Measurement

Vectors and trigonometry are interconnected concepts that play a crucial role in mathematics, physics, engineering and various other fields. Vectors represent quantities that have both magnitude and direction and are essential for describing motion, forces and other physical quantities. Understanding vectors involves applying trigonometric ratios and functions, which relate the angles and sides of right-angle triangles. Trigonometry helps us calculate unknown angles and side lengths, serving as a tool for solving real-world problems involving distances, velocities, and forces. Learning about vectors and trigonometry enhances your spatial reasoning skills and prepares you for advanced studies in mathematics.

These concepts are also closely linked to other subjects such as geography, where bearings are used to navigate and physics, where vectors are used to describe motion. Mastery of these concepts opens doors to a wide range of career paths in science, engineering, and technology.

At the end of this section, you will be able to:

• Recognise a vector as a quantity with both magnitude and direction, and identify, gather, and interpret information about real-world applications of vectors.

• Represent a vector in two-space geometrically as a directed line segment, with directions expressed in different ways (e.g., 320°, N40°W) and algebraically; then recognise vectors with the same magnitude and direction but different positions as equal vectors.

• Investigate the three basic trigonometric ratios (tangent, sine and cosine) of an acute angle in degrees.

• Find the trigonometric functions of special angles 30⁰, 45⁰and 60⁰, including using the calculator to determine the values of sine, cosine and tangent of angles up to 360⁰.

• Solve problems, using the three primary trigonometric ratios for angles from 0° to 360° in standard position.

Key Ideas

• Vector: It is a quantity that has both magnitude (length) and direction.

• Magnitude: In vectors, magnitude refers to its length or size.

• Direction: The direction of a vector refers to the line along which it points.

• Trigonometry is the study of the relation between the sides and angles of triangles, particularly right triangles.

• The three basic trigonometric ratios are: Sine (sin), Cosine (cos) and tangent (tan).

• Trigonometric functions of special angles are: 30⁰, 45⁰and 90⁰.

In Junior High School, you were taught bearings and vectors. Now is the time to remind yourselves of true bearings, forward bearings and back-bearings and bearings with vectors. Have a look back at what you have done previously, or look in textbooks or go online to find out all.

Vectors define the movement of objects from one point to another. Vectors carry a point A to point B. The length of the line between the two points A and B is called the magnitude of the vector and the direction of the displacement of point A to point B is called the direction of the vector AB.

Figure 1: Representation of a vector Real life Applications of Vectors Vectors play an important role in physics. For instance, velocity, displacement, acceleration and forces are all vector quantities that have a magnitude as well as a direction.

Real-life uses of vectors

• Vectors can be used in finding the direction in which the force is applied to move an object.

• The concept of vectors aids in understanding how gravity uses a force of attraction on an object to work.

• Vectors can be used in obtaining the motion of a body which is confined to a plane.

• Vectors help in defining the force applied on a body simultaneously in the three dimensions.

• In the field of Engineering, for a structure not to collapse, vectors are used where the force is much stronger than the structure will sustain.

• Vectors are used in various oscillators.

Types of vectors Zero Vectors Unit Vectors Position Vectors Equal Vectors Negative Vector Parallel Vectors Orthogonal Vectors Co-initial Vectors Parallel and Collinear Vectors Parallel vectors are vectors that have the same or opposite direction. Collinear vectors are vectors that lie on the same line or have the same line of action. A vector parallel to another vector is also collinear with it because both vectors share the same direction (or opposite directions if one is negative), meaning they are aligned along the same line.

Zero vector does not have a defined direction because it does not point anywhere;

it is essentially the point at the origin with no direction.

Co-initial vectors: Vectors that have the same initial point but different endpoints are called “co-initial vectors”. Co-initial vectors are useful in various contexts, such as in physics to represent different forces or motions originating from the same point.

Start Point (Initial Point): The point where the vector begins.

Endpoint (Terminal Point): The point where the vector ends.

In vector notation, v is a vector with an initial point (x₁, y₁) and a terminal point (x₂, y₂), then the start point is (x₁, y₁) and the endpoint is (x₂, y₂).

Vectors in 2-D

A 2D vector is a mathematical entity that has both a magnitude (length) and a direction in a two-dimensional space. It is represented as an ordered pair of numbers, typically (x, y).

I hope you are familiar with the standard (x, y) Cartesian coordinate system in the plane. That is, each point P in the plane is identified with its x and y components:

P (p₁, p₂).

To determine the coordinates of a vector, a, in the plane, the first step is to translate the vector so that its tail is at the origin of the coordinate system. Then, the head of the vector will be at some point (a₁, a₂) in the plane. We call (a₁, a₂) the coordinates or the components of the vector a. We often write a R²to denote that it can be described by two real coordinates.

Figure 2: Vector in 2D

Using the Pythagorean Theorem, we can obtain an expression for the magnitude of a vector in terms of its components. Given a vector a = (a₁, a₂), the vector is the hypotenuse of a right triangle whose legs are length a₁and a₂. Hence, the length of the vector a is |a| = √(a₁)²+ (a₂)²Example 1 Consider the vector a represented by the line segment which goes from the point (1, 2) to the point (4, 6). Calculate the components and the length of this vector?

Solution

To find the components, translate the line segment one unit left and two units down. Now, the line segment begins at the origin and ends at (4−1, 6−2) = (3, 4).

Therefore, a = (3, 4). The length of a is |a| = √(3)²+ (4)²= √9 + 16 = √25 = 5 units Equal Vectors Equal vectors in mathematics refer to vectors that have the same magnitude and direction. Here are some key points about equal vectors. That is, two vectors are equal if they have the same magnitude and the same direction. Equal vectors are important in various mathematical operations, such as vector addition, subtraction and comparison. When vectors have the same magnitude and direction, they exhibit similar characteristics and can be treated as equivalent in many mathematical contexts.

Magnitude: The magnitude of a vector refers to its length or size. If the magnitudes of two vectors are equal, they have the same length. The magnitude of a zero vector is 0, as it has no length.

