Question

In a circle of radius 'r' and center 'O', P and Q are points on the circle such that OP $\perp$ OQ.

Suppose a = $\frac{1}{\sqrt{2}}$. Find PQ in terms of a and r.

• A. $ar$
• B. $2a$
• C. $2r$
• D. $2ar$

Answer 1

The correct option is D $2ar$

Refer to the following figure. Since P and Q are points on the circle,

OP = OQ = r. Since OPQ is a right angled triangle,

$P{Q}^2$ = $Q{P}^2$ + $O{Q}^2$ = $2r^2$

$\Rightarrow$ PQ = $r\sqrt{2}$

Since a = $\frac{1}{\sqrt{2}}$,

2a = $\sqrt{2}$

So, PQ = r(2a) = 2ar