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Grade 10 Mathematics Teachers’ Book

INNODEMS

Subsection 2.5.2 Area of Quadrilaterals

Subsubsection 2.5.2.1 Area of Squares and Rectangles

The area of a two-dimensional shape is the amount of surface it covers, measured in square units (e.g., cm², m²). A quadrilateral with all right angles and opposite sides parallel is called a rectangle. When all four sides are equal, then it becomes a square.
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Learner Experience 2.5.5.

Work in groups: Form groups of \(2\) or \(3\) students.
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Materials: Grid/graph paper and a ruler
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Instructions:
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Consider the rectangle below:
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  1. Count the number of unit squares inside the rectangle.
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  2. Verify your answer using multiplication.
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Now consider the square below:
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  1. Count the number of unit squares inside the square.
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  2. Verify your answer using multiplication.
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What relationship do you observe between the side lengths and the number of square units?
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Key Takeaway 2.5.10.

The area of a rectangle is found by multiplying its length and width:
\begin{equation*} \textbf{Area of rectangle} = \text{length} \times \text{width} \end{equation*}
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A square is a special rectangle where all sides are equal, so:
\begin{equation*} \textbf{Area of square} = \text{side} \times \text{side} = \text{side}^2 \end{equation*}
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In both cases, the area represents the number of square units that fit inside the shape.
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Example 2.5.11.

Find the area of the following:
  1. A square with side \(9 \, cm\)
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  2. A rectangle with length \(12 \, m\) and width \(7 \, m\)
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Solution.
  1. Area of square =
    \begin{equation*} 9 \times 9 = 81 \, \text{cm}^2 \end{equation*}
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  2. Area of rectangle =
    \begin{equation*} 12 \times 7 = 84 \, \text{m}^2 \end{equation*}
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Example 2.5.12.

A rectangular field has length \(120 \;m\) and width \(75 \;m\text{.}\) Find the area in \(m^2\text{.}\)
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Solution.
Area of rectangle \(=\) length \(\times\) width
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Area \(= 120 \times 75\)
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The field covers \(9,000 \;m^2\)
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Example 2.5.13.

The area of a square is \(196\; m^2\text{.}\) Find the length of its side.
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Solution.
Area of square \(= \text{side}^2\)
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\(196 = \text{side}^2 \)
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\(\text{side} = \sqrt{196} = 14 \;m\)
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The side length is \(14 \;m\)
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Example 2.5.14.

The area of a rectangle is \(150 \;cm^2\) and its width is \(10 \;cm\text{.}\) Find the length?
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Solution.
Area of rectangle \(=\) length \(\times\) width
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\(150 = \text{length} \times 10\)
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length \(= 150 \div 10 = 15 \;cm\)
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The length is \(15 \;cm\)
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Example 2.5.15.

Students work independently
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  1. Find the area of a square with side \(8 \;cm\)
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  2. Find the area of a rectangle with length \(15 \;m\) and width \(9 \;m\)
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  3. A square garden has area \(144 \;m^2\text{.}\) Find the length of one side
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  4. A rectangle has area \(200 \;cm^2\) and width \(8 \;cm\text{.}\) Find the length
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Exercises Exercises

