RD Sharma 2017 Solutions for Class 7 Math Chapter 21 Mensuration II are provided here with simple step-by-step explanations. These solutions for Mensuration II are extremely popular among class 7 students for Math Mensuration II Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2017 Book of class 7 Math Chapter 21 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2017 Solutions. All RD Sharma 2017 Solutions for class 7 Math are prepared by experts and are 100% accurate.

Page No 21.15:

Question 1:

Find the area of a circle whose radius is
(i) 7 cm
(ii) 2.1 m
(iii) 7 km

Answer:

(i) We know that the area A of a circle of radius is given by Aπr2.
     Here, r  = 7 cm
A = 227×72 cm2.
A = 227×7×7 cm2 = 22×7 cm2 = 154 cm2.
(ii) ​We know that the area A of a circle of radius r is given by Aπr2.
     Here, r  = 2.1 m
∴ A = 227×2.12 m2.
A = 227×2.1×2.1 m2 = 22×0.3×2.1 m2 = 13.86 m2.
(iii) ​We know that the area A of a circle of radius r is given by A =  πr2.
     Here, r  = 7 km
∴ A = 227×72 km2.
A = 227×7×7 km2 = 22×7 km2 = 154 km2.

Page No 21.15:

Question 2:

Find the area of a circle whose diameter is
(i) 8.4 cm
(ii) 5.6 m
(iii) 7 km

Answer:

(i) Let r be the radius of the circle. Then, r = 8.4 ÷ 2 = 4.2 cm.
∴ Area of the circle  = πr2
    A = 227×4.22 cm2A =227×4.2 ×4.2cm2 = 22×0.6×4.2 cm2 = 55.44 cm2..

(ii) Let r be the radius of the circle. Then, r = 5.6 ÷ 2 = 2.8 m.
 Area of the circle  = πr2
    A = 227×2.82 m2A =227×2.8 ×2.8 m2 = 22×0.4×2.8 m2 = 24.64 m2.
(iii) ​​Let r be the radius of the circle. Then, r = ÷ 2 = 3.5 km.
     Area of the circle  = πr2
    A = 227×3.52 km2A =227×3.5 ×3.5 km2 = 22×0.5×3.5 km2 = 38.5 km2.

Page No 21.15:

Question 3:

The area of a circle is 154 cm2. Find the radius of the circle.

Answer:

Let the radius of the circle be r cm. 
Area of the circle (A) = 154 cm2
    154 = 227×r2 cm2r2 =154×722  r2 =107822  r2 =49 r =7 cm.

Hence, the radius of the circle is 7 cm.

Page No 21.15:

Question 4:

Find the radius of a circle, if its area is
(i) 4 π cm2
(ii) 55.44 m2
(iii) 1.54 km2

Answer:

(i) ​Let the radius of the circle be r cm. 

∴ Area of the circle (A) = 4π cm2
    4π = π×r2 cm2r2 =4ππ  = 4  r =2 cm.

(ii) ​Let the radius of the circle be r cm. 

∴ Area of the circle (A) = 55.44 m2
    55.44 = 227×r2 m2r2 =55.44×722  r2 =5.04×72  r2 =17.64  r =4.2 m

(iii) ​Let the radius of the circle be r cm. 

∴ Area of the circle (A) = 1.54 km2
    1.54 = 227×r2 km2r2 =1.54×722  r2 =10.7822  r2 =0.49  r =0.7 km = 700 m

Page No 21.15:

Question 5:

The circumference of a circle is 3.14 m, find its area.

Answer:

We have :

Circumference of the circle = 3.14  ​m = 2πr
 3.14 m =2×227 x r mr=3.14×72×22 m = 12 m.

Area of the circle (A) = πr2 
    A = 227×122 m2A =227×12×12 m2  = 2228 m2 = 0.785 m2.

Page No 21.15:

Question 6:

If the area of a circle is 50.24 m2, find its circumference.

Answer:

We have :
Area of the circle (A) = ​πr2 = 50.24 m2
50.24 m2= 227×r2 r2 =50.24×722 m2  = 351.6822 m2 = 15.985 m2r = 3.998 m.


Circumference of circle (C) = 2πr
 C =2×227 x 3.998 mC=44×0.571 m = 25.12 m. 

Page No 21.15:

Question 7:

A horse is tied to a pole with 28 m long string. Find the area where the horse can graze. (Take π = 22/7).

Answer:

We have :
Length of the string = ​28 m
The area over which the horse can graze is the same as the area of a circle of radius 28 m.
Hence, required area = π​r2227×28×28m2 = 22×4×28×m2 = 2464 m2.

