Let A B C D be a square and E be a point outside A B C D such tha… - Vidyadip

MCQ

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

A
$8$
✓
$10$
C
$\sqrt{120}$
D
$\sqrt{125}$

✓

Answer

Correct option: B.

$10$

b
(b)

Given, area of $\triangle E A B=$ area of square $A B C D$

$E B=E D=\sqrt{130}$

Let side of square $=x$

$B M=\frac{x}{\sqrt{2}}=A M$

Area of $\triangle A E B=$ Area of $\triangle B E M$ - area of

$=\frac{1}{2} E M \times B M-\frac{1}{2} A M \times B M$

$=\frac{1}{2} B M(E M-A M)$

$=\frac{1}{2} \frac{x}{\sqrt{2}}\left(\sqrt{\left.130-\frac{x^2}{2}-\frac{x}{\sqrt{2}}\right)}\right.$

$=\frac{1}{2}+\frac{x}{\sqrt{2}}\left(\sqrt{\left.130-\frac{x^2}{2}-\frac{x}{\sqrt{2}}\right)=x^2}\right.$

$=\sqrt{130-\frac{x^2}{2}}=2 \sqrt{2} x+\frac{x}{\sqrt{2}}$

$130-\frac{x^2}{2}=\left(\begin{array}{c}5 x \\ \sqrt{2}\end{array}\right)^2$

$130-\frac{x^2}{2}=\begin{array}{c}25 x^2 \\ 2\end{array}$

$13 x^2=130 \Rightarrow x^2=10$

$\therefore$ Area of square $=10$

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