Diameter, d = 7 cm, therefore r = \(\frac{7}{2}\) cm
Area of the shaded region = Area of the square – 4 x Area of the circular quadrant
\(={ a }^{ 2 }-4\times \frac { 1 }{ 4 } { \pi r }^{ 2 }\)
\(=(7\times 7)-\left( 4\times \frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \right) \)
= 49 – 38.5 = 10.5 sq.cm. (approx.)

Area of the dining table of Kamalesh = πr² sq. units
\(=\frac{22}{7}\times70\times70\)
A = 15400 sq.cm (approximately.)

Area of the circular quadrant dining table of Tharun
\(=\frac{1}{4}\pi r^2=\frac{1}{4}\times \frac{22}{7}\times140 \times140\)
A = 15400 sq.cm (approximately.)
We find that, the area of the dining tables of both of them have the same area.

(i) The length of the iron frame required = length of the arc + 4r
= πr + 4r
\(=\left( \frac { 22 }{ 7 } \times 49 \right) +(\times 49)\)
= 154 + 196
= 350 cm (approximately)
(ii) Area of each of the mirror sectors
\(=\frac { { \theta }^{ 0 } }{ { 360 }^{ 0 } } \times { \pi r }^{ 2 }\)
\(=\frac { { 60 }^{ 0 } }{ { 360 }^{ 0 } } \times \frac { 22 }{ 7 } \times 49\times 49\)
= 1257.67 sq.cm (approximately)

Since, the sectors are cut from an equilateral triangle, the central angle of each of them is 60°.
∴ Area of the shape formed, \(A=3\times \left( \frac { { \theta }^{ 0 } }{ { 360 }^{ 0 } } \times { \pi r }^{ 2 } \right) \)
\(=3\times \frac { { 60 }^{ 0 } }{ { 360 }^{ 0 } } \times \pi \times 5\times 5\)
\(=3\times \frac { 1 }{ 6 } \times \pi \times 5\times 5\)
= 12.5 π sq.cm

Radius, r = 7.5 cm and n = 6
Area of each of the sectors, A = \(\frac{1}{n}\times \pi r^2\) sq. units
\(\frac{1}{6}\pi \times 7.5\times 7.5\)
= 9.375 x \(\frac{22}{7}\)