In a quadrilateral $A B C D$, a circle of radius 2 is inscribed. The angle $\angle D A B$ is a right angle. Side $A B$ is 5, and side $B C$ is 6. Find the area of the quadrilateral $A B C D$.
Find the trigonometric functions of the angle $\angle ABC$; extend until the intersection of sides $AD$ and $BC$.

## Solution

Let the circle with center $O$ touch the sides $AB, BC, CD$, and $AD$ of the given quadrilateral at points $M, N, K$, and $L$, respectively. Then $AMOL$ is a square. Therefore,

$$
BN = BM = AB - AM = 5 - 2 = 3, \quad CK = CN = BC - BN = 6 - 3 = 3.
$$

Extend sides $AD$ and $BC$ until they intersect at point $Q$. Let $\angle OBM = \angle OBN = \alpha$, and $\angle AQB = \beta$. From the right triangle $OBM$, we find that $\operatorname{tg} \alpha = \frac{OM}{BM} = \frac{2}{3}$. Then

$$
\operatorname{tg} \angle ABQ = \operatorname{tg} 2\alpha = \frac{2 \operatorname{tg} \alpha}{1 - \operatorname{tg}^2 \alpha} = \frac{12}{5} \\
\operatorname{tg} \beta = \operatorname{tg} \angle AQB = \operatorname{ctg} 2\alpha = \frac{5}{12}, \quad \cos \beta = \frac{12}{13} \\
AQ = AB \operatorname{tg} 2\alpha = 5 \cdot \frac{12}{5} = 12, \quad QL = AQ - AL = 12 - 2 = 10
$$

$$
BQ = \frac{AQ}{\cos \beta} = 12 : \frac{12}{13} = 13, \quad CQ = BQ - BC = 13 - 6 = 7
$$

Let $DK = DL = t$. Then $CD = DK + KC = t + 3$, and $DQ = QL - DL = 10 - t$. Apply the cosine theorem to triangle $CDQ$:

$$
CD^2 = DQ^2 + CQ^2 - 2 \cdot DQ \cdot CQ \cdot \cos \beta
$$

or

$$
(t + 3)^2 = 49 + (10 - t)^2 - 2 \cdot 7(10 - t) \cdot \frac{12}{13}.
$$

From this equation, we find that $t = \frac{14}{17}$. Therefore,

$$
S_{\mathrm{ABCD}} = \frac{AB + BC + CK + KD + DL + AL}{2} \cdot OM = \frac{5 + 6 + 3 + \frac{14}{17} + \frac{14}{17} + 2}{2} \cdot 2 = \frac{300}{17} = 17 \frac{11}{17}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-40.jpg?height=429&width=949&top_left_y=843&top_left_x=560)

## Answer

$17 \frac{11}{17}$.