Area of Polygons

Section 2.5 Area of Polygons

Why polygons?

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The

area of a polygon

refers to the amount of space enclosed within its boundaries. It is measured in square units e.g., cm², m², in² etc.

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Activity 2.5.1.

Materials needed:

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Sticks (e.g. matchsticks, toothpicks, broom straws, twigs)
Thread, string, rubber bands, or yarn
Recycled cardboard, cereal boxes, or packaging paper
Bottle caps or buttons (as vertices)
Scissors
Glue or tape $(d)$
Ruler and protractor (if available)

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Have you seen shapes with straight sides around you?
What do you think makes a shape a polygon?

In pairs or small groups, create physical models of different polygons using the available materials.
Using sticks $(a)$ or straws for sides and caps $(c)$ or buttons for vertices

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Use sticks or straws for sides and caps or buttons for vertices as shown alongside.
(a) Construct at least:

(b) A triangle

(c) A quadrilateral

(d) A pentagon

(e) A hexagon
Present what type of polygon each model is and how many sides/vertices it has.
“What’s common in all these shapes?”
“What’s the minimum number of sides a polygon can have?”

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Key Takeaway

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♦ A

polygon

is a closed two dimension formed by straight lines that connect end to end at a point called a

vertex .

Examples of polygons include triangles, quadrilaterals, pentagons, hexagons, heptagons,octagons, nonagons and decagons.

Their sides do not curve or overlap.

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Note that most of the interior angles of a polygon are smaller than exterior angles.

$Types of Polygons$

We have different types of polygons but we are going to focus on only two in this chapter Regular polygons and irregular polygons.

🔹 Regular polygons – All sides and angles are equal (e.g., square, equilateral triangle).

🔹 Irregular polygons – Sides and angles are not equal (e.g., a scalene triangle, a random quadrilateral).

🔹 Convex polygons – All interior angles are less than 180°.

🔹 Concave polygons – At least one interior angle is greater than 180°.

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Why is Area Important?

🔗

Would you wish to pursue a career some day? Have you ever seen an Architect?

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So finding area is Essential in design and architecture as well as construction of houses for it helps to determine the spacing of rooms in a house in relation to the effect desired.

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Subsection 2.5.1 Area of Triangles.

Activity 2.5.2.

Building a Triangle Garden.

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Working im groups of groups of 3–5 students.

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Materials needed:

🔗

Cereal boxes or recycled cardboard
String or yarn
Pencil and Scissors
Chalk (if outdoors) or masking tape (if indoors)
Manila paper or newspaper
A few small stones or bottle caps (for weights or markers)

🔗

Preferably conuct this activity outside on sand or cement, or inside with tape/string.
- Ensure you have a rectangle (or square) drawn with chalk/tape—e.g. 60 cm × 40 cm.
- Can you use string to form a triangle that perfectly fits inside this rectangle?
- Can you make the triangle cover exactly half of the space?
- Use a string from corner to corner, or make triangles by connecting midpoints.
- Take a manila paper or cardboard. Draw a rectangle (e.g. 10 cm by 6 cm).
- Cut along a diagonal to get two right triangles. Rearrange the two triangles to form a rectangle again.
- Now try : Drawing a triangle with the same base and height, but different shape (scalene, obtuse). Cut and rearrange it against the original rectangle.
On the board or chart, formulate the rule:
- If the area of a rectangle is base × height, and the triangle is always half of it, What will be the formula of a triangle?
- Write it down together in your groups.

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Review of properties of a triangle .

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A triangle is a three-sided polygon with three angles which add up to $180^{\circ}$ and three vertices.
$Key Properties of a Triangle$

♦ A triangle has three Sides and Three Angles.

♦ A triangle has three edges and three vertices (corner points).

♦ Angle Sum Property - The sum of all three interior angles of a triangle adds up to 180°.
$Types of Triangles (by Sides).$

🔹 $Equilateral Triangle$ - All sides of the triangle are equal and have an angle of 60° each.
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🔹 $Isosceles Triangle$ - Two sides of the triangle are equal, and the angles opposite these sides are also equal.
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🔹 $Scalene Triangle$ - All three sides of the triangle have different lengths, and all angles are different.
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$Types of Triangles (by Angles).$

🔹 $Acute-Angled Triangle (T_{1})$ - All three angles are less than 90°.

🔹 $Right-Angled Triangle (T_{2})$ - One angle is exactly 90°.

🔹 $Obtuse-Angled Triangle (T_{3})$ - One angle is greater than 90° but less than $180 \circ .$

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Example 2.5.1.

Calculate the unknown variables in each of the following figures

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Solution.

(a) An isosceles triangle has two angles that are equal. Sum of interior angles of triangle is

180^{\circ}

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\begin{aligned} = & 180^{\circ} - 36^{\circ} \\ = & 144^{\circ} \\ = & \frac{144^{\circ}}{2} \\ = & 72^{\circ} \end{aligned}

Therefore

x = 72^{\circ}

y = 72^{\circ}

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(b) To find

x

we use pythagorean relationship.

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a^{2} + b^{2} = c^{2}

In our case our

c

x

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20^{2} + 15^{2} = c^{2}

400 + 225 = 625

= \sqrt{625}

= 25 cm

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Exercise

🔗

Determine the unknown angles and sides in the figures below

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Subsubsection 2.5.1.1 Area of a Triangle using Heron’s Formula

Activity 2.5.3.

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Pair up students and give each pair a triangular surface to measure. This could be:

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A triangular classroom table
A triangular signboard
A triangular cutout from paper
A drawn triangle on graph paper

$Materials needed.$
Ruler
Graph paper
Calculator
String or measuring tape
Pre-cut paper triangles (optional)
1. Recall the Standard Triangle Area Formula
  - How do we find the area of a triangle?
    
    Expected answer ; $a = \frac{1}{2} \times base \times height$
  - This method works only when we know the height.
  - $What if we don’t know the height?$ Today, they will discover how to find the area $only using side lengths .$
2. Construct a Triangle and Introduce Semi-Perimeter
  - Draw a triangle with sides labeled as $a, b and c .$ Using a ruler or measuring tape, measure the $three sides of the triangle$ and record the lengths.
  - Measure and calculate the semi-perimeter using:
    
    $\begin{aligned} S = & \frac{a + b + c}{2} \end{aligned}$
    
    This helps divide the triangle into manageable parts for the area calculation.
3. Express the $Area$ in Terms of the $Semi-Perimeter.$
  - Use the right-angle criteria.i.e:
    
    $Drop a perpendicular$ from $one vertex to the opposite side$ (splitting the triangle into two right-angled triangles).
    
    Let’s call the base $b$ and split it into two parts using the perpendicular.
  - Use the $Pythagorean Theorem$ to express the height $h$ in terms of $a, b, c .$
    
    Instead of solving fully, guide students to realize that the height can be found using algebraic manipulation.
  - Introduce the $squared form$ and take the $square root :$
    
    $\begin{aligned} A = & \sqrt{S (S - a) (S - b) (S - c)} \end{aligned}$
  - This is the final area using Heron’s Formula.
  - In conclusion, Heron’s Formula is finding the area of any triangle using only its sides.

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Key Takeaway

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Given three unequal sides of a triangle, we can calculate the area using a formula called Heron’s Formula.