1.
Find the area of a square with side \(15 \, \text{cm}\)
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Answer.
\(225 \, \text{cm}^2\)
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2.
Find the area of a rectangle with length \(8 \, \text{m}\) and width \(5 \, \text{m}\text{.}\)
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Answer.
\(40 \, \text{m}^2\)
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3.
A rectangular field has length \(120 \, m\) and width \(75 \, m\text{.}\) Find the area in \(m^2\text{.}\)
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Answer.
\(9000 \, \text{m}^2\)
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4.
The area of a square is \(196 \, m^2\text{.}\) Find the length of its side.
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Answer.
\(14 \, m\)
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5.
The area of a rectangle is \(150 \, cm^2\) and its width is \(10 \, cm\text{.}\) Find the length.
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Answer.
\(15 \, cm\)
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6.
Which has a greater area:
  • a square with side \(14 \, cm\text{,}\) or
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  • a rectangle with length \(18 \, cm\) and width \(10 \, cm\text{?}\)
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Show your working.
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Answer.
Area of square = \(14 \times 14 = 196 \, \text{cm}^2\) Area of rectangle = \(18 \times 10 = 180 \, \text{cm}^2\) Therefore, the square has a greater area.
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7.
Find the area of a rectangle with length \((2x + 3) \text{cm}\) and width \((x - 1) \text{cm}\text{.}\)
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Answer.
\((2x + 3)(x - 1) = 2x^2 + x - 3 \, \text{cm}^2\)
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8.
A square garden has area \(256 \, m^2\text{.}\) One side is increased by \(2 \, m\text{.}\) Find the new area.
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Answer.
\(324 \, m^2\)
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9.
A classroom floor measures \(9 \, m\) by \(7 \, m\text{.}\) If each floor tile covers \(1 \, m^2\text{,}\) how many tiles are required?
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Answer.
\(63 \, \text{tiles}\)
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10.
A rectangular wall is \(12 \, m\) long and \(3 \, m\) high. If one litre of paint covers \(6 \, m^2\text{,}\) how many litres of paint are needed?
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Answer.
\(6 \, \text{litres}\)
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11.
A farmer wants to fence a rectangular plot measuring \(50 \, m\) by \(30 \, m\text{.}\) Calculate:
  • (i) the area of the plot,
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  • (ii) the perimeter of the plot.
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Answer.
(i) Area = \(1500 \, \text{m}^2\) (ii) Perimeter = \(160 \, \text{m}\)
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12.
A square piece of land has side \(x \, m\text{.}\) Write an expression for its area.
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Answer.
\(x^2 \, \text{m}^2\)
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13.
A rectangle has length \((x + 4)\) metres and width \(6\) metres. Form an algebraic expression for its area.
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Answer.
\(6(x + 4) = 6x + 24 \,\text{m}^2\)
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14.
A rectangular field has length \((2x - 1)\) metres and width \((x + 3)\) metres. Form an expression for its area and simplify.
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Answer.
\((2x - 1)(x + 3) = 2x^2 + 5x - 3 \, \text{m}^2\)
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15.
A square has side \((x + 3)\) cm. Form and simplify an expression for its area.
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Answer.
\((x + 3)^2 = x^2 + 6x + 9 \, \text{cm}^2\)
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16.
The length of a rectangle is twice its width. If the width is \(x\) metres, form an expression for its area.
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Answer.
\(2x \times x = 2x^2 \, \text{m}^2\)
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17.
A rectangular playground is \(40 \, m\) long and \(25 \, m\) wide. A path of width \(2 \, m\) is built all around the outside. Find the area covered by the path.
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Answer.
Area of playground = \(40 \times 25 = 1000 \,\text{m}^2\) Area of playground + path = \((40 + 4)(25 + 4) = 44 \times 29 = 1276 \, \text{m}^2\) Area of path = \(1276 - 1000 = 276 \, \text{m}^2\)
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18.
A square metal sheet has side \(20 \, cm\text{.}\) A smaller square of side \(8 \, cm\) is cut from one corner. Find the remaining area.
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Answer.
\(336 \, \text{cm}^2\)
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19.
A rectangular garden is \(30 \, m\) long and \(18 \, m\) wide. A square pond of side \(6 \, m\) is built inside the garden. Find the remaining area of the garden.
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Answer.
Area of garden = \(30 \times 18 = 540 \, \text{m}^2\) Area of pond = \(6 \times 6 = 36 \, \text{m}^2\) Remaining area = \(540 - 36 = 504 \, \text{m}^2\)
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20.
A square tile has an area of \(64 \, \text{cm}^2\text{.}\) Find the length of one side of the tile.
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Answer.
\(8 \, \text{cm}\)
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21.
The length of a rectangle exceeds its width by \(4 \, \text{m}\text{.}\) If the area is \(105 \, \text{m}^2\text{,}\) find the dimensions of the rectangle and its perimeter.
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Answer.
Let the width be \(x\,\text{m}\text{,}\) so the length is \(x+4\,\text{m}\text{.}\) Area equation:
\begin{equation*} x(x+4)=105 \end{equation*}
gives
\begin{equation*} x^2+4x-105=0. \end{equation*}
Factorising,
\begin{equation*} (x+15)(x-7)=0\text{,} \end{equation*}
so the positive solution is \(x=7\,\text{m}\text{.}\) Thus the length is \(11\,\text{m}\text{.}\) Dimensions: \(7\,\text{m}\) by \(11\,\text{m}\text{.}\) Perimeter:
\begin{equation*} 2(7+11)=36\,\text{m}\text{.} \end{equation*}
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Subsubsection 2.5.2.2 Area of Parallelograms

Learner Experience 2.5.6.

\(\textbf{Exploring the Area of a parallelogram}.\)
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Materials needed.
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Steps for the activity.
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  1. Draw a parallelogram.
    1. On graph paper, draw a parallelogram with a given base and height. Label the vertices A, B, C and D.
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    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.
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    3. Use a ruler to measure and compare opposite sides (AB and CD, BC and AD. )
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Key Takeaway 2.5.17.