Page No 21.15:

Question 8:

A steel wire when bent in the form of a square encloses an area of 121 cm2. If the same wire is bent in the form of a circle, find the area of the circle.

Answer:

We have :
Area of the square = 121 cm2 
⇒ (side)2 = (11)2 cm2 
⇒ side =  11 cm.
So, the perimeter of the square = 4(side) = (4 x 11) cm = 44 cm.

Let r be the radius of the circle. Then,
Circumference of the circle = Perimeter of the square
  ⇒     2πr =  44
  ⇒     2 x 227x r =  44
 ⇒ r = 7 cm.
∴ Area of the circle = π​r2  = 227×7×7 = 22×7 = 154 cm2 .

Page No 21.15:

Question 9:

A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of of road.

Answer:

We have :
Circumference of the circular park = 2πr = 352 m
  ​  ⇒     2 x 227r =  352
   ⇒ r = 56 m.
Radius of the path including the 7m wide road = (r +7) =  56 +7 = 63 m.
∴ Area of the road :
 =π632 - π562 m2=π632 - 562 m2=π63 +5663 - 56 m2=π1197=227×119×7 m2 = 22×119 m2 = 2618 m2.

 

Page No 21.15:

Question 10:

Prove that the area of a circular path of uniform width h surrounding a circular region of radius r is π h (2r + h).

Answer:

Radius  of the circular region = r
​Radius of the circular path of uniform width h surrounding the circular region of radius r = (rh).
​∴ Area of the path
 =πr+h2-πr2=πr+h2-r2=πr2 + 2rh +h2 -r2=π2rh+h2=πh2r+h



Page No 21.16:

Question 11:

The perimeter of a circle is 4πr cm. What is the area of the circle?

Answer:

We have :
    Given perimeter of the circle = 4πr cm = 2​π (2r) cm

    We know that, the perimeter of a circle = 2πr

    ∴ Radius of the circle = 2r cm
    Area of the circle= πr2   = ​π (2r)2 = 4​πr2 .

Page No 21.16:

Question 12:

A wire of 5024 m length is in the form of a square. It is cut and made a circle. Find the ratio of the area of the square to that of the circle.

Answer:

We have:
    Perimeter of the square = 5024 m = Circumference of the circle
⇒​ 4 x Side of the square = 5024
∴ Side of the square = 50244 = 1256 m.
Let the area of the square be A1 and the area of the circle be A2.
Area of the square (A1)= side x side =50244 ×50244m2 .
Circumference of the circle = 5024 m
    ⇒ 2 πr = 5024 m

  2×227×r=5024 m  r = 5024×72×22.
 
Area of the circle (A2)= πr2   = ​227×5024×72×22×5024×72×22=5024×5024××72×2×22m2.

 A1:A2= 50244×50244 : 5024×5024××72×2×22 A1A2= 50244×50244 ÷ 5024×5024××72×2×22  A1A2=50244×50244×2×2×225024×5024××7A1A2=1114.A1:A2 = 11:14.
Hence, the ratio of the area of the square to the area of the circle is 11:14.

Page No 21.16:

Question 13:

The radius of a circle is 14 cm. Find the radius of the circle whose area is double of the area of the circle.

Answer:

Let the area of the circle whose radius is 14 cm be A1.
Let the radius and area of the circle, whose area is twice the area of the circle A1 , be r2 and A2, respectively.

Thus,
A1πr2 = π142 = 227×14×14 cm2=44×14 cm2= 616 cm2.
A2 = 2 × A1 = 2 × 616 = 1232 cm2
A2 = πr22 = 1232 cm2
 227×r22 = 1232 cm2r22 = 1232×722 = (56×7) cm2r2 = 56×7 = 7×8×7 =7×7×4×2  = 142 cm

Hence, the radius of the circle A2 is 142 cm.

Page No 21.16:

Question 14:

The radius of one circluar field is 20 m and that of another is 48 m. Find the radius of the third circular field whose area is equal to the sum of the areas of two fields.

Answer:

Let the area of the circle whose radius is 20 m be A1 , and the area of the circle whose radius
is 48 m be A2. Let A3 be the area of a circle that is equal to the sum of the areas of the two fields, with the radius of its field being r cm.
A3 =  A1 + A2

 A1 =  π202 = 227×20×20 m2=400π m2A2 = π (48)2 = 227×48×48 m2=2304π m2A3 = A1+A2  = 400π +2304π = π400 +2304 m2A3 = πr2 =  π400 +2304 m2r2 = 400 +2304 m2r = 2704 m = 52 m

Page No 21.16:

Question 15:

The radius of one circular field is 5 m and that of the other is 13 m. Find the radius of the circular field whose area is the difference of the areas of first and second field.

