNIT 1

MEASUREMENT, SHAPE AND SPACE

INFORMAL GEOMETRY AND SPATIAL SENSE

Spatial sense is an intuitive feel for shape and space. It involves the concepts of traditional geometry, including an ability to recognize, visualize, represent, and transform geometric shapes. It also involves other, less formal ways of looking at two- and three-dimensional space, such as paper-folding, transformations, tessellations, and projections. Geometry is all around us in art, nature, and the things we make. Students of geometry can apply their spatial sense and knowledge of the properties of shapes and space to the real world.

Meaning and Importance

Geometry is the study of spatial relationships. It is connected to every strand in the mathematics curriculum and to a multitude of situations in real life. Geometric figures and relationships have played an important role in society’s sense of what is aesthetically pleasing. From the Greek discovery and architectural use of the golden ratio to M. C. Escher’s use of tessellations to produce some of the world’s most recognizable works of art, geometry and the visual arts have had strong connections. Well-constructed diagrams allow us to apply knowledge of geometry, geometric reasoning, and intuition to arithmetic and algebra problems. The use of a rectangular array to model the multiplication of two quantities, for instance, has long been known as an effective strategy to aid in the visualization of the operation of multiplication. Other mathematical concepts which run very deeply through modern mathematics and technology, such as symmetry, are most easily introduced in a geometric context. Whether one is designing an electronic circuit board, a building, a dress, an airport, a bookshelf, or a newspaper page, an understanding of geometric principles is required.

DEVELOPMENT AND EMPHASES

Traditionally, elementary school geometry instruction has focused on the categorization of shapes; at the secondary level, it has been taught as the prime example of a formal deductive system. While these perspectives of the content are important, they are also limiting. In order to develop spatial sense, students should be exposed to a broader range of geometric activities at all grade levels.

By virtue of living in a three-dimensional world, having dealt with space for five years, children enter school with a remarkable amount of intuitive geometric knowledge. The geometry curriculum should take advantage of this intuition while expanding and formalizing the students’ knowledge. In early elementary school, a rich, qualitative, hands-on study of geometric objects helps young children develop spatial sense and a strong intuitive grasp of geometric properties and relationships. Eventually they develop a comfortable vocabulary of appropriate geometric terminology. In the middle school years, students should begin to use their knowledge in a more analytical manner to solve problems, make conjectures, and look for patterns and generalizations. Gradually they develop the ability to make inferences and logical deductions based on geometric relationships and to use spatial intuition to develop more generic mathematical problem-solving skills. In high school, the study of geometry expands to address coordinate, vector, and transformational viewpoints which utilize both inductive and deductive reasoning. Geometry instruction at the high school level should not be limited to formal deductive proof and simple measurement activities, but should include the study of geometric transformations, analytic geometry, topology, and connections of geometry with algebra and other areas of mathematics.

At all grade levels, the study of geometry should make abundant use of experiences that require active student involvement. Constructing models, folding paper cutouts, using mirrors, pattern blocks, geoboards, and tangrams, and creating geometric computer graphics all provide opportunities for students to learn by doing, to reflect upon their actions, and to communicate their observations and conclusions. These activities and others of the same type should be used to achieve the goals in the seven specific areas of study that constitute this standard and which are described below.

In their study of spatial relationships, young students should make regular use of concrete materials in hands-on activities designed to develop their understanding of objects in space. The early focus should be the description of the location and orientation of objects in relation to other objects. Additionally, students can begin an exploration of symmetry, congruence, and similarity. Older students should study the two dimensional representations of three-dimensional objects by sketching shadows, projections, and perspectives.

In the study of properties of geometric figures, students deal explicitly with the identification and classification of standard geometric objects by the number of edges and vertices, the number and shapes of the faces, the acuteness of the angles, and so on. Cut-and-paste constructions of paper models, combining shapes to form new shapes and decomposing complex shapes into simpler ones are useful exercises to aid in exploring shapes and their properties. As their studies continue, older students should be able to understand and perform classic constructions with straight edges and compasses as well as with appropriate computer software. Formulating good mathematical definitions for geometric shapes should eventually lead to the ability to make hypotheses concerning relationships and to use deductive arguments to show that the relationships exist.

The standard geometric transformations include translation, rotation, reflection, and scaling. They are central to the study of geometry and its applications in that these movements offer the most natural approach to understanding congruence, similarity, symmetry, and other geometric relationships. Younger children should have a great deal of experience with flips, slides, and turns of concrete objects, figures made on geoboards, and drawn figures. Older students should be able to use more formal terminology and procedures for determining the results of the standard transformations. An added benefit of experience gained with simple and composite transformations is the mathematical connection that older students can make to functions and function composition.

Coordinate geometry provides an important connection between geometry and algebra. Students can work informally with coordinates in the primary grades by finding locations in the room, and by studying simple maps of the school and neighborhood. In later elementary grades, they can learn to plot figures on a coordinate plane, and still later, study the effects of various transformations on the coordinates of the points of two- and three-dimensional figures. High-school students should be able to represent geometric transformations algebraically and interpret algebraic equations geometrically.

Measurement and geometry are interrelated, and an understanding of the geometry of measurement is necessary for the understanding of measurement. In elementary school, students should learn the meaning of such geometric measures as length, area, volume and angle measure and should be actively involved in the measurement of those attributes for all kinds of two- and three-dimensional objects, not simply the standard ones. Throughout school, they should use measurement activities to reinforce their understanding of geometric properties. All students should use these experiences to help them understand such principles as the quadratic change in area and cubic change in volume that occurs with a linear change of scale.

Trigonometry and its use in making indirect measurements provides students with another view of the interrelationships between geometry and measurement.

Geometric modeling is a powerful problem-solving skill and should be used frequently by both teachers and students. A simple diagram, such as a pie-shaped graph, a force diagram in physics, or a dot-and-line illustration of a network, can illuminate the essence of a problem and allow geometric intuition to aid in the approach to a solution. Visualization skills and understanding of concepts will both improve as students are encouraged to make such models.

The relationship between geometry and deductive reasoning originated with the ancient Greek philosophers, and remains an important part of the study of geometry. A key ingredient of deductive reasoning is being able to recognize which statements have been justified and which have been assumed without proof. This is an ability which all students should develop in all areas, not just geometry, or even just mathematics! At first, deductive reasoning is informal, with students inferring new properties or relationships from those already established, without detailed explanations at every step. Later, deduction becomes more formal as students learn to use all permissible assumptions in a proof and as all statements are systematically justified from what has been assumed or proved before. The idea of deductive proof should not be confused with the specific two-column format of proof found in most geometry textbooks. The reason for studying deductive proof is to develop reasoning skills, not to write out arguments in a particular arrangement. Note that proof by mathematical induction is another deductive method that should not be neglected.

Much of the current thinking about the development of geometric thinking in students comes from the work of a pair of Dutch researchers, Pierre van Hiele and Dina van Hiele-Geldof. Their model of geometric thinking identifies five levels of development through which students pass when assisted by appropriate instruction.

·Visual recognition of shapes by their appearances as a whole (level 0)

·Analysis and description of shapes in terms of their properties (level 1)

·Higher “theoretical” levels involving informal deduction (level 2)

·Formal deduction involving axioms and theorems (level 3)

·Work with abstract geometric systems (level 4).

(Geddes & Fortunato, 1993)

Although the levels are not completely separate and the transitions are complex, the model is very useful for characterizing levels of students’ thinking. Consistently, the research shows that appropriately targeted instruction is critical to children’s movement through these levels. Stagnation at early levels is the frequent result of a geometry curriculum that never moves beyond identification of shapes and their properties. The discussion in this K-12 Overview draws on this van-Hiele model of geometric thinking.

IN SUMMARY, students of all ages should recognize and be aware of the presence of geometry in nature, in art, and in human-built structures. They should realize that geometry and geometric applications are all around them and, through study of those applications, come to better understand and appreciate the role of geometry in life. Carpenters use triangles for structural support, scientists use geometric models of molecules to provide clues to understanding their chemical and physical properties, and merchants use traffic-flow diagrams to plan the placement of their stock and special displays. These and many, many more examples should leave no doubt in students’ minds as to the importance of the study of geometry.

NOTE: Although each content standard is discussed in a separate chapter, it is not the intention that each be treated separately in the classroom. Indeed, as noted in the Introduction to this Framework, an effective curriculum is one that successfully integrates these areas to present students with rich and meaningful cross-strand experiences.

THE DEVELOPMENT OF GEOMETRIC THINKING

The van Hiele Levels of Geometric Thought

There is some well-established research that has been influencing school curriculum development internationally for many years now, but the practical details are still unknown to most teachers. This research began in the 1950's with a husband and wife team in the Netherlands, Pierre and Dina van Hiele. Pierre van Hiele has continued to develop the theory over the years, and many other researchers around the world have investigated its basis and application in various ways. The main theory emphasises that despite some natural development of spatial thinking, deliberate instruction is needed to move children through several levels of geometric understanding and reasoning skill. It is based on the firm belief that it is inappropriate to teach children Euclidean geometry following the same logical construction of axioms, definitions, theorems and proofs that Euclid used to construct the system. Children don't think on a formal deductive level, and therefore can only memorise geometric facts and 'rules', but not understand the relationships between the ideas, if taught using this approach.

The van Hiele theory puts forward a hierarchy of levels of thinking spanning the ages of about five years through to academic adults. Originally there were five levels, that have been adapted and renamed by various researchers, but now van Hiele concentrates on the three levels that cover the normal period of schooling. The main content focus is on two-dimensional (plane) shape.

Level 1: Visual

This level begins with 'nonverbal thinking'. Shapes are judged by their appearance and generally viewed as 'a whole', rather than by distinguishing parts. Although children begin using basic shape names, they usually offer no explanation or associate the shapes with familiar objects. For example, a child might say, "It's a square because it looks like one", or "I know it's a rectangle because it looks like a box". This could be likened to young children's ability to recognise some words by sight, before they understand the individual letter sounds and how they blend together to form words.

Level 2: Descriptive

At this level, children can identify and describe the component parts and properties of shapes. For example, an equilateral triangle can be distinguished from other triangles because of its three equal sides, equal angles and symmetries. Children need to develop appropriate language to go with the new specific concepts. However, at this stage the properties are not 'logically ordered', which means that the children do not perceive the essential relationships between the properties. So, with the equilateral triangle for example, they do not understand that if a triangle has three equal sides it must have three equal angles.

Level 3: Informal Deduction

In this level, the properties of shapes are logically ordered. Students are able to see that one property precedes or follows from another, and can therefore deduce one property from another. They are able to apply what they already know to explain the relationships between shapes, and to formulate definitions. For example, they could explain why all squares are rectangles. Although informal deduction such as this forms the basis of formal deduction, the role of axioms, definitions, theorems and their converses, is not understood.

ACTIVITY PHASES TO DEVELOP GEOMETRIC THINKING

Included in van Hiele's theory is a sequence of five phases of activity types that are designed to promote the movement of children's thinking from one level to the next. There are a variety of ways to use these phases when planning lessons. For example, a teacher might choose to deal with the topic of squares, and so design all the tasks to involve the manipulation, construction, discussion and so on, of squares. Another way to use the phases is to take a more general approach and explore a number of different shapes simultaneously. Cycling through these phases using materials such as tangrams and various mosaic puzzles enables children to build a rich background in visual and descriptive thinking about shapes and their properties.

Activity Examples:Traditional seven-piece tangram sets are commonly available and so have been used as the basis of activities to illustrate each phase. The topic is not 'Tangrams', but the 'Properties of 2D Shapes'. The activities are not really age specific, but are probably best suited to children in the range of 7 to 10 years old. It is desirable for each child to have his or her own tangram, so multiple copies could be made from card and stored in envelopes. The pieces in the example below have been numbered for easy reference in this article, but it is also useful to do this (or colour-code) with the children's sets. (Find or make a tangram set and use it to try out the activities as you read. Instructions for folding a tangram can be foundhereor one can easily be drawn using grid paper as shown below).

Phase 1: Inquiry

Learning begins with play! The children should be encouraged to freely explore the materials and hence discover some properties and structures. While the children are playing, the teacher has the opportunity to observe and informally assess the children's thinking and language.

Example:Give each child a tangram set and simply ask 'What can you do with these pieces?' Encourage the children to share and talk about the shapes and pictures they have made. Allow plenty of time for children to freely explore the pieces. During this play the children will become familiar with the size and shape of the pieces, and they begin to see how they fit together. In other words, they begin to discover the properties and relationships.

Phase 2: Direct Orientation

Activities are presented in such a way that children's attention is focussed on particular characteristics of the shapes or puzzle pieces. Ideas for directed activities will come from watching the children at play, as well some pre-planned tasks.

Example:In free play a child may have used pieces 3 and 5 to make piece 6. So ask the children to find out whether all the pieces can be made from two smaller pieces. They might do this by fitting the two pieces together on top of the larger piece. Other questions to explore are:

·Which pieces can be made from three other pieces? Have the children record what they find by tracing around the shapes.

·One activity will often lead to another; How many different ways can the largest triangle (1) be made using the other pieces?

·Choose two shapes. How many different shapes can you make with them? Draw them all and give them names. Use three shapes to make a new shape (not like one of the other pieces). Trace around it. How many ways can you make this shape? Can you make it with and without the numbers showing? Draw all the solutions.