A parallelogram is a quadrilateral whose opposite sides are equal and parallel. Which other quadrilateral has similar characteristics like these?
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🔹 The area of a parallelogram is given by: \(Base \times Height\text{:}\)
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\begin{align*} \text{A } = \amp \, b \times h \end{align*}
where \(b\) is the base and \(h \) the perpendicular distance between the given pair of parallel sides.
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Make sure you use the perpendicular height \(h\text{,}\) not the slanted edge of the parallelogram.
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Example 2.5.18.

  • A parallelogram PQRS is of sides 28cm and 7cm. If \(\angle\) QRS is \(75^\circ\text{.}\)
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    a. Find the height the parallelogram using sine rule.
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    b. Find the area of the parallelogram.
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Solution.
From Q drop a perpendicular to meet RS at T considering \(\angle QRS\text{.}\)
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\(\text{Height (QT)} = \text{Hypotenuse(BC)} \times \text{sin} \theta\)
\begin{align*} QT = \amp 7 \, \text{sin} 75^\circ \\ = \amp 6.76 \, \text{cm} \end{align*}
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b.
\begin{align*} \text{Area of PQRS} = \amp RS \times PT \\ = \amp 28 \times 7 \, \text{sin} \, 75^\circ \\ = \amp 189.32 \, \text{cm}^2\\ \amp \end{align*}
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The area of the parallelogram is \(189.32 \, \text{cm}^2.\)
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Example 2.5.19.

A parallelogram \(\mathbf{ABCD}\) has sides \(18\; cm\) and \(12 \;cm\text{.}\) Given that \(\angle \mathbf{BAD} = 38^\circ\text{.}\)
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  1. Find the height of the parallelogram relative to base \(\mathbf{AB}\text{.}\)
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  2. Find the area of the parallelogram.
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  3. Find the area if side \(\mathbf{AD}\) is taken as the base instead.
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Solution.
  1. Drop a perpendicular from point \(\mathbf{D}\) to meet \(\mathbf{AB}\) at point \(\mathbf{E}\)
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    In right triangle \(\mathbf{ADE}\text{:}\)
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    Height\(=\mathbf{AD\;sin\;(\angle BAD)}\)
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    \(\mathbf{DE} = 12 \times sin \;38^\circ\)
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    \(\mathbf{DE} = 7.39 \; cm\)
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  2. \(\text{Area} = \text{Base} \times \text{Height}\)
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    \(\text{Area} = 18 \times 7.39 = 133.02\; cm^2\)
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  3. Now the height relative to base AD comes from side AB
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    Height \(=\) \(AB \; sin \;38^\circ\)
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    Height \(= 11.08 \; cm\)
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    Area \(= 12 \times 11.08 = 132.96 \; cm^2\)
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Exercises Exercises

1.
Find the area of the parallelogram given below.
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Answer.
Area \(= 4 \times 3 = 12 \, \text{cm}^2\text{.}\)
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2.
Find the missing height of the parallelogram ABCD given the area is 24 cm² and base is 6 cm.
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Solution.
\(= \text{Area} \div \text{base} = 24 \div 6 = 4 \, \text{cm}\text{.}\)
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3.
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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Solution.
Area \(= 10 \times 6 = 60 \, \text{m}^2\text{.}\)
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4.
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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Solution.
Area \(= 12 \times 5 = 60 \, \text{ft}^2\)
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5.
A solar panel is shaped like a parallelogram with a base of 8 meters and an inclined height of 4 meters. Find its surface area.
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Solution.
Area \(= 8 \times 4 = 32 \, \text{m}^2\text{.}\)
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Subsubsection 2.5.2.3 Area of Rhombuses

Learner Experience 2.5.7.

\(\textbf{Materials Needed:}\)
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Ruler, Protractor, Pencil, Graph paper, Scissors (optional) and String or thread (optional)
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  • 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.
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  • Measure the Sides using a ruler to confirm that all four sides are of equal length.
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  • Measure the angles using a protractor to measure each of the interior angles. Observe: Opposite angles should be equal.
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  • Draw the Diagonals:
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    Draw diagonal AC and diagonal BD.
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    Measure their lengths and angles at the intersection point.
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    Observe: Diagonals bisect each other at 90°, and they are not equal.
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  • Cut and Fold (Optional):
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    Cut out the rhombus and fold it along both diagonals.
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    Observe the symmetry and how the diagonals act as lines of symmetry.
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\(\textbf{Discussion Questions}\)
  1. What do you notice about the sides and angles of the rhombus?
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  2. How do the diagonals interact with each other?
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  3. What makes a rhombus different from a square or a general parallelogram?
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Key Takeaway 2.5.21.

