### 10.393 An isosceles trapezoid is circumscribed around a circle with radius $R=1$ cm, and the area of the trapezoid is $5 \mathrm{~cm}^{2}$. Find the area of the quadrilateral whose vertices are the points of tangency of the circle and the trapezoid.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-239.jpg?height=278&width=351&top_left_y=1236&top_left_x=160)

A
Let $O$ be the center of the circle circumscribed around the given isosceles trapezoid $ABCD$, and let $K, L, M, N$ be the points of tangency with the circle. $\angle BAO = \angle OAK = \angle MKO$, since $MK \perp AO$ (AO is the bisector and, therefore, the height of the isosceles $\triangle AMK$). Additionally, $\angle NMK = 90^\circ$ because $NK$ is the diameter (Theorem 6), $\angle AOB = 90^\circ$ (Theorem 9) $\Rightarrow \triangle MNK \sim \triangle AOB \Rightarrow \left(\frac{AB}{NK}\right)^2 = \frac{S_{AOB}}{S_{MNK}}$ (Theorem 7). But $NK = 2R = 2$ (by condition), $AB = \frac{AD + BC}{2}$ (Theorem 2) $= \frac{S_{ABCD}}{NK} = \frac{5}{2}$, $S_{AOB} = \frac{AB \cdot OM}{2} = \frac{5}{4}$. Therefore, $S_{MNK} = 0.8$, $S_{MNKL} = 2S_{MNK} = 1.6$.

Answer: $1.6 \text{ cm}^2$.