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Follow the simple guidelines below and let’s explore on how to get to the formula.

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Consider the triangle below.

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Perimeter refers to adding all round.

$P = a + b + c$
Let’s find the Semi perimeter of the triangle by the perimeter diving by half.

$S = \frac{a + b + c}{2}$
Then find the area using Heron’s formula:

$A = \sqrt{S (S - a) (S - b) (S - c)}$

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Calculate $S$ (half of the triangle’s perimeter):
Then calculate the Area:

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Example 2.5.2.

If the length of the sides of a triangle

A B C

are

4

inches,

3

inches and

5

inches. Calculate its area using the heron’s formula.

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$triangle$

Solution.

To find: Area of the triangle

A B C .

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Given that AB

= 4

in and BC

= 3

in, AC

= 5

in.

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Using Heron’s Formula,

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\begin{aligned} A & = \sqrt{(s (s - a) (s - b) (s - c))} \end{aligned}

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\begin{aligned} s & = \frac{(a + b + c)}{2} \end{aligned}

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\begin{aligned} s & = \frac{(4 + 3 + 5)}{2} \end{aligned}

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\begin{aligned} = 6 units \end{aligned}

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Substitute in the values,

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\begin{aligned} A & = \sqrt{(6 (6 - 4) (6 - 3) (6 - 5))} \end{aligned}

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\begin{aligned} A & = \sqrt{(6 (2) \times (3) \times (1))} \end{aligned}

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\begin{aligned} A & = \sqrt{36} \end{aligned}

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\begin{aligned} = 6 {in}^{2} \end{aligned}

The area of the triangle is

6 {in}^{2} .

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Exercise

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1. In the diagram below

∠ A B C

is right-angled at B. Complete the table below.

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Table 2.5.3. Trigonometry question test

Trigonometic ratios	Solution
sin C	=
=	$\frac{B C}{A C}$
tan $θ$	=
Value of AC using pythagorean relationship	=
Area given a, b and sin $θ$	=

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2.The area of a triangle with sides 5 cm, 12 cm, and

x

cm is 30 cm². Find the possible values of

x .

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3. A right-angled triangle has sides 9 cm, 12 cm and 15 cm.

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(a) Verify that the triangle satisfies the Pythagorean relationship.

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(b) Find its area using both the standard formula and Heron’s formula.

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4. A triangular plot of land has sides measuring 50 m, 65 m and 75 m. Find the area of the land using Heron’s formula.

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5. An aeroplane at

J

is flying directly over a point

D

on the ground at a height of 5 kilometres. It is heading to land at point

K .

The angle of depression from

J

K

8^{\circ} .

S

is a point along the route from

D

K .

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I) What’s the size of

∠ J K D ?

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II) Calculate the distance

D K,

correct to the nearest metre

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III) If the distance

S K

8

kilometres, calculate the distance

D S .

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IV) Calculate the area of

△ J K D .

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Subsubsection 2.5.1.2 Area of a triangle given two sides and an included angle

Activity 2.5.4.

What you require:
- A tall object (flagpole, tree or building)
- Measuring tape or ruler.
- A sunny day ☀
$Identify the Tall Object .$
- Choose a tree, flagpole, or lamp post as the vertical height (like $A$ in the diagram below).
- The ground acts as the base $(B C) .$
$Find the Shadow Length$
- Measure the length of the shadow cast by the object on the ground ( $B C$ ).
- This represents the long horizontal base in the diagram.
$Measure the Angle of Elevation$
- Stand some distance away and use a protractor or a phone app to measure the angle of elevation from your eyes to the top of the object.
- This represents the $30^{\circ}$ angle at $B$ in the diagram.
- If no protractor is available, use similar triangles by measuring the shadow of a known object (like a stick) and comparing proportions.
Discuss your findings with your group members.
- $Apply Trigonometry Using Sine$
  
  Using the sine function to calculate the height of the object $A N .$
  
  Drop a perpendicular line from Point A to meet line BC at N .
- Formula:
  
  $\begin{aligned} H e i g h t (A N) = & Hypotenuse(AB) \times sin 30^{\circ} \end{aligned}$
- AN is therefore the height for triangle ABC.
- In a triangle $sin θ = \frac{Opposite}{Hypotenuse} .$
  
  From the figure shown above, Opposite = AN, the height of triangle ABN whereas the hypotenuse i $A B$ is the longest side of a triangle.
  
  Where $A N$ = $h$ and $B A$ = $a$
  
  $\begin{aligned} In triangle ABN, sin 30^{\circ} = & \frac{A N}{a} \\ sin 30^{\circ} = & \frac{h}{a} \\ height of ∠ ABC h = & a sin 30^{\circ} \\ Since sin 30^{\circ} = 1 / 2 a \end{aligned}$
- Area of Triangle ABC will be:
  
  $\begin{aligned} △ A B C = & \frac{1}{2} \times Base (b) \times Height (h) = a sin θ \\ = & \frac{1}{2} ab sin θ \end{aligned}$

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♦ Generally, if the lengths of the two sides and an included angle of a triangle are given, then the area of the triangle is

A = \frac{1}{2} ab sin θ .

Where

b is the base and h height.

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Exploration 2.5.5.

Deriving the Formula: $\frac{1}{2} a b sin C$
- Step 1: Let’s recall the Basic Formula for finding area of a triangle.
- Area = $\frac{1}{2} \times Base \times Height$
$Step 2: Consider a Triangle with an Angle .$
- Let’s take a triangle ABC with sides $a,$ $b$ and included angle $C .$
- Side $a$ and Side $b$ form the triangle.
- The height $h$ is perpendicular from the top vertex to the base.
$Step 3: Express Height in Terms of Sin$
- Using trigonometry, we know that in a right-angled triangle:
- $sin θ = \frac{Hypotenuse}{Opposite}$
- In our case, the height $h$ is the $opposite side$ of angle $C,$ and side $b$ is the $hypotenuse .$
- So, we can write it as:
  
  $h = b sin C$
$Step 4: Substitute into the Area Formula$
- Now, take the basic Formula for finding area of a triangle;
- Substituting $Base$ and $Height = b sin C$ .
  
  $\begin{aligned} Area = & \frac{1}{2} \times a \times (b sin C) \\ Area = & \frac{1}{2} a b sin C \end{aligned}$
$NOTE:$
- $\begin{aligned} Area = & \frac{1}{2} a b sin C \end{aligned}$
- This formula is useful when we know two sides of a triangle and the angle between them instead of the height..

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Example 2.5.4.

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1.A triangle HFG has sides

10 cm, 7 cm and 9 cm.

Find:

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(a) Its area.

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(b) The sizes of its angles.

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2. The area of triangle ABC is

28.1 {cm}^{2}

. It’s side AB

= 7.2 cm

and

∠

ABC

= {48.6}^{\circ} .

Find:

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(a) The length of the perpendicular from A to BC.

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(b) The length of BC.

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Solution 1.

(a) Find the area using Heron’s formula.

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\begin{aligned} A & = \sqrt{(s (s - a) (s - b) (s - c))} \end{aligned}

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\begin{aligned} s & = \frac{(a + b + c)}{2} \end{aligned}

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\begin{aligned} s & = \frac{(10 + 7 + 9)}{2} c m \end{aligned}

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\begin{aligned} = 13 cm \end{aligned}

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Substitute in the values.

