The circle constructed on the larger lateral side $AB$ of the right trapezoid $ABCD$ as its diameter intersects the base $AD$ at its midpoint. It is known that $AB=10, CD=6$. Find the midline of the trapezoid.
Let $M$ be the midpoint of $A D$. Then $\angle A M B=90^{\circ}$, so the quadrilateral $M B C D$ is a rectangle, $B M=C D=6$. From the right triangle $A B M$ we find that

$$
A M=\sqrt{A B^{2}-B M^{2}}=\sqrt{100-36}=8
$$

Then $B C=M D=A M=8$. Therefore, the midline of the trapezoid $A B C D$ is

$$
\frac{1}{2}(A D+B C)=\frac{1}{2}(16+8)=12
$$

## Answer

12.00