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The diagonal of a rectangular trapezoid and its lateral side are equal.

Find the midline of the trapezoid if the height of the trapezoid is 2 and the lateral side is 4.

#
The height of the given trapezoid, drawn from the vertex of the obtuse angle, passes through the midpoint of the larger base.

## Solution

Let $BC$ be the smaller base of trapezoid $ABCD$, $AB=2$ be the smaller lateral side, $CD$ be the larger lateral side, and $AC=CD=4$. Drop a perpendicular $CK$ to the base $AD$. In the right triangle $CKD$, the leg $CK$ is half the hypotenuse $CD$. Therefore, $\angle CDK=30^{\circ}$, and

$DK=2\sqrt{3}$. Since $K$ is the midpoint of $AD$, the midline of the trapezoid is $1/2(AD+BC)=3/2 DK$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d3cd76ad0dc53ae543bdg-34.jpg?height=329&width=880&top_left_y=1745&top_left_x=589)

## Answer

$3\sqrt{3}$