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\begin{aligned} A & = \sqrt{(13 (13 - 10) (13 - 7) (13 - 9))} \end{aligned}

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\begin{aligned} A & = \sqrt{(13 (6) \times (3) \times (4))} \end{aligned}

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\begin{aligned} A & = \sqrt{936} \end{aligned}

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\begin{aligned} = 30.6 {cm}^{2} \end{aligned}

The area of the triangle is

30.6 {cm}^{2} .

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(b) Find the angles using the sine area formula.

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A = \frac{1}{2} ab sin C

A = 30.6 {cm}^{2}

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To find

∠ H F G

Using sides

a = 10

and

b = 7

and area

= 30.6 {cm}^{2}

🔗

\begin{aligned} 30.6 {cm}^{2} = & \frac{1}{2} \times 10 5 cm \times 7 cm \times sin A \\ Sin A = & \frac{30.6 {cm}^{2} cm}{35 cm} \\ sin A = & 0.8743 \\ A = & s i n^{- 1} 0.8743 \\ = & {60.96}^{\circ} \end{aligned}

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To find

∠ F G H

Using sides

a = 10

and

b = 9

and area

= 30.6 {cm}^{2}

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\begin{aligned} 30.6 {cm}^{2} = & \frac{1}{2} \times 10 5 cm \times 9 cm \times sin C \\ Sin C = & \frac{30.6 {cm}^{2} cm}{45 cm} \\ sin C = & 0.68 \\ C = & {sin}^{- 1} 0.68 \\ = & {42.84}^{\circ} \end{aligned}

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To find the

∠ G H F

we use the angle sum property.

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\begin{aligned} ∠ G H F = & 180^{\circ} - (∠ F G H + ∠ H F G) \\ = & (180 - 103.80)^{\circ} \\ = & {76.2}^{\circ} \end{aligned}

Similarly we can use the Sin A rule which states that :

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\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}

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\begin{aligned} \frac{sin B}{b} & = \frac{sin C}{c} \\ \frac{sin B}{10 cm} = & \frac{sin 0.8743}{9} \\ sin B = & 0.9714 \\ = & {sin}^{- 1} 0.9714 \\ B = & {76.2}^{\circ} \end{aligned}

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Solution 2.

(a) Find the length of the perpendicular from A to BC.

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The formula for area using base and height.

🔗

\begin{aligned} A = & \frac{1}{2} \times b a s e \times h e i g h t \\ 28.1 {cm}^{2} = & \frac{1}{2} \times b \times h \\ h = & \frac{2 \times 28.1}{B C} \end{aligned}

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(b) Finding BC using sine rule

🔗

\begin{aligned} A = & \frac{1}{2} \times A B \times B C \times sin θ \\ 28.1 {cm}^{2} = & \frac{1}{2} \times 7.2 cm \times B C \times sin {48.6}^{\circ} \\ sin {48.6}^{\circ} = & 0.7501 \\ 28.1 {cm}^{2} = & \frac{1}{2} \times 7.2 cm \times B C \times sin 0.7501 \\ B C = & \frac{28.1 \times 2}{7.2 \times 0.7501} \\ = & 10.4 cm \end{aligned}

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Now substituting BC into the height equation :

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\begin{aligned} h = & \frac{2 \times 28.1}{10.4} \\ = & 5.4 cm \end{aligned}

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Exercise2

🔗

1.

In a triangle

△ QRS,

QR

= 10 cm,

RS

= 24 cm and

QS

= 26 cm. Find the length of the perpendicular from vertex

Q

to side

RS

.Therefore find it’s area.

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2.

△ ABC,

∠ BAC = 40^{\circ},

∠ ABC = 65^{\circ},

and side

BC = 8

cm. Find it’s area.

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3. △ PQR is isosceles with PQ = PR = 10

cm. The base angle is

48^{\circ} . Find its area .

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4. Triangle XYZ

has side

XY = 15

cm,

YZ = 20

cm, and

XZ = 25 c m . Show that this is a right-angled triangle and find it's area.

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5. Triangle MNO

\, has

∠ NMO = 50^{\circ}

∠ MNO = 60^{\circ}

, and

MO = 18

cm. Find the length of

MN using the sine rule thus find the triangle's area.

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6. △ PQR

has

∠ RPQ = 35^{\circ}

∠ PQR = 75^{\circ}

, and

QR = 15

cm. Find the area of triangle PQR.\)

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Subsection 2.5.2 Area of Quadrilaterals.

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Activity 2.5.6.

$Materials needed:$
Grid paper or geoboards
Markers or colored pencils for highlighting shapes
Printable worksheets (with grid paper and templates for quadrilaterals)
RulersPencils and erasers
Protractors (for measuring angles)
Scissors (if cutting shapes from paper)
1. Prepare the worksheets with grid paper or geoboard templates and space for calculations.
2. Prepare cut-out shapes of different quadrilaterals (square, rectangle, parallelogram, rhombus, trapezoid, etc.), if using physical materials.
3. Set up a projector or board to display instructions, the properties of quadrilaterals, and the step-by-step breakdown.
Take that cut-out shapes or access to digital templates of various quadrilaterals. Sort the shapes based on their properties:

Number of parallel sides

Number of equal sides

Angles (right angles or acute/obtuse angles)

Symmetry
For each quadrilateral, discuss with your patner and record:

1. The type of quadrilateral.

2. Whether it has parallel sides or right angles.

3. The number of sides of equal length.

Learners should be able to classify quadrilaterals by their properties.

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Key Takeaway

🔗

Properties of Quadrilaterals.

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🔹 Below are the key types of quadrilaterals and their properties:

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Quadrilateral	Properties
Parallelogram	Opposite sides are parallel and equal. Opposite angles are equal. Diagonals bisect each other. 🔗
Rectangle	Opposite sides are equal and parallel. All angles are $90^{\circ} .$ Diagonals are equal and bisect each other. 🔗
Square	All sides are equal. All angles are $90^{\circ} .$ Diagonals are equal, bisect each other at $90^{\circ},$ and are perpendicular. 🔗
Rhombus	All sides are equal. Opposite angles are equal. Diagonals bisect each other at $90^{\circ}$ . 🔗
Trapezium (Trapezoid)	One pair of opposite sides is parallel. Non-parallel sides are called legs. If legs are equal, it is called an isosceles trapezium. 🔗
Kite	Two pairs of adjacent sides are equal. One pair of opposite angles is equal. Diagonals intersect at $90^{\circ}$ , and one diagonal bisects the other. 🔗

♦ Can you remember the formula for finding the area of any Quadrilateral in the table above?

🔗

♦ If yes which one and what is the formula?

🔗

♦ Kindly write it down exchange your answers with your deskmate and see if their answers are correct.

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Example 2.5.5.

Find the area of a rectangular classroom measuring 6m by 4m .

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Solution.

Area of a rectangle = L e n g t h \times W i d t h

= 6 m \times 4 m

= 24 m^{2}

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Subsubsection 2.5.2.1 Area of a Parallelogram

Activity 2.5.7.

Exploring the Area of a parallelogram .

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Materials needed.

