If two opposite vertices of a square are (5, | vedclass

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(A) Let the opposite vertices of the square be $A(5, -4)$ and $C(-3, 2)$.The length of the diagonal $d$ is the distance between these two points:$d =

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(A) Let the opposite vertices of the square be $A(5, -4)$ and $C(-3, 2)$.The length of the diagonal $d$ is the distance between these two points:$d = \sqrt{(5 - (-3))^2 + (-4 - 2)^2}$$d = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$.For a square with side length $s$,the diagonal $d = s\sqrt{2}$.Thus,$s\sqrt{2} = 10 \implies s = \frac{10}{\sqrt{2}} = 5\sqrt{2}$.The area of the square is $s^2 = (5\sqrt{2})^2 = 25 	imes 2 = 50$ square units.

If two opposite vertices of a square are $(5, -4)$ and $(-3, 2)$,find its area.

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