A \(\textbf{rhombus}\) is an equilateral quadrilateral. All its sides are of \(\textbf{equal length}\text{,}\) vut all it’s angles are not. All its sides are equal and the diagnols bisect at \(90 ^\circ\text{.}\) Which other quadrilateral has all its sides equal?
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Compare the rhombus with the rectangle below.
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Area of a Rhombus.
For a Rhombus with diagonals \(d_1\) and \(d_2\text{,}\) we have:
\begin{equation*} \text{Area} = \tfrac{1}{2} \times d_1 \times d_2 \end{equation*}
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We can also find the area using the side length \(s\) and the perpendicular height \(h\text{:}\)
\begin{equation*} \text{Area} = s \times h \end{equation*}
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\(\textbf{ Properties of a Rhombus}\)
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  • All sides are equal in length.
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  • Opposite angles are equal.
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  • Diagonals bisect each other at right angles.
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  • Diagonals bisect the interior angles.
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  • Some examples of a rhombus are Tiles or patterns in flooring, or The shape of playing cards (diamonds suit).
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Example 2.5.22.

A rhombus has a diagonal of 10 cm and another diagonal of 24 cm. Find:
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i) The area of the rhombus.
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ii).The length of one side of the rhombus.
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Solution.
i.
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One diagonal = 10 cm
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Another diagonal = 24 cm
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\begin{align*} \text{Area} = \amp \frac{1}{2 }\times d_1 \times d_2\\ = \amp \frac{1}{2} \times 10\, \text{cm} \times 24\, \text{cm} \\ = \amp 120 \, \text{cm}^2 \end{align*}
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ii. Length of One Side of the Rhombus
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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.
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Let’s find the length of one side using the Pythagorean Theorem.
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Each side of the rhombus is the hypotenuse of a right triangle with legs:
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\begin{align*} \text{Half of AC } = \amp \frac{10}{2} = 5\, \text{cm}\\ \text{Half of BD } = \amp \frac{24}{2} = 12\, \text{cm}\\ \text{Side} = \amp \sqrt{(5)^2 + (12)^2} \\ = \amp \sqrt{25\,\text{cm} + 144\,\text{cm}} \\ = \amp \sqrt{169 \,\text{cm}}\\ = \amp 13\,\text{cm} \end{align*}
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Example 2.5.23.

A rhombus has sides of length \(16 \; cm\) and one of its angles is \(60^\circ\text{.}\) Find:
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  1. The perpendicular height \(h\) of the rhombus.
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  2. The area of the rhombus (by using \(A = s \times h\))
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  3. The length of the long diagonal.
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  4. The length of the short diagonal (remember that we also have \(A = \frac{1}{2}\times d_1 \times d_2\))
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Solution.
  1. We drop a perpendicular to \(AB\) at \(D\) to create \(DE\text{,}\) which has lenght \(h\text{:}\)
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    Given that \(AD = 16\) cm, we can use trigonometry to get:
    \begin{align*} \sin 60^\circ \amp = \frac{h}{AD} \\ h \amp = 16 \times \frac{\sqrt{3}}{2} \\ \amp = 8\sqrt{3} \text{ cm} \end{align*}
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  2. As \(A = s \times h\text{,}\) we have \(A = 16 \times 8 \sqrt{3} = 128\sqrt{3}\text{ cm}^2\)
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  3. We can move the perpendicular to \(C\) to get another right-angled triangle:
    As discussed in the Key Takeaway, the diagonals of a rhombus bisect the angles, so we have a \(30^\circ\) angle, and so:
    \begin{align*} \sin 30^\circ \amp = \frac{CF}{AC} \\ AC \amp = \frac{h}{\frac{1}{2}}\\ \amp = 16 \sqrt{3} \text{ cm} \end{align*}
    Thus the long diagonal \(AC\) has length \(16 \sqrt{3} \text{ cm}\)
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  4. Since \(AC\) and \(BD\) are the diagonals, we can use the results from (b) and (c) to get:
    \begin{align*} A \amp = \frac{1}{2} \times AC \times BD \\ 128\sqrt{3} \amp = \frac{1}{2} \times 16 \sqrt{3} \times BD \\ BD \amp = 16 \text{ cm} \end{align*}
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Checkpoint 2.5.24.

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Checkpoint 2.5.25.

