Question

Let $ABCD$ be a square and $E$ be a point outside $ABCD$ such that $E,A,C$ are collinear in that order. Suppose $EB=ED=\sqrt{130}$ and the areas of triangle $EAB$ and square $ABCD$ are equal. Then the area of square $ABCD$ is?

• A. $8$
• B. $10$
• C. $\sqrt{120}$
• D. $\sqrt{125}$

Answer 1

The correct option is B $10$

Let the side length of square be $a$

Since $E,A,C$ are collinear$,\angle EAB={135}^{\circ }$

Let $\angle AEB=\theta$ , then, $\angle ABE={45}^{\circ }-\theta$

Applying Sine rule in $△EAB,$

$\frac{\sqrt{130}}{sin\left(\angle EAB\right)}=\frac{AB}{sin\left(\angle AEB\right)}$

$\Rightarrow \frac{\sqrt{130}}{1/\sqrt{2}}=\frac{a}{sin\theta }$

$\Rightarrow a=\sqrt{260}sin\theta$ $-------\left(1\right)$

Area of $△EAB=\frac{1}{2}\left(AB\right)\left(EB\right)sin\left(\angle ABE\right)=\frac{1}{2}\left(a\right)\left(\sqrt{130}\right)sin({45}^{\circ }-\theta )$

Area of square $ABCD=a^2$

Since, the areas of $△EAB$ and square $ABCD$ are equal,

$\frac{1}{2}\left(a\right)\left(\sqrt{130}\right)sin({45}^{\circ }-\theta )=a^2$

$\Rightarrow a=\frac{\sqrt{130}}{2}sin({45}^{\circ }-\theta )$ $-------\left(2\right)$

From $\left(1\right)$ and $\left(2\right),$

$\sqrt{260}sin\theta =\frac{\sqrt{130}}{2}sin({45}^{\circ }-\theta )$

$\Rightarrow 2\sqrt{2}sin\theta =\frac{cos\theta -sin\theta }{\sqrt{2}}$

$\Rightarrow tan\theta =\frac{1}{5}$

$\Rightarrow sin\theta =\frac{1}{\sqrt{26}}$

So, $a=\sqrt{260}\times \frac{1}{\sqrt{26}}=\sqrt{10}$

Area of square $=a^2=10$