 Circle inscribed in an angle  Area of a trapezoid

A circle with a radius of 3 is inscribed in a right trapezoid, the smaller base of which is 4. Find the larger base of the trapezoid.
Let $O$ be the center of a circle with radius 3, inscribed in trapezoid $ABCD$ with bases $AD$ and $BC=4$, $\angle C = \angle D = 90^\circ$, and the circle touching sides $CD$, $BC$, $AB$, and $AD$ at points $K$, $L$, $M$, and $N$ respectively. The quadrilaterals $ONDK$ and $OKCL$ are squares, so

$$
DN = OK = 3, \quad CL = OK = 3, \quad BM = BL = BC - CL = 4 - 3 = 1.
$$

The rays $AO$ and $BO$ are the angle bisectors of the angles at the lateral side of the trapezoid. The sum of these angles is $180^\circ$, and the sum of their halves is $90^\circ$. Therefore, $\angle AOB = 90^\circ$, which means $OM = 3$ is the altitude of the right triangle $AOB$ drawn from the vertex of the right angle, so $OM^2 = BM \cdot AM$, from which we find that

$$
\begin{gathered}
AM = \frac{OM^2}{BM} = \frac{9}{1} = 9 \\
\text{Then} \\
AN = AM = 9, \quad AD = AN + DN = 9 + 3 = 12
\end{gathered}
$$

## Answer

12.00

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