Let ABCD be a square and E be a point outside | vedclass

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(B) Let the side length of the square $ABCD$ be $x$. The area of the square is $x^2$.Let $M$ be the intersection of the diagonals $AC$ and $BD$. Since

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$10$

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(B) Let the side length of the square $ABCD$ be $x$. The area of the square is $x^2$.Let $M$ be the intersection of the diagonals $AC$ and $BD$. Since $ABCD$ is a square,the diagonals bisect each other at right angles,so $AM = MC = BM = MD = \frac{x}{\sqrt{2}}$.Since $E, A, C$ are collinear,$EA$ lies on the diagonal $AC$. In $	riangle EBD$,$EB = ED = \sqrt{130}$ and $BD$ is the base. The altitude from $E$ to $BD$ is $EM$. Since $	riangle EBD$ is isosceles,$EM \perp BD$.In $	riangle EBM$,by the Pythagorean theorem,$EM^2 + BM^2 = EB^2$.$EM^2 + (\frac{x}{\sqrt{2}})^2 = 130$ $\Rightarrow EM^2 = 130 - \frac{x^2}{2}$ $\Rightarrow EM = \sqrt{130 - \frac{x^2}{2}}$.The area of $	riangle EAB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AB 	imes EA 	imes \sin(\angle EAB)$. Since $E, A, C$ are collinear,$\angle EAB = 135^\circ$. Thus,$	ext{Area} = \frac{1}{2} x \cdot EA \cdot \sin(135^\circ) = \frac{1}{2} x \cdot EA \cdot \frac{1}{\sqrt{2}}$.Alternatively,using the coordinates with $A$ at $(0,0)$,$B$ at $(x,0)$,$D$ at $(0,x)$,and $C$ at $(x,x)$,the line $AC$ is $y=x$. Point $E$ lies on $y=x$ with $x_E $EB^2 = (x+a)^2 + a^2 = 130$ and $ED^2 = a^2 + (x+a)^2 = 130$. Area of $	riangle EAB = \frac{1}{2} |x_A(y_B-y_E) + x_B(y_E-y_A) + x_E(y_A-y_B)| = \frac{1}{2} |0 + x(-a-0) + (-a)(0-0)| = \frac{ax}{2}$.Given $\frac{ax}{2} = x^2 \Rightarrow a = 2x$.Substitute $a=2x$ into $EB^2 = (x+2x)^2 + (2x)^2 = 130$ $\Rightarrow 9x^2 + 4x^2 = 130$ $\Rightarrow 13x^2 = 130$ $\Rightarrow x^2 = 10$.The area of the square is $x^2 = 10$.

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

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