Ex.12.3 Q12 Areas Related to Circles Solution - NCERT Maths Class 10

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Question

In Figure, \({OACB}\) is a quadrant of a circle with centre \({O}\) and radius \(3.5\; \rm{cm.}\) If \(OD = 2 \,\rm{cm,}\) find the area of the

(i) Quadrant \({OACB,}\) 

(ii) Shaded region.

Text Solution

What is known?

\({OACB}\) is a quadrant of a circle with centre \({O} \) and radius \((r) = \rm{3.5 cm}\)

What is unknown?

(i) Area of the quadrant \({OACB}\) 

(ii) Area of the shaded region

Reasoning:

(i) Since quadrant mean \(\begin{align}\frac{1}{4}\rm{th}\end{align}\) part.

Therefore  angle  at the centre of a quadrant of a circle\(\begin{align} = \frac{{360}}{4} = {90^ \circ }\end{align}\)

Area of the quadrant OACB \(\begin{align} = \frac{1}{4}\pi {r^2}\end{align}\)

We can get area of quadrant \({OACB}\) with radius  \(r = \text{3.5 cm}\) 

 (ii) Visually from the figure it is clear that

Area of shaded region \(=\) Area of quadrant \(OACB\; –\) Area of \(\Delta {BDO}\)

Since \(\angle {BOD} = {90^ \circ }\)

\(\therefore\;\) For side OB of \(\Delta {BDO, OD}\)  is the 

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

We can find Area of \(\Delta {BDO}\) with base \(= OB = \text{3.5 cm}\) (radius of quadrant) height \(= OD = \text{2 cm}\)

Steps:

(i)  Since \({OACB}\) is a quadrant, it will subtend \(\begin{align} \text{} \,\, \theta = \frac{{{{360}^\circ }}}{4} = {90^\circ }\end{align}\) angle at \({O}\) 

Radius \((r) = \text{3.5 cm}\) 

Area of quadrant \({OACB}\) \(\begin{align}= \frac{1}{4}\pi {r^2}\end{align}\)

\[\begin{align}&= \frac{1}{4} \times \frac{{22}}{7} \times {\left( {3.5cm} \right)^2}\\&= \frac{1}{4} \times \frac{{22}}{7} \times \frac{7}{2} \times \frac{7}{2}c{m^2}\\&= \frac{{77}}{8}c{m^2}\end{align}\]

\(\begin{align}{\rm{In }}\Delta BDO,{\rm{ }}OB = r = 3.5cm = \frac{7}{2}cm\,OD = 2cm\,\angle BOD = {90^{\rm{\circ}}}\end{align}\)

\[\begin{align}{\rm{Area of }}\Delta BDO &= \frac{1}{2} \times {\rm{ }}base{\rm{ }} \times height\\&= \frac{1}{2} \times OB \times OD\\& = \frac{1}{2} \times \frac{7}{2}cm \times 2cm\\&= \frac{7}{2}c{m^2}\end{align}\]

From figure, it is observed that:

Area of shaded region \(=\) Area of Quadrant \({OACB} \,-\) Area of \(\Delta {BDO}\)

\[\begin{align}&= \frac{{77}}{8} - 3.5\,{\text{c}}{{\text{m}}^2}\\&={ \frac{{77}}{8} - \frac{7}{2}{\text{c}}{{\text{m}}^2}}\\&= {\frac{{77 - 28}}{8}{\text{c}}{{\text{m}}^2}}\\&= {\frac{{49}}{8}{\text{c}}{{\text{m}}^2}}\end{align}\]