(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

(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): Since \(\Delta\)ABC is a right angled triangle
\(\therefore\)  Area of triangle ABC \(=\frac{1}{2} \times 7 \times 24=84 \mathrm{~m}^{2}\)
(ii) (a): Area of region grazed by the cow
\(=\frac{\angle A}{360^{\circ}} \times \pi \times(3.5)^{2}\)
(iii) (d): Area of the region grazed by the buffalo and the horse
\(=\frac{\angle B}{360^{\circ}} \times \pi \times(3.5)^{2}+\frac{\angle C}{360^{\circ}} \times \pi \times(3.5)^{2} \)
\(=\frac{(\angle B+\angle C)}{360^{\circ}} \times \pi \times(3.5)^{2}\)
(iv) (d): Total area grazed by the cow, the horse and the buffalo
\(=\frac{\angle A}{360^{\circ}} \times \pi \times(3.5)^{2}+\frac{(\angle B+\angle C)}{360^{\circ}} \times \pi \times(3.5)^{2}\)
\(=\frac{(\angle A+\angle B+\angle C)}{360^{\circ}} \times \pi \times(3.5)^{2} \)
\(=\frac{22}{7} \times \frac{180^{\circ}}{360^{\circ}} \times 3.5 \times 3.5\)
\(\left(\because\right. sum of interior angles of a triangle is 180^{\circ} )\)
\(=\frac{77}{4}=19.25 \mathrm{~m}^{2}\)
(v) (b): Area of the field that cannot be grazed
= Area of MBC - Area of region grazed by all the three animals
= 84 - 19.25 = 64.75 m²

(i) (a): Since BOC is the diameter and \(\angle\)BAC = 90°
\(\therefore\) BC² =AB² +AC²
= 72 + 242 = 625
\(\Rightarrow\) BC = 25 cm
\(\therefore\) Radius of circle \(=\frac{25}{2} \mathrm{~cm}=12.5 \mathrm{cn}\)
(ii) (c): Area of circle \(=\pi(12.5)^{2}=\frac{22}{7} \times 12.5 \times 12.5\)
= 491.07 cm²
(iii) (b): Clearly, \(\angle\)COD = 90°
[\(\therefore\)\(\angle\)COB = 180° and equal arcs subtends equal angles at the centre]
Area of region COD \(=\frac{90^{\circ}}{360^{\circ}} \times \pi r^{2}\)
\(=\frac{1}{4}(491.07)=122.76 \mathrm{~cm}^{2}\)
(iv) (d):  \(\text { Area of } \Delta B A C=\frac{1}{2} \times A B \times A C \)
\(=\frac{1}{2} \times 7 \times 24=84 \mathrm{~cm}^{2}\)
(v) (b): Area of the polluted region = Area of circle - Area of sector COD - Area of \(\Delta\)ABC
= 491.07 - 122.76 - 84
= 284.31 cm²