Properties and characteristics of the tangent [ Trigonometric ratios in a right-angled triangle ]

Given a rhombus $A B C D$. A circle of radius $R$ is tangent to the lines $A B$ and $A D$ at points $B$ and $D$ respectively, and intersects the side $B C$ at point $L$, such that $4 B L=B C$. Find the area of the rhombus.
Drop a perpendicular from the center of the circle to side $B C$.

## Solution

Let $B C=4 x$. Then $B L=x, C L=3 x$. Drop the perpendicular $O P$ from the center $O$ of the given circle to the side $B C$ of the rhombus. Then

$$
C P=C L+L P=3 x+\frac{x}{2}=\frac{7 x}{2}
$$

If $\angle B C D=\alpha$, then

$$
O P=P C \tan \frac{\alpha}{2}=P C \cdot \frac{O B}{A B}=\frac{7 x}{2} \cdot \frac{R}{4 x}=\frac{7 R}{8} .
$$

By the Pythagorean theorem,

$$
O B^{2}=O P^{2}+B P^{2} \text {, or } R^{2}=\frac{49}{64} R^{2}+\frac{x^{2}}{4} .
$$

From this, we find that $x=\frac{R \sqrt{15}}{4}$. Then

$$
A B=4 x=R \sqrt{15}, \tan \frac{\alpha}{2}=\frac{R}{4 x}=\frac{1}{\sqrt{15}}, \sin \alpha=\frac{2 \tan \frac{\alpha}{2}}{1+\tan^{2} \frac{\alpha}{2}}=\frac{\sqrt{15}}{8}
$$

$$
S_{\mathrm{ABCD}}=A B \cdot A D \sin \alpha=\frac{15 R^{2} \sqrt{15}}{8}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1b164f828743b46f847cg-35.jpg?height=335&width=855&top_left_y=227&top_left_x=607)

## Answer

$\frac{15 R^{2} \sqrt{15}}{8}$