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

ANSWER:
∆ ADL is a right angle triangle.
So, DL = 52 - 42 = 9 = 3 cm
Similarly, in ∆ BMC, we have:
MC = 52 - 42 = 9 = 3 cm
∴ DC = DL + LM + MC = 3 + 7 + 3 = 13 cm
Now, ar(trapezium. ABCD) = 1/2⨯ (sum of parallel sides) ⨯ (distance between them)
=1/2 ⨯ (7 + 13) ⨯ 4
= 40 cm 2
Hence, DC = 13 cm and area of trapezium = 40 cm 2

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