[ The Law of Cosines [ Area of a Triangle (through two sides and the angle between them)

A circle touches the side $AD$ of quadrilateral $ABCD$ at point $D$, and side $BC$ at its midpoint $M$. The diagonal $AC$ intersects the circle at points $K$ and $L$ (with $AK < AL$). It is known that $AK=5$, $KL=4$, and $LC=1$. The rays $AD$ and $BC$ intersect at point $S$, and $\angle ASB=120^{\circ}$. Find the radius of the circle and the area of quadrilateral $ABCD$.
Let $O$ be the center of the circle, $R$ be its radius. Since the circle is inscribed in the angle $A S B$, its center lies on the bisector of this angle, so $\angle O S D=60^{\circ}$. From the right triangle $O D S$ we find that $S D=O D \cot$

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-18.jpg?height=91&width=841&top_left_y=2267&top_left_x=14)

$$
A D=\sqrt{A K \cdot A L}=\sqrt{5 \cdot 9}=3 \sqrt{5}, C M=\sqrt{C K \cdot C L}=\sqrt{5 \cdot 1}=\sqrt{5}
$$

Then

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-18.jpg?height=103&width=1072&top_left_y=2481&top_left_x=493)

By the cosine rule $A C 2=S A 2+S C 2-2 S A \cdot S C \cos 120^{\circ}$, or

$$
100=\left(\frac{R}{(\sqrt{3}}+3 \sqrt{5}\right) 2+\left(\frac{R 2}{\sqrt{3}} \sqrt{5}\right) 2+\left(\frac{R}{(\sqrt{3}}+3 \sqrt{5}\right)\left(\frac{R}{\sqrt{3}}-\sqrt{5}\right) .
$$

From this equation, we find that $R=4 \sqrt{5}-\sqrt{15}$. Then

$$
S A=A D+S D=3 \sqrt{5}+\frac{12}{\sqrt{3}}=3 \sqrt{5}+\frac{4 \sqrt{5}-\sqrt{15}}{\sqrt{3}}=\frac{2 \sqrt{15}+4 \sqrt{5}}{\sqrt{3}},
$$

$$
S B=S M+M B=\frac{R}{\sqrt{3}}+\sqrt{5}=\frac{4 \sqrt{5}-\sqrt{15}}{\sqrt{3}}+\sqrt{5}=\frac{4 \sqrt{5}}{\sqrt{3}}
$$

$$
S_{\triangle A S B}=\frac{1}{2} S A \cdot S B \sin 120^{\circ}=\frac{1}{2} \cdot \frac{2 \sqrt{15}+\mathbf{4} \sqrt{5}}{\sqrt{3}} \cdot \frac{4 \sqrt{5}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2}=\frac{10 \sqrt{3}+20}{\sqrt{3}}
$$

$$
S D=\frac{R^{2}}{\sqrt{3}}=\frac{\mathbf{4} \sqrt{5}-\sqrt{15}}{\sqrt{3}}, S C=\frac{\frac{R}{\sqrt{3}}}{\sqrt{5}} \sqrt{5}=\frac{\mathbf{4} \sqrt{5}-\sqrt{15}}{\sqrt{3}}-\sqrt{5}=\frac{4 \sqrt{5}-2 \sqrt{15}}{\sqrt{3}},
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-19.jpg?height=123&width=1114&top_left_y=699&top_left_x=472)

Therefore,

$$
S_{A B C D}=S_{\triangle A S B}-S_{\triangle D S C}=\frac{10 \sqrt{3}+20}{\sqrt{3}}-\frac{55-30 \sqrt{3}}{2 \sqrt{3}}=\frac{50 \sqrt{3}-15}{2 \sqrt{3}}=\frac{50-5 \sqrt{3}}{2} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-19.jpg?height=1501&width=1884&top_left_y=1093&top_left_x=127)

## Answer

$$
R=4 \sqrt{5}-\sqrt{15}, S_{A B C D}=\frac{50-5 \sqrt{3}}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-20.jpg?height=121&width=2038&top_left_y=111&top_left_x=13)
[ Area of a triangle (using two sides and the angle between them).}

The circle touches the side $A D$ of the quadrilateral $A B C D$ at point $D$, and the side $B C$ at its midpoint $M$. The diagonal $A C$ intersects the circle at points $K$ and $L$ (with $A K<A L$). It is known that $A K=3, K L=5, L C=1$. The rays $A D$ and $B C$ intersect at point $S$, and $\angle A S B=60^{\circ}$. Find the radius of the circle and the area of the quadrilateral $A B C D$.

## Answer

$R=\frac{3 \sqrt{6}-\sqrt{2}}{2}, S_{A B C D}=27$