Direction: The direction of a vector refers to the line along which it points. If two vectors have the same direction, they are parallel and point in the same line or path.

Notation: In mathematical notation, equal vectors are typically denoted by placing an arrow on top of the vector symbols. For example, if vector a and vector b are equal, it is written as →a = → b

Example 2

Consider two displacement vectors in a coordinate system. If vector A represents a displacement of 5 meters to the east, and vector B represents a displacement of 5 meters to the east, then vector A and vector B are equal. They have the same magnitude (5 meters) and the same direction (east).

Example 3

Imagine two velocity vectors of moving objects. If vector C represents a velocity of 30 kilometres per hour north, and vector D represents a velocity of 30 kilometres per hour north, then vector C and vector D are equal. They have the same magnitude (30 kilometres per hour) and the same direction (north).

Figure 3: Demonstrating magnitude and direction of a vector

Activity 1: Study the following scenario Scenario: The local government in Ghana wants to establish a new market area in a region between two existing village, Village A and Village B. They need to determine a suitable location for the new market that is equidistant from both villages and strategically positioned.

Given Information

• Village A is located at coordinates (2, 3) in kilometers.

• Village B is located at coordinates (7, 8) in kilometers.

• The new market area should be located such that it is equidistant from both Village A and Village B.

Task: Find the coordinates of the new market area that is equidistant from both Village A and Village B and the distance from each village to the market.

Step-by-Step Solution

Step One: Calculate the Midpoint

Step Two: Use the distance formula to calculate the distance between two points, (x₁, y₁) and (x₂,y₂).

Step Three: locate the new market area Step Four: Report the findings and coordinate with local authorities to finalise the location.

Example 4

A ship is sailing with a bearing of 060° at a speed of 20 km/h. What direction is it heading?

Solution

Step One: A bearing of 045° means the ship is heading 45° clockwise from True North.

Step Two: Interpret the direction:

• North is 0°

• East is 90°

• South is 180°

• West is 270° ∴ Bearing 045° means the ship is moving in a direction that is 45° clockwise from True North. This is exactly half way between North and East. So the ship is heading North-East

Activity 2

Indicate the types of vectors in the following Scenario Card Examples:

• A car parked in front of the school?

• A direction arrow on a map?

• A student walking from the classroom to the library?

• A thrown ball flying through the air?

• A speeding car accelerating down the highway?

Activity 3

Perform the following task in class.

• Take a task sheet with vectors written in component form (e.g., (2 3))

• Identify equal vectors for each given vector.

• Work in pairs to match the vectors with their equal pairs.

• Discuss and explain your reasons to your friend.

I hope you are all feeling more confident now with your use of vectors. Remember that they are simply quantities which have direction as well as magnitude and they determine the position of one point in space, relative to another.

It is now time to delve into the wonderful world of Trigonometry and all its applications.

What is Trigonometry?

Trigonometry is the study of the relation between the sides and angles of triangles, particularly right-angled triangles. It thus helps in finding the measure of unknown dimensions of a triangle using formulas and identities based on this relationship.

Figure 4: Trignometry

Trigonometric Ratios

A ratio is a statement of a mathematical relationship between two objects, often represented as a fraction. If we consider two sides of a right-angled triangle with respect to a given internal angle, the ratio of the two sides has a special relationship to the angle. Take a look at the three sides of the triangle.

In relation to the angle (θ) using ideas from the concept of Pythagoras’ theorem, we can deduce the basic ratios in trigonometry that help in establishing a relationship between the ratios of sides of a right-angled triangle with the angle.

Figure 5: Right-angled

triangle

Figure 6: Relationship between the ratios of sides of a right-angled triangle with the angle The three common trigonometric ratios we see above are Sine, Cosine and Tangent, shortened to become sin, cos and tan respectively. The trigonometric ratios enable us to determine the ratio of two sides of a right-angled triangle given an internal angle or find an angle given the ratio of two sides of a right-angled triangle. The three trigonometric ratios are defined as follows:

For example, If θ is the angle in a right-angled triangle formed between the adjacent and hypotenuse, then sinθ = Opposite_________ Hypotenuse cosθ = Adjacent_________ Hypotenuse tanθ = Opposite_______ Adjacent Let us now try some examples using these special ratios.

Example 5

James is standing 31 metres away from the base of a Harbour Centre. He looks up to the top of the building at a 78° angle. How tall is the Harbour Centre?

Figure 7: Diagram of the scenario

Solution

From the right-angled triangle XYZ, tan(θ) = Opposite_______ Adjacent From the diagram, we can substitute the given side and angle.

Thus: tan(78°) = ⌊YZ⌋____ 31 This implies, ⌊YZ⌋ = 31 × tan(78°) ⌊YZ⌋ = 31 × 4.704 ≈145.824 Therefore, the height of the harbour centre is approximately 146 m.

Example 6

Thomas is standing at the top of a building that is 45 metres high and looks at his friend who is standing on the ground, 22 metres from the base of the building.

What is the angle of depression?

Figure 8: Diagram of the scenario

Solution

From right-angled triangle XYZ, tan(θ) = Opposite_______ Adjacent From the diagram, we can substitute the given sides. Thus: tan(θ) = 45/22 This implies, tan(θ) = 2.04545454545… θ = tan⁻¹(2.045454545…) = 63.94650479… Therefore, the angle of depression from Thomas to his friend is approximately 64° Trigonometric functions of special angles 30⁰, 45⁰ and 60⁰Use a unit circle to calculate the values of basic trigonometric functions- sine, cosine and tangent.

The following diagram shows how trigonometric ratios sine and cosine can be represented in a unit circle.

Figure 9: Unit circle Derive the trigonometric ratios of 30°, 45° and 60° from the 30-60-90 and 45-45- 90 special triangles.

Figure 10: Trigonometric ratios

Activity 4

Perform the following activities by considering the triangle in the figure. The

solution is below – but have a go by yourself before looking!