From these activities the children gain a more specific understanding of the properties of the shapes. They will notice the lengths of sides, some that are equal and others that are half or double the length of others. They will visualise how the angles will fit together and how to flip and turn the shapes to fit various spaces.

Phase 3: Explication

This involves tasks and games that deliberately develop the vocabulary associated with the ideas that have been encountered so far, while continuing to explore the properties of the shapes. The teacher clarifies terms the children are already using and introduces new terms. Activities that encourage the children to use the vocabulary as they talk and write about their experiences should be utilised.

Example:During some workshops, the teacher names the shapes; square, isosceles triangle, and parallelogram, as well as other vocabulary associated with the properties; similar, equal sides, equal angles, right angle, symmetry and parallel sides. Questions like the following will provide opportunities for the terms to use as well as clarifying the associated concepts:

·Which shapes have a right angle? Match them up in a pile. (What size are the other angles?)

·Trace around each shape on paper and cut it out. How many lines of symmetry does each shape have?

·What is the same about all the triangles?

·Which shapes have parallel sides?

·Which shapes have sides the same length within themselves? Which shapes have sides the same length as other shapes? half/double the length?

·Play 'Mystery Bag' where a child feels a hidden shape and describes it to the rest of the class, who try to name the shape.

The children should be encouraged to explain and demonstrate what they find out, perhaps using pieces on an overhead projector, or making a display poster.

Phase 4: Free Orientation

The children engage in activities and problem-solving tasks that are open-ended or can be completed in different ways. The aim is for children to utilise what they have learned and become more proficient.

Example:The children should engage in more challenging tasks, that draw on the knowledge and skills previously developed; such as:

·How many ways can you make a square from some or all of the pieces?

·Complete classic tangram puzzles of outlines of birds and animals.

·Draw a completed tangram square (like the one illustrated above) on an 8x8 square grid, examine the pieces carefully in relation to the grid, then work out a way to enlarge all the pieces.

Phase 5: Integration

Opportunities are given for the children to pull together their new knowledge and reflect on it as a whole. They should be able to express or summarise what they have learned in some way.

Example: Small groups of children can design charts, class books and displays that present what they have learned about the tangram shapes. Pairs of children could prepare "What shape am I?" cards which could be used by individuals or in team games. Class games like "Twenty Questions" will help children to logically organise their knowledge of properties of the shapes. Individual assessment might be carried out by giving children open-ended tasks, like 'Choose a shape and write everything you can about it', and 'Choose another shape and design a What shape am I? Card'.

Outcomes of the Five Phases of Activities

As a consequence of engaging in a series of activities such as those described above, the majority of children will have progressed from simply recognising some shapes, to being able to discuss the shapes in terms of specific geometric properties and perhaps make some comparisons between shapes. In other words, they will have moved from Level 1 Visual thinking to Level 2 Descriptive thinking in regards to the shapes treated in the activities. Although the knowledge and vocabulary developed in respect to this specific set of shapes will assist in the study of different shapes, many more similar experiences will be needed to develop geometric thinking about different shapes. For example, the puzzle shapes presented below would provide good opportunities for the children to explore types of triangles and quadrilaterals. (Note that it is easily copied using equilateral-triangle dot or grid paper).

EXPLORING SHAPES AND THEIR PROPERTIES

In geometry,2d shapesand 3d shapes are explained widely to make you understand the different types of objects you come across in real life. These shapes have their own pattern and properties.Depending on many factors, such as angle, sides, length, height, width, area, volume, etc., the shapes can vary.These 2D and3D shapeshave been taught to us since our primary classes. Let us learn various types of two-dimensional shapes here, in this article.

2D SHAPES

In maths, 2d shapes can be defined as the plane figures that can be drawn on a flat (or plane) surface or a piece of paper. All the 2d shapes have various parameters such asarea and perimeter. Some of the 2d shapes contain sides and corners, whereas some have curved boundaries.

2D Shapes Names: Circle, Triangle, Square, Rectangle, Pentagon, Octagon.

The basic types of 2d shapes are a circle, triangle, square, rectangle, pentagon, quadrilateral, hexagon, octagon, etc. Apart from the circle, all the shapes are considered as polygons,whichhavesides. A polygon which has all the sides and angles as equal is called a regular polygon. Includingthe circle, an ellipse is also a non-polygon shape. Both circle and ellipse have acurvedshape, whereas the polygons have a closed structure with sides. Now let us discuss some shapes one by one.

Circle

A circle is a closed 2d figure in which the set of all the points in the plane is equidistant from a given point called “center”. The distance from the center to the outer line of the circle is called a radius. The example of the circle in real life are wheels, pizzas, orbit, etc.

Triangle

A triangle is a three-sided polygon (2d Shape) which has three edges and three vertices. The sum of all the three angles of a triangle is equal to 180°. Pyramids are the best example of a triangle shape. You can also learn theproperties of trianglehere.

Square

A square is a four-sided polygon (2d Shape), whose four sides are equal in length and all the angles are equal to 90°. It is considered to be a two-dimensional regular quadrilateral. The diagonals of the square also bisect each other at 90°. A wall or a table where all the sides are equal are the examples of square shape.

Rectangle

A rectangle is a 2d shape which has four sides, where the opposite sides are equal and parallel to each other. All the angles of a rectangle are equal to 90°. A brick, TV, cardboard, which has length and breadth are examples of the rectangle.

Pentagon

A pentagon is a five-sided polygon (2d Shape), and it can be regular or irregular. In the case of a regular pentagon, each interior angle is equal to 108°, and each exterior angle measures 72°. It has five diagonals. The Pentagon building, which is the headquarters of the US Department of Defense, is a great example of the pentagon shape.

Octagon

An octagon is an eight-sided polygon which can be either regular or irregular. It is a 2d shape which has eight angles. The sum of all the interior angles of an octagon is 1080°. The stop sign board has an octagon shape, which you can see on the roadside.

2D Shapes Properties

Go through the below to learn all the properties of 2D shapes.

Area and Perimeter of 2D Shapes

The area is the region covered by a 2d shape on a plane. The areas for different shapes are given below:

2d Shapes and 3d Shapes

We know that 2d shapes are flat figures and 3d shapes are solid figures. Below are the few comparisons of these two types of shapes.

3D SHAPES

In Geometry, 3D shapes are known as three-dimensional shapes or solids. 3D shapes have three different measures such as length, width, and height as its dimensions. The only difference between 2D shape and 3D shapes is that 2D shapes do not have a thickness or depth.

Usually, 3D shapes are obtained from the rotation of the 2D shapes. The faces of the solid shapes are the 2D shapes. Some examples of the 3D shapes are a cube, cuboid, cone, cylinder, sphere, prism and so on.

TYPES OF 3D SHAPES

The 3D shapes consist of both curved shaped solid and the straight-sided polygon called the polyhedron. The polyhedrons are also called the polyhedra, which are based on the 2D shapes with straight sides. Now, let us discuss the details about the polyhedrons and curved solids.

Polyhedrons

Polyhedrons are 3D shapes. As discussed earlier, polyhedra are straight-sided solids, which has the following properties:

·Polyhedrons should have straight edges.

·It should have flat sides are called the faces

·It must have the corners, called vertices

Like polygons in two-dimensional shapes, polyhedrons are also classified into regular and irregular polyhedrons and convex and concave polyhedrons.

The most common examples of polyhedra are:

·Cube: It has 6 square faces, 8 vertices and 12 edges

·Cuboid: It has 6 rectangular faces, 8 vertices and 12 edges

·Pyramid: It has a polygon base, straight edges, flat faces and one vertex

·Prism: It has identical polygon ends and flat parallelogram sides

Some other examples of regular polyhedrons are tetrahedrons, octahedrons, dodecahedrons, icosahedrons, and so on. These regular polyhedrons are also known as platonic solids, whose faces are identical to each face.

For example, the most commonly used example of a polyhedron is a cube, which has 6 faces, 8 vertices, and 12 edges.

Curved Solids

The 3D shapes that have curved surfaces are called curved solids. The examples of curved solids are:

·Sphere: It is a round shape, having all the points on the surface equidistant from center

·Cone: It has a circular base and a single vertex

·Cylinder: It has parallel circular bases, connected through curved surface

Faces Edges and Vertices

Faces, edges and vertices are three important measures of 3D shapes, that defines their properties.

·Faces – A face is a curve or flat surface on the 3D shapes

·Edges – An edge is a line segment between the faces

·Vertices – A vertex is a point where the two edges meet

PROPERTIES OF 3D SHAPES

As we already discussed above the properties of 3D shapes are based on their faces, edges and vertices. Thus, we can have a brief of all the properties here in the table.

SURFACE AREA AND VOLUME OF 3D SHAPES

The two different measures used for measuring the 3D shapes are:

·Surface Area

·Volume

Surface Areais defined as the total area of the surface of the two-dimensional object. The surface area is measured in terms of square units, and it is denoted as “SA”. The surface area can be classified into three different types. They are:

·Curved Surface Area (CSA) – Area of all the curved regions

·Lateral Surface Area (LSA) – Area of all the curved regions and all the flat surfaces excluding base areas

·Total Surface Area (TSA) – Area of all the surfaces including the base of a 3D object

Volumeis defined as the total space occupied by the three-dimensional shape or solid. It is measured in terms of cubic units and it is denoted by “V”.

3D SHAPES FORMULAS

The formulas of 3D shapes related to surface areas and volumes are:

EXPLORING NETS OF 3D SHAPES

A net is a flattened out three-dimensional solid. It is the basic skeleton outline in two dimensions, which can be folded and glued together to obtain the 3D structure. Nets are used for making 3D shapes. Let us have a look at nets for different solids and its surface area and volume formula.

Cuboid

A cuboid is also known as a rectangular prism. The faces of the cuboid are rectangular. All the angle measures are 90 degrees.

Take a matchbox. Cut along the edges and flatten out the box. This is the net for the cuboid. Now if you fold it back and glue it together similarly as you opened it, you get the cuboid.

Cube

A cube is defined as a three-dimensional square with 6 equal sides. All the faces of the cube have equal dimension.

Take a cheese cube box and cut it out along the edges to make the net for a cube.

Cone

A cone is a solid object that has a circular base and has a single vertex. It is a geometrical shape that tapers smoothly from the circular flat base to a point called the apex.

Take a birthday cap which is conical. When you cut a slit along its slant surface, you get a net for cone.

Cylinder

A cylinder is a solid geometrical figure, that has two parallel circular bases connected by a curved surface.

When you cut along the curved surface of any cylindrical jar, you get a net for the cylinder. The net consists of two circles for the base and the top and a rectangle for the curved surface.

Pyramid

A pyramid, also known as a polyhedron. A pyramid can be any polygon, such as a square, triangle and so on. It has three or more triangular faces that connect at a common point is called the apex.

The net for a pyramid with a square base consists of a square with triangles along its four edges.

HAND SKETCHING OF COMMON SOLIDS

EXPLORING RELATIONSHIP AMONG FACES, EDGES AND VERTICES

A 3D shape or an object is made up of a combination of certain parts. Most of the solid figures consist of polygonal regions. These regions are- faces, edges, and vertices. Solidgeometric shapeswhich have faces, edges and vertices are known as polyhedrons.

Faces of 3D Shapes

·The flat surface of a polyhedron is its face. Solid shapescan have more than one face. The cube shown below has 6 faces viz. ABCD, EFGH, ADHE, DHGC, BFGC, and AEFB.

·Cubes and cuboids have 6 faces.

·Cones have a flat face and a curved face.

·Cylinders have 2 flat faces and a curved face.

·A sphere has a curved face.

Edges of 3D Shapes

·The faces meet each other at edges. Edges are straight lines which serve as the junction of two faces. The cube shown below has 12 edges namely AB, BF, EF, AE, AD, DH, EH, HG, FG, BC, CG, and CD.

·Cubes and cuboids have 12 edges.

·Cones have 1 edge.

·Cylinders have 2 edges.

·A sphere has no edge.

Vertices of 3D Shapes

·The points of intersection of edges denote the vertices. Vertices are represented by points. In the cube shown below A, B, C, D, E, F, G, and H are the 8 vertices of the cube.

·Cubes and cuboids have 8 vertices.

·Cones have 1 vertex.

·Cylinders have no vertex.

·Spheres have no vertex (the surface is a curve).

Now that, we are familiar with polyhedrons let’s move onto to their types.

TYPES OF POLYHEDRON

·Convex Polyhedron:If the surface of a polyhedron (which consists of its faces, edges, and vertices) does not intersect itself and the line segment connecting any two points of the polyhedron lies within its interior part or surface then such a polyhedron is a convex polyhedron.

·Concave Polyhedron:A non-convex polyhedron is termed as a concave polyhedron.

Euler’s Formula

According toEuler’s formulafor any convex polyhedron, the number of Faces (F) and vertices (V) added together is exactly two more than the number of edges (E).

F + V = 2 + E

A polyhedron is known as a regular polyhedron if all its faces constitute regular polygons and at each vertex the same number of faces intersect.

Fig A is a regular polyhedron as all the faces are regular polygons and B is an irregular polygon.