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Exercises Exercises

1.
A rhombus has diagonals measuring 16 cm and 30 cm.
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  1. Find the area of the rhombus.
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  2. Find the side length of the rhombus.
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Solution.
  1. \begin{align*} \text{Area} \amp = \frac{1}{2} \times d_1 \times d_2\\ \amp = \frac{1}{2} \times 16 \times 30\\ \amp = 240 \, \text{cm}^2 \end{align*}
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  2. The diagonals of a rhombus bisect each other at right angles. The half diagonals are \(8\) cm and \(15\) cm.
    \begin{align*} \text{Side} \amp = \sqrt{8^2 + 15^2}\\ \amp = \sqrt{64 + 225}\\ \amp = \sqrt{289}\\ \amp = 17 \, \text{cm} \end{align*}
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2.
A diamond-shaped road sign is a rhombus with a diagonal of 40 cm and another diagonal of 60 cm. Find:
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  1. The area of the sign.
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  2. The length of one side of the sign.
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Solution.
  1. \begin{align*} \text{Area} \amp = \frac{1}{2} \times d_1 \times d_2\\ \amp = \frac{1}{2} \times 40 \times 60\\ \amp = 1{,}200 \, \text{cm}^2 \end{align*}
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  2. The half diagonals are \(20\) cm and \(30\) cm.
    \begin{align*} \text{Side} \amp = \sqrt{20^2 + 30^2}\\ \amp = \sqrt{400 + 900}\\ \amp = \sqrt{1300}\\ \amp \approx 36.06 \, \text{cm} \end{align*}
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3.
The perimeter of a rhombus is 48 cm. Find the length of one side.
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Solution.
Since all four sides of a rhombus are equal:
\begin{align*} \text{Side} \amp = \frac{\text{Perimeter}}{4}\\ \amp = \frac{48}{4}\\ \amp = 12 \, \text{cm} \end{align*}
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5.
A rhombus has one of its angles measuring \(60°\text{.}\) Find the other three angles.
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Solution.
In a rhombus, opposite angles are equal and consecutive angles are supplementary (sum to \(180^\circ\)).
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The angle opposite to \(60°\) is also \(60°\text{.}\)
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Each of the other two angles is \(180° - 60° = 120°\text{.}\)
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Therefore the four angles are: \(60°, \; 120°, \; 60°, \; 120°\text{.}\)
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6.
A tiling design on a floor is made of rhombus-shaped tiles. Each tile has diagonals of 18 cm and 24 cm.
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  1. Find the area of one tile.
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  2. If 20 such tiles cover a portion of the floor, what is the total area covered?
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Solution.
  1. \begin{align*} \text{Area of one tile} \amp = \frac{1}{2} \times d_1 \times d_2\\ \amp = \frac{1}{2} \times 18 \times 24\\ \amp = 216 \, \text{cm}^2 \end{align*}
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  2. \begin{align*} \text{Total area} \amp = 20 \times 216\\ \amp = 4{,}320 \, \text{cm}^2 \end{align*}
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Subsubsection 2.5.2.4 Area of Trapeziums

Learner Experience 2.5.8.

\(\textbf{materials needed: }\)
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Materials: Ruler, Protractor, Pencil, Graph paper, Scissors (optional) and Calculator (optional)
  • Draw a Trapezium:
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    On graph paper, draw a quadrilateral with one pair of opposite sides parallel (e.g., AB ∥ CD).
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    Make sure the other pair (AD and BC) are not parallel.
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    Label the trapezium as ABCD.
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  • Identify the Parts:
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    Label the bases (parallel sides).
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    Label the legs (non-parallel sides).
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    Draw the height (perpendicular distance between the two bases).
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  • Now, let’s break it down algebraically:
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    Imagine the Area of a Rectangle: If we could transform the trapezium into a rectangle, the area of that rectangle would be the \(\textbf{average of the two bases times the height}\) . This works because the average length of the two parallel sides is a \(\textbf{"representative" length for the trapezium,}\) and multiplying it by the \(\textbf{height}\) gives us the \(\textbf{correct area}\text{.}\)
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  • Mathematical Expression:
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    The average length of the two parallel sides is given by
    \begin{align*} \text{Average of bases}= \amp \frac{\text{Base}_1 + \text{Base}_2}{2} \end{align*}
    Therefore, the area of the trapezium is:
    \begin{align*} Area =\amp \text{Average of bases} \times \text {height} \\ = \amp \frac{\text{Base}_1 + \text{Base}_2}{2} \times \text{height} \end{align*}
    This is the formula for the area of a trapezium.
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  • \(\displaystyle \textbf{Understanding the Formula}\)
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  • The bases are the lengths of the parallel sides.
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    The height is the perpendicular distance between the two bases.
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    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.
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Key Takeaway 2.5.27.

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.
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Properties of a trapezium.
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  • A trapezium looks like a bridge. Suggest other three real life examples of trapeziums.
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  • One pair of sides are parallel (they never meet).
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  • The height is the straight-up distance between the parallel sides.
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  • The angles inside add up to 360°.
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The area of a trapezium is given by the formula:
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\begin{align*} A =\amp \frac{{a+b}}{2} \times {h} \end{align*}
where\(a \) and \(b\) are the two parallel sides and \(h \) height.
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\(\textbf{Study Questions.}\)
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  1. Define a trapezium in your own words.
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  2. Identify which of the following shapes are trapeziums giving reasons why:
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    \(a.\) A square
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    \(b.\) A shape with one pair of parallel sides
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    \(c\) A rectangle
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    • In the images provided below read the question and tick the appropriate answer.
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  3. Draw a trapezium and label its bases, legs, and height.
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  4. How many angles does a trapezium have? What is the sum of all interior angles?
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  5. What is the difference between a trapezium and a parallelogram?
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Example 2.5.28.

