Question : In the given figure, the area of isosceles triangle $\mathrm{ABE}$ is $72\;\mathrm{cm^2}$ and $\mathrm{BE = AB}$ and $\mathrm{AB = 2 AD}$, $\mathrm{AE \parallel DC}$, then what is the area (in$\;\mathrm{cm^2}$) of the trapezium $\mathrm{ABCD}$?
Option 1: 108
Option 2: 124
Option 3: 136
Option 4: 144
Correct Answer: 144
Solution :
Given that the area of triangle $\mathrm{ABE}$ is $72\;\mathrm{cm^2}$.
The triangle is an isosceles.
$\mathrm{AB} = \mathrm{BE} = \sqrt{2 \times 72} = 12\;\mathrm{cm}$
$\mathrm{AB = 2 AD}$
⇒ $\mathrm{AD} = \frac{\mathrm{AB}}{2} = 6\;\mathrm{cm}$
Since $\mathrm{AE} \parallel \mathrm{DC}$
$\mathrm{CE} = \mathrm{AD} = 6\;\mathrm{cm}$, and $\mathrm{BC} = \mathrm{BE} + \mathrm{CE} = 12 + 6 = 18\;\mathrm{cm}$
The area of trapezium $\mathrm{ABCD}$ $=\frac{1}{2} \times (\mathrm{sum\; of\; parallel\; sides}) \times \mathrm{height} = \frac{1}{2} \times (18 + 6) \times 12 = 144\;\mathrm{cm^2}$
Hence, the correct answer is 144.
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