Question

A rod bent into an angle is placed with its vertexr at the centre of a circle and it is found that $\frac{1}{10}$ of the circle lies within it. If it is placed with its vertex on the circumference, what part of the circle would be within it ?

• A. $\frac{1}{10}$
• B. $\frac{1}{6}$
• C. $\frac{1}{5}$
• D. $\frac{3}{10}$

Answer 1

The correct option is C

$\frac{1}{5}$

Let the rod make an angle x when placed at the centre.

Now, when placed on the circumference.

$\angle APB=\angle AOB=x^{\circ }$

In the second figure,

$\angle A{O}^{\prime }B=2\angle APB$

$\angle A{O}^{\prime }B=2x^{\circ }$

Therefore, $\angle AOB=\frac{1}{2}\angle A{O}^{\prime }B$

$\therefore$ Part of second circle is 2$\times \frac{1}{10}=\frac{1}{5}$ of the circle.