🔗

Graph paper
Scissors
Ruler
Colored pencils
A printed or drawn parallelogram template.

🔗

Steps for the activity.

🔗

Draw a parallelogram.
1. On graph paper, draw a parallelogram with a given base and height. Label the vertices A, B, C and D.
2. Use a pair of campus to drop a perpendicular. Lable the intersection point on line CD as H representing the height of the parallelogram.
3. Use a ruler to measure and compare opposite sides (AB and CD, BC and AD. )

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Key Takeaway

🔗

A parallelogram is a quadrilateral whose opposite sides are equal and parallel. Which other quadrilateral has similar characteristics like these?

🔗

🔹 The area of a parallelogram is given by:

B a s e \times H e i g h t :

🔗

\begin{aligned} A = & b \times h \end{aligned}

where

b

is the base and

h

the perpendicular distance between the given pair of parallel sides.

🔗

🔹 The area of a parallelogram given the base is 28cm and the height 8cm is:

🔗

\begin{aligned} A = & b \times h \\ = & 28 cm \times 8 cm \\ = & 224 {cm}^{2} \end{aligned}

🔗

Example 2.5.6.

A parallelogram PQRS is of sides 28cm and 7cm. If $∠$ QRS is $75^{\circ} .$

a. Find the height the parallelogram using sine rule.

b. Find the area of the parallelogram.

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Solution.

🔗

From Q drop a perpendicular to meet RS at T considering

∠ Q R S .

🔗

Height (QT) = Hypotenuse(BC) \times sin θ

🔗

\begin{aligned} Q T = & 7 sin 75^{\circ} \\ = & 6.76 cm \end{aligned}

🔗

\begin{aligned} Area of PQRS = & R S \times P T \\ = & 28 \times 7 sin 75^{\circ} \\ = & 189.32 {cm}^{2} \end{aligned}

🔗

The area of the parallelogram is

189.32 {cm}^{2} .

🔗

Exercise

🔗

1. Find the area of the parallelogram given below.

🔗

2. The parallelogram shown below has diagonals. Find the length of the diagonal.

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3. Find the missing height of the parallelogram ABCD given the area is 24 cm² and base is 6 cm.

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4. A construction company is building a parallelogram-shaped floor with a base of 10 meters and height of 6 meters. Find the area of the floor.

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5. A billboard has a parallelogram shape with a base of 12 feet and a height of 5 feet. Find the area of the billboard.

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6. A solar panel is shaped like a parallelogram with a base of 8 meters and an inclined height of 4 meters. Find it’s surface area.

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Subsubsection 2.5.2.2 Area of a Rhombus

Activity 2.5.8.

Materials Needed:

🔗

Ruler, Protractor, Pencil, Graph paper, Scissors (optional) and String or thread (optional)

🔗

Draw a Rhombus Using a ruler, draw a quadrilateral with all sides equal in length. Make sure the opposite angles are equal. Label the vertices as A, B, C, D in order.
Measure the Sides using a ruler to confirm that all four sides are of equal length.
Measure the angles using a protractor to measure each of the interior angles. Observe: Opposite angles should be equal.
Draw the Diagonals:

Draw diagonal AC and diagonal BD.

Measure their lengths and angles at the intersection point.

Observe: Diagonals bisect each other at 90°, and they are not equal.
Cut and Fold (Optional):

Cut out the rhombus and fold it along both diagonals.

Observe the symmetry and how the diagonals act as lines of symmetry.

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Discussion Questions

🔗

What do you notice about the sides and angles of the rhombus?
How do the diagonals interact with each other?
What makes a rhombus different from a square or a general parallelogram?

🔗

Key Takeaway

🔗

rhombus

is an equilateral quadrilateral. All it’s sides are of

equal length,

But all it’s angles are not. All its sides are equal and the diagnols bisect at

90^{\circ} .

Which other quadrilateral has all its sides equal?

🔗

Compare the rhombus with the rectangle below.

🔗

Properties of a Rhombus

🔗

All sides are equal in length.
Opposite angles are equal.
Diagonals bisect each other at right angles.
Diagonals bisect the interior angles.
Some examples of a rhombus are Tiles or patterns in flooring, or The shape of playing cards (diamonds suit).

🔗

Example 2.5.7.

A rhombus has a diagonal of 10 cm and another diagonal of 24 cm. Find:

🔗

i) The area of the rhombus.

🔗

ii).The length of one side of the rhombus.

🔗

Solution.

🔗

One diagonal = 10 cm

🔗

Another diagonal = 24 cm

🔗

\begin{aligned} Area = & \frac{1}{2} \times d_{1} \times d_{2} \\ = & \frac{1}{2} \times 10 cm \times 24 cm \\ = & 120 {cm}^{2} \end{aligned}

🔗

ii. Length of One Side of the Rhombus

🔗

In a rhombus, the diagonals bisect each other at 90°. That means each side of the rhombus forms a right-angled triangle with half of each diagonal.

🔗

Let’s find the length of one side using the Pythagorean Theorem.

🔗

Each side of the rhombus is the hypotenuse of a right triangle with legs:

🔗

\begin{aligned} Half of AC = & \frac{10}{2} = 5 cm \\ Half of BD = & \frac{24}{2} = 12 cm \\ Side = & \sqrt{(5)^{2} + (12)^{2}} \\ = & \sqrt{25 cm + 144 cm} \\ = & \sqrt{169 cm} \\ = & 13 cm \end{aligned}

🔗

Exercise

🔗

A rhombus has diagonals measuring 16 cm and 30 cm.

a) Find the area of the rhombus.

b) Find the side length of the rhombus.
A diamond-shaped road sign is a rhombus with a diagonal of 40 cm and another diagonal of 60 cm. Find:

(a) The area of the sign.

(b) The length of one side of the sign.
The perimeter of a rhombus is 48 cm. Find the length of one side.
A car logo is shaped like a rhombus. The diagonals measure 14 cm and 10 cm. Find its area.
A rhombus has one of its angles measuring 60°. Find the other three angles.
A tiling design on a floor is made of rhombus-shaped tiles. Each tile has diagonals of 18 cm and 24 cm.

(a) Find the area of one tile.

(b) If 20 such tiles cover a portion of the floor, what is the total area covered?

🔗

Subsubsection 2.5.2.3 Area of a Trapezium

🔗

Activity 2.5.9.

🔗

materials needed:

🔗

Materials: Ruler, Protractor, Pencil, Graph paper, Scissors (optional) and Calculator (optional)

🔗

Draw a Trapezium:

On graph paper, draw a quadrilateral with one pair of opposite sides parallel (e.g., AB ∥ CD).

Make sure the other pair (AD and BC) are not parallel.

Label the trapezium as ABCD.
Identify the Parts:

Label the bases (parallel sides).

Label the legs (non-parallel sides).

Draw the height (perpendicular distance between the two bases).
Now, let’s break it down algebraically:

Imagine the Area of a Rectangle: If we could transform the trapezium into a rectangle, the area of that rectangle would be the $average of the two bases times the height$ . This works because the average length of the two parallel sides is a $"representative" length for the trapezium,$ and multiplying it by the $height$ gives us the $correct area .$
Mathematical Expression:

The average length of the two parallel sides is given by

$\begin{aligned} Average of bases = & \frac{B a s e_{1} + B a s e_{2}}{2} \end{aligned}$

Therefore, the area of the trapezium is:

$\begin{aligned} A r e a = & Average of bases \times height \\ = & \frac{B a s e_{1} + B a s e_{2}}{2} \times height \end{aligned}$

This is the formula for the area of a trapezium.
$Understanding the Formula$
The bases are the lengths of the parallel sides.

