2012-II-15

Ans: A
In $\Delta ADC$,

$\begin{array}{rcl}
AC & = & \sqrt{8^2 + 6^2} \\
& = & 10 \mbox{ cm}
\end{array}$

In $\Delta ABC$,

$\begin{array}{rcl}
AB & = & \sqrt{26^2 – 10^2} \\
& = & 24 \mbox{ cm}
\end{array}$

Therefore, the area of quadrilateral $ABCD$

$\begin{array}{cl}
= & \dfrac{6\times 8 }{2}+ \dfrac{10\times 24}{2} \\
= & 144 \mbox{ cm}^2
\end{array}$