How to solve this semi-circle problem?

Question

The figure above shows a semi-circle. $\angle BAP$ is $\alpha$ radian. The area of semi-circle is bisected by $AP$. Prove that $$2\alpha+\sin 2\alpha = \frac{\pi}{2}$$

I have simply no clue to solve this problem. Can anyone help me? Thanks so much.

Answer 1

Area of semi-circle $= \frac{1}{2} \pi r^{2}$.

Let $O$ be the centre. Then

$\angle BOP = 2\alpha$ and $\angle AOP = \pi-2\alpha$

Area of $ABP=$ Area of sector $OBP +$ Area of $\Delta OAP$

$\implies \frac{1}{4} \pi r^{2} = \frac{1}{2} r^{2}(2\alpha)+\frac{1}{2} r^{2} \sin (\pi-2\alpha)$

$\implies \frac{1}{4} \pi r^{2} = r^{2}\alpha+\frac{1}{2} r^{2} \sin 2\alpha$

$\implies \frac{1}{2} \pi = 2\alpha+ \sin 2\alpha$

(QED)

Note that the radius $r$ will be automatically cancelled out.

P.S.: The word "bisect" is not good. May say "$AP$ divides the semi-circle into $2$ equal areas".

Answer 2

Let $r$ be the radius & $O$ be the center of the semi-circle.

Area of the segment ABP of semi-circle $$(\text{area of sector OBP with angle of aperture } 2\alpha)+(\text{area of isosceles}\ \triangle AOP)$$

$$=\frac{1}{2}r^2(2\alpha)+\frac{1}{2}r^2\sin(\pi-2\alpha)=\color{blue}{\frac{1}{2}r^2(2\alpha)+\frac{1}{2}r^2\sin(2\alpha)}$$ But the area of semi-circle is bisected by $AP$ hence, Area of the segment ABP of semi-circle $$=\text{half the area of semi-circle}=\frac 12\frac{\pi r^2}{2}=\color{blue}{\frac{\pi r^2}{4}}$$ Equating both the values, $$\frac{1}{2}r^2(2\alpha)+\frac{1}{2}r^2\sin(2\alpha)=\frac{\pi r^2}{4}$$ $$\color{red}{2\alpha+\sin(2\alpha)=\frac{\pi}{2}}$$