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      ⇒     R= (8)+ (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 = πr2Area 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 = 142 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 = 82 cm

Draw BM ⊥ AC.