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 $10$


$EB=ED=\sqrt{130}$
Let the side of square be $x$
Area of square $ABCD$ = area of triangle $EAB$,
$\therefore x^2=\frac{1}{2}\times x\times \sqrt{130}\times \sin \theta \Rightarrow \sin \theta =\frac{2x}{\sqrt{130}}.......\left(i\right)$

Area of the
$△EBD=△EAB+△EAD+△ABD\Rightarrow \frac{1}{2}\times \sqrt{130}\times \sqrt{130}\times \sin ({90}^{\circ }-2\theta )=x^2+x^2+\frac{x^2}{2}\Rightarrow 65\cos 2\theta =\frac{5x^2}{2}\Rightarrow 1-2{\sin }^2\theta =\frac{5x^2}{130}\Rightarrow 1-2\times \frac{4x^2}{130}=\frac{5x^2}{130}\Rightarrow 1=\frac{13x^2}{130}\Rightarrow x^2=10$

Hence the area of the square $=10$