Figure 11: Right-angled triangle If Cosθ = 6/10

a. Which angle is represented by θ? How do you know?

b. Find the numerical value of sin θ. Express your answer as a simplified fraction.

c. Find the numerical value of tan θ. Express your answer as a simplified fraction.

Solution

a. Since Cos θ = Adjacent_________ Hypotenuse, then the angle θ is BCA.

b. Sin θ = Opposite_________ Hypotenuse = x/10 Find the length of the adjacent side, x, by using the Pythagoras’ theorem, C²= A²+ B².

Substitute C = 10 and A = 6 into the Pythagoras’ theorem, then square root the expression to find the value of x.

Sin θ = √10²− 6²_ 10 = √100 − 36_ 10 = √64_ 10 = 8/10 Sin θ = 4/5

c. In right-angled triangles, recall that the tan of an angle is equal to the ratio of the side length opposite that angle and the adjacent length.

That is, tan θ = Opposite_______ Adjacent = x/6 We previously found that x = 8. Therefore, substitute the value into the trigonometric identity.

tan θ = 8/6 Simplify the fraction.

tan θ = 4/3 Application of the three primary trigonometric ratios to solve Real-life Problems The practical applications of these ratios are vast and diverse. Architects and engineers employ trigonometry to design structures, surveyors rely on it to measure land, astronomers utilise it to calculate distances between celestial bodies, and pilots use it for navigation purposes. Trigonometry can be used to measure the height of a building or mountains. Trigonometry truly lies at the heart of many scientific and technical fields.

Trigonometry in video games: Have you ever played the game, Mario? When you see him so smoothly glide over the road blocks, trigonometry helps Mario jump over these obstacles.

Trigonometry in flight engineering: Flight engineers have to take into account speed, distance and direction along with the speed and direction of the wind.

The wind plays an important role in how and when a plane will arrive wherever needed. This is solved using vectors to create a triangle using trigonometry to solve. For example, if a plane is travelling at 234 mph, 45 degrees N of E, and there is a wind blowing due south at 20 mph. Trigonometry will help to solve for that third side of your triangle which will lead the plane in the right direction. The plane will travel with the force of wind added on to its course.

Trigonometry in physics: In physics, trigonometry is used to find the components of vectors, model the mechanics of waves (both physical and electromagnetic) and oscillations, sum the strength of fields and use dot and cross products. Even in projectile motion you have a lot of applications of trigonometry.

Other real-life Applications of Trigonometry are;

• archaeology

• criminology

• marine biology

• marine engineering

• navigation

Activity 5:

Consider the following scenarios and answer the questions that follow. You will need a scientific calculator. Do not round your answers in the middle of the solution as this tends to lead to rounding errors. Try and keep using the full calculator display. Also, ensure that all measurements are in the same units before calculating. Write clear and concise solutions for each problem, including the sketch, steps taken and the final answer.

Scenario 1 A boy is standing 30m from the base of a tree. To see the top of the tree he must look up at an angle of 45⁰. He looks up at the tree and wonders, how tall is the tree?

Figure 12: Diagram of the scenario Step One: Identify which side forms an angle of 90⁰with the horizontal.

Step Two: Based on step one, identify the triangle shown above. A quick sketch always helps in these scenarios.

Step Three: Identify which of the three basic ratios can be used to calculate the height of the tree.

Step Four: We know the distance to the base of the tree is 30m and the angle formed is 45 degrees, so we can calculate the height of the tree.

(Hint: Height = 30m, so if you found that you are doing brilliantly!)

Scenario 2 You are standing at a point on the ground and looking up at the top of a building. The angle of elevation from where you are standing to the top of the building is 30°. You are standing 50 metres away from the base of the building.

i) Identify the right-angled triangle and do a quick sketch to represent the scenario.

ii) Which of the three basic trigonometric ratios can be used to calculate the height of the building?

iii) Use the ratio identified in (ii) to find the height of the building.

Scenario 3 A person on top of a cliff looks down at a boat on the sea. The angle of depression from the top of the cliff to the boat is 25° and the horizontal distance from the base of the cliff to the boat is 200 meters.

i) Identify the right-angled triangle and do a quick sketch to represent the scenario.

ii) Which of the three basic trigonometric ratios can be used to calculate the height of the cliff?

iii) Use the ratio identified in (ii) to find the height of the cliff.

Scenario 4 You are standing on one bank of a river and need to determine the width of the river. You know that the angle of depression from the top of a tall tree (20 meters high) to a point on the opposite bank directly across is 45°.

i) Sketch a right-angled triangle based on the scenario. Remember that a sketch always helps.

ii) Use the appropriate trigonometric ratio (sine, cosine, or tangent) based on the given information to find the width of the river.

Review Questions 6.1

1. The vector u has an initial point at (−2, 1) and an endpoint at (4, −2).

What is the vector’s length?

2. Determine if the vectors →a = (3, −4) and → b = (−6, 8) are equal.

3. Determine whether the two vectors are equal.

• Vector P: Magnitude = 5, Direction = 30 degrees above the positive x-axis.

• Vector Q: Magnitude = 5, Direction = 60 degrees above the positive x-axis.

4. Explain the following concepts as used in vectors:

i) Why might a Parallel Vector also be considered Collinear Vectors?

ii) What is the magnitude and direction of a Zero Vector?

iii) Vectors that have the same initial point are known as…?

iv) What is the start and endpoint of a vector called?

5. Find the length of the vector a = (− 3 4 ).

6. Using the magnitude formula, find the magnitude of the vector with u = (− 6 8 ).

7. Given the vectors →u = (2 5)and →v = ( 4 10), find a scalar k such that u = kv.

Review Questions 6.2

1. Given a right-angle triangle ABC with angle A as the right-angle and side lengths as follows: AB = 5 cm, BC = 13 cm. Calculate the values of the trigonometric ratios for angle B (sine, cosine and tangent). Keep your answers as fractions.

2. State the exact values of the trigonometric functions (sine, cosine and tangent) for the special angles 30 degrees, 45 degrees, and 60 degrees.

