Area of sector with central \(\angle\)P
\(=\frac{\angle P}{360^{\circ}} \times \pi \times(14)^2\)
Area of sector with central \(\angle\)Q
\(=\frac{\angle Q}{360^{\circ}} \times \pi \times(14)^2\)
and area of sector with central \(\angle\)R
\(=\frac{\angle R}{360^{\circ}} \times \pi \times(14)^2\)
Now, area of shaded region
\(=(14)^2 \times \pi \frac{(\angle P+\angle Q+\angle R)}{360^{\circ}}\)
\(=196 \times \pi \times \frac{180^{\circ}}{360^{\circ}}\) [by angle sum property of triangle]
308 cm²

1386 cm²

(75.36 - \(36\sqrt { 3 } \)) cm²

42 cm²

Let r₁=radius of the inscribed circle=\(\frac {4}{2}\)=2 cm
and r₂=radius of the circumscribed circle

Now, in triangle OBA,
(OA)²=(OB)²+(BA)²
[by Pythagoras theorem].
\(\Rightarrow \quad { r }^{ 2 }={ r }_{ 1 }^{ 2 }+{ r }_{ 1 }^{ 2 }\)
\(\Rightarrow { r }_{ 2 }=\sqrt { 2{ r }_{ 1 }^{ 2 } } \)
\(\Rightarrow { r }_{ 2 }=\sqrt { 2\times { 2 }^{ 2 } } =2\sqrt { 2 } cm\)
Now, area of inscribed circle
\(=\pi { r }_{ 1 }^{ 2 }=\pi { (2) }^{ 2 }=4\pi \quad { cm }^{ 2 }\)     ..(i)
and area of circumscribed circle=\(\pi { r }_{ 2 }^{ 2 }=\pi { (2\sqrt { 2 } ) }^{ 2 }=8\pi \quad { cm }^{ 2 }\)
Area of circumscribed circle=\(2(4\pi )\)
=2 x (Area of inscribed circle) [from Eq. (i)]