Answer: A(□ABCD) = 192 sq. units
In right-angled \(\triangle ADB\) (\(\angle ADB = 90^\circ\)):
\[BD^2 = 25^2 - 15^2 = 400,\quad BD = 20\]Draw DE ⊥ AB. The altitude on the hypotenuse:
\[DE = \frac{AD \times BD}{AB} = \frac{15 \times 20}{25} = 12\] \[AE = \sqrt{15^2 - 12^2} = 9\]Similarly draw CF ⊥ AB; \(\triangle ACB\) is right angled at C with BC = 15, so in the same way FB = 9 and CF = 12.
□DEFC is a rectangle, so DC = EF = AB − AE − FB = 25 − 9 − 9 = 7.
\[A = \frac{1}{2}(AB + DC) \times DE\] \[= \frac{1}{2}(25 + 7) \times 12 = 192\]