Q.18. ABCD is trapezium in which AB = 7 cm, AD = 5 cm, BC = 5 cm, DC = p cm and distance between AB and CD is 1 less than BC. Find the value of 'p' and its area.

Dear Student,

∆ADL is a right angle triangle.
So, DL = $\sqrt{\left(5^2 - 4^2 \right)} = \sqrt{9} = 3 \mathrm{cm}$
Similarly, in ∆BMC, we have:
MC = $\sqrt{\left(5^2 - 4^2 \right)} = \sqrt{9} = 3 \mathrm{cm}$
∴ 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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