The sunshade of a window is in the form of isoceles trapezium whose p

1.	The sunshade of a window is in the form of isoceles 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 Rs 2 per sq. cm.
Answer» Given the parallel sides a = 81 cm; b = 64 cm Distance between ‘a’ and ‘b’ is height h = 6 cm Area of the trapezium = \(\frac{1}{2}\) x h(a + b) sq. units = \(\frac{1}{2}\) x 6 x (81 + 64) = 3 x 145 cm² = 435 cm² Cost of painting 1 cm² = Rs 2 Cost of painting 435 cm² = Rs 435 x 2 = Rs 870 Cost of painting = Rs 870

The sunshade of a window is in the form of isoceles 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 Rs 2 per sq. cm.

Answer»

Given the parallel sides a = 81 cm; b = 64 cm

Distance between ‘a’ and ‘b’ is height h = 6 cm

Area of the trapezium = \(\frac{1}{2}\) x h(a + b) sq. units

= \(\frac{1}{2}\) x 6 x (81 + 64) = 3 x 145 cm² = 435 cm²

Cost of painting 1 cm² = Rs 2

Cost of painting 435 cm² = Rs 435 x 2 = Rs 870

Cost of painting = Rs 870

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The sunshade of a window is in the form of isoceles 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 Rs 2 per sq. cm.

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