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

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)]

There are four equally semi-circles and JKLM formed a square
\(\therefore\)  FH=14-(3+3)=8 cm
Let the side of square JKLM be x can
Then, FH=\(\frac { x }{ 2 } +x+\frac { x }{ 2 } \)
\(\Rightarrow\) 8=2x \(\Rightarrow\)  x=4
So, the side of square should be 4 cm and radius of semi-circle of both ends are 2 cm each.
\(\therefore\) Area of square JKLm=(4)²=16 cm²
Area of semi-circle JHM=\(\frac { \pi { r }^{ 2 } }{ 2 } \)
\(=\frac { \pi \times (2)^{ 2 } }{ 2 } =2\pi \quad { cm }^{ 2 }\)
\(\therefore\) Area of four semi-circles=\(4\times 2\pi \)
Now, area of square ABCD=(14)²=196 cm²
\(\therefore\) Area of shaded region=Area of square ABCD-(Area of four semi-circles + Area of square JKLM)
\(=196-(8\pi +16)=196-16-8\pi \)
\(=(180-8\pi ){ cm }^{ 2 }\)
Hence, the required of the shaded region is \((180-8\pi ){ cm }^{ 2 }\)