1. Let R be the radius of the circle which has circumference equal to the sum of circumferences of the two circles, then according to question,
27πR = 2π(19) + 2π(9) ⇒ R = 19 + 9 ⇒ R = 28 cm
2. Let R be the radius of the circle which has area equal to the sum of areas of the two circles, then
According to the question,
πR2 = π(8)2 + π(6)2 ⇒ R2 = (8)2 + (6)2 ⇒ R2 = 64 + 36
⇒ R2 = 100 ⇒ R = 10 cm
3. Gold: Diameter = 21 cm 
Area of gold scoring region 
Red: Area of red scoring region 
Blue: Area of blue scoring region 
Black: Area of black scoring region 
White: Area of white scoring region 
4. Diameter of wheel = 80 cm ⇒ Radius of wheel (r) = 40 cm
Distance covered by wheel in one revolution 
∵ Distance covered by wheel in 1 hour = 66 km = 66000 m = 6600000 cm
∴ Distance covered by wheel in 10 minutes 
∴ No. of revolutions = 
5. (A) Circumference = Area ⇒ 2πr = πr2 ⇒ r = 2 units
6. Here, r = 6 cm and θ = 600
7. Given, 
We know that for quadrant of circle, θ = 900
∴ Area of quadrant 
8. Here, 
∴ Area swept 
9. (i) Here, r = 10 cm and θ = 900
∴ Area of minor segment = Area of minor sector – Area of Δ OAB
= 78.5 – 50 = 28.5 cm2
(ii) For major sector, radius = 10 cm and θ = 3600 – 900 = 2700
∴ Area of major sector 
10. Given, r = 21 cm and θ = 600
⇒ Area of segment = 231 – Area of Δ OAB …(i)
In right angled triangle OMA and OMB,
OM = OB [Radii of the same circle]
OM = OM [Common]
∴ Δ OMA ≅ Δ OMB [ RHS congruency]
∴ AM = BM [By CPCT]
∴ M is the mid-point of AB and ∠ AOM = ∠ BOM
Therefore, in right angled triangle OMA,
Using eq. (i),
Area of segment formed by corresponding chord 
11. Here, r = 15 cm and θ = 600
For, Area of Δ AOB,
Draw OM ⊥ AB.
In right triangles OMA and OMB,
OA = OB [Radii of same circle]
OM = OM [Common]
∴ Δ OMA ≅ Δ OMB [RHS congruency]
∴ AM = BM [By CPCT]
In right angled triangle OMA, 
Area of minor segment = Area of minor sector – Area of Δ AOB
= 117.75 – 97.3125 = 20.4375 cm2
And, Area of major segment = πr2 — Area of minor segment
= 706.5 – 20.4375 = 686.0625 cm2
12. Here, r = 15 cm and θ =1200
For, Area of Δ AOB,
Draw OM ⊥ AB.
In right triangles OMA and OMB,
OA = OB [Radii of same circle]
OM = OM [Common]
∴ Δ OMA ≅ Δ OMB [RHS congruency]
∴ AM = BM [By CPCT]
In right angled triangle OMA, 
∴ Area of corresponding segment = Area of corresponding sector – Area of Δ AOB
= 150.72 – 62.28 = 88.44 cm2
13. (i) Area of quadrant with 5 m rope 
(ii) Area of quadrant with 10 m rope 
∴ The increase in grazing area = 78.5 – 19.625 = 58.875 m2
14. (i) Diameter = 35 mm 
Circumference 
Length of 5 diameters = 35 x 5 = 175 mm …..(ii)
∴ Total length of the silver wire required = 110 + 175 = 285 mm
(ii) 
The area of each sector of the brooch 
15. Here, r = 45 cm and 
Area between two consecutive ribs of the umbrella 
16. Here, r = 25 cm and θ =1150
The total area cleaned at each sweep of the blades = 
17. Here, r = 16.5 km and θ = 800
The area of sea over which the ships are warned = 
18. r = 28 cm and 
For, Area of Δ AOB,
Draw OM ⊥ AB.
In right triangles OMA and OMB,
OA = OB [Radii of same circle]
OM = OM [Common]
∴ Δ OMA ≅ Δ OMB [RHS congruency]
∴ AM = BM [By CPCT]
∴ Area of minor segment = Area of minor sector – Area of Δ AOB = 410.67 – 333.2 = 77.47 cm2
∴ Area of one design = 77.47 cm2
∴ Area of six designs = 77.47 x 6 = 464.82 cm2
Cost of making designs = 464.82 x 0.35 = Rs. 162.68
19. (D) Given, r = R and θ = p
20. ∠ RPQ = 900 [Angle in semi circle is 900]
∴ RQ2 = PR2 + PQ2 = (7)2 + (24)2 = 49 + 576 = 625
⇒ RQ = 25 cm ⇒ Diameter of the circle = 25 cm
∴ Radius of the circle = 
Area of the semicircle = 
Area of right triangle RPQ 
Area of shaded region = Area of semicircle – Area of right triangle RPQ
21. Area of shaded region = Area of sector OAC – Area of sector OBD
22. Area of shaded region
= Area of square ABCD – (Area of semicircle APD + Area of semicircle BPC)
23. Area of shaded region = Area of circle + Area of equilateral triangle OAB – Area common to the circle and the triangle
24. Area of remaining portion of the square = Area of square – (4 x Area of a quadrant + Area of a circle)
25. Area of design = Area of circular table cover – Area of the equilateral triangle ABC
∵ G is the centroid of the equilateral triangle.
∴ radius of the circumscribed circle 
26. Area of shaded region = Area of square – 4 x Area of sector
27. (i) Distance around the track along its inner edge 
28. Area of shaded region = Area of circle + Area of semicircle ACB – Area of Δ ACB
29. Area of equilateral triangle 
Area of shaded region = Area of Δ ABC 
= 17320.5 – 15700 = 1620.5 cm2
30. Area of remaining portion of handkerchief = Area of square ABCD – Area of 9 circular designs
31. (i) Area of quadrant OACB 
(ii) Area of shaded region = Area of quadrant OACB – Area of Δ OBD
32. 
Area of shaded region = Area of quadrant OPBQ – Area of square OABC
33. Area of shaded region = Area of sector OAB – Area of sector OCD
34. In right triangle BAC, BC2 = AB2 + AC2 [Pythagoras theorem]
⇒ BC2 = (14)2 + (14)2 = 2(14)2 ⇒ BC = 14√2 cm
∴ Radius of the semicircle 
∴ Required area = Area BPCQB
35. In right triangle ADC, AC2 = AD2 + CD2 [Pythagoras theorem]
⇒ AC2 = (8)2 + (8)2 = 2(8)2 AC = √128 = 8√2 cm
Draw BM ⊥ AC.
