In the adjoining figure, ABCD is a trapezium in which AB || DC; AB = 7 cm; AD = BC = 5 cm and the distance between AB and DC is 4 cm. Find the length of DC and hence, find the area of trap. ABCD.

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Solution

∆ADL is a right angle triangle.
So, DL = $\sqrt{(5^{2} - 4^{2})} = \sqrt{9} = 3 cm$
Similarly, in ∆BMC, we have:
MC = $\sqrt{(5^{2} - 4^{2})} = \sqrt{9} = 3 cm$
∴ DC = DL + LM + MC = 3 + 7 + 3 = 13 cm
Now, ar(trapezium. ABCD) = $\frac{1}{2}$ ⨯ (sum of parallel sides) ⨯ (distance between them)
= $\frac{1}{2}$ ⨯ (7 + 13) ⨯ 4
= 40 cm²
Hence, DC = 13 cm and area of trapezium = 40 cm²

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