A trapezium has parallel sides of lengths \(14 \, \text{cm}\) and \(8 \, \text{cm}\text{.}\) The non-parallel sides are each \(5 \, \text{cm}\text{.}\)
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Calculate the area of the trapezium.
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Solution.
  1. Sketch and analyse the trapezium.
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    Since the non-parallel sides are equal (each \(5 \, \text{cm}\)), the trapezium is isosceles.
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    The difference between the parallel sides is:
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    \(14 - 8 = 6 \, \text{cm}\)
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    This difference is shared equally on both sides:
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    \(\frac{6}{2} = 3 \, \text{cm}\)
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  2. Find the height using Pythagoras’ Theorem.
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    Consider one of the right triangles formed.
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    \begin{gather*} h^2 + 3^2 = 5^2\\ h^2 + 9 = 25 \\ h^2 = 16 \\ h = 4 \, \text{cm} \end{gather*}
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  3. Apply the trapezium area formula.
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    \begin{gather*} A = \frac{1}{2} (a + b)h \\ = \frac{1}{2} (14 + 8) \times 4 \\ = \frac{1}{2} (22) \times 4 \\ = 44 \, \text{cm}^2 \end{gather*}
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Example 2.5.29.

Calculate the area of the figure below.
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Solution.
The given figure is a trapezium with the following vertices:
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  • \(\displaystyle A(-2,3)\)
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  • \(\displaystyle B(2,3)\)
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  • \(\displaystyle C(3,-1)\)
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  • \(D(-3,-1)\)
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    From the diagram, we observe:
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  • The two parallel bases are ABAB and DCDC.
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  • The height is given as 4 cm, which is the perpendicular distance between the parallel bases.
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Step 1: Determine the lengths of the bases
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Length of \(AB:\)
\begin{align*} AB = |x_1-x_2| \amp = |2-(-2)| = |2+2| = 4 \, \text{cm} \end{align*}
Length of \(DC:\)
\begin{align*} DC = |x_1-x_2| \amp = |3-(-3)| = |3+3| = 6 \, \text{cm} \end{align*}
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Step 2: Use the trapezium area formula.
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The area of a trapezium is given by:
\begin{align*} \text{Area} = \amp \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height} \end{align*}
Substituting the values:
\begin{align*} A = \amp \frac{1}{2} \times (4+6) \, \text{cm} \times 4 \, \text{cm}\\ = \amp \frac{1}{2} \times 10 \, \text{cm} \times 4 \, \text{cm} \\ = \amp 20 \, \text{cm}^2 \end{align*}
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THe area of the trapezium is \(20 \, \text{cm}^2\)
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Exercises Exercises

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.
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Solution.
\begin{align*} \text{Area} \amp = \frac{1}{2}(a + b) \times h\\ \amp = \frac{1}{2}(8 + 14) \times 5\\ \amp = \frac{1}{2} \times 22 \times 5\\ \amp = 55 \, \text{cm}^2 \end{align*}
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2.
The bases of a trapezium are 12 cm and 6 cm and its area is \(45 \, \text{cm}^2\text{.}\) Find the height.
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Solution.
\begin{align*} 45 \amp = \frac{1}{2}(12 + 6) \times h\\ 45 \amp = \frac{1}{2} \times 18 \times h\\ 45 \amp = 9h\\ h \amp = 5 \, \text{cm} \end{align*}
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3.
Find the missing angle in a trapezium where three angles are \(70^\circ\text{,}\) \(85^\circ\) and \(95^\circ\text{.}\)
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Solution.
The sum of interior angles of any quadrilateral is \(360^\circ\text{.}\)
\begin{align*} \text{Missing angle} \amp = 360^\circ - (70^\circ + 85^\circ + 95^\circ)\\ \amp = 360^\circ - 250^\circ\\ \amp = 110^\circ \end{align*}
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4.
A trapezium has equal non-parallel sides (legs). What special type of trapezium is this?
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Solution.
A trapezium with equal non-parallel sides (legs) is called an isosceles trapezium. Its base angles are equal and its diagonals are equal in length.
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5.
The parallel sides of a trapezium are 20 m and 30 m and its height is 12 m. Find its area.
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Solution.
\begin{align*} \text{Area} \amp = \frac{1}{2}(a + b) \times h\\ \amp = \frac{1}{2}(20 + 30) \times 12\\ \amp = \frac{1}{2} \times 50 \times 12\\ \amp = 300 \, \text{m}^2 \end{align*}
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6.
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.
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Solution.
\begin{align*} \text{Area} \amp = \frac{1}{2}(a + b) \times h\\ \amp = \frac{1}{2}(150 + 300) \times 200\\ \amp = \frac{1}{2} \times 450 \times 200\\ \amp = 45{,}000 \, \text{m}^2 \end{align*}
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7.
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.
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Solution.
\begin{align*} \text{Area} \amp = \frac{1}{2}(a + b) \times h\\ \amp = \frac{1}{2}(60 + 80) \times 50\\ \amp = \frac{1}{2} \times 140 \times 50\\ \amp = 3{,}500 \, \text{cm}^2 \end{align*}
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8.
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?
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Solution.
\begin{align*} \text{Area} \amp = \frac{1}{2}(a + b) \times h\\ \amp = \frac{1}{2}(1.2 + 1.8) \times 0.75\\ \amp = \frac{1}{2} \times 3.0 \times 0.75\\ \amp = 1.125 \, \text{m}^2 \end{align*}
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Subsubsection 2.5.2.5 Area of Kites

