(i) (c): Radius of circle representing red region
\(=\frac{22}{2}=11 \mathrm{~cm}\)  [\(\because\) Diameter = 22 cm (Given)]
(ii) (a): Area of red region = \(\pi r^{2}\)
\(=\frac{22}{7} \times 11 \times 11=380.28 \mathrm{~cm}^{2}\)
(iii) (b): Radius of circle formed by combining red and silver region = Radius of red region + width of silver sign
= (11 + 10.5) cm = 21.5 cm
(iv) (d): Area of silver region
= Area of combined region - Area of red region
\(=\frac{22}{7} \times 21.5 \times 21.5-380.28\)
= 1452.78 - 380.28 = 1072.50 crrr'
(v) (b): Area of circular path formed by two concentric circles = \(\pi\left(r_{1}^{2}-r_{2}^{2}\right)\) sq. units

Let r and R be the radii of each smaller circle and larger circle respectively.
We have, \(d=\frac{1}{4} D\)
\(\Rightarrow r=\frac{1}{4} R \Rightarrow r=\frac{1}{4} \times 16 \Rightarrow r=4 \mathrm{~cm}\)
(i) (c): Area of smaller circle = \(\pi r^{2}\)
\(=\frac{22}{7} \dot{\times} 4 \times 4=50.28 \mathrm{~cm}^{2}\)
(ii) (a): Area oflarger circle = \(\pi R^{2}\)
\(=\frac{22}{7} \times 16 \times 16=\frac{5632}{7}=804.57 \mathrm{~cm}^{2}\)
(iii) (b): Area of the black colour region = Area of larger circle - Area of 4 smaller circles
= 804.57 - 4 x 50.28 = 603.45 cm²
(iv) (d): Area of quadrant of a smaller circle
\(=\frac{1}{4} \times 50.28=12.57 \mathrm{~cm}^{2}\)
(v) (c): Area between two concentric circles
\(=\pi\left(R^{2}-r^{2}\right)=\frac{22}{7}\left(5^{2}-2^{2}\right) \)
\(=\frac{22}{7}(25-4)=\frac{22}{7} \times 21=66 \mathrm{~cm}^{2}\)

(i) (b): Side of square PQRS = 27 cm
\(\therefore\) Area of square PQRS = 27 x 27 = 729 cm²
(ii) (c): Area of rectangle left for car parking is area of
region PSUT = 27 x 3 = 81 cm²
(iii) (a) : Diameter of semi circle = \(P V=\frac{P S}{2}=\frac{27}{2}\)
= 13.5 cm
\(\therefore\) Radius of semi circle \(=\frac{13.5}{2}=6.75 \mathrm{~cm}\)
(iv) (d): Area of a semi -circle \(=\frac{1}{2} \pi r^{2}\)
\(=\frac{1}{2} \times \frac{22}{7} \times 6.75 \times 6.75=71.59 \mathrm{~cm}^{2}\)
(v) (b): Area of shaded region = area of rectangular plot QRUT - area of two semi-circles
= 30 x 27 - 2 x 71.59 = 666.82 cm²

(i) (b):Area of square ABCD = 42 x 42
= 1764 cm2
(ii) (c): Area of quadrant BCD
\(=\frac{1}{4} \times \frac{22}{7} \times 42 \times 42=1386 \mathrm{~cm}^{2}\)
(iii) (a) : Area of \(\Delta C E F=\frac{1}{2} \times C E \times C F\)
\(=\frac{1}{2} \times 7 \times 7=24.5 \mathrm{~cm}^{2}\)
(iv) (d): Area of shaded region = Area of quadrant BCD + Area of \(\Delta\)CEF
= 1386 + 24.5 = 1410.5 cm²
(v) (b): Area of the unshaded region = Area of square ABCD - Area of quadrant BCD
= 1764 - 1386 = 378 cm²

(i) (b): Volume of the earth taken out
\(=\pi\left(\frac{7}{2}\right)^{2} \times 12=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12=462 \mathrm{~m}^{3}\)
(ii) (d): Area of the rectangular field = 30 x 15 = 450 m²
(iii) (a): Area of top of the pit = \(=\pi\left(\frac{7}{2}\right)^{2} =\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \)
\(=\frac{77}{2}=38.5 \mathrm{~m}^{2}\)
(iv) (d): Area of the remaining field = Area of rectangular field - area of top of pit
= 450 - 38.5 = 411.5 m²
(v) (c): The rise in the level of field = \(=\frac{462}{411.5}=1.12 \mathrm{~m}\)