In a rectangular trapezoid, the smaller base is equal to the height, and the larger base is equal to $a$. Find the lateral sides of the trapezoid, given that one of them is tangent to a circle passing through the ends of the smaller base and tangent to the larger base.
Let $BC$ be the smaller base, $BA = x$ be the smaller lateral side of trapezoid $ABCD$. Then the point $M$ of tangency of the given circle with the lower base $AD$ lies on the perpendicular bisector of the base $BC$. Therefore, $AM = BC / 2 = x / 2, CD = MD = a - x / 2$.

Let $K$ be the projection of vertex $C$ onto $AD$. Then $KD = a - x, CK = x$. By the Pythagorean theorem, $(a - x / 2)^2 = x^2 + (a - x)^2$, from which $x = 4a / 7, CD = a - x / 2 = 5a / 7$.

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-20.jpg?height=335&width=395&top_left_y=553&top_left_x=840)

## Answer

$4a / 7, 5a / 7$.