Learner Experience 2.5.9. Constructing a Kite Using Geometry Tools.

How to construct a kite accurately using a ruler, compass, and protractor.
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  • Materials needed:
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  • Graph paper or plain paper
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  • Ruler , Compass, protractor, pencil and eraser
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  • Follow this steps.
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    \(\textbf{Step 1: Drawing the diagnol.}\)
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  • Draw a vertical line of length 10 cm (this will be the longer diagonal, \(d_1\)).
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  • Label the midpoint of this line as O.
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    \(\textbf{Step 2: Draw the Perpendicular Diagonal}\)
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  • At O, use a protractor to draw a perpendicular line.
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  • Mark \(4 \, \text{cm} \) on each side of \(O\) (total shorter diagonal \(d_2 = \)\(8cm\text{.}\)
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    \(\textbf{Step 3: Mark the Kite's Vertices}\)
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  • Label the four points where the diagonals intersect as A, B, C, and D.
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  • Connect \(\textbf{A}\) to \(\textbf{B}\text{,}\) \(\textbf{B}\) to \(\textbf{C}\text{,}\) \(\textbf{C}\) to \(\textbf{D}\) and \(\textbf{D}\) to \(\textbf{A}\text{.}\)
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    \(\textbf{Step 3:Check Properties }\)
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  • Measure adjacent sides to ensure two pairs are equal.
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  • Verify opposite angles (one pair should be equal)
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  • Confirm diagonals are perpendicular.
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Design your own kite patterns on graph paper and justify why their design would be aerodynamic and stable in the air.
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Key Takeaway 2.5.31.

A kite is a quadrilateral with two pairs of adjacent sides equal in length and one pair of opposite angles equal.
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A kite can be analyzed using coordinate geometry when its vertices are given on a Cartesian plane.
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Since the diagonals are perpendicular, right-angled triangles are formed.
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We can use trigonometric ratios (sine, cosine, tangent) to find missing angles and side lengths.
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Area of a Kite.
If a kite has diagonals \(d_1\) and \(d_2\text{,}\) then the area is half the product of the diagonals:
\begin{align*} \text{Area} \amp = \frac{1}{2} \times d_1 \times d_2 \end{align*}
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Example 2.5.32.