3. Label the sides of the triangle and find the hypotenuse, opposite and adjacent with regards to θ.

4. For the given triangle find the sine, cosine and tangent ratio.

5. For the given triangle find the sine, cosine and tangent ratios in relation to angle θ.

6. Sakumonor Skating Club has a beginner ski slope which makes a 15⁰angle with the ground. How far does a skier travel over a horizontal distance of 120 meters?

7. A ramp is to be built from the ground to the back of a semi-truck. The bed of the truck is 1.3 meters above the ground. The ramp makes a 22⁰angle with the ground. How long will the ramp be?

8. The angle of elevation from a boat to the top of a 48-meter lighthouse is 25⁰. How far is the boat from the base of the lighthouse?

9 A prince is standing 100 meters from the base of a tower where a beautiful princess is being held captive. The princess sees the prince at a 30⁰angle of depression. How high will the prince have to climb to rescue the princess?

Perimeter, area and volume are important concepts in geometry that help us measure and compare the sizes of 2D and 3D shapes. Understanding these ideas is useful for solving everyday problems, such as figuring out how much fencing is needed for a garden, how much paint is needed to cover a wall or how much liquid a container can hold. These concepts are connected and are the basis for learning more advanced topics in geometry and calculus. They are also important in subjects like physics and engineering, where measuring shapes and volumes is needed for designing and analysing structures At the end of this section, you will be able to:

• Solve problems that involve identifying and comparing referents for SI and imperial area measurements of regular, composite and irregular 2-D shapes including decimal and fractional measurements and verify the solutions.

• Estimate the perimeter and area of a given regular, composite or irregular 2-D shapes.

• Solve a contextual problem that involves the perimeter and area of a regular, a composite or an irregular 2-D shape.

• Solve problems that involve SI and imperial units in volume of prisms.

• Solve real world problems that involves the volume of prisms.

Key ideas

• Perimeter: refers to the total distance around the boundary of a two- dimensional shape.

• Area: refers to the amount of space enclosed within the boundaries of a two-dimensional shape.

• Volume: refers to the amount of space occupied by an object, such as a box, a swimming pool, or a water tank.

Measurement is a key part of understanding the world and referents help by providing clear reference points for accurate measurement. Referents are standards or examples that we compare other quantities to, so everyone can use the same system to measure things. They help us make fair comparisons, understand results and keep measurements consistent in different situations. We will explore the importance of referents in measurement, looking at natural, man-made and relative referents. We will also discuss how choosing or changing referents can affect how we see things. By learning about referents, we can better understand how measurement works and why it is important.

Example 1

Investigate to validate the following referents in the tables.

Table 1: Referents for Imperial Linear Measurement

Imperial Measurement

Referent Inch Thumb length, from the tip to the first knuckle, or the thickness of a hockey puck Foot Standard floor tile in a classroom Yard Arm span from tip of nose, yard stick, length of a guitar Mile Distance walked in 20 minutes

Table 2: Referents for SI Linear Measurement

SI Measurement Referent

Millimetres Thickness of a dime, or a fingernail Centimetres Width of a fingernail, black keys on a piano, crayon, paper clip, or AA battery Metre Distance from a doorknob to the floor, width of a volleyball net, metre stick, waist height Kilometre Distance walked in 15 minutes

Table 3: Referents for Area

Measurement Referent

≈1 ft²Area of a floor tile ≈1 in²Area of a postage stamp ≈1 cm²Area of a fingernail ≈ 2 m²Area of an exterior house door ≈ 93.5in²or 600cm²Area of exercise notebook ≈ 1500 m²or 17 000 ft²Area of an ice rink surface ≈ 32 ft²or 3 m²Area of a sheet of plywood Project work I would like you to do the project below by estimating the area of the referents in the table and submitting the final project to your teacher.

Table 4: Referents for Area

Referent Measurement

Area of your television set Area of the marker board in the classroom Area of the classroom Area of the school field Area of the head teacher’s office Perimeter and Area of 2-D Shapes In Primary and Junior High school, you explored how to find the area and perimeter of basic shapes like squares, rectangles, triangles, and circles. We will now build on that knowledge by learning how to calculate the area and perimeter of more complex shapes—specifically, trapeziums, kites and parallelograms.

Before we dive into the new lesson, let’s refresh our memory with a quick revision.

Please watch the video below, which covers the methods we used to calculate the area and perimeter of squares, rectangles, triangles, and circles. This will help you recall the key concepts from the previous lesson and prepare you for the new material we’ll cover this week.

A video on Area and Perimeter of Rectangle, Square, Triangle, Circle (https:// youtu.be/9KvIQD2DjVg) Perimeter and area of shapes Perimeter is the total distance around a shape. To find the perimeter, you add up the lengths of all sides. Area is the measure of the space enclosed by a 2D shape.

Area is measured in square units.

Investigating the area and perimeter of shapes using graphs/geodots We can use graph sheets to investigate the perimeter and area of shapes. Look at the pictures below. A trapezium has been drawn on a graph sheet to help determine its area.

Figure 1: A trapezium on a grid We know that area is a measure of how many square units will fit inside a shape.

So, how many squares are inside our trapezium? There are 24 full squares plus eight half squares, which means the area of the trapezium is 28 square units.

Take a look at the rhombus drawn on the graph sheet below:

Figure 2: A rhombus on a graph We can determine the perimeter of the shape by counting the number of squares covered by the line at S1 or S3. Each of these two sides cover approximately 12 squares. Now, since all the sides of a rhombus are equal, we can say that S2 and S4 are 12 squares as well. Assuming the side of a square on the graph is 1cm, then we can calculate the perimeter as S₁+ S₂+ S₃+ S₄= 12cm + 12cm + 12cm + 12cm = 48cm.

Now, let us undertake activities to help us understand how to explore the perimeter and area of kites, parallelograms, rhombuses, and trapeziums using the graph sheet or geodot.

