Question

Two parallel chords are drawn on the same side of the centre of a circle of radius 20. It is found that they subtend ${60}^0$ and ${120}^0$ angles at the centre of the circle. Then the perpendicular distance between the chords is:

• A. $5(\sqrt{3}-1)$
• B. $10(\sqrt{3}-1)$
• C. $10(\sqrt{2}-1)$
• D. $5(\sqrt{3}+1)$

Answer 1

Answer: (B)

$10(\sqrt{3}-1)$

Let OF and OE be the perpendicular at AB and CD respectively

Consider triangle AOB

In triangle$\angle AOB$$={120}^o$

Therefore , $\angle AOF={60}^o$

Then , $cos{60}^o=\frac{OF}{AO}$

$\frac{1}{2}=\frac{OF}{20}$

$OF=10unit$

Now consider triangleCOD,

In triangle $\angle COD={60}^o$

Therefore $\angle COE={30}^o$

Then, $cos{30}^o=\frac{OE}{CO}$

$\frac{\sqrt{3}}{2}=\frac{OE}{20}$

$OE=10\sqrt{3}unit$

Hence the perpendicular distance between the two chords $=10\sqrt{3}-10$

$=10(\sqrt{3}-1)$