It is said that in 1750, Euler derived the well known formula V + F – E = 2 to describe polyhedrons.At first glance, Euler’s formula seems fairly trivial.Edges, faces and vertices are considered by most people to be the characteristic elements of polyhedron.Surprisingly however, concise labelling of such characteristics was not presented until the 1700’s.Leonhard Euler, recognizing the deficiency, began his investigation of general polyhedra, and the relationship between their elements.

Euler emphasized five major components of a polyhedron in an attempt to find the relationship between them.These five components were vertices, (a location where 2 or more edges meet), faces (contained and defined by 3 or more edges), edges (defined as the “ridge or sharp edge”[2]of a polyhedron), sides (used to refer to the sides of each face), and plane angles (the angle found at a vertex, contained by 2 sides).These definitions, contrasted with the features thatEuclidhad relied on previously, right angles, and bases, led to far more possible relationships between characteristics.

From above, green denotes the edge of the cube; red denotes the plane angle;

the yellow sphere occurs at a vertex; and all faces are shaded purple.

Sides are in black, they surround the purple faces.

In the remainder, let:-V be the number of vertices,

-F be the number of faces,

-E be the number of edges,

-S be the number of sides, and

-P be the number of plane angles.

By naming each component, Euler observed some general relationships that occur for all polyhedron.For example, the definition of a side leads to the equation 2S = E, where S is the number of sides and E is the number of edges.It can then be stated that the number of edges must be an even number, for if this were not the case, it would be possible to have a half of a side, which contradicts the definition of S.

By looking at the condition to form a face, we can derive another formula.Since a face is a surface enclosed by at least 3 sides, we can state that S ≥ 3F.Furthermore, there must be at least 3 faces surrounding every vertex ( F ≥ 3V ).Additionally, it can be stated that the number of plane angles equals the number of sides, P = S, that 2E ≥ 3F, and that 2E ≥ 3V

In summary:2S = E

S ≥ 3F

F ≥ 3V

P = S

2E ≥ 3F

2E ≥ 3V

The Definition of a Polyhedron

There are many definitions of a polyhedron.The definition used in this project given by P. Cromwell in his book entitled “Polyhedra” is the following:

“Apolyhedronis the union of a finite set of polygons such that

i.Any pair of polygons meet only at their sides or corners.

ii.Each side of each polygon meets exactly one other polygon along an edge.

iii. It is possible to travel from the interior of any polygon to the interior of another.

iv.Let V be any vertex and lef F₁, F₂, …, F_nbe the n polygons which meet at V.It is possible to travel over the polygons F_ifrom one to any other without passing through V.”

Euler’s Own Proof

1. Explanation

Although Euler presented the formula, he was unable to prove his result absolutely.His proof is based on the principle that polyhedrons can be truncated.Euler proceeds by starting with a polyhedron consisting of a large number of vertices, faces, and edges.By removing a vertex, you remove at least 3 faces (while exposing a new face), and at least 3 edges.As you continue, more vertices are removed, until eventually you will find that Euler’s proof degenerates into an object that is not a polyhedron.A polyhedron must consist of at least 4 vertices.If there are less than 4 vertices present, a degenerate result will occur, and Euler’s formula fails.While the proof fails to prove his formula, it does show that truncated and augmented platonic polyhedra satisfy the equation, (thereby including classes such as the Archimedean solids, and pyramids).

2. Proof in Pictures

Click here to see an example ofEuler’s proofusing an octahedron.

LEGENDRE’S PROOF USING RADIAL PROJECTIONS

I. Explanation

Another interesting proofthat has wider applications, is a proof credited to Adrien Marie Legendre.This proof inscribes a convex polyhedron inside a sphere, where the centre is found in the region enclosed by the polyhedron, and the polyhedron is fully inscribed.We then proceed to project light rays from the origin through each of the vertices on the polyhedron.By finding where these rays intersect the sphere, and connecting the points of intersection by the arcs that characterize the shortest distance between two points along the sphere, we produce a radial projection of the polyhedron.Vertices are now the points where the arcs meet, edges are now arcs, and faces are now spherical polygons.By combining the characteristics of a sphere, spherical polygons, and the summarized characteristics of vertices, edges, and faces (from above), Euler’s formula can be derived.

Main facts:

-If you sum all of the angles surrounding a vertex, the resulting angle will be 2π.It follows that the total angle sum around all the vertices will be 2πV.

-From above, 2S = E.

-Each side has an angle of π.

-Each face has an angle of 2π.

-The surface area of a sphere of R=1 is 4 π.

Let’s now establish the equation for the surface area of a sphere.Let us assume the radius of the sphere is 1.

Area of a Spherical Polygon** = angle sum – sides on single face (π) + 2

Surface Area =(# of spherical polygons mapped on the sphere) (Area of Spherical Polygon)

Surface Area = V(angle sum) - Sπ +2πF

4 π = 2 πV – 2E π + 2πF

If we divide both sides of the equation by 2π, we get the formula:

2 = V – E + F

**This formula is from Girard’s Theorem for spherical polygons, although no further discussion of this theorem is provided in this project.

II.Proof in Pictures

The following demonstrates elements of Legendre’s proof of Euler’s formula through pictures, by inscribing a tetrahedron inside a unit sphere.Clickhereto view it.

III.Difficulties

At first glance, it seemed like an easy task to solve for the path along the surface of a sphere that connected two points.However, upon attempting to code such a procedure, it turns out there are some interesting questions that need to be answered.

Let us first look at how to parameterize a unit sphere of form x²+ y²+ z²= 1.Let u, v be the parameters we introduce to describe this curve.This being said, any point on the sphere satisfies the equation

(cos(u)cos(v), cos(v)sin(u), sin(v) )where 0^o≤ u, v ≤ 360^o.

However, knowing how to parameterize the unit sphere is unnecessary when trying to derive the direct path from any two points, A and B, because there is an additional condition that must be satisfied.To find the appropriate arc, we must find the plane that contains A, B, and the origin of the sphere.This plane will also contain all of the points that describe the path between the two points.

After looking at different cases, I found the easiest way to find the arc along the great circle was to calculate the midpoint between the two points we wish to connect.Let C be the centre of the sphere.

Let A = (x_A, y_A, z_A), B = (x_B, y_B, z_B), and C = (0, 0, 0).

The direction of the midpoint is therefore M, where

M = [ ((x_A+ x_B)/2), ((y_A+ y_B)/2),((z_A+ z_B)/2) ].

To find the midpoint along the arc of the great circle, we must therefore find where the vector of the midpoint intersects x²+ y²+ z²= 1.

To do accomplish this, you can find the unit length M, and divide M by it.This will result in the midpoint.Therefore, the midpoint along a great arc, m, is:

m = M / ||M||

A PROOF BASED PRINCIPALLY ON ILLUSTRATIONS

i.Explanation

I found Von Staudt’s proof to be one of the nicer proofs for Euler’s formula.Von Staudt’s proof begins by picking any vertex on a convex polyhedron.From this vertex, we look for a vertex that has never been coloured, and connect the vertices by colouring along the edge that connects them.Now at the new vertex, we look for another vertex that has never been coloured.If we find one, we connect the vertices and colour along the edge that connects them again.We proceed in this fashion until there are no more vertices to be visited.

If there are no more vertices to colour, then all vertices have been coloured.When traversing a polyhedron in this manner, we can see that this proof requires that there is always a path that touches all vertices only once.

Around the same time that this proof was being published, an Irish mathematician Sir William Rowan Hamilton (1805 – 1865) was examining the problem of finding a path in a graph, or along a solid, where each vertex is traversed only once.This path is called a Hamiltonian circuit, and finding whether or not a circuit exists in a figure is quite a challenge.To solve the problem on an arbitrary graph is known to be intractable; no efficient algorithm is known.Hamilton actually created this idea as a sort of puzzle, find the path to connect every vertex on a dodecahedron, visiting every vertex only once, and sold his dodecahedron puzzle to an Irish merchant who began to sell it.It turns out that it is quite a difficult task to identify whether or not a solid has a Hamiltonian circuit or not, and it may require a proof by exhaustion to determine whether or not a path is possible.However, it has been proven that all Platonic solids, Archimedean solids, and planar-4-connected graphs have Hamiltonian circuits.

As we begin our proof, we initially connect 2 vertices, and for every other edge we shade (let’s use red), we find another vertex that we previously had not visited.Therefore, by counting the number of edges that we shade red we can determine the number of vertices.The formula is:

Red Edges = V – 1

Furthermore, Von Staudt states that all the vertices have been coloured.To prove this, assume this is not the case.Then there would be a vertex that is not coloured, which means we were not done colouring our edges red.

Next we will begin by examining faces.Pick a face and shade it green, as well as any edges that have not yet been coloured red.Proceed by finding another face who hasonly onegreen side, and shade green as before.Continue until there are either no more faces to colour that satisfy the condition.Since it is impossible that a face exists with all four sides coloured, all faces must be shaded green.The relationship between green edges and faces can be described as:

Green Edges = F – 1

The total number of edges, E, will equal the number of green edges and red edges.

E = Red Edges + Green Edges

E = V – 1 + F – 1

Therefore,

E = V + F – 2

Proof in Pictures

I have illustratedVon Staudt’s proof for a cubeinitially to clearly demonstrate how the proof is to proceed.Next, the same proof is illustrated on adodecahedron.

CONCLUSION

For over two centuries people have discussed, proven and disproved Euler’s formula.The sample proofs I have provided by no means represent the sole representations of Euler’s formula.It is possible to find proofs based on electrical charge that will prove the formula.There are others that useJordancurves and planar graphs, and there are other proofs that rely on topology.Furthermore, by carefully declaring the case and family of solids you wish to explain, the formulaV + F – 2 = E can be generalized (by changing the 2) to explain more topologically complicated solids.(L’Huilier, a Swiss mathematician, investigated such figures and discovered that this was possible.)It is this variety of proofs and applications of Euler’s formula that makes it such an interesting topic to study.

CONCEPT OF MEASUREMENT

MEASUREMENT

Measurement is a process of finding a number that shows the amount of something.

Measurement, the process of associating numbers with physical quantities and phenomena. Measurement is fundamental to the sciences; toengineering, construction, and other technical fields; and to almost all everyday activities. For that reason, the elements, conditions, limitations, and theoretical foundations of measurement have been much studied. See alsomeasurement systemfor a comparison of different systems and the history of their development.

Measurements may be made by unaided human senses, in which case they are often called estimates, or, more commonly, by the use of instruments, which may range in complexity from simple rules for measuring lengths to highly sophisticated systems designed to detect and measure quantities entirely beyond the capabilities of the senses, such as radio waves from a distant star or the magnetic moment of asubatomic particle.

Measurement begins with a definition of the quantity that is to be measured, and it always involves a comparison with some known quantity of the same kind. If the object or quantity to be measured is not accessible for direct comparison, it is converted or “transduced” into ananalogousmeasurement signal. Since measurement always involves some interaction between the object and the observer or observing instrument, there is always an exchange of energy, which, although in everyday applications is negligible, can become considerable in some types of measurement and thereby limit accuracy.

MEASUREMENT INSTRUMENTS AND SYSTEMS

In general, measuring systemscomprisea number of functional elements. One element is required to discriminate the object and sense its dimensions or frequency. This information is then transmitted throughout the system by physical signals. If the object is itself active, such as water flow, it maypowerthe signal; if passive, it must trigger the signal by interaction either with an energetic probe, such as a light source orX-ray tube, or with a carrier signal. Eventually the physical signal is compared with a reference signal of known quantity that has been subdivided or multiplied to suit the range of measurement required. The reference signal is derived from objects of known quantity by a process calledcalibration. The comparison may be ananalogprocess in which signals in a continuous dimension are brought to equality. Analternativecomparison process isquantizationby counting, i.e., dividing the signal into parts of equal and known size and adding up the number of parts.

Other functions of measurement systemsfacilitatethe basic process described above. Amplification ensures that the physical signal is strong enough to complete the measurement. In order to reducedegradationof the measurement as it progresses through the system, the signal may be converted to coded or digital form. Magnification, enlarging the measurement signal without increasing its power, is often necessary to match the output of one element of the system with the input of another, such as matching the size of the readoutmeterwith the discerning power of thehuman eye.

One important type of measurement is the analysis ofresonance, or the frequency of variation within a physical system. This is determined byharmonic analysis, commonly exhibited in the sorting of signals by a radio receiver.Computationis another important measurement process, in which measurement signals are manipulated mathematically, typically by some form of analog ordigital computer. Computers may also provide a control function in monitoring system performance.

Measuring systems may also include devices for transmitting signals over great distances (seetelemetry). All measuring systems, even highly automated ones, include some method of displaying the signal to an observer. Visual display systems may comprise acalibratedchart and a pointer, anintegrateddisplay on acathode-ray tube, or a digital readout. Measurement systems often include elements for recording. A common type utilizes a writing stylus that records measurements on a moving chart. Electrical recorders may include feedback reading devices for greater accuracy.

The actual performance of measuring instruments is affected by numerous external and internal factors. Among external factors are noise and interference, both of which tend to mask or distort the measurement signal. Internal factors include linearity, resolution, precision, and accuracy, all of which are characteristic of a given instrument or system, anddynamicresponse, drift, andhysteresis, which are effects produced in the process of measurement itself. The general question of error in measurement raises the topic of measurement theory.

