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

Question

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

Team Careers360 · Accepted Answer

Correct Answer: 144

Solution : Given that the area of triangle $\mathrm{ABE}$ is $72\;\mathrm{cm^2}$.
The triangle is an isosceles.
$\mathrm{AB} = \mathrm{BE} = \sqrt{2 imes 72} = 12\;\mathrm{cm}$
$\mathrm{AB = 2 AD}$
⇒ $\mathrm{AD} = \frac{\mathrm{AB}}{2} = 6\;\mathrm{cm}$
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 $\mathrm{ABCD}$ $=\frac{1}{2} imes (\mathrm{sum\; of\; parallel\; sides}) imes \mathrm{height} = \frac{1}{2} imes (18 + 6) imes 12 = 144\;\mathrm{cm^2}$
Hence, the correct answer is 144.

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Question : In the given figure, the area of isosceles triangle ABE is 72cm2 and BE=AB and AB=2AD, AE∥DC, then what is the area (incm2) of the trapezium ABCD? Option 1: 108Option 2: 124Option 3: 136Option 4: 144

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Question : In the given figure, the area of isosceles triangle $ABE$ is $72 {cm}^{2}$ and $BE = AB$ and $AB = 2 AD$ , $AE ∥ DC$ , then what is the area (in ${cm}^{2}$ ) of the trapezium $ABCD$ ?

Option 1: 108
Option 2: 124
Option 3: 136
Option 4: 144