The height is the perpendicular distance between the two bases.

The average of the bases gives us a "typical length" for the trapezium, and when multiplied by the height, it gives the area, just like how you would calculate the area of a rectangle.

🔗

Key Takeaway

🔗

A trapezium is a four-sided polygon (quadrilateral) that has one pair of opposite sides that are parallel. These parallel sides are called the bases of the trapezium. The other two sides are not parallel and are called the legs.

🔗

Proterties of a trapezium .

🔗

A trapezium looks like a bridge. Suggest other three real life examples of trapeziums.
One pair of sides are parallel (they never meet).
The height is the straight-up distance between the parallel sides.
The angles inside add up to 360°.

🔗

Study Questions.

🔗

Define a trapezium in your own words.
Identify which of the following shapes are trapeziums giving reasons why:

$a .$ A square

$b .$ A shape with one pair of parallel sides

$c$ A rectangle
- In the images provided below read the question and tick the appropriate answer.
Draw a trapezium and label its bases, legs, and height.
How many angles does a trapezium have? What is the sum of all interior angles?
What is the difference between a trapezium and a parallelogram?

🔗

The area of a trapezium is given by the formula:

🔗

\begin{aligned} A = & \frac{a + b}{2} \times h \end{aligned}

where

a

and

b

are the two parallel sides and

h

height.

🔗

Example 2.5.8.

Calculate the area of the figure below.

🔗

Solution.

The given figure is a trapezium with the following vertices:

🔗

$A (- 2, 3)$
$B (2, 3)$
$C (3, - 1)$
$D (- 3, - 1)$

From the diagram, we observe:
The two parallel bases are ABAB and DCDC.
The height is given as 4 cm, which is the perpendicular distance between the parallel bases.

🔗

Step 1: Determine the lengths of the bases

🔗

Length of

A B :

🔗

\begin{aligned} A B = | x_{1} - x_{2} | & = | 2 - (- 2 | = | 2 + 2 | = 4 cm \end{aligned}

Length of

D C :

🔗

\begin{aligned} D C = | x_{1} - x_{2} | & = | 3 - (- 3 | = | 3 + 3 | = 6 cm \end{aligned}

🔗

Step 2: Use the trapezium area formula.

🔗

The area of a trapezium is given by:

🔗

\begin{aligned} Area = & \frac{1}{2} \times ({Base}_{1} + {Base}_{2}) \times Height \end{aligned}

Substituting the values:

🔗

\begin{aligned} A = & \frac{1}{2} \times (4 + 6) cm \times 4 cm \\ = & \frac{1}{2} \times 10 cm \times 4 cm \\ = & 20 {cm}^{2} \end{aligned}

🔗

THe area of the trapezium is

20 {cm}^{2}

🔗

Exercise

🔗

1.

A trapezium has a height of 5 cm, the top base is 8 cm and the bottom base is 14 cm. Find its area.

🔗

2. a)

The bases of a trapezium are 12 cm and 6 cm and its area is 45 cm². Find the height.

🔗

b) Find the missing angle in a trapezium where three angles are 70°, 85° and 95°.

🔗

c) A trapezium has equal non-parallel sides (legs). What special type of trapezium is this?

🔗

d) The parallel sides of a trapezium are 20 m and 30 m and its height is 12 m. Find its area.

🔗

3.

A trapezium-shaped farm has a shorter base of 150 m, a longer base of 300 m and a height of 200 m. Find the area of the farm.

🔗

4.

A car window is shaped like a trapezium with a height of 50 cm, the top base is 60 cm and the bottom base is 80 cm. Find the area of the window.

🔗

5.

A trapezoidal table has a top base of 1.2 m, a bottom base of 1.8 m and a height of 0.75 m. What is its surface area?

🔗

Subsubsection 2.5.2.4 Area of a Kite

Activity 2.5.10. Constructing a Kite Using Geometry Tools.

How to construct a kite accurately using a ruler, compass, and protractor.

🔗

Materials needed:
Graph paper or plain paper
Ruler , Compass, protractor, pencil and eraser
Follow this steps.

$Step 1: Drawing the diagnol.$
Draw a vertical line of length 10 cm (this will be the longer diagonal, $d_{1}$ ).
Label the midpoint of this line as O.

$Step 2: Draw the Perpendicular Diagonal$
At O, use a protractor to draw a perpendicular line.
Mark $4 cm$ on each side of $O$ (total shorter diagonal $d_{2} =$ $8 c m .$

$Step 3: Mark the Kite's Vertices$
Label the four points where the diagonals intersect as A, B, C, and D.
Connect $A$ to $B,$ $B$ to $C,$ $C$ to $D$ and $D$ to $A .$

$Step 3:Check Properties$
Measure adjacent sides to ensure two pairs are equal.
Verify opposite angles (one pair should be equal)
Confirm diagonals are perpendicular.

🔗

Design your own kite patterns on graph paper and justify why their design would be aerodynamic and stable in the air.

🔗

$\textbf{Key Takeaway$

🔗

A kite is a quadrilateral with two pairs of adjacent sides equal in length and one pair of opposite angles equal.

🔗

A kite can be analyzed using coordinate geometry when its vertices are given on a Cartesian plane.

🔗

Since the diagonals are perpendicular, right-angled triangles are formed.

🔗

We can use trigonometric ratios (sine, cosine, tangent) to find missing angles and side lengths.

🔗

Finding Angles Using Trigonometry

🔗

If a kite has diagonals

d_{2}

and

d_{2}

we can find interior angles using:

🔗

\begin{aligned} sin θ = & \frac{opposite side}{hypotenuse} \\ cos θ = & \frac{adjacent side}{hypotenuse} \\ tan θ = & \frac{opposite side}{adjacent side} \end{aligned}

🔗

If a kite has diagonals 14 cm and 10 cm, then each right-angled triangle within the kite has:

🔗

\begin{aligned} Base = \frac{d_{1}}{2} = & \frac{14}{2} \\ = & 7 cm \\ Height = \frac{d_{2}}{2} = & \frac{10}{2} \\ = & 5 cm \end{aligned}

The angle θθ between the diagonal and a side can be found using:

🔗

\begin{aligned} tan θ = & \frac{5}{7} \\ θ = & {tan}^{- 1} 0.7143 \\ = & {35.5}^{\circ} \end{aligned}

🔗

Example 2.5.9.

A kite whose has the diagonals are 16 cm and 12 cm.

🔗

a) Find the area of the a kite

🔗

b) If one pair of adjacent sides is 10 cm, find the perimeter of the kite.

🔗

c) Find the angles of the kite using trigonometry.

🔗

Solution.

