Ex.12.3 Q1 Areas Related to Circles Solution - NCERT Maths Class 10

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Question

Find the area of the shaded region in the figure, if \(PQ = 24\,\rm{cm},\) \(PR = 7\rm\,{cm}\) and \(O\) is the centre of the circle.

Text Solution

 

What is known?

\(PQ = 24\, {\rm{cm}}, PR = 7\,{\rm{cm}},\, O\) is the center the circle.

Use \(\begin{align}\pi = \frac{{22}}{7}\end{align}\)

What is unknown?

Area of the shaded region in figure.

Reasoning:

(i) Visually it’s clear that

Area of the shaded region \(=\)  Area of semicircle \(RPQ \,-\)  Areaof \(\Delta RPQ\)

\[\begin{align}= {\frac{1}{2}\pi = {{(OR)}^2} - {\text{Area of }}\Delta \rm{RPQ}}\end{align}\]

Since we don’t know \(RQ\) (diameter) \(OR\) and \(PQ\) (radii of circle), we are unable to find either area of semicircle and we don’t know \(RQ\) so we can’t find area of \(\Delta RPQ\) using heron’s formula (since either \(2\) sides are known) or by\(\begin{align} \frac{1}{2} \times {\text{base}} \times {\text{height}}\end{align}\) as we have only \(2\) sides and that their respective heights.

So, we need to find \(RQ.\) 

(ii) Using the knowledge:

Angle subtended by an arc at any point on the circle is half of the angle subtended by it at the center.

\(\begin{align} \therefore \angle ROQ &= {{180}^\circ } \\ \therefore \angle RPQ &= \frac{1}{2} \times {{180}^\circ = {{90}^\circ }} \end{align}\)

(\(\rm{OR}\)) angle in a semicircle \( = {90^ \circ }\)

We get \(\Delta RQP\) is a right angle triangle.

Using Pythagoras theorem

\(\begin{align} RQ^2 &= PR^2 + PQ^2 \end{align}\)

We get \[\begin{align} RQ &= \sqrt PR^2 + {PQ^2} \\\therefore \text{radius} &= \frac{1}{2}\,RQ.\end{align}\]

We can find the area of semicircle.

And since \(\angle RPQ = {90^ \circ }\)

\(\therefore\; RP\) is the height for \(PQ\) or vice versa

So using the formula.

Area of \(\begin{align} \Delta = \frac{1}{2} \times {\text{base}} \times {\text{height}}\end{align}\)

Area of \(\begin{align} \Delta \,RPQ = \frac{1}{2} \times PQ \times RP \end{align}\)

Steps:

\( PQ = 24\,{\rm{cm}}\,\,, PR = 7\,\rm{cm}\)

Hence the angle in a semicircle is a right angle

\(\therefore \;LRPQ = {90^\circ }\)

\( \Rightarrow \Delta {RQP}\) is a right angled triangle.

\(\therefore \;\) Using Pythagoras theorem.

\[\begin{align}{P{Q^2}} &= {P{R^2}} + {P{Q^2}}\\{RQ }&= \sqrt {{7^2} + {{24}^2}} \\ &= \sqrt {49 + 576} \\ &= \sqrt {625} \\ &= 25\,{\rm{cm}}\end{align}\]

\(\therefore\;\)Radius (\(r\))  \(\begin{align}= \frac{{25}}{2}{\text{cm}}\end{align}\)

Area of shaded region\( = \)Area of semicircle \(RPQ\; –\) Area of \( \Delta RQP\)

\[\begin{align}&= \frac{1}{2} \times \pi {{\text{r}}^{\text{2}}} - \frac{{\text{1}}}{{\text{2}}} \times {{PQ}} \times {RP}\\ &= \frac{1}{2} \times \left[ {\frac{{22}}{7} \times \frac{{25}}{2} \times \frac{{25}}{2} - 24 \times 7} \right]\\ &= \frac{1}{2} \times \left[ {\frac{{6875}}{{14}} - 168} \right]\\ &= \frac{1}{2} \times \left[ {\frac{{6875 - 2352}}{{14}}} \right] = \frac{1}{2} \times \frac{{4523}}{{14}}\\ &= \frac{{4523}}{{28}}\,{\text{cm}^2}\\& = 161.535\,\,{\text{cm}^2}\\ &= 161.54\,\,{\text{cm}^2}\,\left( {{\text{approximately}}} \right)\end{align}\]