8,9

The projection of the lateral side of an isosceles trapezoid on the larger base is $a$, the midline of the trapezoid is $b$, and the acute angle at the base is $45^{\circ}$. Find the radius of the circle circumscribed around the trapezoid.
Let $ABCD$ be an isosceles trapezoid with bases $AD > BC$, $\angle ADC = 45^{\circ}$, $CH$ be the height of the trapezoid, and $R$ be the radius of the circumscribed circle around the trapezoid. It is known that the projection of the diagonal of the isosceles trapezoid onto the larger base is equal to the semi-sum of the bases, i.e., the midline of the trapezoid. Then,

$$
CH = DH = a, \quad AH = \frac{1}{2}(AD + BC) = b, \quad AC = \sqrt{CH^2 + AH^2} = \sqrt{a^2 + b^2}
$$

The circumscribed circle around the trapezoid coincides with the circumscribed circle around triangle $ACD$. By the Law of Sines,

$$
R = \frac{AC}{2 \sin \angle ADC} = \frac{AC}{2 \sin 45^{\circ}} = \frac{\sqrt{a^2 + b^2}}{2 \cdot \frac{\sqrt{2}}{2}} = \sqrt{\frac{a^2 + b^2}{2}}
$$

## Answer

$\sqrt{\frac{a^2 + b^2}{2}}$.