Find the Area of the Kite

Let

d_{1} = 16 cm

and

d_{2} = 12 cm

🔗

\begin{aligned} Area = & \frac{1}{2} \times d_{1} \times d_{2} \end{aligned}

substitute d_{1} = 16 cm and d_{2} = 12 cm

🔗

\begin{aligned} Area = & \frac{1}{2} \times 16 cm \times 12 cm \\ = & \frac{192}{2} {cm}^{2} \\ = & 96 {cm}^{2} \end{aligned}

🔗

(b) Find the Perimeter of the Kite. The diagonals bisect each other at right angles, so each half-diagonal forms a right-angled triangle.

\frac{d_{1}}{2} = 6 cm

and

\frac{d_{2}}{2} = 8 cm

Using the Pythagoras Theorem:

🔗

\begin{aligned} a^{2} + b^{2} = c^{2} \\ = & 6^{2} + 8^{2} \\ c^{2} = & 36 + 64 \\ \sqrt{c^{2}} = & \sqrt{100} \\ c = & 10 cm \end{aligned}

Since a kite has two pairs of equal sides, the perimeter is:

🔗

\begin{aligned} P = & 2 (a + b) \\ = & 2 (10 + 10) cm \\ = & 40 cm \end{aligned}

🔗

\begin{aligned} tan θ = & \frac{Opposite}{adjacent} \\ = & \frac{6}{8} \\ θ = & {tan}^{- 1} 0.75 \\ θ = & {36.87}^{\circ} \end{aligned}

🔗

A kite is a quadrilateral, meaning it has four angles.

🔗

The sum of the interior angles of any quadrilateral is given by the formula:

🔗

Sum of Interior Angles=

(n - 2) \times 180^{\circ}

🔗

where

n = 4

(since a kite has 4 sides)

(4 - 2) \times 180^{\circ}

= 2 \times 180^{\circ}

= 360^{\circ}

🔗

Since the kite is symmetric, the larger angle (

α

)is

🔗

\begin{aligned} = & 360^{\circ} - 2 θ \\ = & 360^{\circ} - 2 (36.87)^{\circ} \\ = & 360^{\circ} - {73.74}^{\circ} \\ 2 (α) = & {286.26}^{\circ} \\ α = & {143.13}^{\circ} \end{aligned}

🔗

The angles of the kite are:

143^{\circ}, {143.13}^{\circ}, {36.87}^{\circ} and {36.87}^{\circ}

🔗

Exercise

🔗

1. Find the area of the kite given below.

🔗

1. A kite has diagonals of length 10 cm and 8 cm. Find its area.

🔗

3. A playground has a kite-shaped design with diagonals measuring 15 meters and 12 meters. Find its area.

🔗

4. A rescue helicopter designates an emergency landing zone in the shape of a kite. The diagonals measure 60 m and 45 m. Calculate the available landing space.

🔗

5. A playground has a kite-shaped design with diagonals measuring 15 meters and 12 meters. Find its area.A playground has a kite-shaped design with diagonals measuring 15 meters and 12 meters. Find its area.

🔗

6. Engineers are designing a kite-shaped solar panel with diagonals of 30 m and 18 m. What is the total solar-collecting area?

🔗

7. A relief team sets up a kite-shaped safe zone for disaster survivors. The diagonals measure 50 m and 40 m. What is the area available for the survivors?

🔗

Subsection 2.5.3 Area of Regular Polygons

Activity 2.5.11.

Materials Needed:

i.Compass,

ii. Ruler

ii. Protractor

iv. Calculator

v. Pencil and paper

vi. Colored pencils (optional)

🔗

Draw a line from the center O perpendicular to a side (e.g., side BC). This line is the apothem (denoted a). Label the apothem clearly.

🔗

$\textbf{Key Takeaway$

🔗

♦ The

area

of a

regular polygon

can be found using this formula, where

P

is the

perimeter

and

a

is the apothem

🔗

The

apothem

is the distance from the center of a regular polygon to

a

side. Where

a

is the half

\frac{1}{2} side.

🔗

You want to find the area of the triangle. You can see it has an apothem of

\sqrt{3}

meters and a side length of 10 meters.

🔗

If you also find the perimeter, P , then you can use the formula A=

\frac{1}{2} P a

to find the area.

🔗

P = length of each side \times a, where a = h

P = 10 m \times 3

= 30, m

Now, use the formula for the area of a regular polygon is :

🔗

\begin{aligned} A = & \frac{1}{2} \times 30 \times \sqrt{3} \\ = & 125.98 m^{2} \end{aligned}

🔗

Example 2.5.10.

A regular hexagon with center h is shown below

🔗

Similarly you can use

A = \frac{1}{2} \times n \times S \times a

🔗

Solution.

Find the perimeter P, Then use the formula A=

\frac{1}{2} P a

to find the area.

🔗

P = length of each side \times a

P = 20 cm \times 6

= 120 {cm}^{2}

Now, use the formula for the area of a regular polygon is :

🔗

\begin{aligned} A = & \frac{1}{2} \times 120 \times 14 cm \\ = & 840 {cm}^{2} \end{aligned}

🔗

Where

n = number of sides, S = length of each side and a = \frac{S}{2}

🔗

\begin{aligned} A = & \frac{1}{2} n \times S \times a \\ = & \frac{1}{2} \times 6 \times 20 cm \times 14 cm \\ = & \frac{1, 680}{2} {cm}^{2} \\ = & 840 {cm}^{2} \end{aligned}

🔗

Subsubsection 2.5.3.1 Area of Heptagon

A heptagon is a seven-sided polygon. It has seven edges and seven vertices.

🔗

Activity 2.5.12.

Objective:

🔗

learners are expected to know how to construct a regular heptagon (7-sided polygon) using a compass, ruler, and protractor.

🔗

Materials Needed:

🔗

Materials Needed:

♦ Compass

♦ Ruler

♦ Protractor

♦ Pencil

♦ Eraser

♦ Graph paper (optional)
Step:1 Draw a Circle.
- Place the compass pointer on the paper and draw a circle of any radius.
- Mark the center (O) of the circle.
Step 2: Draw a Horizontal Diameter
- Use the ruler to draw a straight line measuring 10cm through the center (O), creating a diameter (A 0 to P1).
- Label the first point P1 on the circumference.
Step 3: Divide the Circle into Seven Equal Parts
- Use a protractor to measure angles of 360° ÷ 7 = 51.43° from point P1.
- Mark each 51.43° interval around the circle to get seven points.
Step 4: Connect the Points as shown below.
- Use the ruler to draw straight lines connecting the seven points in sequence.
- The heptagon is now complete!

🔗

Extended Activity.

🔗

🔹 Try drawing a heptagon using only a compass (without a protractor)

🔗

🔹 Shade the inside of the heptagon with different colors to make a pattern.

🔗

🔹 Find the sum of the interior angles of the heptagon. (

H i n t : (n - 2) \times 180^{\circ}

)

🔗

Discussion Questions

🔗

1. What is the sum of all interior angles of a heptagon?

🔗

2. How do we calculate the measure of one interior angle of a regular heptagon?

🔗

3. Can you find a heptagon in real life (architecture, logos, etc.)?

🔗

$\textbf{Key Takeaway$

🔗

Properties of a Regular Heptagon. .

🔗

♦ Sum of interior angles =

(n - 2) \times 180^{\circ} .

Where n is the number of sides.

🔗

♦ Example for a Heptagon (7-sided polygon).