MEASUREMENT THEORY

Measurement theory is the study of how numbers are assigned to objects and phenomena, and its concerns include the kinds of things that can be measured, how different measures relate to each other, and the problem oferrorin the measurement process. Any general theory of measurement must come to grips with three basic problems: error; representation, which is the justification of number assignment; and uniqueness, which is the degree to which the kind of representation chosen approaches being the only one possible for the object or phenomenon in question.

Various systems ofaxioms, or basic rules and assumptions, have been formulated as a basis for measurement theory. Some of the most important types of axioms include axioms of order, axioms of extension, axioms of difference, axioms of conjointness, and axioms ofgeometry. Axioms of order ensure that the order imposed on objects by the assignment of numbers is the same order attained in actual observation or measurement. Axioms of extension deal with the representation of such attributes as time duration, length, and mass, which can be combined, or concatenated, for multiple objects exhibiting the attribute in question. Axioms of difference govern the measuring of intervals. Axioms of conjointness postulate that attributes that cannot be measured empirically (for example, loudness, intelligence, or hunger) can be measured by observing the way their component dimensions change in relation to each other. Axioms of geometry govern the representation of dimensionally complex attributes by pairs of numbers, triples of numbers, or evenn-tuples of numbers.

The problem of error is one of the central concerns of measurement theory. At one time it was believed that errors of measurement could eventually be eliminated through the refinement of scientific principles and equipment. This belief is no longer held by most scientists, and almost all physical measurements reported today are accompanied by some indication of the limitation of accuracy or the probable degree of error. Among the various types of error that must be taken into account are errors of observation (which include instrumental errors, personal errors, systematic errors, and random errors), errors of sampling, and direct and indirect errors (in which oneerroneousmeasurement is used in computing other measurements).

Measurement theory dates back to the 4th centuryBC, when a theory of magnitudes developed by the Greek mathematiciansEudoxus of Cnidusand Thaeatetus was included inEuclid’sElements. The first systematic work on observational error was produced by the English mathematician Thomas Simpson in 1757, but the fundamental work on error theory was done by two 18th-century French astronomers,Joseph-Louis LagrangeandPierre-Simon Laplace. The first attempt to incorporate measurement theory into the social sciences also occurred in the 18th century, whenJeremy Bentham, a British utilitarian moralist, attempted to create a theory for the measurement of value. Modernaxiomatictheories of measurement derive from the work of two German scientists,Hermann von Helmholtzand Otto Hölder, and contemporary work on the application of measurement theory to psychology and economics derives in large part from the work ofOskar MorgensternandJohn von Neumann.

Since most social theories are speculative in nature, attempts to establish standard measuring sequences or techniques for them have met with limited success. Some of the problems involved in social measurement include the lack of universally accepted theoretical frameworks and thus of quantifiable measures, sampling errors, problems associated with the intrusion of the measurer on the object being measured, and the subjective nature of the information received from human subjects. Economics is probably thesocial sciencethat has had the most success in adopting measurement theories, primarily because many economic variables (like price and quantity) can be measured easily and objectively. Demography has successfully employed measurement techniques as well, particularly in the area of mortality tables.

EXAMPLES OF MEASUREMENT

Time

The ongoing sequence of events is time. We can measure time in seconds, minutes, hours, days, weeks, months, and years.

A clock and a calendar help us to measure time.

Weight

The amount of matter a thing consists of is called its weight. Measuring weight means to measure the heaviness of a thing.

Weight can be measured in grams, kilograms, and pounds.

Length

The amount of something that is measured from one end to the other along the longest side is called its length.

Length is measured in centimeters, meters, kilometers, feet, and miles

Capacity

Capacity is a measure of how much quantity a thing can hold.

Capacity is measured in liters and gallons

Temperature

The temperature of a thing is the measurement of how hot or cold it is.

Temperature is measured in Celsius, Fahrenheit, and Kelvin.

We can convert from one unit of measurement to another.

There are two measurement systems:438834709

The Metric System:

This system is based on the meter, liter, and gram as units of length (distance), capacity (volume), and weight (mass) respectively.

The US Standard Units:

This system uses inches, feet, yards, and miles to measure length or distance.

Capacity or volume is measured in fluid ounces, cups, pints, quarts or gallons.

Weight or mass is measured in ounces, pounds and tons.

USING NON-STANDARD AND STANDARD UNITS OF MEASUREMENT;

Standard units are common units of measurement such as centimetres, grams and litres. Non-standard measurement units might include cups, cubes or sweets.

Standard units are what we usually used to measure things like weight, length and volume. Standard units that would be introduced in primary school are grams, kilograms, meters, kilometres, millilitres and litres.

Non-standard units are used by children in EYFS and Year 1 to introduce without having to use scales of any kind as this can make it seem more complicated.

Although standard measurements are an essential part of the 2014 National Curriculum for Maths, non-standard measurements are often used with EYFS and Year one students to help them grasp the concept of measurements.

HOW ARE NON-STANDARD UNITS USED IN THE CLASSROOM?

Children are first introduced to the concept of measurement during the EYFS stage of their learning. However, children at this stage don't read any scales - instead, they usually begin to measure everyday items on how they feel (light or heavy). You could ask children to compare items by asking them: do you think the pencil or sharpener is heavier? This is a great way to get them thinking about measurement and how we determine the difference between items.

Once children reaching KS1 students are prompted to measure using centimetres or kilograms, it is common for teachers to introduce them to the concept and skills of measuring using non-standard units.

These are often practical, easy to visualise examples which can be demonstrated in the classroom. For instance, how many cups of sand are needed to fill a bucket or how many biscuits will fit into a tin.

This will allow students to develop their counting skills as well as get used to the language around measurements, such as “x units got into y unit”. This creates a strong foundation of knowledge which will help them as they begin to learn standard units of measurement.

USING STANDARD UNITS OF MEASUREMENT

Children begin to use standard units in Year 2. They'll begin to learn and understand the different equipment needed to measure different items. For example, they'd need to know whether to measure items using centimetres, kilograms or metres.

Once children reaching Year 3 they'll carry on working on everything they've learnt in Year 2, but at more advanced levels. Year 2 children will begin to problem-solve using practical ways such as scales. They will also begin to understand the different conversion measurements and should be able to recall the following:

·1 litre = 1000 millilitres.

·1 metre = 100 centimetres.

·1 kilogram = 1000 grams.

In year 4 children will start to convert and should be able to determine different measurements. For example, children should know that 1.4 litres is the same as 1400 ml. This will help them with problem-solving when sometimes children must find out a measurement and convert it to a standard unit.

In Year 5, children will solve problems using measurement further but will also use addition, subtraction, multiplication and division to achieve the answer.

Once children reach the last year of Primary education (Year 6) they will continue problem solving using the four operations, whilst converting between units using decimals.

EXAMPLES OF STANDARD AND NON-STANDARD UNITS

The following are all standard units of measurement. Though they may not be common, they are still accepted units of measurement across the world.

·Centimetres (1 cm = 0.39 Inches)

·Feet (1 Foot = Approx. 30 cm)

·Kilograms (1 Kilogram = 2.2. Pounds)

·Cups (1 Cup = 10 Fluid Ounces)

·Hands (1 Hand = 4 inches)

These units of measurement, however, are non-standard and not generally accepted or known by other people. They are likely only used in a classroom situation or in a casual context. They may give only a very approximate measurement - more like an estimate than a fixed measurement.

·Heads (e.g. Megan is a head taller than her sister)

·Sweets (e.g. 20 sweets in a bag)

·Squares (e.g. 15 squares of chocolate in a bar)

·Phone Book (e.g. This is heavier than a stack of Phone Books!)

·Stone’s Throw (e.g. It’s not far - it’s only a stone's throw away.)

HOW CAN I TEACH STANDARD AND NON-STANDARD UNITS TO MY CLASS?

There are a number of National Curriculum aims relating to measurements when it comes to KS1 Maths. It is important that students have a grounding in both standard and non-standard units of measurement.

TYPES OF ANGLES

When two lines intersect, at the point of their intersection an angle is formed. Learning about angles is important, as they form the base of geometry. The two rays that form the angle are known as the sides of the angle. Also, it is not necessary that an angle is formed by intersection of two straight lines; it can be formed by intersection of two curved lines too.

1. Acute Angle

An angle which measures less than 90° is called an acute angle. The measure between 0° to 90°. In the picture below, the angle formed by the intersection of PQ and QR at Q forms an angle PQR which measures 45°. Thus, PQR is called an acute angle.

2. Right Angle

An angle which measures exactly 90° is called a right angle. It is generally formed when two lines are perpendicular to each other. In the figure below, line AB intersects line BC at B and form an angle ABC which measures 90°.

3. Obtuse Angle

An angle that measures greater than 90° is known as the obtuse angle. The angle measure ranges from 90° to 180°. An obtuse angle can also be found out if we have the measure of the acute angle.

Obtuse Angle Measure = (180 - acute angle measure)

In the picture above, line segment DO intersects line segment OQ at point O and forms an angle DOQ measuring 120°. Thus, it is an obtuse angle.

Also, if we extend line OQ to OP then we can find a measure of the acute angle.

DOP = 180° - DOQ = 180° - 120° = 60°

1. Straight Angle

The angle which measures exactly 180° is called a straight angle. This is similar to a straight line, thus the name straight angle.

A straight angle is nothing but a mixture of an obtuse angle and acute angle on a line.

2. Reflex Angle

The angle which measures greater than 180° and less than 360° is known as the reflex angle. The reflex angle can be calculated if the measure of the acute angle is given, as it is complementary to the acute angle on the other side of the line.

Using the reflex angle, we can find the measure of the acute angle.

A Measure of Acute Angle = 360° – a Measure of Reflex Angle

Complementary & SupplementaryAngle

1. Complementary Angle

If two angles add up to measure 90° then they are known as complementary angles. The angles don't have to be adjacent to each other to be known as complementary. As long as they add up to 90° they will be known as complementary angles.

In the figure, a and b the angles are present adjacent to each other and add up to 90° and thus are known as complementary angles. In figure c and d, the angles are not adjacent to each other, but they add up to 90° and thus they are known as complementary angles.

2. Supplementary Angles

When two angles add up to 180° then they are known as supplementary angles. There are various types of supplementary angles.

Vertical Angles

Angles which have a common vertex and the sides of the angle are formed by the same lines are known as vertical angles. Vertical angles are equal to each other.

In the above figure, 1 and 3, 2 and 4, 6 and 8 and 5 and 7 are vertical angles. Also, 3, 4,5, 6 are known as interior angles and 1,2,7,8 are known as exterior angles.

b.Alternate Interior Angles

These are a pair of interior angles present on the opposite side of the transversal. The easiest way to spot alternate interior angles is to identify a "Z” on the interior side.

In the above figure, 3 and 5, 4 and 6 are interior angles. The interior angles equal to one another.

c.Alternate Exterior Angles

This is similar to alternate interior angles; just that it is present on the exterior side. In the above figure, 1 and 7, 2 and 8 are the pair of alternate exterior angles. Similar to alternate interior angles, even alternate exterior angles are equal to one another.

d.Corresponding Angles

Angles which are present in a similar position are known as corresponding angles.

Adjacent angles have the same vertex and arm.

e.Adjacent angleshave the same vertex and arm.

f.Zero Angle

A zero angle (0°) is an angle formed when both the angle’s arms are at the same position.

∠RPQ = 0°(zero angle)

g.Complete Angle

A complete angle is equal to 360°. 1 revolution is equal to 360°.

Positive and Negative Angles

When measuring from a line:

·apositiveangle goescounterclockwise(opposite direction that clocks go)

·anegativeangle goes clockwise

Example: −67°

Summary

The major basis of geometry is angles. Angles finds its application in nearly all types of questions, be it trigonometry to closed shapes. Understanding angles and angle types will help in solving a lot of tricky questions. Thus, make sure that you understand it well.

PARTS OF AN ANGLE

The corner point of an angle is called thevertex

And the two straight sides are calledarms

The angle is theamount of turnbetween each arm.

HOW TO LABEL ANGLES

There are two main ways to label angles:

1. give the angle a name, usually a lower-case letter likeaorb, or sometimes a Greek letter likeα(alpha) orθ(theta)

2. or by the three letters on the shape that define the angle, with the middle letter being where the angle actually is (its vertex).

Example angle "a" is "BAC", and angle "θ" is "BCD"

FINDING PERIMETER AND AREAS OF TRIANGULAR

PERIMETER OF A TRIANGLE

The perimeter of any two-dimensional figure refers to a distance around the figure. We can compute the perimeter of any closed figure by just summing up the length of all the three sides. In this section, you will learn how to calculate the perimeter of different types of triangles when there lengths are known.

Formula

The formula for finding the perimeter of a triangle is given below:

Perimeter = Sum of three sides

The standard unit of perimeter is meter or centimeter.

