Question

A chord of a circle subtends an angle of $\theta$ at the centre of the circle. The area of minor segment cut off by the chord is one eighth of the area of the circle. Prove that $8\sin \frac{\theta }{2}\cos \frac{\theta }{2}+\pi =\frac{\pi \theta }{45}.$

Answer 1

$\Rightarrow$ Chord $AB$ subtend angle $\theta$ at center $O$.

$\Rightarrow$ Let radius of the center circle be $r$.

$\Rightarrow$ Area of the minor segment cut off by the chord $AB$ = Area of sector $AOB$ - Area of triangle $OAB$

$\therefore$ Area of the minor segment cut off by the chord $AB=\frac{\theta }{{360}^o}\times \pi r^2-\frac{1}{2}\times r^2\times \sin \theta$

$\Rightarrow$ Since, area of this minor segment $=\frac{1}{8}\times$ Area of circle.

$\Rightarrow$ $\frac{\theta }{{360}^o}\times \pi r^2-\frac{1}{2}\times r^2\times \sin \theta =\frac{1}{8}\times \pi r^2$

$\Rightarrow$ $\frac{\pi \theta }{{360}^o}-\frac{\sin \theta }{2}=\frac{\pi }{8}$

$\Rightarrow$ $\frac{\pi \theta }{{45}^o}-4\sin \theta =\pi$

$\Rightarrow$ $\pi +4\sin \theta =\frac{\pi \theta }{{45}^o}$

$\Rightarrow$ $\pi +4\times 2\sin \frac{\theta }{2}.cos\frac{\theta }{2}=\frac{\pi \theta }{{45}^o}$

$\therefore$ $8\sin \frac{\theta }{2}.\cos \frac{\theta }{2}+\pi =\frac{\pi \theta }{{45}^o}$

Hence Proved