Question

A circle with area A $c{m}^2$ is contained in the interior of a larger circle with area (A +B) $c{m}^2$ and the radius of the larger circle is 4 cm. If A, B, A+B are in arithmetic progression, then the diameter (in cm) of the smaller circle is

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{4\sqrt{3}}{3}$
• C. $\frac{8\sqrt{3}}{3}$
• D. $2\sqrt{3}$

Answer 1

The correct option is C $\frac{8\sqrt{3}}{3}$

Given :

Area of smaller circle $=$ A $c{m}^2$

Area of larger circle $=$ A+B $c{m}^2=\pi r^2=16\pi$

$\therefore$ A $=16\pi -$ B

A, B, A+B i.e $16\pi -$ B, B, $16\pi$ are in arithmetic progression

$\therefore$ $(16\pi -B)-B=B-16\pi$

$⟹$ $32\pi =3B$

$⟹$ $B=\frac{32}{3}\pi$

$\therefore$ $A=16\pi -\frac{32}{3}\pi$

$⟹A=\frac{48\pi -32\pi }{3}$

$=\frac{16\pi }{3}$

But A=Area of smaller circle $=\pi \times radiu{s}^2=\frac{16\pi }{3}$

$⟹$ $radius=\frac{4}{\sqrt{3}}$

$\therefore$ Diameter $=\frac{8}{\sqrt{3}}=\frac{8\sqrt{3}}{3}$.

Hence, option C is correct.