The diagonals of a trapezoid are perpendicular to each other. One of them is equal to 6, and the second one forms an angle of $30^{\circ}$ with the base. Find the midline of the trapezoid.
Through the vertex of the trapezoid, draw a line parallel to the first diagonal.

## Solution

Let the diagonal $AC$ of trapezoid $ABCD$ be 6, and the diagonal $BD$ forms an angle of $30^{\circ}$ with the larger base $AD$. Let the line drawn through vertex $B$ parallel to $AC$ intersect the extension of base $AD$ at point $K$. Then $KBD$ is a right triangle with the leg $BK$ opposite the $30^{\circ}$ angle. Therefore, $BC + AD = AK + AD = 2 BK = 2 AC = 12$, and the midline of the trapezoid is half the sum of the bases.

Answer

## 6.

Find the height of an isosceles triangle drawn to the lateral side if the base is $a$ and the lateral side is $b$.

## Hint

The product of the desired height and the lateral side is equal to the product of the base and the height drawn to it.

## Solution

The height dropped to the base is $\frac{\sqrt{4 b^{2} - a^{2}}}{2}$ (see problem $\underline{\text{54199}}$). Therefore, twice the area of the triangle is $\frac{a \sqrt{4 b^{2} - a^{2}}}{2}$. Dividing it by the lateral side, we get the height drawn to it.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-43.jpg?height=389&width=366&top_left_y=812&top_left_x=846)

## Answer

$\frac{a \sqrt{4 b^{2} - a^{2}}}{2 b}$