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📐 Chapter 6: Measuring Space: Perimeter and Area • Exercise Set 6.2

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Exercise Set 6.1 4Exercise Set 6.2 4Exercise Set 6.3 3End-of-Chapter Exercises 5All Exercises 16
Q2

The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.

Answer: This is an isosceles trapezium. Let the parallel sides be $a = 20$ cm and $b = 40$ cm, and non-parallel sides be $c = 26$ cm.
Drop perpendiculars from the ends of the shorter parallel side to the longer parallel side. This forms a rectangle in the middle (width 20) and two identical right triangles on the ends.
Base of each right triangle = $\frac{40 - 20}{2} = 10$ cm.
Using Pythagoras to find height $h$: $h^2 + 10^2 = 26^2 \Rightarrow h^2 = 676 - 100 = 576 \Rightarrow h = 24$ cm.
Area of trapezium = $\frac{1}{2}(a + b)h = \frac{1}{2}(20 + 40) \times 24 = \frac{1}{2}(60) \times 24 = 30 \times 24 =$ 720 cm².

20 cm 10 cm 20 cm 10 cm 26 cm 26 cm h = 24 cm
Isosceles trapezium decomposed into a 20×24 rectangle and two 10-24-26 right triangles
Q3

Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.

Answer: Let sides be $a=8$, $b=11$, $c$. Perimeter $= a+b+c = 32 \Rightarrow 8+11+c = 32 \Rightarrow c = 13$ cm.
Semi-perimeter $s = \frac{32}{2} = 16$ cm.
Using Heron's formula: Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{16(16-8)(16-11)(16-13)}$
Area $= \sqrt{16 \times 8 \times 5 \times 3} = \sqrt{16 \times 4 \times 2 \times 15} = 4 \times 2 \sqrt{30} =$ $8\sqrt{30}$ cm².

Q4

The sides of a triangular plot are in the ratio 3:5:7; its perimeter is 300 m. Find its area.

Answer: Let sides be $3x, 5x, 7x$.
$3x + 5x + 7x = 300 \Rightarrow 15x = 300 \Rightarrow x = 20$.
Sides are $a=60$ m, $b=100$ m, $c=140$ m. Semi-perimeter $s = 150$ m.
Heron's Formula: Area $= \sqrt{150(150-60)(150-100)(150-140)}$
Area $= \sqrt{150 \times 90 \times 50 \times 10} = \sqrt{6,750,000} =$ $1500\sqrt{3}$ m².

Q5

One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm², find the length of the shorter diagonal.

Answer: Let the shorter diagonal be $d_1 = x$. Then the longer diagonal $d_2 = 2x$.
Area of a rhombus = $\frac{1}{2} \times d_1 \times d_2$.
$\frac{1}{2} \times x \times 2x = 128 \Rightarrow x^2 = 128$
$x = \sqrt{128} = $ $8\sqrt{2}$ cm.