# Square 2

Points D[10,-8] and B[4,5] are opposed vertices of the square ABCD. Calculate area of the square ABCD.

Result

S =  102.5

#### Solution:

$a = \dfrac{ \sqrt{ (10-4)^2+(-8-5)^2 }}{ \sqrt{2} } \ \\ S = a^2 = \dfrac{(10-4)^2+(-8-5)^2}{2} = 102.5$

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