$\left[\begin{array}{l}{[\text { Transfer of sides, diagonals, etc. ] }} \\ {[\text { Area of a trapezoid }}\end{array}\right]$

Find the area of a trapezoid with bases 11 and 4 and diagonals 9 and 12.
Through the vertex of the trapezoid, draw a line parallel to the diagonal.

## Solution

Through vertex $C$ of the smaller base $BC$ of trapezoid $ABCD (BC=4, AD=11, AC=9, BD=12)$, draw a line parallel to diagonal $BD$, intersecting line $AD$ at point $K$. In triangle $ACK$,

$$
AC=9, CK=BD=12, AK=AD+DK=AD+BC=11+4=15.
$$

Since $AK^2=AC^2+CK^2$, triangle $ACK$ is a right triangle. Its area is half the product of the legs, i.e.,

$$
S_{\triangle \mathrm{ACK}}=\frac{1}{2} AC \cdot CK=54
$$

The area of trapezoid $ABCD$ is equal to the area of this triangle, since triangles $ABC$ and $CDK (BC=DK$, and the heights dropped to these sides are equal to the height of the trapezoid) are equal in area.

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-43.jpg?height=371&width=757&top_left_y=1584&top_left_x=656)

## Answer

54.