NCERT Solutions for Class 10 Maths Chapter 12: Areas Related to Circles || CBSE Class 10 Maths Chapter 12 Constructions solutions Ex 12.2

Radius \( (r) \) of circle \( =21 \mathrm{~cm} \)
The angle subtended by the given arc \( =60^{\circ} \)
Length of an arc of a sector of angle \( \theta=\frac{\theta}{360^{\circ}} \times 2 \pi r \)
(i) Length of arc \( \mathrm{ACB}=\frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21 \)
\(=\frac{1}{6} \times 2 \times 22 \times 3\)
\(=22 \mathrm{~cm}\)
(ii) Area of sector \( \mathrm{OACB}=\frac{\theta}{360^{\circ}} \pi r^{2} \)
\(=\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 21 \times 21\)
\(=\frac{1}{6} \times \frac{22}{7} \times 21 \times 21\)
\(=231 \mathrm{~cm}^{2}\)
(iii) In \( \triangle \mathrm{OAB} \),
As \( \mathrm{OA}=\mathrm{OB} \)
\(\Rightarrow \angle \mathrm{OAB}=\angle \mathrm{OBA} \text { (Angles opposite to equal sides are equal) }\)
\(\Rightarrow \angle \mathrm{OAB}+\angle \mathrm{AOB}+\angle \mathrm{OBA}=180^{\circ}\)
\(\Rightarrow 2 \angle \mathrm{OAB}+60^{\circ}=180^{\circ}\)
\(\Rightarrow 2 \angle \mathrm{OAB}=180^{\circ}-60^{\circ}\)
\( \Rightarrow \angle \mathrm{OAB}=60^{\circ} \)
Hence,
\( \triangle \mathrm{OAB} \) is an equilateral triangle
Area of equilateral triangle \( =\frac{\sqrt{3}}{4}(S i d e)^{2} \)
\( \Rightarrow \) Area of \( \triangle \mathrm{OAB}=\frac{\sqrt{3}}{4}( Side)^{2} \)
\( \frac{\sqrt{3}}{4} \times(21)^{2} \)
\( =\frac{441 \sqrt{3}}{4} \mathrm{~cm}^{2} \)
Area of segment \( \mathrm{ACB}= \) Area of sector \( \mathrm{OACB}\) - Area of \( \triangle \mathrm{OAB} \) \( =\left(231-\frac{441 \sqrt{3}}{4}\right) \mathrm{cm}^{2} \)

Radius \( (r) \) of circle \( =15 \mathrm{~cm} \)
Area of sector \( \mathrm{OPRQ}=(\frac{ 60 }{ 360 }) \times \pi r^{2} \)
\(=\frac{ 1 }{ 6 } \times 3.14 \times(15)^{2}\)
\(=117.75 \mathrm{~cm}^{2}\)
In \( \triangle \mathrm{OPQ} \),
\(\angle \mathrm{OPQ}=\angle \mathrm{OQP}\) (As \(\mathrm{OP}=\mathrm{OQ})\)
\(\angle \mathrm{OPQ}+\angle \mathrm{OQP}+\angle \mathrm{POQ}=180^{\circ}\)
\(2 \angle \mathrm{OPQ}=120^{\circ}\)
\(\angle \mathrm{OPQ}=60^{\circ}\)
\( \Delta \mathrm{OPQ} \) is an equilateral triangle.
\(\text { Area of } \Delta \mathrm{OPQ}=\frac{\sqrt{3}}{4} \times(Side)^{2}\)
\(=\frac{\sqrt{3}}{4} \times(15)^{2}\)
\(=\frac{225 \sqrt{3}}{4}\)
\(=56.25 \times \sqrt{3}\)
\(=97.3125 \mathrm{~cm}^{2}\)
Area of segment \( \mathrm{PRQ}= \) Area of sector \( \mathrm{OPRQ}\) - Area of \( \triangle \mathrm{OPQ} \)
\(=117.75-97.3125\)
\(=20.4375 \mathrm{~cm}^{2}\)
Area of major segment \( \mathrm{PSQ}\) = Area of circle - Area of segment \( \mathrm{PRQ}\)
\(=\pi(15)^{2}-20.4375\)
\(=3.14 \times 225-20.4375\)
\(=706.5-20.4375\)
\(=686.0625 \mathrm{~cm}^{2}\)

Let us draw a perpendicular \( \mathrm{OV}\) on chord \( \mathrm{ST}\). It will bisect the chord \( \mathrm{ST}\) and the angle \( \mathrm{O}\).
\(\mathrm{SV}=\mathrm{VT}\)
In \( \triangle \mathrm{OVS} \),
\(\text { As, } \cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}\)
\(\frac{O V}{O S}=\cos 60^{\circ}\)
\(\frac{O V}{12}=1\)
\(\mathrm{OV}=6 \mathrm{~cm}\)
\(\frac{s v}{s o}=\sin 60^{\circ}\)
\(\frac{s v}{12}=\frac{\sqrt{3}}{2}\)
\(\mathrm{SV}=6 {\sqrt{3}} \mathrm{~cm}\)
\(\mathrm{ST}=2 \times \mathrm{SV}\)
\(=2 \times 6{\sqrt{3}}\)
\(=12 {\sqrt{3}} \mathrm{~cm}\)
Area of \( \Delta \mathrm{OST}=\frac{1}{2} \times 12 {\sqrt{3}} \times 6 \)
\(=36 {\sqrt{3}}\)
\(=36 \times 1.73\)
\(=62.28 \mathrm{~cm}^{2}\)
Area of sector \( \mathrm{OUST} =\frac{120}{360} \times \pi \times(12)^{2} \)
\(=\frac{1}{3} \times 3.14 \times 144\)
\(=150.72 \mathrm{~cm}^{2}\)
Area of segment \( \mathrm{SUT} = \) Area of sector \( \mathrm{OUST}\) - Area of \( \triangle \mathrm{OST}\)
\(=150.72-62.28\)
\(=88.44 \mathrm{~cm}^{2}\)