Measuring Space: Perimeter and Area | Exercise 6.2

Question 2

The parallel sides of a trapezium are 40 cm40\text{ cm} and 20 cm20\text{ cm}. If its non-parallel sides are both equal, each being 26 cm26\text{ cm}, find the area of the trapezium.

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Solution
Understand the Question
  • An isosceles trapezium has equal non-parallel sides (26 cm26\text{ cm} each) and parallel sides of lengths 40 cm40\text{ cm} and 20 cm20\text{ cm}.
  • By drawing perpendicular heights from the vertices of the shorter parallel side to the longer base, the trapezium is split into a central rectangle and two congruent right-angled triangles.
  • We first find the base of these right-angled triangles, use the Pythagoras theorem to compute the vertical height hh, and then apply the formula: Area=12×(sum of parallel sides)×height\text{Area} = \dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}

Step 1 · Find the Base of the Right Triangle

Draw perpendiculars from the vertices of the shorter base (20 cm20\text{ cm}) to the longer base (40 cm40\text{ cm}).Diagram 1

Since the non-parallel sides are equal, the two right-angled triangles formed on either side are congruent.

Difference between parallel sides=40 cm−20 cm=20 cm\begin{aligned} \text{Difference between parallel sides} &= 40\text{ cm} - 20\text{ cm} \\[0.6em] &= 20\text{ cm} \end{aligned} Base of each right triangle=20 cm2=10 cm\begin{aligned} \text{Base of each right triangle} &= \dfrac{20\text{ cm}}{2} \\[0.6em] &= 10\text{ cm} \end{aligned}

Step 2 · Calculate the Height

In one of the right-angled triangles, using the Pythagoras theorem where the hypotenuse is 26 cm26\text{ cm} and the base is 10 cm10\text{ cm}:

h2=(hypotenuse)2−(base)2=(26 cm)2−(10 cm)2=676 cm2−100 cm2=576 cm2h=576 cm2=24 cm\begin{aligned} h^2 &= (\text{hypotenuse})^2 - (\text{base})^2 \\[0.6em] &= (26\text{ cm})^2 - (10\text{ cm})^2 \\[0.6em] &= 676\text{ cm}^2 - 100\text{ cm}^2 \\[0.6em] &= 576\text{ cm}^2 \\[0.6em] h &= \sqrt{576\text{ cm}^2} \\[0.6em] &= 24\text{ cm} \end{aligned}

Step 3 · Calculate the Area of the Trapezium

Using the area formula for a trapezium:

Area=12×(sum of parallel sides)×height=12×(40 cm+20 cm)×24 cm=12×60 cm×24 cm=30 cm×24 cm=720 cm2\begin{aligned} \text{Area} &= \dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} \\[0.6em] &= \dfrac{1}{2} \times (40\text{ cm} + 20\text{ cm}) \times 24\text{ cm} \\[0.6em] &= \dfrac{1}{2} \times 60\text{ cm} \times 24\text{ cm} \\[0.6em] &= 30\text{ cm} \times 24\text{ cm} \\[0.6em] &= 720\text{ cm}^2 \end{aligned}
Answer

720 cm2720\text{ cm}^2

Common Mistakes
  • Using Slant Height as Vertical Height: Directly multiplying 12×(40+20)×26\dfrac{1}{2} \times (40 + 20) \times 26 instead of computing the perpendicular height h=24 cmh = 24\text{ cm}.
  • Base Calculation Error: Forgetting to divide the difference of the bases (20 cm20\text{ cm}) by 22, erroneously using 20 cm20\text{ cm} as the triangle base instead of 10 cm10\text{ cm}.

More questions in Exercise 6.2

Q1

Find the area of triangle ADE in Fig. 6.31.

Q2

The parallel sides of a trapezium are 40 cm40\text{ cm} and 20 cm20\text{ cm}. If its non-parallel sides are both equal, each being 26 cm26\text{ cm}, find the area of the trapezium.

Q3

Find the area of a triangle, given that its sides are 8 cm8\text{ cm} and 11 cm11\text{ cm} long, and its perimeter is 32 cm32\text{ cm}.

Q4

The sides of a triangular plot are in the ratio 3:5:73 : 5 : 7; its perimeter is 300 m300\text{ m}. Find its area.

Q5

One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm2128\text{ cm}^2, find the length of the shorter diagonal.

Q6

ABCDABCD is a parallelogram. PP and QQ are any two points on side ABAB. What can you say about the ratio area(ΔPCD):area(ΔQCD)\text{area}(\Delta \text{PCD}) : \text{area}(\Delta \text{QCD})?

Q7

OO is any point on the diagonal PRPR of a parallelogram PQRSPQRS. Prove that the areas of triangles PSOPSO and PQOPQO are equal.

Q8

If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.)

Q9

In ΔABC\Delta\text{ABC}, the midpoint of BC\text{BC} is D\text{D} (Fig. 6.32). Median AD\text{AD} is drawn. P\text{P} is any point on AD\text{AD}. Show that area(ΔABP)=area(ΔACP)\text{area}(\Delta\text{ABP}) = \text{area}(\Delta\text{ACP}).

Q10

Given a square ABCDABCD, let PP be a point within it. Join PAPA, PBPB, PCPC, PDPD (Fig. 6.33). What is the ratio of the areas of the red region (ΔPAB\Delta PAB and ΔPCD\Delta PCD) and the green region (ΔPBC\Delta PBC and ΔPDA\Delta PDA)?

Q11

In ΔABC\Delta\text{ABC}, DD is the midpoint of AB\text{AB}. PP is any point on BC\text{BC}, and QQ is a point on AB\text{AB} such that CQ∥PD\text{CQ} \parallel \text{PD}. PQ\text{PQ} is joined (Fig. 6.34). Prove that

Area(ΔBPQ)=12Area(ΔABC)\text{Area}(\Delta\text{BPQ}) = \dfrac{1}{2} \text{Area}(\Delta\text{ABC})

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