A kite whose has the diagonals are 16 cm and 12 cm.
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a) Find the area of the a kite
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b) If one pair of adjacent sides is 10 cm, find the perimeter of the kite.
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c) Find the angles of the kite using trigonometry.
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Solution.
\(\textbf{ Find the Area of the Kite}\) Let \(d_1 = 16 \, \text{cm}\) and \(d_2 = 12 \, \text{cm}\)
\begin{align*} \textbf{Area} = \amp \frac{1}{2} \times d_1 \times d_2 \end{align*}
\(\text{substitute} \, d_1 = 16 \, \text{cm and } d_2 = 12 \, \text{cm}\)
\begin{align*} \text{Area} = \amp \frac{1}{2} \times 16 \, \text{cm } \times 12 \, \text{cm} \\ = \amp \frac{192}{2} \, \text{cm}^2 \\ = \amp 96 \, \text{cm}^2 \end{align*}
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(b) Find the Perimeter of the Kite. The diagonals bisect each other at right angles, so each half-diagonal forms a right-angled triangle. \(\frac{d_1}{2} = 6\text { cm }\) and \(\frac{d_2}{2} = 8 \text{cm}\) Using the Pythagoras Theorem:
\begin{align*} \amp a^2 + b^2 = c^2\\ = \amp 6^2 + 8^2\\ c^2 = \amp 36 + 64\\ \sqrt{c^2} = \amp \sqrt{100} \\ c = \amp 10 \, \text{cm} \end{align*}
Since a kite has two pairs of equal sides, the perimeter is:
\begin{align*} P = \amp 2(a+b) \\ = \amp 2(10 + 10) \, \text{cm}\\ = \amp 40 \, \text{cm} \end{align*}
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(c) Finding the Angles Using Trigonometry. Using trigonometry in the right-angled triangle:
\begin{align*} \text{tan} \, \theta = \amp \frac{\text{Opposite}}{\text{adjacent}} \\ = \amp \frac{6}{8} \\ \theta = \amp \text{tan}^{-1} \, 0.75 \\ \amp \\ \theta = \amp 36.87^\circ \end{align*}
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A kite is a quadrilateral, meaning it has four angles.
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The sum of the interior angles of any quadrilateral is given by the formula:
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Sum of Interior Angles=\((n-2) \times 180^\circ\)
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where \(n = 4\)(since a kite has 4 sides) \((4-2) \times 180^\circ \) \(= 2 \times 180^\circ\) \(= 360^\circ\)
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Since the kite is symmetric, the larger angle ( \(\alpha \) )is
\begin{align*} = \amp 360^\circ - 2 \theta \\ = \amp 360^\circ - 2 (36.87)^\circ \\ = \amp 360^\circ - 73.74 ^\circ \\ 2(\alpha)= \amp 286.26 ^\circ\\ \alpha = \amp 143.13^\circ \end{align*}
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The angles of the kite are:\(143^\circ, \, 143.13^\circ, \, 36.87^\circ \, \text{and} \, 36.87^\circ\)
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Example 2.5.33.

A kite has diagonals of lengths \(18\) cm and \(14\) cm.
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  1. Find the area of the kite.
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  2. Determine all interior angles of the kite using trigonometry.
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Solution.
  1. Area of the kite:
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    Let \(d_1 = 18 \; \text{cm}\) and \(d_2 = 14\; \text{cm}\)
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    \begin{align*} \text{Area} = \amp \frac{1}{2} \times d_1 \times d_2\\ = \amp \frac{1}{2} \times 18 \times 14\\ = \amp 126 \text{ cm}^2 \end{align*}
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  2. Interior angles using trigonometry:
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    Consider the triangle formed by half-diagonals:
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    \(\tan \theta = \frac{d_1/2}{d_2/2} = \frac{9}{7}\)
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    \(\theta = \tan^{-1}(\frac{9}{7})\)
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    \(\theta \approx 52.13^\circ\)
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    The other angles of the kite:
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    \(\alpha = 180^\circ - 2 \theta = 180 - 2(52.13) \approx 75.74^\circ\)
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    Therefore, the angles of the kite are approximately:
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    \(75.74^\circ, 75.74^\circ, 104.26^\circ, 104.26^\circ\)
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Exercises Exercises

1.
Find the area of the kite given below.
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Solution.
The diagonals are \(d_1 = 6\) cm and \(d_2 = 4\) cm.
\begin{align*} \text{Area} \amp = \frac{1}{2} \times d_1 \times d_2\\ \amp = \frac{1}{2} \times 6 \times 4\\ \amp = 12 \, \text{cm}^2 \end{align*}
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2.
A kite has diagonals of length 10 cm and 8 cm. Find its area.
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Solution.
\begin{align*} \text{Area} \amp = \frac{1}{2} \times d_1 \times d_2\\ \amp = \frac{1}{2} \times 10 \times 8\\ \amp = 40 \, \text{cm}^2 \end{align*}
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3.
A playground has a kite-shaped design with diagonals measuring 15 meters and 12 meters. Find its area.
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Solution.
\begin{align*} \text{Area} \amp = \frac{1}{2} \times d_1 \times d_2\\ \amp = \frac{1}{2} \times 15 \times 12\\ \amp = 90 \, \text{m}^2 \end{align*}
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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.
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Solution.
\begin{align*} \text{Area} \amp = \frac{1}{2} \times d_1 \times d_2\\ \amp = \frac{1}{2} \times 60 \times 45\\ \amp = 1{,}350 \, \text{m}^2 \end{align*}
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5.
Engineers are designing a kite-shaped solar panel with diagonals of 30 m and 18 m. What is the total solar-collecting area?
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Solution.
\begin{align*} \text{Area} \amp = \frac{1}{2} \times d_1 \times d_2\\ \amp = \frac{1}{2} \times 30 \times 18\\ \amp = 270 \, \text{m}^2 \end{align*}
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6.
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?
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Solution.
\begin{align*} \text{Area} \amp = \frac{1}{2} \times d_1 \times d_2\\ \amp = \frac{1}{2} \times 50 \times 40\\ \amp = 1{,}000 \, \text{m}^2 \end{align*}
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