Question

Find the area of the circle in which a chord of length $$\sqrt{2}$$ makes an angle $${\displaystyle \frac{\pi }{2}}$$ at the centre.
(A) $${\displaystyle \frac{\pi }{2}}$$ sq. units
(B) $$2\pi$$ sq. units
(C) $$\pi$$ sq. units
(D) $${\displaystyle \frac{\pi }{4}}$$ sq. units

Answer 1

Hint: Assume a circle with centre O having a chord PQ of length $$\sqrt{2}$$ which makes an angle of $${\displaystyle \frac{\pi }{2}}$$ at the centre. Draw a line OR perpendicular to the chord PQ. We can say that $$\mathrm{\Delta }ORP$$ and $$\mathrm{\Delta }ORQ$$ are congruent to each other from RHS criteria. We can say that PR=RQ and $$\mathrm{\angle }POR=\mathrm{\angle }QOR={45}^0$$ . In the $$\mathrm{\Delta }ORP$$ , apply $$\sin {45}^0$$ and then OP which is radius. Now, solve further and then find the area of the circle.

Complete step-by-step answer:

Let us assume a circle with centre O which has a chord of length $$\sqrt{2}$$ which makes an angle of $${\displaystyle \frac{\pi }{2}}$$ at the centre.
Now, Draw a line OR perpendicular to the chord PQ.
In the $$\mathrm{\Delta }ORP$$ and $$\mathrm{\Delta }ORQ$$ , we have
$$\mathrm{\angle }ORP=\mathrm{\angle }ORQ={90}^0$$
OP=OQ (radius of the circle)
OR=OR (common)
$$\mathrm{\Delta }ORP\cong \mathrm{\Delta }ORQ$$
Thus, $$\mathrm{\Delta }ORP$$ and $$\mathrm{\Delta }ORQ$$ are congruent with each other.
It means the sides of $$\mathrm{\Delta }ORP$$ and $$\mathrm{\Delta }ORQ$$ are equal to each other.
So, PR=RQ and also $$\mathrm{\angle }POR=\mathrm{\angle }QOR$$ .
According to the question, we have PQ= $$\sqrt{2}$$ .
Now,
$$\begin{array}{rl}& PR+RQ=PQ\\ & \Rightarrow 2PR=\sqrt{2}\end{array}$$
$$\Rightarrow PR={\displaystyle \frac{1}{\sqrt{2}}}$$ ………………….(1)
It is given that, $$\mathrm{\angle }POQ={90}^0$$ ………………(2)
We also have, $$\mathrm{\angle }POQ=\mathrm{\angle }POR+\mathrm{\angle }QOR$$ ……………………………(3)
From equation (2) and equation (3), we get
$$\begin{array}{rl}& \mathrm{\angle }POQ=\mathrm{\angle }POR+\mathrm{\angle }QOR\\ & \Rightarrow {90}^0=2\mathrm{\angle }POR\\ & \Rightarrow {45}^0=\mathrm{\angle }POR\end{array}$$
In the $$\mathrm{\Delta }ORP$$ , we have
$$\mathrm{\angle }POR={45}^0$$
$$\sin {45}^0={\displaystyle \frac{PR}{OP}}$$ ……………(4)
$$\sin {45}^0={\displaystyle \frac{1}{\sqrt{2}}}$$ ……………..(5)
From equation (1), equation (4), and equation (5), we get
$$\begin{array}{rl}& \sin {45}^0={\displaystyle \frac{PR}{OP}}\\ & \Rightarrow {\displaystyle \frac{1}{\sqrt{2}}}={\displaystyle \frac{{\displaystyle \frac{1}{\sqrt{2}}}}{OP}}\\ & \Rightarrow OP=1\end{array}$$
OP is the radius of the circle. We have got the radius of the circle.
Area = $$\pi {(radius)}^2$$
Area = $$\pi {(1)}^2=\pi$$ sq. units.
Hence, option (C) is correct.

Note: In this question, we have to find the area and for the area of the circle we need radius of the circle. So, one can apply $$\sin {\displaystyle \frac{\pi }{2}}$$ in the $$\mathrm{\Delta }POQ$$ . We know that in a right-angled triangle the side opposite to the sine angle is taken as height. If we do so then the side PQ will become height but the side PQ is hypotenuse in $$\mathrm{\Delta }POQ$$ which is a contradiction. So, we cannot directly apply $$\sin {\displaystyle \frac{\pi }{2}}$$ in the $$\mathrm{\Delta }POQ$$ .