Let AB = x = CD , AD = y = BC , AC = z

$z^{2}=x^{2}+y^{2}$........1

Given z + x = 5y ( as AC + AB = 5 AD)

z = 5y - x ..........2

squaring both sides in equation 2 we get

$z^{2}=25y^{2}+x^{2}-10\ast x \ast y$........3

Now Subtracting 1 from 3 we get

$0 = 24y^{2} - 10\ast x \ast y$

x / y = 24 / 10 = 12 / 5

z - y = 8 (as AC - AD = 8) $\Rightarrow$ z = 8 + 5 = 13

we will get the area of rectangle ABCD as = x * y = 60

**Option C will be the Correct Answer**.