The following table shows the formula for computing the sides of three types of triangle:

·It shows that the perimeter of an equilateral triangle is 3 times its length

·The perimeter of an isosceles triangle is the sum of 2 times length of the equal sides and base

·The perimeter of the scalene triangle is equal to the sum of the lengths of all three sides of the triangle

Consider the following triangle whose lengths of three sides are given:

The length of the sides are 12cm 9 cm and 8 cm respectively. To calculate the perimeter, substitute these values in the formula below:

Perimeter = a + b + c

= 12 + 9 + 8 = 29 cm

AREA OF A TRIANGLE

An area of a triangle is defined as follows:

The total region that is enclosed by three sides of a triangle is known as an area of that triangle

This area of a triangle is equal to the half of the product of base and height of a triangle. Mathematically, we can write the formula of area of a triangle like this:

Area =

It means that to calculate the area of a triangle, we should know the base and height of the triangle. It should be kept in mind that the base and height of a triangle are perpendicular to each other. Sometimes, the height of a triangle is unknown. To find the area of such a triangle, we need to know the height first. In these situations, we different formulas to compute the areas of triangle. All these scenarios are discussed below with examples.

AREA OF A TRIANGLE WHEN HEIGHT IS KNOWN

In this section, we will see how to calculate the area of a triangle with known height.

Find the area of the following triangle:

In the above triangle, height and base of the triangle is given, so we will use the general formula for finding the area of the triangle.

Area =

Area =

Area =

AREA OF AN EQUILATERAL TRIANGLE (HEIGHT UNKNOWN)

All sides of an equilateral triangle are equal. The formula to calculate the area of an equilateral triangle is given below:

Area =

For example, consider an equilateral triangle below whose measurement of one side is given:

We will substitute 8 in the above formula to calculate the area of an equilateral triangle:

Area =

Area =

Area =

AREA OF AN ISOSCELES TRIANGLE WITH KNOWN SIDES ONLY

Isosceles triangle is a triangle whose length of two sides are equal. We can have a situation in which the sides of an isosceles triangle are known, but the height is unknown. In this situation, we can use the following formula to compute the area of a triangle:

Area =

In the above triangle, the lengths of two sides are equal, hence it is an isosceles triangle. We do not know the height of the triangle, so we will use the above discussed formula to compute its area.

Area =

Area =

Area =

We know that remembering all the above formulas for equilateral or isosceles can be difficult. When sides are known and height is unknown, you can simply use the Heron's formula which is discussed below to calculate the area of a triangle. This is true for all types of triangles whether they are equilateral, isosceles, right angle or squalene.

HERON'S FORMULA TO CALCULATE THE AREA WHEN SIDES ARE KNOWN ONLY

We can also use the Heron's formula to calculate the area of a triangle when the length of three sides are known. To employ this formula, we should know the perimeter of a triangle which refers to the distance covered around the triangle and is computed by summing up the length of all three sides of the triangle. Use the following steps to calculate the area using Heron's formula:

·In the first step of the process, find the semi perimeter of the triangle by summing up the three sides and dividing the figure by 2. Semi perimeter is simply half of the perimeter of a triangle.

·In the second step, apply the value obtained in the first step of the triangle in the main formula which is known as Heron's formula.

The Heron's formula to calculate the area of a triangle is given below:

Area =

Here, s is the semi perimeter and a, b and c represent the sides of the triangle.

Consider the following example:

Find the area of the following triangle:

The values of sides a, b and c are 12 cm, 9 cm and 8 cm. First we will calculate the semi perimeter by substituting these values in the below formula:

Semi perimeter =

Semi perimeter =

Semi perimeter =

Now, we will substitute this value of semi perimeter in the below formula to find the area:

Area =

Area =

Area =

Consider another example below

Find the area of the following triangle:

The values of sides a, b and c are 18 cm, 15 cm and 9 cm. First we will calculate the semi perimeter by substituting these values in the below formula:

Semi perimeter =

Semi perimeter =

Semi perimeter =

Now, we will substitute this value of semi perimeter in the below formula to find the area:

Area =

Area =

Area =

Area =

CIRCUMFERENCE AND AREA OF CIRCULAR REGIONS

WHAT IS THE CIRCUMFERENCE OF A CIRCLE?

Thecircumferenceof a circle is the distance around the outside of the circle. It is like theperimeterof other shapes like squares. You can think of it as the line that defines the shape. For shapes made of straight edges this line is called theperimeterbut for circles this defining line is called thecircumference.

This diagram shows the circumference of a circle.

There are two other important distances on a circle, theradius (r)and thediameter(d).The radius, the diameter, and the circumference are the three defining aspects of every circle. Given the radius or diameter and pi you can calculate the circumference. The diameter is the distance from one side of the circle to the other at its widest points. The diameter will always pass through the center of the circle. The radius is half of this distance. You can also think of the radius as the distance between the center of the circle and its edge.

This diagram shows the circumference, diameter, center, and radius on a circle.

HOW CAN YOU CALCULATE THE CIRCUMFERENCE OF A CIRCLE?

If you know the diameter or radius of a circle, you can work out the circumference. To begin with, remember that pi is an irrational number written with the symbolπ. π is roughly equal to 3.14.

The formula for working out the circumference of a circle is:

CIRCUMFERENCE OF CIRCLE =Π X DIAMETER OF CIRCLE

This is typically written as C =πd. This tells us that the circumference of the circle is three “and a bit” times as long as the diameter. We can see this on the graphic below:

You can also work out the circumference of a circle if you know its radius. Remember that the diameter is double the length of the radius. We already know that C =πd. If r is the radius of the circle, then d = 2r. So, C = 2πr.

Example 1

If a circle has a diameter of 10cm, what is its circumference?

Answer

We know that C = πd. Since the diameter is 10cm, we know that C = π x 10cm = 31.42cm (to 2 decimal places).

Example 2

If a circle has a radius of 3m, what is its circumference?

Answer

We know that C = 2πr. Since the radius is 3m, we know that C = π x 6m = C= 18.84m (to 2 decimal places).

Example 3

Find the missing length (marked with a ?) on the diagram below:

Answer

The missing length is the circumference. Using the knowledge that the diameter is 4.3m on the diagram and knowing thatC = πd, we can calculate the circumference. With a little thinking we can easily figure out that, C = π x 4.3m = 13.51m (to 2 decimal places). The missing length is 13.51m.

HOW TO CALCULATE THE CIRCUMFERENCE OF THE EARTH?

Have you ever wondered how big the earth is? Well, using pi it’s possible to work out the circumference of the Earth! Scientists have discovered that the diameter of the Earth is 12,742km. Given this information, what is the circumference of the Earth? Get out a piece of paper and a calculator and see if you can work it out on your own.

Again, we know that C = πd, and that the diameter of the earth is 12,742km. Using this information, we can calculate the circumference of the Earth as C = π x 12,742km = 40,030km.

AREA OF A CIRCLE

The area of a circle is the space occupied by the circle in a two-dimensional plane. Alternatively, the space occupied within the boundary/circumference of a circle is called the area of the circle. The formula for the area of a circle is A = πr², where r is the radius of the circle. The unit of area is the square unit, for example, m², cm², in², etc.Area of Circle = πr²or πd²/4 in square units, where(Pi) π= 22/7 or 3.14.Pi (π) is the ratio of circumference to diameter of any circle. It is a special mathematical constant.

The area of a circle formula is useful for measuring the region occupied by a circular field or a plot. Suppose, if you have a circular table, then the area formula will help us to know how much cloth is needed to cover it completely. The area formula will also help us to know the boundary length i.e., the circumference of the circle. Does a circle have volume? No, a circle doesn't have a volume. A circle is atwo-dimensional shape, it does not have volume. A circle only has an area and perimeter/circumference. Let us learn in detail about the area of a circle, surface area, and its circumference with examples.

Circle and Parts of a Circle

A circle is a collection of points that are at a fixed distance from the center of the circle. A circle is a closed geometric shape. We see circles in everyday life such as a wheel, pizzas, a circular ground, etc. The measure of the space or region enclosed inside the circle is known as the area of the circle.

Radius:The distance from the center to a point on the boundary is called theradius of a circle. It is represented by the letter 'r' or 'R'. Radius plays an important role in the formula for the area and circumference of a circle, which we will learn later.

Diameter:A line that passes through the center and its endpoints lie on the circle is called thediameter of a circle. It is represented by the letter 'd' or 'D'.

Diameter formula:The diameter formula of a circle is twice its radius. Diameter = 2 × Radius

d = 2r or D = 2R

If the diameter of a circle is known, its radius can be calculated as:

r = d/2 or R = D/2

Circumference:The circumference of the circle is equal to the length of its boundary. This means that the perimeter of a circle is equal to its circumference. The length of the rope that wraps around the circle's boundary perfectly will be equal to its circumference. The below-given figure helps you visualize the same. The circumference can be measured by using the given formula:

where 'r' is the radius of the circle and π is the mathematical constant whose value is approximated to 3.14 or 22/7. The circumference of a circle can be used to find the area of that circle.

For a circle with radius ‘r’ and circumference ‘C’:

·π = Circumference/Diameter

·π = C/2r = C/d

·C = 2πr

Let us understand the different parts of a circle using the following real-life example.

Consider a circular-shaped park as shown in the figure below. We can identify the various parts of a circle with the help of the figure and table given below.

WHAT IS THE AREA OF CIRCLE?

The area of a circle is the amount of space enclosed within the boundary of a circle. The region within the boundary of the circle is the area occupied by the circle. It may also be referred to as the total number of square units inside that circle.

AREA OF CIRCLE FORMULAS

The area of a circle can be calculated in intermediate steps from the diameter, and the circumference of a circle. From the diameter and the circumference, we can find the radius and then find the area of a circle. But these formulae provide the shortest method to find the area of a circle. Suppose a circle has a radius 'r' then the area of circle = πr²or πd²/4 in square units, where π = 22/7 or 3.14, and d is the diameter.

Area of a circle, A = πr²square units

Circumference / Perimeter = 2πr units

Area of a circle can be calculated by using the formulas:

·Area = π × r², where 'r' is the radius.

·Area = (π/4) × d², where 'd' is the diameter.

·Area = C²/4π, where 'C' is the circumference.

EXAMPLES USING AREA OF CIRCLE FORMULA

Let us consider the following illustrations based on the area of circle formula.

Example1:If the length of the radius of a circle is 4 units. Calculate its area.

Solution:
Radius(r) = 4 units(given)
Using the formula for the circle's area,
Area of a Circle = πr²
Put the values,
A = π4²
A =π × 16
A = 16π ≈ 50.27

Answer:The area of the circle is 50.27 squared units.

Example 2:The length of the largest chord of a circle is 12 units. Find the area of the circle.

Solution:
Diameter(d) = 12 units(given)
Using the formula for the circle's area,
Area of a Circle = (π/4)×d²
Put the values,
A = (π/4) × 12²
A = (π/4) × 144
A = 36π ≈ 113.1

Answer:The area of the circle is 113.1 square units.

AREA OF A CIRCLE USING DIAMETER

The area of the circle formula in terms of thediameteris: Area of a Circle = πd²/4. Here 'd' is the diameter of the circle. The diameter of the circle is twice the radius of the circle. d = 2r. Generally from the diameter, we need to first find the radius of the circle and then find the area of the circle. With this formula, we can directly find the area of the circle, from the measure of the diameter of the circle.

AREA OF A CIRCLE USING CIRCUMFERENCE

The area of a circle formula in terms of the circumference is given by the formula(Circumference)24π(Circumference)24π. There are two simple steps to find the area of a circle from the given circumference of a circle. The circumference of a circle is first used to find the radius of the circle. This radius is further helpful to find the area of a circle. But in this formulae, we will be able to directly find the area of a circle from the circumference of the circle.

AREA OF A CIRCLE-CALCULATION

The area of the circle can be conveniently calculated either from the radius, diameter, or circumference of the circle. The constant used in the calculation of the area of a circle is pi, and it has a fractional numeric value of 22/7 or adecimalvalue of 3.14. Any of the values of pi can be used based on the requirement and the need of the equations. The below table shows the list of formulae if we know the radius, the diameter, or the circumference of a circle.

DERIVATION OF AREA OF A CIRCLE

Why is the area of the circle is πr²? To understand this, let's first understand how the formula for the area of a circle is derived.

Observe the above figure carefully, if we split up the circle into smaller sections and arrange them systematically it forms a shape of a parallelogram. When the circle is divided into even smaller sectors, it gradually becomes the shape of a rectangle. The more the number of sections it has more it tends to have a shape of a rectangle as shown above.

The area of a rectangle is = length × breadth

The breadth of a rectangle = radius of a circle (r)

When we compare the length of a rectangle and the circumference of a circle we can see that the length is = ½ the circumference of a circle

Area of circle = Area of rectangle formed = ½ (2πr) × r

Therefore, the area of the circle is πr², where r, is the radius of the circle and the value of π is 22/7 or 3.14.

SURFACE AREA OF CIRCLE FORMULA

The surface area of a circle is the same as the area of a circle. In fact, when we say the area of a circle, we mean nothing but its total surface area. Surface area is the area occupied by the surface of a 3-D shape. The surface of a sphere will be spherical in shape but a circle is a simple plane 2-dimensional shape.

If the length of the radius or diameter or even the circumference of the circle is given, then we can find out the surface area. It is represented in square units. The surface area of circle formula = πr²where 'r' is the radius of the circle and the value of π is approximately 3.14 or 22/7.

REAL-WORLD EXAMPLE ON AREA OF CIRCLE

Ron and his friends ordered a pizza on Friday night. Each slice was 15 cm in length.

