Question

Find the area of the shaded region in figure, if $PQ = 24cm$, $PR = 7cm$ and O is the centre of the circle.

Answer 1

Hint:Above question is based on properties of diameter of circle. We will apply Thale’s theorem and then find the required dimensions by using pythagoras theorem.
As in the above question the chord RQ is going through centre O, therefore we will take RQ as diameter of circle.
Formula and theorems used
1. Thales Theorem – It states that any diameter of a circle subtends a right angle to any point on the circle.
2. Pythagoras theorem – \[Hypotenuse{e^2} = \] sum of square of legs
3. Area of circle $ = \pi {r^2}$


Complete step by step solution :
It is given that a circle with centre O.
Chord $PQ = 24cm$
$PR = 7cm$

We need to find an area of shaded portion.
From above figure, we can say that RQ is a diameter so circle is divided in semi circles
And area of shaded portion $ = $ area of semi-circle – area of triangle QRP.
Let’s solve the question.
We know that the angle made by diameter in circumference of a circle is always $90^\circ $.
So, $\angle RPQ = 90^\circ $
Therefore QR will be the hypotenuse of $\Delta RPQ$.
So in $\Delta RPQ$
$R{Q^2} = R{P^2} + P{Q^2}$
$R{Q^2} = {(7)^2} + {(24)^2}$
$R{Q^2} = 49 + 576$
$ = 625$
$RQ = 25cm$
So, diameter of circle is, $d = 25cm$ then its radius $r = \dfrac{d}{2} = \dfrac{{25}}{2}cm$
(i) Area of circle $ = \pi {r^2}$
$ = \dfrac{{22}}{7} \times {\left( {\dfrac{{25}}{2}} \right)^2}$
$ = 491.071c{m^2}$
Area of semi-circle $ = $ area of circle $ = \dfrac{{491.07}}{2} = 245.5cm$
(ii) Area of $\Delta RPQ = \dfrac{1}{2}(base \times height)$
$ = \dfrac{1}{2}(24 \times 7)$
$ = 84c{m^2}$
(iii) Therefore area of shaded region $ = $ Area of semi-circle – area of $\Delta RPQ$
$ = (245.5 - 84)$
$ = 161.5c{m^2}$

Note : You should be familiar with all circle theorems to the point where
You can identify when they should be used.
Like we have used Thales’s theorem in the above question.