Understanding the Shapes

Kite: A quadrilateral with two pairs of adjacent sides that are equal. The diagonals intersect at right angles.

Parallelogram: A quadrilateral with opposite sides parallel and equal in length.

Rhombus: A parallelogram where all four sides are equal and diagonals bisect each other at right angles.

Activity 1

To help you explore the perimeter and area of kites, parallelograms, rhombuses and trapeziums using a geodot or graph sheet, here is a breakdown of hands- on activities to guide you.

Materials Needed:

• Geodot or graph sheets

• Rulers

• Pens or pencils

• Formula sheets for perimeter and area Step-by-Step Process:

Kite:

1. Using the geodot or graph sheet, plot the vertices of a kite. Remember that adjacent sides are equal in length and the diagonals intersect at right angles.

2. Count the dots or use the graph scale to determine the lengths of the sides.

3. Calculate the perimeter by adding the lengths of all four sides.

4. To find the area, measure the lengths of the diagonals (from vertex to vertex) and use the formula:

Area = 1/2 × diagonal 1 × diagonal 2 Parallelogram:

1. Draw a parallelogram on the geodot or sheet. Remember that the opposite sides are parallel. (Like a rectangle that has been sat on and pushed over.)

2. Measure the lengths of the base and height (perpendicular distance between the bases).

3. Perimeter is calculated as: Perimeter = 2 × (Base + Side )

4. Area is calculated using: Area = Base × Perpendicular Height Click on the links below to watch how to calculate the area of parallelogram using graph sheet:

https://youtu.be/nZcOo0HMSBg https://youtu.be/pfamBrBf8BQ Rhombus:

1. Construct a rhombus on the graph sheet. Remember it is a shape with four equal length sides. (Like a square that has been sat on and pushed over.)

2. Measure the diagonals.

3. Calculate the perimeter by multiplying the side length by 4:

Perimeter = 4 × Side

4. For the area, use the diagonals:

Area = 1/2 × diagonal 1 × diagonal 2 Click on the link below to watch how to calculate the area of rhombus using graph sheet: https://youtu.be/EM7b-IIbtTQ Trapezium:

1. Draw a trapezium on graph paper. Remember it is a shape with only one pair of parallel sides of length

2. Measure and record the lengths of the sides and the perpendicular height between the parallel sides.

Perimeter = sum of all sides Area using the formula:

Area = (__

2) × (sum of parallel sides) × perpendicular height Click on the link below to watch how to calculate the area of Trapezium using graph sheet: https://youtu.be/NLEiOsPPmwk Calculating the perimeter and area of shapes using simple formulae Take a look at the formulae for the following shapes:

Formula Perimeter (P) = 2(a + b) here, a and b are the 2 adjacent sides In ABCD, a = AB = AD b = BC = CD

Figure 3: A kite Area of a kite = 1/2 × (d₁) × (d₂) square units Perimeter = c +c + a + b Area of an isosceles trapezoid = 1/2 (a + b) h square units

Figure 4: A trapezium Perimeter = AB + AD + BC + CD Area of a rhombus = 1/2 ×(d₁) × (d₂) square units

Figure 5: A rhombus Perimeter = 2(a + b) or (a + a + b + b) Area of parallelogram = base × perpendicular height = b × h units²Figure 6: A parallelogram For trapeziums, investigate the different types:

• Isosceles Trapezium

• Scalene Trapezium

• Right Trapezium

Example 2

The perimeter and area for the following shapes can be determined as;

Figure 7: A kite. Figure 8: A rhombus Perimeter (P) = 2(a + b), ∴ P = 2(51 + 64) = 230m Area = 1/2 × 72 × 89 = 3204 m²Perimeter = AB + BC + CD + DA = 6 + 6 + 6 + 6 = 24cm Area of one triangle = 1/2 × 3 × 5 = 7.5 cm² So, area of rhombus = 4 × 7.5 = 30 cm²

Figure 9: A parallelogram Figure 10: A trapezium Perimeter = 8 + 8 + 12 + 12 = 40 cm Area = 12 × 6 = 72 cm²Perimeter = 4.1 + 3 + 5 + 6.8 = 18.9 in Area = 1/2(3 + 6.8) × 3.8 = 18.62 in²Example 3 Find the perimeter of the shape below

Figure 11: A parallelogram

Solution:

Step 1: Identify the shape if it is not given In this case you will see that the shape is a parallelogram.

Step 2: Identify the lengths of the sides A parallelogram has two pairs of equal sides.

In this case, you are given:

o Base = 12 cm o Side = 8 cm

Step 3: Write the perimeter formula for a parallelogram The perimeter is the total distance around the parallelogram, calculated as Perimeter = 2 × (Base + Side)

Step 4: Substitute the given values into the formula P = 2 × (12 cm + 8 cm)

Step 5: Add the length and width Add the values inside the parentheses/Bracket:

P = 2 × 20 cm

Step 6: Multiply the result by 2 Finally, multiply by 2 to account for both pairs of sides: P = 40 cm The perimeter of the parallelogram is 40 cm

Example 4

Find the perimeter of the shape below

Figure 12: A trapezium

Solution:

Step 1: Identify the shape This is a trapezium, which has four sides of varying lengths.

Step 2: Write the formula for the perimeter The perimeter of a trapezium is the total distance around the shape, which is calculated by adding all four sides together:

P = side 1 + side 2 + side 3 + side 4

Step 3: Identify the side lengths From the question, the given side lengths of the trapezium are:

o Side 1 = 12 mm o Side 2 = 13 mm o Side 3 = 14 mm o Side 4 = 18 mm

Step 4: Substitute the values into the formula P = 12 mm + 13 mm + 14 mm + 18 mm

Step 5: Add the side lengths Add the values: P=12+13+14+18=57 mm The perimeter of the trapezium is 57 mm.

Example 5

Calculate the area of the parallelogram below

Figure 13: A parallelogram

Solution:

Step 1: Identify the base and perpendicular height The base of the parallelogram is given as 10 cm.