Calculate the area of the pizza that was ordered by Ron. You can assume that the length of the pizza slice is equal to the pizza’s radius.

Solution:

A pizza is circular in shape. So we can use the area of a circle formula to calculate the area of the pizza.

Radius is 15 cm

Area of Circle formula = πr²= 3.14 × 15 × 15 = 706.5

Area of the Pizza = 706.5 sq. cm.

WORKED EXAMPLES

1.Find the circumference and area of radius 7 cm.

Solution:

Circumference of circle = 2πr

= 2 × 22/7 × 7

= 44 cm

Area of circle = πr²

= 22/7 × 7 × 7 cm²

= 154 cm²
2.A race track is in the form of a ring whose inner circumference is 220 m and outer circumference is 308 m. Find the width of the track.

Solution:

Let r₁and r₂be the outer and inner radii of ring.

Then 2πr₁= 308

2 × 22/7 r₁= 308

⇒r₁= (308 ×7)/(2 ×22)

⇒r₁= 49 m

2πr₂= 220

⇒2 ×22/7 ×r₂= 220

⇒r₂= (220 ×7)/(2 ×22)

⇒r₂= 35 m

Therefore, width of the track = (49 - 35) m = 14 m

3.The area of a circle is 616 cm². Find its circumference.

Solution:

We know area of circle = πr²

⇒22/7 ×r²= 616

⇒r²= (616 ×7)/22

⇒r²= 28 ×7

⇒r = √(28 ×7)

⇒r = √(2 ×2 ×7 ×7)

⇒r = 2 ×7

⇒r = 14 cm

Therefore, circumference of circle = 2πr

= 2 × 22/7 × 14

= 88 cm

4.Find the area of the circle if its circumference is 132 cm.

Solution:

We know that the circumference of circle = 2πr

Area of circle = πr²

Circumference = 2πr = 132

⇒2 ×22/7 ×r = 132

⇒r = (7 ×132)/(2 ×22)

⇒r = 21 cm

Therefore, area of circle = πr²

= 22/7 × 21 × 21

= 1386 cm²

5.The ratio of areas of two wheels is 25 : 49. Find the ratio of their radii.

Solution:

If A₁and A₂are the area of wheels,

A₁/A₂=25/49

⇒(πr₁²)/(πr₂²) = 25/49

⇒(r₁²)/(r₂²) = 25/49

⇒r₁/r₂= √(25/49)

⇒r₁/r₂= 5/7

Therefore, ratio of their radii is 5 : 7.

6.The diameter of a wheel of a motorcycle is 63 cm. How many revolutions will it make to travel 99 km?

Solution:

The diameter of the wheel of a motorcycle = 63 cm

Therefore, circumference of the wheel of motorcycle = πd

= 22/7 × 63

= 198 cm

Total distance travelled by motorcycle = 99 km

= 99 × 1000

= 99 × 1000 × 100 cm

Therefore, number of revolutions = (99 × 1000 × 100)/198 = 50000
7.The diameter of a wheel of cycle is 21 cm. It moves slowly along a road. How far will it go in 500 revolutions?

Solution:

In revolution, distance that wheel covers = circumference of wheel Diameter of wheel = 21 cm

Therefore, circumference of wheel = πd

= 22/7 × 21

= 66 cm

So, in 1 revolution distance covered = 66 cm

In 500 revolution distance covered = 66 × 500 cm

= 33000 cm

= 33000/100 m

= 330 m
8.The circumference of a circle exceeds the diameter by 20 cm. Find the radius of the circle.

Solution:

Let the radius of circle of = r m.

Then circumference = 2 πr

Since, circumference exceeds diameter by 20

Therefore, according to question;

2 πr = d + 20

⇒2 πr = 2r + 20

⇒2 ×(22/7) ×r = 2r + 20

⇒44r/7 - 2r = 20

⇒(44r - 14r)/7 = 20

⇒30r/7 = 20

⇒r = (7 ×20)/30

⇒r =14/3

So, the radius of circle = 14/3 cm = 42/3 cm

9.A piece of wire in the form of rectangle 40 cm long and 26 cm wide is again bent to form a circle. Find the radius of the circle.

Solution:

Length of wire = Perimeter of rectangle

= 2(l + b)

= 2(40 + 26)

= 2 × 66

= 132 cm

When it is again bent to form a circle, then

Perimeter of circle = Perimeter of rectangle

2 πr = 132 cm

⇒2 ×22/7 ×r = 132

⇒r = (132 ×7)/(2 ×22)

⇒r = 21 cm

SURFACE AREA AND VOLUMES OF PRISMS AND PYRAMIDS

Both prism and pyramid are basically 3D shapes. Even though we have different formulas to find surface area of prism and pyramid, the basic idea of finding surface area is to add the areas of all the faces.

First, let us look at, how to find surface area of a prism.

Surface Area of Prism

Let us consider the rectangle prism given below.

Here is the basic idea to find surface area of the above rectangular prism.

Surface Area = Sum of areas of all six faces

Let us find the area of each face separately.

Area of the front face (red colored) = l x h

Area of the back face (blue colored) = l x h

Area of the left side face (green colored) = w x h

Area of the right side face (green colored) = w x h

Area of the top portion (purple colored) = l x w

Area of thebase (purple colored) = l x w

Now,

Surface area = lh + lh + wh + wh + lw + lw

Surface area = 2lh + 2wh + 2lw

Surface area = 2(lh + wh + lw)

This is the formula to find surface area of a rectangular prism.

Note:

Rectangular prism is also known as cuboid.

We can apply the above explained basic idea to find surface area of any prism without remembering the formulas.

Let us find surface area of the cube given below.

We know that the shape of each face of a cube is a square.

In the above cube, the side length of each face is "a".

So, area of each face (square)= a x a = a²

Therefore,

Surface area of cube = 6 x area of each face

Surface area of cube = 6a²

Now, let us find surface area of the triangular prism given below.

In the above triangular prism, there are five faces. The shape of the base and the two slanting faces is rectangle. The shape of two faces on the left side and right side is triangle.

For the given triangular prism,

Area of the base = Lb.

Area of the first slanting face = Ls

Area of the other slanting face = Ls

Area of the front face = (1/2)bh

Area of the back face = (1/2)bh

So,

surface area = sum of the area of 5 faces

surface area = Lb + 2Ls + 2 x (1/2)bh

Surface area of triangular prism = Lb+2Ls+bh

Now let us look at, how to find surface area of a prism.

Surface Area of Pyramid

Let us consider the pyramid with square base given below.

Here is the basic idea to find surface area of the above pyramid.

Surface Area = Sum of areas of all five faces (Including the base)

For any pyramid, if the shape of the base is square, then we will have four side walls. The shape of each side wall will be a triangle with equal area.

In the above pyramid, the base is a square with side length "a" and each wall is a triangle with base "a" and height "h"

Let us find the area of each face separately.

Area of the base = a x a = a²

Area of each side wall = (1/2) ah

Area of all four side walls = 4 x (1/2) ah = 2ah

Surface area of the above pyramid is

= a²+ 2ah

This is the formula to find surface area of a pyramid with square base.

We can apply the above explained basic idea to find surface area of a pyramid with triangular base.

Let us find surface area of a pyramid with triangular base.

For any pyramid, if the shape of the base is equilateraltriangle, then we will have three side walls. The shape of each side wall will be a triangle with equal area.

In the above pyramid, the base is an equilateral triangle with side length "a".

And each wall is a triangle with base "a" and height "h"

Let us find the area of each face separately.

Area of the base = (√3/4)a²

Area of each side wall = (1/2)ah

Area of all 3 side walls = 3 x (1/2)ah = (3/2)ah

Surface area of the above pyramid is

=(√3/4)a²+ (3/2)ah

This is the formula to find surface area of a pyramid with equilateral triangle base.

Note:

If the base is not equilateral triangle and it is either scalene triangle or isosceles triangle, then the area of side walls will not be equal. We have to find area of each side wall separately.

Practice Problems

Problem 1:

Find the surface area of the cuboid shown below.

Solution:

Surface area of cuboid is

= Sum of areas of all six faces

In cuboid, each face is a rectangle. So we can use area of rectangle formula to get area of each face.

Area of the front face = 8 x 12 = 96cm²

Area of the back face = 8 x 12 = 96cm²

Area of the left side face = 4 x 8 = 32cm²

Area of the right side face = 4 x 8 = 32cm²

Area of the top portion = 4 x 12 = 48cm²

Area of thebase = 4 x 12 = 48cm²

Surface area of the above cuboid is

= Sum of areas of all six faces

=96 + 96 + 32 + 32 + 48 + 48

= 96 + 96 + 32 + 32 + 48 + 48

= 352 cm²

Alternative Method:

We can use the formula given below to find surface area of cuboid.

Formula for surface area of cuboid is

= 2(lh + wh + lw)

Substitute l = 12, w = 4 and h = 8.

= 2(12x8 + 4x8 + 12x4)

= 2(96 + 32 + 48)

= 2(176)

= 352 cm²

Problem 2:

Find the surface area of the cube shown below.

Solution:

We know that the shape of each face of a cube is a square.

In the above cube, the side length of each face is "8".

So, area of each face (square) is

= 8 x 8

= 64 cm²

Therefore,surface area of the cube is

= 6 x area of each face

= 6 x 64

= 384 sq.cm

Problem 3:

Find the surface area of the triangular prism shown below.

Solution:

In the above triangular prism, there are five faces. The shape of the base, vertical face and slanting face is rectangle. The shape of two faces on the left side and right side is triangle.

For the given triangular prism,

Area of the base = 7 x 4 = 28cm²

Area of the vertical face = 3 x 7 =21cm²

Area of the slanting face = 5 x 7 = 35cm²

Area of the front face = (1/2) x 4 x 3 = 6cm²

Area of the back face = (1/2) x 4 x 3 = 6cm²

So,surface area of the above triangular prism is

= sum of the area of 5 faces

= 28 + 21 + 35 + 6 + 6

= 96 cm²

Problem 4:

Find e surface area of the triangular prism shown below.

Solution:

In the above triangular prism, there are five faces. The shape of the base and the two slanting faces is rectangle. The shape of two faces on the left side and right side is triangle.

For the given triangular prism,

Area of the base = 8 x 12 = 96cm²

Area of the first slanting face = 12 x 5 = 60cm²

Area of the other slanting face = 12 x 5 = 60cm²

Area of the front face = (1/2) x 8 x 3 = 12cm²

Area of the back face = (1/2) x 8 x 3 = 12cm²

So,surface area of the abovethe triangular prism is

= sum of the area of 5 faces

= 96 + 60 + 60 + 12 + 12

= 240 cm²

Problem 5:

Find the surface area of the pyramid shown below.

Solution:

Surface area of the pyramid is

= Sum of areas of all 5 faces

In the above pyramid, the base is a square with side length 5 cm and each wall is a triangle with base 5 cm and height 8 cm.

Let us find the area of each face separately.

Area of the base = 5 x 5 = 25 sq.cm

Area of each side wall = (1/2) x 5 x 8 = 20 sq.cm

Area of all 4 side walls = 4 x 20 = 80 sq.cm

Surface area of the above pyramid is

= 25 + 80

=105 sq.cm

The surface area is the area that describes the material that will be used to cover a geometric solid. When we determine the surface areas of a geometric solid we take the sum of the area for each geometric form within the solid.

The volume is a measure of how much a figure can hold and is measured in cubic units. The volume tells us something about the capacity of a figure.

A prism is a solid figure that has two parallel congruent sides that are called bases that are connected by the lateral faces that are parallelograms. There are both rectangular and triangular prisms.

To find the surface area of a prism (or any other geometric solid) we open the solid like a carton box and flatten it out to find all included geometric forms.

To find the volume of a prism (it doesn't matter if it is rectangular or triangular) we multiply the area of the base, called the base area B, by the height h.

V=B⋅hV=B⋅h

A cylinder is a tube and is composed of two parallel congruent circles and a rectangle which base is the circumference of the circle.

Example

The area of one circle is:

A=πr2A=πr2

A=π⋅22A=π⋅22

A=π⋅4A=π⋅4

A≈12.6A≈12.6

The circumference of a circle:

C=πdC=πd

C=π⋅4C=π⋅4

C≈12.6C≈12.6

The area of the rectangle:

A=C⋅hA=C⋅h

A=12.6⋅6A=12.6⋅6

A≈75.6A≈75.6

The surface area of the whole cylinder:

A=75.6+12.6+12.6=100.8units2A=75.6+12.6+12.6=100.8units2

To find the volume of a cylinder we multiply the base area (which is a circle) and the height h.

V=πr2⋅hV=πr2⋅h

A pyramid consists of three or four triangular lateral surfaces and a three or four sided surface, respectively, at its base. When we calculate the surface area of the pyramid below we take the sum of the areas of the 4 triangles area and the base square. The height of a triangle within a pyramid is called the slant height.

The volume of a pyramid is one third of the volume of a prism.

V=13⋅B⋅hV=13⋅B⋅h

The base of a cone is a circle and that is easy to see. The lateral surface of a cone is a parallelogram with a base that is half the circumference of the cone and with the slant height as the height. This can be a little bit trickier to see, but if you cut the lateral surface of the cone into sections and lay them next to each other it's easily seen.

