Question:
The area of trapezium ABCD in the given figure is
(a) $62 \mathrm{~cm}^{2}$
(b) $93 \mathrm{~cm}^{2}$
(c) $124 \mathrm{~cm}^{2}$
(d) $155 \mathrm{~cm}^{2}$
Solution:
(c) $124 \mathrm{~cm}^{2}$
In the right angle triangle BEC, we have:
$E C=\sqrt{17^{2}-15^{2}}=\sqrt{289-225}=\sqrt{64}=8 \mathrm{~cm}$
$\operatorname{ar}($ trapez $. A B C D)=\frac{1}{2} \times($ sum of parallel sides $) \times$ distance between them $=\frac{1}{2} \times 31 \times 8=124 \mathrm{~cm}^{2}$
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