The perpendicular height is the shortest distance from the base to the opposite side, which is given as 5 cm.

Step 2: Write the formula for the area of a parallelogram The formula is: Area = base × perpendicular height

Step 3: Substitute the given values Base = 10 cm Perpendicular height = 5 cm Area=10 cm × 5 cm

Step 4: Multiply the base by the perpendicular height Area = 50cm²The area of the parallelogram is 50 cm².

Solve real-life problems on perimeter and area of 2-D shapes

Example 6

The perimeter of a parallelogram is equal to 48cm. Two of its sides are 16cm each.

How long is each of the other side?

Solution:

We are given the Perimeter (P) = 48 cm and side a = 16 cm.

To find the length of the other side b of the parallelogram, we will use the formula of the perimeter of a parallelogram P = 2(a + b).

P = 2(a + b) ⇒ 48 = 2 (16 + b) ⇒ 16 + b = 48/2 ⇒ b = 24 – 16 ⇒ b = 8 cm Answer: The other side of the parallelogram is 8 cm.

Example 7

A garden in the shape of a parallelogram has dimensions 12 metres and 9 metres and a perpendicular height of 8 metres. If the owner wants to put a fence around the garden, how much fencing material is needed? Also, what is the area of the garden?

Figure 14: A parallelogram

Solution

To find the perimeter of the parallelogram, we need to add the lengths of all four sides. Since a parallelogram has opposite sides of equal length, the perimeter is:

Perimeter = 2(12 + 9) = 2(21) = 42 metres To find the area of the parallelogram, we use the formula:

Area = base × perpendicular height = 12 × 8 = 96 square metres Therefore, the owner needs 42 m of fencing material and the area of the garden is 96 m².

Example 8

Gina is designing a rectangular banner to promote her school’s art exhibition.

She wants to include a large rhombus-shaped logo in the middle of the banner.

The diagonal dimensions of the logo rhombus are 96 cm and 60 cm. The banner dimensions are 1.5 m × 1 m.

Gina has a few design questions:

i. What is the height and the width of the rhombus logo?

ii. What is the area of the rhombus logo?

iii. Will her logo rhombus fit inside the 1.5 m × 1 m banner dimensions if she aligns the edges parallel?

iv. How much blank space will there be at the top/bottom or sides once the rhombus logo is centered aligned in the banner.

Solution

i. Height = Shorter diagonal = 60 cm Width = Longer diagonal = 96 cm

ii. Area = 1/2 × Height × Width = 1/2 × 60 × 96 = 2 880 cm²iii. Yes, aligned vertically or horizontally, the rhombus height 60 cm and width 96 cm will fit within the 150 cm × 100 cm banner with room to spare.

iv. Top/Bottom blank space = (Banner height − Rhombus height)________________________ 2 = 150 − 60/2 = 45cm Side blank space= (Banner width − Rhombus width)________________________ 2 = 100 − 96/2 = 2cm So, when centred, there would be 45 cm blank space above and below the rhombus logo in the banner and 2 cm on either side.

Example 9

Nagbija Isaac who is a builder is designing a window with a parallelogram-shaped glass pane. The base of the glass pane is 50 cm, and the perpendicular height is 40 cm. What is the area of the glass pane?

Solution:

To find the area of the parallelogram-shaped glass pane designed by Nagbija Isaac, we follow these steps:

Step 1: Write down the formula for the area of a parallelogram The formula to calculate the area of a parallelogram is:

Area = base × perpendicular height

Step 2: Identify the values for the base and height Base (b) = 50 cm Perpendicular Height (h) = 40 cm

Step 3: Substitute the values into the formula Area = 50 cm × 40 cm

Step 4: Multiply the base by the height Area = 2000 cm² The area of the parallelogram-shaped glass pane is 2 000 cm².

Example 10

A park has a rhombus-shaped flower bed with diagonals measuring 12 metres and 8 metres. What is the area of the flower bed?

Solution

To find the area of the rhombus-shaped flower bed in the park follow these steps:

Step 1: Write down the formula for the area of a rhombus The formula to calculate the area (A) of a rhombus using its diagonals is:

A = 1/2 × d₁× d₂

Step 2: Identify the values for the diagonals Diagonal 1 (d₁) = 12 metres Diagonal 2 (d₂) = 8 metres

Step 3: Substitute the values into the formula A = 1/2 × 12 m × 8 m

Step 4: Multiply the diagonals A = 1/2 × 96 m²Step 5: Divide by 2 to get the final area A = 48m² The area of the rhombus-shaped flower bed is 48 m².

Example 11

A contractor is designing a trapezium-shaped wheelchair ramp for Mr. Otonko.

The parallel sides are 2 metres and 5 metres, and the perpendicular height of the ramp is 1.5 metres. What is the area of the ramp?

Solution

To find the area of the trapezium-shaped wheelchair ramp, follow these steps:

Step 1: Write down the formula for the area of a trapezium The formula to calculate the area of a trapezium is:

Area = 1/2 × (a + b) × perpendicular h

Step 2: Identify the values for the parallel sides and height Parallel side 1 (a) = 2 metres Parallel side 2 (b) = 5 metres Perpendicular Height (h) = 1.5 metres

Step 3: Substitute the values into the formula Area = 1/2 × (2 m + 5 m) × 1.5 m

Step 4: Add the lengths of the parallel sides Area = 1/2 × 7 m × 1.5 m

Step 5: Complete the calculation to find the area Area = 5.25 m² The area of the trapezium-shaped wheelchair ramp is 5.25 m².

Example 12

You want to make a large decorative kite at a local Guinea Corn festival, celebrated by the Konkombas. The diagonals of the kite are 2.5 metres and 1.5 metres. How much material do you need to make the kite?

Solution

To determine how much material is needed to make the kite, we need to calculate the area of the kite. A kite’s area can be found using its diagonals.

