Question

In Figure , $AB=BC$. If the area of $△$ $ABE$ is $x$, calculate the area of $△$ $ACD$.

• A. $x\sqrt{2}$
• B. $x\sqrt{3}$
• C. $2x$
• D. $3x$
• E. $4x$

Answer 1

The correct option is E $4x$

$AB=BC$ ....Given

So, $AB:AC=1:2$ ...(I)

$\angle BED+\angle BEA={180}^o$ ...Angles in linear pair

$\therefore {118}^o+\angle BEA={180}^o$

$\therefore \angle BEA={62}^o=\angle CDA$ ...(II)

In $△ABE$ and $△ACD$

$\angle A$ is the common angle.

$\therefore \angle BEA=\angle CDA$ ... from (II)

$△ABE\sim △ACD$ ....AAA test of similarity

So, $\frac{A\left(△ABE\right)}{A\left(△ACD\right)}={\left(\frac{AB}{AC}\right)}^2$ ....Theorem on ratio of areas of similar triangles

$\Rightarrow \frac{A\left(△ABE\right)}{A\left(△ACD\right)}={\left(\frac{1}{2}\right)}^2$

$\Rightarrow \frac{x}{A\left(△ACD\right)}=\frac{1}{4}$

$A\left(△ACD\right)=4x$ sq. units