The surface area of a cone is thus the sum of the areas of the base and the lateral surface:

Abase=πr2andALS=πrlAbase=πr2andALS=πrl

A=πr2+πrlA=πr2+πrl

Example

Abase=πr2Abase=π⋅32Abase≈28.3andALS=πrlALS=π⋅3⋅9ALS≈84.8Abase=πr2andALS=πrlAbase=π⋅32ALS=π⋅3⋅9Abase≈28.3ALS≈84.8

A=πr2+πrl=28.3+84.8=113.1units2A=πr2+πrl=28.3+84.8=113.1units2

The volume of a cone is one third of the volume of a cylinder.

V=13π⋅r2⋅hV=13π⋅r2⋅h

Example

Find the volume of a prism that has the base 5 and the height 3.

B=3⋅5=15B=3⋅5=15

V=15⋅3=45units3

CONGRUENCE, SYMMETRY AND OTHER PROPERTIES OF GIVEN SHAPES.

Congruenttriangles have both the same shape and the same size. In the figure below, trianglesABC andDEF are congruent; they have the same angle measures and the same side lengths.

Similartriangles have the same shape, but not necessarily the same size. In the figure below, trianglesABCandXYZare similar, but not congruent; they have the same angle measures, but not the same side lengths.

Note:If two objects are congruent, then they are also similar.

What skills are tested?

·Determining whether two triangles are congruent

·Determining whether two triangles are similar

·Using similarity to find a missing side length

WHAT ARE THE TRIANGLE CONGRUENCE CRITERIA?

Two triangles are congruent if they meet one of the following criteria.

SSS

: All three pairs of corresponding sides are equal.

SAS

: Two pairs of corresponding sides and the corresponding angles between them are equal.

ASA

: Two pairs of corresponding angles and the corresponding sides between them are equal.

AAS

: Two pairs of corresponding angles and one pair of corresponding sides (not between the angles) are equal.

HL

: The pair ofhypotenusesand another pair of corresponding sides are equal in two right triangles.

WHAT ARE THE TRIANGLE SIMILARITY CRITERIA?

Two triangles are similar if they meet one of the following criteria.

AA

: Two pairs of corresponding angles are equal.

SSS

: Three pairs of corresponding sides are proportional.

SAS

: Two pairs of corresponding sides are proportional and the corresponding angles between them are equal.

FINDING MISSING SIDE LENGTHS IN SIMILAR TRIANGLES

The SSS similarity criterion allows us to calculate missing side lengths in similar triangles. For similar trianglesABCABCA, B, CandXYZXYZX, Y, Zshown below:

TO CALCULATE A MISSING SIDE LENGTH, WE:

1.Write a proportional relationship using two pairs of corresponding sides.

2.Plug in known side lengths. We need to know333of the444side lengths to solve for the missing side length.

3.Solve for the missing side length.

THINGS TO REMEMBER

Congruent triangles have the same corresponding angle measures and side lengths. The triangle congruence criteria are:

·SSS (Side-Side-Side)

·SAS (Side-Angle-Side)

·ASA (Angle-Side-Angle)

·AAS (Angle-Angle-Side)

·HL (Hypotenuse-Leg, right triangle only)

Similar triangles have the same corresponding angle measures and proportional side lengths. The triangle similarity criteria are:

·AA (Angle-Angle)

·SSS (Side-Side-Side)

·SAS (Side-Angle-Side)

If trianglesABCandXYZ are similar, then their corresponding side lengths have the same ratio:

UNIT 2

CONSTRUCTION

CONSTRUCTIONS

Here you will learn to create formal geometric constructions by hand and with dynamic geometry software. You will learn to both copy and bisect segments and angles. You will use your knowledge of angles to construct parallel and perpendicular lines. You will use these basic constructions together with your knowledge of the properties of regular polygons to construct regular polygons.

You learned that a drawing is a rough sketch while a construction is a step-by-step process for creating an accurate geometric figure. A compass and straightedge or folded paper can help when performing constructions by hand.

You mastered copying and bisecting line segments and angles. You used your knowledge of corresponding angles to copy an angle in order to create parallel lines. You learned how to construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle by hand.

Finally, you learned how to use dynamic geometry software such asGeogebrato create a specific polygon using the properties of that polygon.

CONSTRUCTING PERPENDICULAR PARALLEL LINES

CONSTRUCTING PERPENDICULAR LINES

A perpendicular line from a given point

Read through the following steps.

Step 1

Place your compass on the given point (point P). Draw an arc across the line on each side of the given point. Do not adjust the compass width when drawing the second arc.

Step 2

From each arc on the line, draw another arc on the opposite side of the line from the given point (P). The two new arcs will intersect.

Step 3

Use your ruler to join the given point (P) to the point where the arcs intersect (Q).

PQ is perpendicular to AB. We also write it like this: PQ ⊥ AB.

Use your compass and ruler to draw a perpendicular line from each given point to the line segment:

A perpendicular line at a given point on a line

Read through the following steps.

Step 1

Place your compass on the given point (P). Draw an arc across the line on each side of the given point. Do not adjust the compass width when drawing the second arc.

Step 2

Open your compass so that it is widerthan the distance from one of the arcs tothe point P. Place the compass on each arc and draw an arc above or below the point P. The two new arcs will intersect.

Step 3

Use your ruler to join the given point (P) and the point where the arcs intersect (Q).

PQ ⊥ AB

Use your compass and ruler to draw a perpendicular at the given point on each line

COPYING AND BISECTING ANGLES AND LINES

BISECTING LINES

When we construct, or draw, geometric figures, we often need to bisect lines or angles.Bisectmeans to cut something into two equal parts. There are different ways to bisect a line segment.

BISECTING A LINE SEGMENT WITH A RULER

Read through the following steps.

Step 1:Draw line segment AB and determine its midpoint.

Step 2:Draw any line segment through the midpoint.

The small marks on AF and FB show that AF and FB are equal.

CD is called abisectorbecause it bisects AB. AF = FB.

Use a ruler to draw and bisect the following line segments: AB = 6 cm and XY = 7 cm.

In Grade 6, you learnt how to use a compass to draw circles, and parts of circles called arcs. We can use arcs to bisect a line segment.

BISECTING A LINE SEGMENT WITH A COMPASS AND RULER

Read through the following steps.

Step 1

Place the compass on one endpoint of the line segment (point A). Draw an arc above and below the line. (Notice that all the points on the arc aboveand below the line are the same distance from point A.)

Step 2

Without changing the compass width, place the compass on point B. Draw an arc above and below the line so that the arcs cross the first two. (The two points where the arcs cross are the same distance away from point A and from point B.)

Step 3

Use a ruler to join the points where the arcsintersect. This line segment (CD) is the bisector of AB.

Intersectmeans to cross or meet.

Aperpendicularis a line that meets another line at an angle of 90°.

Notice that CD is alsoperpendicularto AB. So it is also called aperpendicular bisector.

Work in your exercise book. Use a compass and a ruler to practise drawing perpendicular bisectors on line segments.

Try this!

Work in your exercise book. Useonly a protractor and rulerto draw a perpendicular bisector on a line segment. (Remember that we use aprotractor to measure angles.)

BISECTING ANGLES

Read through the following steps.

Step 1

Place the compass on the vertex of the angle (point B). Draw an arc across each arm of the angle.

Step 2

Place the compass on the point where one arc crosses an arm and draw an arc inside the angle. Without changing the compass width, repeat for the other arm so that the two arcs cross.

Step 3

Use a ruler to join the vertex to the point where the arcs intersect (D).

DB is the bisector of \(\hat{ABC}\).

Use your compass and ruler to bisect the angles below.

You could measure each of the angles with a protractor to check if you have bisected the given angle correctly.

COPYING LINES AND ANGLES

1.Compass:A device that allows you to create a circle with a given radius. Not only can compasses help you to create circles, but also they can help you to copy distances.

2.Straightedge:Anything that allows you to produce a straight line. A straightedge should not be able to measure distances. An index card works well as a straightedge. You can also use a ruler as a straightedge, as long as you only use it to draw straight lines and not to measure.

3.Paper:When a geometric figure is on a piece of paper, the paper itself can be folded in order to construct new lines.

Let's take a look at some example problems.

1. Use a straightedge to draw a line segment on your paper like the one shown below. Then, use your straightedge and compass to copy the line segment exactly.

[Figure 2]

First use your straightedge and pencil to create a new ray.

[Figure 3]

Now, you have one endpoint of your line segment. Your job is to figure out where the other endpoint should go on the ray. Use your compass to measure the width of the original line segment.

Now, move the compass so that the tip is on the endpoint of the ray.

[Figure 5]

You can now see where the endpoint of the segment should lie on the ray. Draw a little arc with the compass to mark where the endpoint should go.

[Figure 6]

You can use your straightedge to draw the copied line segment in a different color if you wish.

[Figure 7]

Note that in this construction, the compass was used to copy a distance. This is one of the primary uses of a compass in constructions.

2. Use a straightedge to draw an angle on your paper like the one shown below. Then, use your straightedge and compass to copy the angle exactly.

[Figure 8]

Keep in mind that what defines the angle is the opening between the two rays. The lengths of the rays are not relevant.

Start by using your straightedge and pencil to draw a new ray. This will be the bottom of the two rays used to create the angle.

[Figure 9]

Next, use your compass to make an arc through the original angle. It does not matter how wide you open your compass for this.

[Figure 10]

Next, leave your compass open to the same width, and make a similar arc through the new ray.

[Figure 11]

3. Now, you know that the second ray necessary to create the new angle will go somewhere through that arc. Measure the width of the arc on the original angle using the compass.

[Figure 12]

Leave the compass open to the same width, and move it to the new angle.

[Figure 13]

Make a mark to show where the pencil on the compass intersects the arc.

[Figure 14]

4. Use a straightedge to draw another ray that passes through the point of intersection of the two compass markings.

[Figure 15]

You have now copied the angle exactly.

[Figure 16]

5. An angle is created from two line segments. Use a straightedge to draw a similar figure on your paper. Then, use the straightedge and compass to copy the figure exactly.

[Figure 17]

To copy this figure, you will need to copy both the line segments and the angle.

Start by copying the line segment on the bottom using the process outlined in #1 (draw a ray, use the compass to measure the width of the line segment, mark off the endpoint on the ray).

[Figure 18]

Next, copy the angle using the process outlined in #2 (draw an arc through the angle and draw the same arc through the new ray, measure the width of the arc, draw a new ray through the intersection of the two markings).

[Figure 19]

Finally, copy the second line segment by measuring its length using the compass and marking off the correct spot for the endpoint.

[Figure 20]

You can now draw the copied segments in a different color for emphasis.

[Figure 21]

Examples

Example 1

Earlier, you were asked to describe two ways how to use a compass and a straightedge to copy a triangle.

To copy a triangle means to create a congruent triangle. There are four triangle congruence criteria that work for any type of triangle:SSS,SAS, AAS, ASA. You can use SSS, SAS, or ASA combined with copying angles and line segments to copy a triangle.

1.SSS: Copy one-line segment. Copy the other two line segments so that their endpoints intersect.This construction will be explored in Guided Practice #1.

2.SAS: Copy one line segment. Copy an angle from one of the endpoints of the line segment. Copy a second line segment onto the ray created by the copied angle. Connect the endpoints to form the triangle.This is very similar to Example C. This construction is explored in Guided Practice #2.

3.ASA: Copy one line segment. Copy two angles, one from each endpoint. The intersection of the angles will produce the third vertex of the triangle.This construction is explored in Guided Practice #3.

You drew a triangle similar to the one below for the concept problem.

[Figure 22]

Example 2

Copy your triangle using SSS.

Start by copying one line segment. Here, the base line segment is copied.

[Figure 23]

Next, use the compass to measure the length of one of the other sides of the triangle.

[Figure 24]

Move the compass to the location of the new triangle and make an arc to mark the length of the second side of the triangle from the correct endpoint.

[Figure 25]

Repeat with the third side of the triangle.

[Figure 26]

The point where the arcs intersect is the third vertex of the triangle. Connect to form the triangle.

[Figure 27]

Note that with this method, you have only used the lengths of the sides of the triangle (as opposed to any angles) to construct the new triangle.

Example 3

Copy your triangle using SAS.

Start by copying one line segment. Here, the base of the triangle is copied.

[Figure 28]

Next, copy the angle at one of the endpoints of the line segment.

[Figure 29]

Copy the second side of the triangle (that creates the angle you copied) onto the ray that you just drew.

[Figure 30]

Connect to form the triangle.

[Figure 31]

Example 4

Copy your triangle using ASA.

Start by copying one line segment.

[Figure 32]

Next, copy the angle at one of the endpoints.

[Figure 33]

Copy the angle at the other endpoint of the line segment.

[Figure 34]

Connect to form the triangle.

[Figure 35]

Review

1.What is the difference between a drawing and a construction?

2.What is the difference between a straightedge and a ruler?

3.Describe the steps for copying a line segment.

4.Describe the steps for copying an angle.

5.When copying an angle, do the lengths of the lines matter? Explain.

6.Explain the connections between copying a triangle and the triangle congruence criteria.

7.Draw a line segment and copy it with a compass and straightedge.

8.Draw another line segment and copy it with a compass and straightedge.

