Find the shaded area of the triangle below

Question

According to the graph, triangle ABC is equilateral, $C$ and $D$ are points of tangency. Calculate the area of the shaded region.(S:$3R^2-\dfrac{143\pi R^2}{360}$)

$cos30^o=\dfrac{AC}{2R\sqrt3} \implies \dfrac{\sqrt3}{2}=\dfrac{AC}{2R\sqrt3}\\ \therefore AC = 3R=BC=BD\\ S_{BCGD} =2S_{\triangle BCG}= 2.\dfrac{R.3R}{2}=3R^2$

How to calculate theta angle?

Answer 1

1. Find the altitude and it is $R\sqrt(3) \times (3/2) = x, say$.

2. Find BC which is given by the ratio $x: BC = \sqrt (3) : 2$.

3. Find $0.5 \theta$ from $\triangle BCG$.

Answer 2

Comment :As can be seen in figure, $\angle CBG=\angle DAC\approx 18.44^o$, that is $\frac{\pi}{10}<\theta<\frac{\pi}9 $. It is about:

$\frac {18.44}{180}\pi\approx \frac{100\pi}{976}$

In fact we have:

$BC=AC=2(R\sqrt 3) \cos 30^o=3R$

$\angle CBG=tan^{-1} \frac R{3R}=18.44^o$

In triangl BCG we have:

$BGC=\frac{\theta}2=\pi-(\frac {\pi}2+\frac{100\pi}{976})=\frac{388 \pi}{976}\approx \frac{129.33\pi}{162.7}$

$\theta\approx \frac {130\pi}{163}$

Area of segment cnter on G is:

$s=\frac 12\theta R^2=\frac{130\pi}{326} R^2$

$A=3R^2-\frac{130\pi}{326} R^2$

We can see that $\frac{130}{326}\approx \frac{143}{360}$. So we may say:

$A=3R^2-\frac{143\pi}{360} R^2$

Answer 3

From the properties of an equilateral triangle we conclude that $BC=3R$. From here $\tan(\theta/2) =3$. As the area of $BCGD$ equals $3R^2$ the area in question equals $$3R^2 - 2\pi R^2\frac{\arctan(3)}{360}=R^2\left(3-\frac{\pi\arctan(3)}{180}\right)\approx1.750955R^2.$$ The result ist somewhat nicer if the angle is measured in rad: $$R^2\bigl(3-\arctan(3)\bigr).$$