Step 1: Write down the formula for the area of a kite The formula to calculate the area of a kite using its diagonals is:

A = 1/2 × d₁× d₂

Step 2: Identify the values for the diagonals Diagonal 1 (d₁) = 2.5 metres Diagonal 2 (d₂) = 1.5 metres

Step 3: Substitute the values into the formula Area = 1/2 × 2.5 m × 1.5 m

Step 4: Complete the calculation to find the area Area = 1.875 m² You will need 1.875m²of material to make the kite.

PRISMS Let us now look into the world of prisms. We need to:

i. define prisms and identify the elements: bases, height/length, lateral faces.

ii. understand the appropriate formula for calculating the volume of prisms.

iii. calculate volume by substituting dimensions into formula.

iv. Solve problems involving prism volumes contextually.

Defining and identifying the elements: bases, height/length, lateral faces of prisms Watch this video (5 minutes) by clicking on the link which will help you identify parts of prisms: https://youtu.be/SBJv3MqlGEA.

You will realise from the video that: A prism is a three-dimensional solid object having two identical and parallel shapes facing each other. Thus, a prism has a constant cross-section. The identical shapes are called the bases. The bases can have any shape of a polygon for example, triangles, squares, rectangles or a pentagon. The diagram below shows a triangular prism.

Parts A prism has bases, lateral faces, edges and vertices.

Figure 15: A prism

a. Base – The base is one of the parallel faces which makes the 2 ends of any prism. These are congruent. The base determines the cross-section of any prism and it remains uniform throughout the shape.

b. Lateral faces – The non-parallel faces which connects the 2 bases. In this

example there are 3 lateral faces.

c. Vertices – The corners of the shape. In this example there are 6 vertices.

d. Edges – Where any 2 faces meet. In this example there are 9 edges.

Table 5: Examples of other prisms include Square-faced cuboid

• 6 faces (2 squares and 4 rectangular)

• 12 edges

• 8 vertices Rectangular-faced cuboid

• 6 faces (all rectangular)

• 12 edges

• 8 vertices Triangular

• 5 faces (2 triangular and 3 rectangular)

• 9 edges

• 6 vertices

Activity 2

Identify other prisms (in your immediate environment) and indicate the type of face, number of faces, edges and vertices. Indicate the properties and nets of the prisms (learned in Junior High school). Tabulate your results on a sheet of paper and show to your teacher and classmates.

Regular and Irregular Prisms

A prism can also be classified into regular or irregular based on the uniformity of its cross-section. It can be right or oblique, depending on the alignment of its bases. The diagram shows the difference between a regular and irregular triangular prism.

Figure 16a: Regular prism Figure 16b: Irregular prism

1. Regular Prism – It has a base which is a regular polygon with equal side lengths. Regular prisms have identical bases and identical lateral faces.

2. Irregular Prism – It has a base which is an irregular polygon with unequal side lengths. Irregular prisms have identical bases. However, the lateral faces are not identical.

Right and Oblique Prisms

The diagram shows the difference between a right and an oblique pentagonal prism.

Figure 17a: Right prism Figure 17b: Oblique prism

1. Right Prism – Its lateral faces are perpendicular to its bases. The 2 bases of a right prism are aligned perfectly over one another.

2. Oblique Prism – It is a slanted prism. This means that its lateral faces are not perpendicular to its bases. The 2 bases are not aligned perfectly over one another.

In this section we will be focusing on right prisms.

In Junior High School, you learned about surface area and volume of solid shapes. Before we delve into today’s lesson, let’s refresh our minds on surface area and volume of solid shapes. Watch this video (17 minutes) by clicking on the link:

https://youtu.be/cS28z1cdAdY.

I hope you enjoyed the video. Let’s now look at the following examples.

Example 13

Calculate the surface area of a cuboid consisting of a square base of length 4m where the height of the cuboid is 10m.

Solution

Cuboid Net

Figure 18: Cuboid Figure 19: Net of a cuboid To calculate the surface area of a cuboid with a square base, you need to calculate the area of all its faces.

Given:

• The side length of the square base L = 4m

• The height of the cuboid, H = 10m The cuboid has:

• 2 square (base) faces (top and bottom) with area L²• 4 rectangular (lateral) faces, each with area L x H

Step 1: Calculate the area of the square (base) faces Area of one square face = L²= 4²= 16m²Area of two square faces = 2 × 16 = 32m²Step 2: Calculate the area of the rectangular (lateral) faces Area of one rectangular face = L × H = 4 × 10 = 40 m²Area of four rectangular faces = 4 × 40 = 160 m²Step 3: Calculate the total surface area Total surface area = Area of square faces + Area of rectangular faces Total surface area = 32 + 160 = 192m²Thus, the surface area of the cuboid is 192 m² If you are still unsure, watch this video (16 minutes) on Surface Area and volumes of solid figures by clicking on the link:

https://youtu.be/eBAq_caikJ4.

Understanding volume helps quantify three-dimensional spaces which allows for practical applications like determining storage capacities, amounts that fit inside containers, calculating dosages and more.

Volume is a measure of the amount of three-dimensional space that an object occupies.

Some key points about volume:

1. Volume measures the space taken up by a 3D object. It quantifies how much a substance or solid material would fill a three-dimensional container.

2. Volume is measured in cubic units such as cubic metres (m³), cubic centimetres (cm³) and cubic inches (in³).

3. The volume of a 3D shape depends on the dimensions of height, width and depth - unlike area which uses just height and width. The standard mathematical formula used is:

Volume = Area of the base x Height Some examples of volumes we encounter:

• Volume of liquid a bottle can contain

• Volume of concrete needed to lay a house’s foundation

• Volume of medicines and dosage instructions based on that.

Understanding and calculating volumes has many practical real-world applications in fields from shipping and storage to calculating health or building metrics to determining carrying capacity of containers.

The basic volume formulas work for standard 3D geometrical shapes. However, for irregular shapes, volumes are measured by instruments or by estimation using simpler building blocks and their volume formulae.

Understanding the appropriate volume formula Watch this video (4 minutes) by clicking on the link:

https://youtu.be/P72Jfnr66Ac and study Table 6.

