A chord of a circle subtends an angle of θ at the centre of the circle.

Question:

A chord of a circle subtends an angle of θθ at the centre of the circle. The area of the minor segment cut off by the chord is one eighth of the area of the circle. Prove that

$8 \sin \frac{\theta}{2} \cos \frac{\theta}{2}+\pi=\frac{\pi \theta}{45}$

Solution:

We know that the area of circle and area of minor segment of angle $\theta$ in a circle of radius $r$ is given by, $A^{\prime}=\pi r^{2}$ and $A=\left\{\frac{\pi \theta}{360^{\circ}}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}\right\} r^{2}$ respectively.

It is given that,

Area of minor segment $=\frac{1}{8} \times$ area of circle

$\left\{\frac{\pi \theta}{360^{\circ}}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}\right\} r^{2}=\frac{\pi r^{2}}{8}$

$\left\{\frac{\pi \theta}{360^{\circ}}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}\right\} \times 8=\pi$

$\frac{8 \pi \theta}{360^{\circ}}-8 \sin \frac{\theta}{2} \cos \frac{\theta}{2}=\pi$

$\frac{\pi \theta}{45^{\circ}}-8 \sin \frac{\theta}{2} \cos \frac{\theta}{2}=\pi$

$\frac{\pi \theta}{45^{\circ}}=8 \sin \frac{\theta}{2} \cos \frac{\theta}{2}+\pi$

 

 

 

 

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