The area of trapezium ABCD in the given figure is


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) $155 \mathrm{~cm}^{2}$



(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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