🔗

(n - 2) \times 180^{\circ}

(7 - 2) \times 180^{\circ} = 900^{\circ}

🔗

♦ Sum of exterior angles of any

polygon(regular or irregular)

is always equl to

360^{\circ}

for both regular and irregular polygons.

🔗

♦ Each Interior and Exterior Angle (Regular Polygon Only)

🔗

Each Interior Angle (for a regular polygon):

I n t e r i o r A n g l e = \frac{(n - 2) \times 180^{\circ}}{n}

Each Exterior Angle (for a regular polygon):

E x t e r i o r A n g l e = \frac{360^{\circ}}{n}

🔗

♦ Example for a Regular Heptagon:

🔗

Each Interior Angle = \frac{900}{7}

= {128.57}^{\circ}

🔗

Each Exterior Angle = \frac{360}{7}

= {51.43}^{\circ}

🔗

Study Questions

🔗

1. What is the sum of all interior angles of a heptagon ?

🔗

2. How do we calculate the measure of one interior angle of a regular heptagon ?

🔗

3. Can you find heptagons in real life?

🔗

Example 2.5.11.

A regular heptagon measures 10 cm as indicated in the figure below, find its area given the sum of it’s interior angles is

900^{\circ} ?

🔗

Solution.

There are 7 triangles since a heptagon has 7 sides. Area of Triangle

P_{0} O P_{1} = \frac{1}{2} a b sin {51.43}^{\circ}

🔗

\begin{aligned} △ P_{0} O P_{1} = & \frac{1}{2} \times 10 \times 10 sin {51.43}^{\circ} \\ = & \frac{1}{2} \times 100 {cm}^{2} \times sin 51, 43^{\circ} \\ = & \frac{1}{2} \times 100 {cm}^{2} \times 0.7818 \\ = & 39.0923 {cm}^{2} \end{aligned}

🔗

Therefore the total area

🔗

\begin{aligned} = & 7 \times 39.0923 {cm}^{2} \\ = & 273.65 {cm}^{2} \end{aligned}

🔗

Subsubsection 2.5.3.2 Area of a an Octagon

A regular octagon is an(8-sided) polygon with eight vertices (corners) and ten edges (sides).

🔗

Activity 2.5.13.

Materials Needed:

🔗

♦ Compass, Ruler, Protractor

🔗

♦ Pencil, Eraser, Graph paper (optional)

🔗

Drawing a Regular Octagon (8-sided Polygon)

🔗

1. Draw a Circle:

🔗

🔹Use a compass to draw a circle of your deired radius.

🔗

🔹Mark the center (O).

🔗

2. Draw the First Diameter:

🔗

🔹Use a ruler to draw a horizontal diameter (P0 to 95) passing through the center.

🔗

3. Divide the Circle into 8 Equal Parts.

🔗

🔹Use a protractor to measure angles of 360° ÷ 8 = 45° from point P0.

🔗

🔹Mark each 45° interval on the circle to get 8 points.

🔗

4. Connect the Points:

🔗

🔹Use a ruler to connect the eight points in order to form the octagon.

🔗

Discussion Questions

🔗

Study Questions

🔗

1. What is the sum of all interior angles of an octagon ?

🔗

2, How do we calculate the measure of one interior angle of a regular octagon?

🔗

3, Have you ever seen octagon-shaped objects in real life?

🔗

Example 2.5.12.

A regular octagon with O as its centre. If OA is 8 cm, find its area.

🔗

Solution.

There are 8 triangles since an octagon has 8 sides. Area of Triangle

A O B = \frac{1}{2} a b sin 45^{\circ}

🔗

\begin{aligned} △ A O B = & \frac{1}{2} \times 8 \times 8 sin 45^{\circ} \\ = & \frac{1}{2} \times 64 {cm}^{2} \times sin 45^{\circ} \\ = & \frac{1}{2} \times 64 {cm}^{2} \times 0.7071 \\ = & 22.6274 {cm}^{2} \end{aligned}

🔗

Therefore the total area

🔗

\begin{aligned} = & 8 \times 22.6274 {cm}^{2} \\ = & 181.02 {cm}^{2} \end{aligned}

🔗

Example 2.5.13.

The figure below represents a regular octagon with O as its centre. If OA is 7cm, find its area

🔗

Solution.

There are 8 triangles since an octagon has 8 sides and the sum of interior angles of one triangle is

360^{\circ} .

Area of Triangle

A O B = \frac{1}{2} a b sin 45^{\circ}

🔗

The angle of

△ A O B = \frac{360}{8} = 45^{\circ}

🔗

\begin{aligned} △ A O B = & \frac{1}{2} \times 7 \times 7 sin 45^{\circ} \\ = & \frac{1}{2} \times 49 {cm}^{2} \times sin 45^{\circ} \\ = & \frac{1}{2} \times 49 {cm}^{2} \times 0.7071 \\ = & 17.3241 {cm}^{2} \end{aligned}

🔗

Therefore the total area

🔗

\begin{aligned} = & 8 \times 17.3241 {cm}^{2} \\ = & 138.59 {cm}^{2} \end{aligned}

🔗

Subsubsection 2.5.3.3 Area of a Nonagon

A nonagon is a nine-sided polygon with nine vertices (corners) and nine edges (sides).

🔗

Activity 2.5.14.

Materials Needed:

🔗

♦ Compass, Ruler, Protractor

🔗

♦ Pencil, Eraser, Graph paper (optional)

🔗

Drawing a Regular nonagon (9-sided Polygon)

🔗

1. Draw a Circle:

🔗

🔹Use a compass to draw a circle of your desired radius.

🔗

🔹Mark the center (O).

🔗

2. Draw the First Diameter:

🔗

🔹Use a ruler to draw a horizontal diameter (P0 to P6) passing through the center.

🔗

3. Divide the Circle into 9 Equal Parts.

🔗

🔹Use a protractor to measure angles of 360° ÷ 9 = 40° from point P0.

🔗

🔹Mark each 40° interval on the circle to get 9 points.

🔗

4. Connect the Points:

🔗

Study Questions

🔗

What is the sum of all interior angles of an nonagon ?
How do we calculate the measure of one interior angle of a regular nonagon?
Have you ever seen nonagon-shaped objects in real life?

🔗

Example 2.5.14.

Nonagon example.

🔗

One of the angles of a regular polygon is

40^{\circ}

and its sides are

15 cm

long. Sketch the polygon and then find the area of the polygon.

🔗

Solution.

🔗

🔹 From the definition of a polygon all interior angle add up to

360^{\circ} .

🔗

🔹 To find the number of sides of a triangle we use the formula

\frac{360}{Interior angle} = Number of sides .

🔗

Where

n

is the number of sides.

🔗

\begin{aligned} \frac{360^{\circ}}{interior angle} = & Number of sides \\ \frac{360^{\circ}}{40} = & 9 sides \end{aligned}

🔗

🔹Finding the area of the nonagon we use the formula:

🔗

Area of a tringle using sine rule

\times the number of sides

Where our

a b

is the radii.

🔗

\begin{aligned} \frac{1}{2} a b sin 40^{\circ} = & \frac{1}{2} \times 15 cm \times 15 cm \times sin 40^{\circ} \\ = & \frac{1}{2} \times 225 \times 0.6428 {cm}^{2} \end{aligned}

🔗

🔹 There are 9 triangles since a nonagon has 9 sides.

