CAT 2001 — QA Question 22

$AC=\sqrt{{24}^2+{23}^2}=40$

Let us draw DE, which is perpendicular to AC.

Since ΔADC is isosceles, therefore DE will be its median and altitude both.

∴ DE = $\sqrt{{25}^2-{20}^2}=15$

∴ Required area = $\left(\frac{1}{2}\times 24\times 32\right)+\left(\frac{1}{2}\times 40\times 15\right)$ = 384 + 300 = 684

Hence, option (d).