Find the area of a quadrilateral one of whose diagonals is 40 cm

Let $A B C D$ be a quadilateral.

$D$ iagonal, $A C=40 \mathrm{~cm}$

$B L \perp A C$, such that $B L=16 \mathrm{~cm}$

$D M \perp A C$, such that $D M=12 \mathrm{~cm}$

$A$ rea of the quadilateral $=(A$ rea of $\Delta D A C)+(A$ rea of $\Delta A C B)$

$=\left[\left(\frac{1}{2} \times A C \times D M\right)+\left(\frac{1}{2} \times A C \times B L\right)\right] \mathrm{cm}^{2}$

$=\left[\left(\frac{1}{2} \times 40 \times 12\right)+\left(\frac{1}{2} \times 40 \times 16\right)\right] \mathrm{cm}^{2}$

$=(240+320) \mathrm{cm}^{2}$

$=560 \mathrm{~cm}^{2}$

Hence, the area of the quadilateral is $560 \mathrm{~cm}^{2}$.