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Question
The sunshade of a window is in the form of isosceles trapezium whose parallel sides are 81 cm and 64 cm and the distance between them is 6 cm. Find the cost of painting the surface at the rate of ₹ 2 per sq.cm
Solution
Given the parallel sides a = 81 cm, b = 64 cm
Distance between ‘a’ and ‘b’ is height h = 6 cm
Area of the trapezium = `1/2` × h(a + b) sq.units
= `1/2` × 6 × (81 + 64)
= 3 × 145 cm2
= 435 cm2
Cost of painting 1 cm2 = ₹ 2
Cost of painting 435 cm2 = ₹ 435 × 2
= ₹ 870
Cost of painting = ₹ 870.
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