Question

Solve. Question no. 15.
Step wise solution

Answer 1

Given an isosceles trapezium, ABCD. AB = 52 cm, DC = 28 cm, AC = BD = 41 cm.

Let the distance between AB and CD = h, AD = BC = y, and <ABD = θ. <BCD = 180 - θ.

In triangle BCD, BD^2 = CD^2+BC^2–2*BD*BC*cos BCD, or

41^2 = y^2+28^2–2*y*28*cos(180 - θ) …(1)

sin θ = h/y and cos θ = 12/y.

Or (1) becomes 41^2 = y^2+28^2+2*y*28*cos θ

1681 - 784 = y^2+2*y*28*12/y, or

897 = y^2+ 672, or

y^2 = 897–672 = 225, or

y = 15 cm.

The equal sides = 15 cm.

An isosceles trapezium is a cyclic quadrilateral, so we can apply Brahmagupta’s formula to get the area as [(s-a)(s-b)(s-c)(s-d)]^0.5.

Here, 2s = 28+15+52+15 = 110, or s = 55.

Area = [(55–28)(55–15)(55–15)(55–52)]^0.5

= [27*40*40*3]^0.5

= 360 sq cm.

Answer Area of trapezium = 360 sq cm.

Check: If area = 360 sq cm, sum of parallel sides = 28+52 = 80 cm. So the distance between AB and CD = 360*2/80 = 9 cm. sin θ = 9/15 so cos θ = 12/15 = 12/y or y = 15 cm. Correct
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