Table 6: Formulae for calculating volume of various prisms Shape Base Volume of Prism = Base area × height Triangular Prism Triangular Volume of triangular prism = Area of triangle × height of the prism Square Prism Square Volume of square prism = Area of square × height of the prism Rectangular Prism Rectangular Volume of rectangular prism = Area of rectangle × height of the prism

Example 14

Find the volume of a triangular prism whose base area is 64 cm²and height is 7 cm.

Solution

As we know, Volume (V) = Base Area × Height V = B × H, here B = 64 cm², H = 7 cm = 64 × 7 = 448 cm³

Example 15

Find the volume of the prism shown below.

Figure 20: Prism

Solution

V = B × H [Volume of a prism formula] For a rectanglular prism we know that the area of the base = Length × Width V = L × W × H [Area of rectangle formula × Height] V = 6 × 8 × 15 [Substitute values of L, W and H] V = 720 The volume of the prism is 720 cubic yards.

Example 16

A box of popcorn holds 7000 cubic centimetres of popcorn. The length and width of the base of the box are 14cm and 20cm, respectively. Find the height of this box of popcorn.

Figure 21: Popcorn box

Solution

V = B × H [Formula for the volume of a prism] 7000 = L × W × H [Formula for the area of a rectangle; replace V with 7000] 7000 = 14 × 20 × H [Substitute values of L and W] 7000 = 280 × H [Solve for H] 25 = H [Divide each side by 280] The height of the popcorn box is 25cm.

Example 17

What is the base area of the prism if the volume of the prism is 324 cubic units and the height of the prism is 9 units?

Solution

The given dimensions are the volume of the prism = 324 cubic units and the height of the prism = 9 units. Let the base area of the prism be “B”.

Substituting the values in the volume of the prism formula:

Volume of prism = V = B × H = 324 cubic units ⇒ 9B = 324 ⇒ B = 36 square units Therefore, the base area of the prism is 36 square units.

Activity 3

Look for solid objects in your immediate environment (for example, a Marker box, Money box) that are prisms and find the volume of each using the appropriate materials.

Record your results on a sheet of paper and show to your teacher or classmates.

Activity 4

Think about a cylinder. Is this a prism?

Can you determine the volume and surface area of a cylinder whose radius is 3m and height 6m?

Hint: i) Multiply the area of the base by the height to get the volume. ii) Use the net of the cylinder to determine the surface area.

Figure 22: Net of a cylinder

1. Calculate the area of the following shapes

2. Find the area for the shapes below.

3. The perimeter of the given trapezium is 104 cm. Find its area.

4. Solve the following problems i.

Miss. Tackie wants to paint the side of her house.

To buy paint, she must know the area.

What is the area of the side of Miss. Tackie’s house?

ii.

Charles needs paper to cover his kite.

Find the area of the kite

5. A field in the shape of a rhombus has an area of 60 square metres. If one of its diagonals is 12 metres long, find the length of the other diagonal.

6. Jamila is building a deck outside her house with a rectangular area that has outer dimensions of 8m by 6m. She wants to have decorative floor tile patterns in a parallelogram shape inside the deck. Each parallelogram tile has one side measuring 1.2m and a perpendicular height of 0.8m against that side.

(a) What is the area of each parallelogram tile?

(b) If Jamila covers the entire inner deck area with these tiles, how many full tiles would she need?

7. A designer wants to use parallelogram-shaped tiles to cover a rectangular floor that is 6 metres long and 4 metres wide. If each tile has a base of 0.5 metres and a height of 0.3 metres, how many tiles will be needed to cover the entire floor?

8. An architect is designing a parallelogram-shaped roof for a house. The sides of the roof are 10 metres and 6 metres. If the perpendicular height of the roof is 4 metres, what is the area of the roof?

9. A construction company is laying out a rhombus-shaped foundation for a building. If one of the diagonals is 30 metres and the other diagonal is 24 metres, how much land will the foundation cover?

10. A farmer has a trapezium-shaped plot of land. The parallel sides measure 40 metres and 70 metres, and the shortest distance between the parallel sides is 30 metres. What is the area of the land?

11. A farmer is designing a kite-shaped plot in their field, with diagonals measuring 10 metres and 6 metres. What is the area of the kite-shaped plot?

12. Find the height of the prism if the volume of the prism is 729 cubic units and the base area is 27 square units.

13. What is the base area of a prism if the volume of the prism is 300 cubic feet and the height of the prism is 6 feet?

14. The base area of a prism is 123 square yards. The height is 9 yards. Find the volume.

15. What is the volume of a triangular prism with dimensions of 12 m, 16 m and 20 m as shown in the figure below?

16. A swimming pool in the shape of a cuboid has a length of 10 metres, a width of 5 metres, and a depth of 2 metres. If the pool is filled with water, how much water does it hold?

17. The height of a square prism is 11 yards. The volume is 99 cubic yards.

What is the length of the side of the square at the base of the prism?

18. A building is in the shape of a triangular prism. The height of the building is 190 metres. The base of the building is in the form of an equilateral triangle whose side length is 45 metres. Find the lateral surface area of the building.

19. Amina’s doll’s tent is in the shape of a triangular prism. How many cubic centimetres of space is in the tent?

GLOSSARY

• Perimeter: The total distance around the boundary of a two-dimensional shape.

• Area: The measure of the amount of space enclosed within the boundaries of a two-dimensional shape.

• Volume: The amount of space occupied by a three-dimensional object or shape.

• Circumference: The perimeter of a circle, calculated using the formula C=2πr where r is the radius of the circle.

• Cuboid: A three-dimensional shape with six rectangular faces.

• Cube: A special type of cuboid where all sides are equal in length, so all the faces are identical squares.

• Parallelogram: A four-sided polygon with the opposite sides parallel and equal in length.

• Trapezium (Trapezoid): A four-sided polygon with only one pair of parallel sides.

• Prism: A three-dimensional solid with two parallel and congruent faces called bases and all other faces are rectangles.