The projection of the diagonal of an isosceles trapezoid on its larger base is $a$, the lateral side is $b$. Find the area of the trapezoid if the angle at its smaller base is $150^{\circ}$.

#
The projection of the diagonal of an isosceles trapezoid on its larger base is equal to the midline of the trapezoid.

## Solution

From vertex $B$ of the smaller base $BC$ of the isosceles trapezoid $ABCD$, drop a perpendicular $BH$ to its larger base $AD$. Then

$$
DH = AD - AH = AD - \frac{1}{2}(AD - BC) = \frac{1}{2}(AD + BC).
$$

Since $BH$ is the leg of the right triangle $ABH$, lying opposite the angle of $30^{\circ}$, then

$$
BH = \frac{1}{2} AB = \frac{b}{2}
$$

Therefore,

$$
S(ABCD) = \frac{1}{2}(AD + BC) \cdot BH = \frac{ab}{2}
$$

## Answer

$\frac{ab}{2}$.