CBSE Solutions for Class 10 Maths Chapter 12: Areas Related to Circles || CBSE Class 10 Maths Chapter 12 Areas Related to Circles solutions Ex 12.3

Radius of the circle "R" \( =32 \mathrm{~cm} \)
Draw a median AD of the triangle passing through the centre of the circle.
\(\Rightarrow \mathrm{BD}=\frac{ \mathrm{AB} }{ 2 }\)
Since, AD is the median of the triangle
\( \therefore \mathrm{AO}= \) Radius of the circle \( =\frac{ 2 }{ 3 } \mathrm{AD} \) [By the property of equilateral triangle inscribed in a circle]
\(\Rightarrow \frac{ 2 }{ 3 } \mathrm{AD}=32 \mathrm{~cm}\)
\( \Rightarrow \mathrm{AD}=48 \mathrm{~cm} \)
In \( \triangle \mathrm{ADB} \)

By Pythagoras theorem,
\(\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}\)
\(\Rightarrow \mathrm{AB}^{2}=48^{2}+(\frac{ \mathrm{AB} }{ 2 })^{2}\)
\(\Rightarrow \mathrm{AB}^{2}=2304+\frac{ \mathrm{AB}^{2} }{ 4 }\)
\(\Rightarrow \frac{ 3 }{ 4 }\left(\mathrm{AB}^{2}\right)=2304\)
\(\Rightarrow \mathrm{AB}^{2}=3072\)
\(\Rightarrow \mathrm{AB}=32
{\sqrt{3}} \mathrm{~cm}\)
Area of \( \triangle \mathrm{ABC}=
\frac{ {\sqrt{3}} }{ 4 } \times(32 \sqrt{3})^{2} \mathrm{~cm}^{2} \)
\(=768
{\sqrt{3}} \mathrm{~cm}^{2}\)
Area of circle \( =\pi \mathrm{R}^{2} \)
\(=\frac{ 22 }{ 7 } \times 32 \times 32=\frac{ 22528 }{ 7 } \mathrm{~cm}^{2}\)
Area of the design \( = \) Area of circle - Area of \( \triangle \mathrm{ABC} \)
\( =(\frac{ 22528 }{ 7 }-768 {\sqrt{3}}) \mathrm{cm}^{2} \)