Question

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

Answer 1

Hint: Assume a circle with centre O having a chord PQ of length \[\sqrt{2}\] which makes an angle of \[\dfrac{\pi }{2}\] at the centre. Draw a line OR perpendicular to the chord PQ. We can say that \[\Delta ORP\] and \[\Delta ORQ\] are congruent to each other from RHS criteria. We can say that PR=RQ and \[\angle POR=\angle QOR={{45}^{0}}\] . In the \[\Delta ORP\] , 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 \[\dfrac{\pi }{2}\] at the centre.
Now, Draw a line OR perpendicular to the chord PQ.
In the \[\Delta ORP\] and \[\Delta ORQ\] , we have
\[\angle ORP=\angle ORQ={{90}^{0}}\]
OP=OQ (radius of the circle)
OR=OR (common)
\[\Delta ORP\cong \Delta ORQ\]
Thus, \[\Delta ORP\] and \[\Delta ORQ\] are congruent with each other.
It means the sides of \[\Delta ORP\] and \[\Delta ORQ\] are equal to each other.
So, PR=RQ and also \[\angle POR=\angle QOR\] .
According to the question, we have PQ= \[\sqrt{2}\] .
Now,
\[\begin{align}
  & PR+RQ=PQ \\
 & \Rightarrow 2PR=\sqrt{2} \\
\end{align}\]
\[\Rightarrow PR=\dfrac{1}{\sqrt{2}}\] ………………….(1)
It is given that, \[\angle POQ={{90}^{0}}\] ………………(2)
We also have, \[\angle POQ=\angle POR+\angle QOR\] ……………………………(3)
From equation (2) and equation (3), we get
\[\begin{align}
  & \angle POQ=\angle POR+\angle QOR \\
 & \Rightarrow {{90}^{0}}=2\angle POR \\
 & \Rightarrow {{45}^{0}}=\angle POR \\
\end{align}\]
In the \[\Delta ORP\] , we have
\[\angle POR={{45}^{0}}\]
\[\sin {{45}^{0}}=\dfrac{PR}{OP}\] ……………(4)
\[\sin {{45}^{0}}=\dfrac{1}{\sqrt{2}}\] ……………..(5)
From equation (1), equation (4), and equation (5), we get
\[\begin{align}
  & \sin {{45}^{0}}=\dfrac{PR}{OP} \\
 & \Rightarrow \dfrac{1}{\sqrt{2}}=\dfrac{\dfrac{1}{\sqrt{2}}}{OP} \\
 & \Rightarrow OP=1 \\
\end{align}\]
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 \[\sin \dfrac{\pi }{2}\] in the \[\Delta POQ\] . 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 \[\Delta POQ\] which is a contradiction. So, we cannot directly apply \[\sin \dfrac{\pi }{2}\] in the \[\Delta POQ\] .