🔗

🔹Therefore, total area is:

🔗

\begin{aligned} = & 9 \times (\frac{1}{2} \times 225 \times 0.6428) \\ = & 9 \times 225 \times 0.3214 \\ = & 73.6006 {cm}^{2} \end{aligned}

🔗

Subsubsection 2.5.3.4 Area of a Decagon

A decagon is a ten-sided polygon with ten vertices (corners) and ten edges (sides).

🔗

Activity 2.5.15.

Materials neede:

🔗

♦ Compass, Ruler, Protractor,

🔗

♦ Pencil, Eraser, Graph paper (optional)

🔗

Drawing a Regular Decagon (10-sided Polygon)

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1. Draw a Circle:

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🔹Use a compass to draw a circle of your desired radius.

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🔹Mark the center (O).

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2. Draw the First Diameter:

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🔹Use a ruler to draw a horizontal diameter (P0 to P6) passing through the center.

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3. Divide the Circle into 10 Equal Parts.

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🔹Use a protractor to measure angles of 360° ÷ 10 = 36° from point A.

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🔹Mark each 36° interval on the circle to get 10 points.

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4. Connect the Points:

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🔹Use a ruler to connect the ten points in order to form the decagon.

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Study Questions

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What is the sum of all interior angles of an decagon ?
How do we calculate the measure of one interior angle of a regular decagon?
Can you find decagons in real life?
What is the sum of all interior angles of a decagon?

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Example 2.5.15.

A regular decagon has an apothem

a

7.5 cm

and a side length

S

5 cm .

Find its area.

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Solution.

Find the perimeter P, Then use the formula A=

\frac{1}{2} P a

to find the area.

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P = length of each side \times a

P = 5 cm \times 10

= 50 cm

Now, use the formula for the area of a regular polygon is :

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\begin{aligned} A = & \frac{1}{2} \times 50 \times 7.5 cm \\ = & \frac{375}{2} {cm}^{2} \\ = & 187.5 {cm}^{2} \end{aligned}

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Where

n = number of sides, S = length of each side and a = \frac{S}{2}

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\begin{aligned} A = & \frac{1}{2} n \times S \times a \\ = & \frac{1}{2} \times 10 \times 5 c m \times 7.5 cm \\ = & \frac{375}{2} {cm}^{2} \\ = & 187.5 {cm}^{2} \end{aligned}

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Exercise

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1. A regular octagon has a side length of

10 cm .

Calculate its area using the formula for the area of a regular polygon.

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2. A regular nonagon is inscribed in a circle of radius

5 cm .

Find its area and perimeter, considering that all sides are equal and the internal angles can be derived using geometric properties.

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3. The perimeter of a regular decagon is

50 cm .

Determine the length of each side and use it to calculate the area of the decagon.

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4. A stop sign is shaped like a regular octagon with a side length of

30 cm .

If the material costs are based on area, determine the total surface area needed for manufacturing

100

signs.

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5. A company designs floor tiles shaped like regular decagons with a side length of

12 cm .

If a room requires

50

tiles, calculate the total area covered by the tiles.

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Subsection 2.5.4 Area of Irregular Polygons

Activity 2.5.16.

Graph Paper (preferably colored)

Scissors

Glue or tape

Rulers

Graph paper (optional)

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Measuring tape (optional)

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A variety of irregular polygons (either printed or hand-drawn)

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work with a printed or drawn irregular polygon. The task is to divide the irregular polygon into smaller shapes whose areas are easier to calculate (like triangles, rectangles, or trapeziums). Tip: Use straight lines to cut along the diagonals or through the middle of the shape to create triangles or rectangles.
For each smaller shape, ask students to measure the necessary dimensions:

For triangles, they need the base and height.

For rectangles, they need the length and width.

For trapeziums, they need the lengths of the two parallel sides and the height.
Calculate the Area of Each Shape: Once the polygon is broken into smaller shapes, students can calculate the area of each shape using appropriate formulas
Add the Areas: Once the area of each smaller shape has been calculated, students should add up all the areas to find the total area of the original irregular polygo

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Regardless of shape, all polygons are made up of the same parts, sides, vertices, interior angles and exterior angles which may varry in size thus describing why we have irregular polygons versus regular polygons.

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An irregular polgon has a set of atleast two sides or angles that are not the same. This heptagon has many different size angles , making it irregular.

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The interior angles of an irregular nenagon (9 sides) add up to

1, 260^{\circ} .

Because angles are different sizes, individual angles cannot be found the sum of the interior angle.

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Exercise

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1. An irregular pentagon has the following side lengths:

5 cm, 7 cm, 6 cm, 8 cm and 4 cm .

If its total area is estimated using triangulation, determine its approximate area.

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2. A garden is shaped like an irregular hexagon with side lengths

4 m, 6 m, 5 m, 7 m, 8 m and 3 m .

Calculate its perimeter.

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3. A farmer’s land is shaped like an irregular quadrilateral with sides measuring

50 m, 60 m, 40 m and 30 m .

If the land is divided into two triangles for calculation, estimate its total area.

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4. An office space has an irregular pentagonal shape with different side lengths and angles. The flooring cost is calculated based on the total area. If the room is divided into three triangles for estimation, find the approximate flooring cost given a rate of $25 per square meter.

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5. A city park is designed in the shape of an irregular hexagon with measured sides of

20 m, 25 m, 30 m, 28 m, 22 m and 18 m .

If the park’s area is estimated by splitting it into smaller triangles, find the total area.

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6. A regular decagon has a side length of

12 cm .

Calculate its perimeter and area using the formula for the area of a regular polygon.

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7. A large conference room has an octagonal shape with a side length of

9 m .

If the flooring material costs

Ksh . 30

per square meter, find the total flooring cost.

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8. A heptagonal garden has side lengths of

5 m, 7 m, 6 m, 8 m, 9 m, 6 m, and 10 m .

The owner wants to fence the garden. Calculate the total length of fencing required.

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9. A nonagonal water tank has a radius of

5 m .

If it is filled with water, determine the total volume assuming the depth is

4 m .

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10. A decorative fountain is designed in the shape of a regular decagon with a side length of

15 m .

If the cost of tiling is $40 per square meter, determine the total cost of tiling the fountain area.

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11. A regular nonagon is inscribed in a circle of radius

7 cm .

Compute its side length and area.

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12. A heptagonal plot of land has side lengths of

8 m, 10 m, 12 m, 9 m, 11 m, 13 m and 7 m .

Find its perimeter. If the area is approximated by dividing it into triangles, estimate its total area.

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Technology 2.5.16.

"We used to measure with rulers, now we measure with tools that reveal more than the eye can see. Technology is here, not to replace thinking but empower it." Kindly use the links below and explore in these interactive exercises.

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Interactive triangles.

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https://www.mathsisfun.com/geometry/triangles-interactive.html by Math is Fun

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Interactive Quadrilaterals.

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https://www.mathsisfun.com/geometry/quadrilaterals-interactive.html by Math is Fun

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Interactive polygons.

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https://www.mathsisfun.com/geometry/polygons-interactive.html by Math is Fun

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