In an isosceles trapezoid, the bases are 40 and 24, and its diagonals are perpendicular to each other. Find the area of the trapezoid.

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Through the vertex of the trapezoid, draw a line parallel to the diagonal.

## Solution

Let the bases $AD$ and $BC$ of the isosceles trapezoid $ABCD$ be 40 and 24, respectively, and the diagonals $AC$ and $BD$ are perpendicular. Draw a line through vertex $C$ parallel to diagonal $BD$. Let this line intersect the extension of base $AD$ at point $K$. Then $\triangle ACK$ is an isosceles right triangle with base $AK = AD + DK = 64$ and height equal to half of $AK$. Therefore,

$$
S_{\mathrm{ABCD}}=S_{\triangle \mathrm{ACK}}=\frac{1}{2} \cdot 64 \cdot 32=32^{2}=1024 .
$$

## Answer

1024.