9.Draw an angle and copy it with a compass and straightedge.

10.Draw another angle and copy it with a compass and straightedge.

11.Use your straightedge to draw a triangle. Copy the triangle usingSSS≅. Describe your steps.

12.Copy the triangle from #11 usingSAS≅. Describe your steps.

13.Copy the triangle from #11 usingASA≅. Describe your steps.

14.Can you copy the triangle from #11 usingAAS≅? Explain.

15.Compare the methods for copying the triangle. Is one method easier than the others? Explain.

Use your compass to construct a circle like the one shown below on a piece of paper. Describe how to fold the paper two times in order to help you construct a square.

[Figure 1]

CONSTRUCTING ANGLES

In construction of angles by using compass we will learn how to construct different angles with the help of ruler and compass.

1. Construction of an Angle of 60° by using Compass

Step of Construction:

(i)Draw a ray OA.

(ii)With O as centre and any suitable radius draw an arc above OA cutting it at a point B.

(iii)With B as centre and the same radius as before, draw another arc to cut the previous arc at C.
(iv)Join OC and produce it to D.

Then∠AOD = 60°.

2. Construction of an Angle of 120° by using Compass

Step of Construction:

(i)Draw a ray OA.

(ii)With O as centre and any suitable radius draw an arc cutting OA at B.

(iii)With B as centre and the same radius cut the arc at C, then with C as centre and same radius cut the arc at D. Join OD and produce it to E.

Then,∠AOE = 120°.

3. Construction of an Angle of 30° by using Compass

Step of Construction:

(i)Construction an angle ∠AOD = 60° as shown.

(ii)Draw the bisector OE of ∠AOD.

Then,∠AOD = 30°.

4. Construction of an Angle of 90° by using Compass

Step of Construction:

(i)Take any ray OA.

(ii)With O as centre and any convenient radius, draw an arc cutting OA at B.

(iii)With B as centre and the same radius, draw an cutting the first arc at C.

(iv)With C as centre and the same radius, cut off an arc cutting again the first arc at D.

(v)With C and D as centre and radius of more than half of CD, draw two arcs cutting each other at E, join OE.

Then,∠EOA = 90°.

5. Construction of an Angle of 75° by using Compass

Step of Construction:

(i)Take a ray OA.

(ii)With O as centre and any convenient radius, draw an arc cutting OA at C.

(iii)With C as centre and the same radius, draw an cutting the first arc at M.

(iv)With M as centre and the same radius, cut off an arc cutting again the first arc at L.

(v)With L and M as centre and radius of more than half of LM, draw two arcs cutting each other at B, join OB which is making 90°.

(vi)Now with N and M as centres again draw two arcs cutting each other at P.

(vii)Join OP.

Then,∠POA = 75°.

6. Construction of an Angle of 105° by using Compass

Step of Construction:

(i)After making 90° angle take L and N as centre and draw two arcs cutting each other at S.

(ii)Join SO.

Then,∠SOA = 105°.

7. Construction of an Angle of 135° by using Compass

Step of Construction:

(i)Construct ∠AOD = 90°

(ii)Produce ∠AO to B.

(iii)Draw OE to bisect ∠DOB.

∠DOE = 45°

∠EOA = 45° + 90° = 135°

Then,∠EOA = 135°.

8. Construction of an Angle of 150° by using Compass

Step of Construction:

(i)Construct ∠AOC = 120°

(ii)Produce ∠AO to B.

(iii)Draw OD to bisect ∠COB.

Now ∠COD = 30°

Therefore, ∠AOD = 120° + 30° = 150°

Then,∠AOD = 150°.

CONSTRUCTING TRIANGLES WITH GIVEN ANGLES AND LINE SEGMENTS;

Constructing triangles when three sides are given

1.Read through the following steps. They describe how to construct \( \triangle ABC\) with side lengths of 3 cm, 5 cm and 7 cm.

Step 1

Draw one side of the triangle using a ruler. It is often easier to start with the longest side.

Step 2

Set the compass width to 5 cm. Draw an arc 5 cm away from point A. The third vertex of the triangle will be somewhere along this arc.

Step 3

Set the compass width to 3 cm. Draw an arc from point B. Note where this arc crosses the first arc. This will be the third vertex of the triangle.

Step 4

Use your ruler to join points A and B to the point where the arcs intersect (C).

2.Work in your exercise book. Follow the steps above to construct the following triangles:

1.\( \triangle ABC\) with sides 6 cm, 7 cm and 4 cm

2.\(\triangle KLM\) with sides 10 cm, 5 cm and 8 cm

3.\(\triangle PQR\) with sides 5 cm, 9 cm and 11 cm

Constructing triangles when certain angles and sides are given

Use the rough sketches in (a) to (c) below to construct accurate triangles, using a ruler, compass and protractor. Do the construction next to each rough sketch.

The dotted lines show where you have to use a compass to measure the length of a side.

Use a protractor to measure the size of the given angles.

Construct \( \triangle ABC\), withtwo angles andone side given.

Construct a \(\triangle KLM\), withtwo sides andan angle given.

Construct right-angled \(\triangle PQR\), with thehypotenuse and one other side given.

Measure the missing angles and sides of each triangle in 3(a) to (c) on the previous page. Write the measurements at your completed constructions.

Compare each of your constructed triangles in 3(a) to (c) with a classmate's triangles. Are the triangles exactly the same?

CONSTRUCTING POLYGONS

REGULAR POLYGONS

Aregular polygonis a polygon that isequiangularandequilateral. This means that all its angles are the same measure and all its sides are the same length.

Constructing regular polygons

We know thataregular polygonis apolygonthathas all sides of equal length and all interior angles of equal measure. In this lesson we’ll learn how to construct them using compass and a ruler.

Square

Example.Construct a square if we know the sidea.

First, we make a sketch to know the disposition of the points and sides.

We will start by creating a line and a pointA. Then we take the length of the sideain the width of the compass and make an arc that intersects with the line we drew first. The intersection is pointB, our second vertex. Now we need to construct a line that’s perpendicular to theAB, with theBbeing the point of intersection. We do this because we want to create the angle∡ABC=90∘.

Now we have a perpenticular line, and we know the vertexCwill be on it. Take the length of the sideain the width of the compass and make an arc that intersects with the perpendicular line – the intersection is our vertexC. What’s left is constructing the vertexD. We do that by creating two arcs of the circlesc(A,a)andc(C,a). Their intersection is the final vertex, vertexD.

*We only constructed one square here, but we could have constructed four of them by following the same process we did in construction of equilateral triangle.

Example.How to construct a square if we know the radius of it’s circumcircle?

First we draw a pointOand a circlec(O,r). Pick a starting pointAanywhere on the circle. Now let’s draw a diameter fromAthroughO. Let the point of intersection of a diameter and a circle be the pointC. The lineABis one diagonal of the square we want to construct. How to get the other one? We know that diagonals in square are perpendicular, so we create a perpendicular line to the diameterAC. Make sure that the point of intersection is the pointO. Now, the points of intersection of perpendicular and a circlec(O,r)are pointsBandC, our two last vertices.

Regular pentagon

First we draw a sketch by hand. It doesn’t have to be perfect since it’s not our final construction, we’ll just use it for planing.

Example.Construct a regular pentagon if we know the sidea.

Make a ray withBbeing its endpoint and then construct pointAso that|AB|=a. We want to create a bisector of|AB|. Take the compass and make sure the width of compass is the length of sidea(IMPORTANT!). Put the needle onBand make two arcs of the circlec(B,a). Repeat the step for arcs ofc(A,a). The arcs intersect in pointsKandL. Join them to get the midpoint betweenAandB, pointM. Again, keep the compass radius the length ofa, put the compass’ needle onMand make an arc that intersects with bisector line, making the pointN. Now, adjust the compass to the length ofAN. Put the needle inAand make an arc that intersects with the ray we made at the beginning, that will give us a pointP.

The distance fromMtoPis very important distance – it will give us the rest of the vertices. Make the compass’ radius equal to the distance betweenMandP. Put the needle onBand make an arc that intersects with one of the arcs we made to get midpoint. Make the second arc that intersects with the bisector line. The intersects will be pointsEandDrespectively. To get the vertexCwe will put the needle inAand repeat the process.

Example.Construct a regular pentagon if we know the radius of the circ*mscribed circle.

Construct the circlec(O,r)and two perpendicular diameters,AA′―andPP′―. Now lets construct the bisector of segmentOP―, the intersection will be pointM. In the width of compass take the length betweenAandM, and put a compass needle onMto create an arc that intersects withPP′―. The intersection isN.

The distance fromAtoNis the length of the sideaof regular pentagon. Now that we know the length ofa, we need to construct vertices. LetDbe our first vertex.

Firstly, we open the compass to the length ofaand put the needle of compass onA. Now make an arc that intersects with the circlec(O,r|)giving us the vertexB. Without changing the width of compass, we put the needle onBand do the same process to get vertexA, and so on. This process will give us the remaining4vertices.

Regular hexagon

Example.Construct a regular hexagon if we know sidea.

We can break regular hexagon into6equilateral triangles with the sidea. The vertexOis the center of inscribed and circ*mscribed circles, and|AO|=|BO|=|CO|=|DO|=|EO|=|FO|. First we construct△ABOfollowing the proces we used in constructing equilateral triangle. Let’s drawc(O,|AO|). SinceOis the center of circ*mscribed circle, we know that hexagon’s vertices will be on the circle. Now we just take the length ofainto the width of a compass and make4arcs on the circle.

Without changing the width of compass we put the needle of compass on theB, make and arc that intersects with the circlec(O,|AO|)giving us the vertexC. Then we put the needle onCand do the same process to get vertexD, and so on. This process will give us the last4vertices. It doesn’t matter if we start fromA, and do it clockwise or fromBlike we did here, the result will be the same.

Important to remember:In regular hexagon△ABOis equilateral triangle, which means that the length of a radius of circ*mscribed circle and the length of a side are always equal.

Example.Construct a regular hexagon if we know the radius of circ*mscribed circle.

This is even easier. We just draw the circlec(O,r)and pick a starting point on it, let it be pointA. In regular hexagon, we know that the radius of circ*mscribed circle is equal to side of polygon, meaningr=|AO|=|AB|. Now that we have the side of a regular hexagon, we construct the remaining5vertices like we did in the last example.

Regular octagon

Example.How to construct a regular octagon if we know radius of circ*mscribed circle?

First we construct a square within the given circle with its diagonals following the process described above. Then we constructangle bisectorsof∡AOB,∡BOC,∡CODand∡DOA. Bisectors instersect with the circ*mscribed circles giving us4new points,E,F,GandH. Those points are the remaining vertices of an octagon.

Regular decagon

Example.How to construct a regular decagon if we know radius of circ*mscribed circle?

First we construct a regular pentagon within the given circle following the process described in construction of pentagon. Then we connect each vertex to the center of circ*mscribed circle to divide pentagon into5congruent triangles. The next step is constructing construct angle bisectors of∡AOB,∡BOC,∡COD,∡DOEand∡EOA.

Bisectors intersect with the circ*mscribed circles giving us5new points,F,G,H,IandJ. Those points are the remaining vertices of an decagon.

Regular dodecagon

Example.How to construct a regular dodecagon if we know radius of circ*mscribed circle?

Just like we did in the last two examples, we can construct regular dodecagon out of a regular hexagon. Angle bisectors of central angles of a hexagon give us remaining vertices of dodecagon.

UNIT 3

ANGLES AND POLYGONS

ANGLES AT A POINT

Anglesaround a point add up to 360°. This fact can be used to calculate missing angles.

HOW TO FIND MISSING ANGLES AROUND A POINT

I know that there are 360˚in a full turn.

I can see that there is already an angle of 128˚and 141˚.

128 + 141 = 269

To find the missing angle I need to find the difference between 269 and 360.

360 – 269 = 91

Angle x = 91˚

WHY DON’T YOU TRY?

Remember that all of the angles should add up to 360˚.

Add together the angles you already have.

Then find the difference between 360.

WHY NOT HAVE A GO AT SOME OF THESE REASONING AND PROBLEM SOLVING QUESTIONS

SUM OF INTERIOR ANGLES IN A POLYGON

IS IT A POLYGON?

Polygons are 2-dimensional shapes. They are made of straight lines, and the shape is "closed" (all the lines connect up).

Polygoncomes from Greek.Poly-means "many" and-gonmeans "angle".

Types of Polygons

Regular or Irregular

Aregularpolygon has all angles equal and all sides equal, otherwise it isirregular

Concave or Convex

Aconvexpolygon has no angles pointing inwards. More precisely, no internal angle can be more than 180°.

If any internal angle is greater than 180° then the polygon isconcave. (Think: concave has a "cave" in it)

Simple or Complex

Asimplepolygon has only one boundary, and it doesn't cross over itself. Acomplexpolygon intersects itself! Many rules about polygons don't work when it is complex.

More Examples

Names of Polygons

INTERIOR ANGLE FORMULA (DEFINITION, EXAMPLES, SUM OF INTERIOR ANGLES)

If you take a look at other geometry lessons on this helpful site, you will see that we have been careful to mention interior angles, not just angles, when discussing polygons. Every polygon has interior angles and exterior angles, but the interior angles are where all the interesting action is.

INTERIOR ANGLE FORMULA