Answer:

Let the area of the circular field whose radius is 5 m be A1 , and the area of the circular field whose radius is 13 m be A2. Let A3 and r cm be the area and radius of the circular field, that is equal to the difference of the areas of the two fields.
∴ A3 = A2 - A1

 A1 =  π52 = 25π m2A2 = π (13)2 = 169 π m2A3 = A2-A1  = 169π -25 π = 144π m2A3 = πr2 =  144 π m2r2 = 144 m2r = 144 m = 12 m

Hence, the radius of the circular field is 12 m.

Page No 21.16:

Question 16:

Two circles are drawn inside a big circle with diameters 23rd and 13rd of the diameter of the big circle as shown in Fig. 18. Find the area of the shaded portion, if the length of the diameter of the circle is 18 cm.

Answer:

Let the left circle be denoted as the 1st circle and the right circle be denoted as the 2nd circle.
Diameter of the big circle = 18 cm
Radius of the big circle = 9 cm
Diameter of the 1st circle = 23×18=12 cm
Radius of the 1st circle = 6 cm
Diameter of the 2nd circle = 13×18=6 cm
Radius of the 2nd circle = 3 cm
Area of the 1st circle = π(6)2=36πcm2
Area of the 2nd circle = π(3)2=π×3×3=9π cm2
Area of the big circle = π(9)2=π×9×9=81π cm2
Area of the shaded portion = Area of the big circle - (Area of the Ist circle + Area of the IInd circle)
Area of the shaded portion = 81π-(36π+9π)=36π cm2.

Page No 21.16:

Question 17:

In Fig. 19, the radius of quarter circular plot taken is 2 m and radius of the flower bed is 2 m. Find the area of the remaining field.

Answer:

Radius of the quarter circular plot = 2 m
Area of the quarter circular plot = π(2)2=227×4==12.57 m2
Radius of each flower bed = 2 m
Area of four flower beds = 4×14×π(2)2=12.57 m2
Area of the rectangular region = Length × Breadth
Area of the rectangular region = 8 × 6 = 48 m2
Area of the remaining field = Area of the rectangular region - (Area of the quarter circle + Area of the four flower beds)

Area of the remaining field = [48 - (12.57 + 12.57)] m2 = 22.86 m2 .

Page No 21.16:

Question 18:

Four equal circles, each of radius 5 cm, touch each other as shown in Fig. 20. Find the area included between them. (Take π = 3.14).

Answer:

Side of the square = 10 cm
Area of the square = Side × Side
Area of the square = 10×10=100 cm2

Area of the four quarter circles = 4×14×227×52=78.57 cm2
Area included in them = Area of the square - Area of the four quarter circles
Area included in them = ( 100 - 78.57 ) cm2 = 21.43 cm2 .

Page No 21.16:

Question 19:

The area of circle is 100 times the area of another circle. What is the ratio of their circumferences?

Answer:

Let the area of the first circle be A1, the circumference be  C1 and the radius be r1.
Let the area of the second circle be A2 , the circumference be C2  and the radius be r2.

Thus,

    C1: C2=2πr1 : 2πr2C1C2 = 2πr12πr2 = r1r2

We know that :
         
   A1 =  100A2
∴ πr12 = 100×πr22 r12 = 100 ×r22r1 = 10 ×r2r1r2= 10

 Substituting the values, we get:
∴​ C1C2=r1r2 = 101C1:C2 =10:1 

Hence, the ratio of their circumferences is 10:1.



Page No 21.17:

Question 1:

The ratio of the perimeter (circumference) and diameter of a circle is

(a) π                              (b) 2π                               (c) π2                               (d) π4

Answer:

Let r be the radius of the circle. Then
Perimeter of circle = 2πr
Diameter of circle = 2r
Now
Perimeter of circleDiameter of circle=2πr2r=π
Thus, the required ratio is π.
Hence, the correct option is (a)

Page No 21.17:

Question 2:

The ratio of the area and circumference of a circle of diameter d is

(a) d                              (b) d2                               (c) d4                               (d) 2d
 

Answer:

Let r and d be respectively the radius and diameter of the circle. Then
d = 2r
Circumference of circle = 2πr = 2π×d2=πd
Area of circle = πr2 = πd22=πd24
Now
Area of circleCircumference of circle=πd24πd=d4
Hence, the correct option is (c)

Page No 21.17:

Question 3:

The cost of fencing a circular garden of radius 21 m at ₹10 per metre is

(a) ₹1320                           (b) ₹132                               (c) ₹1200                               (d) ₹660

Answer:

Radius (r) = 21 m
Cost per metre = ₹10
Circumference of circle = 2πr=2×227×21=132 m
Cost of fencing = Circumference × Cost per metre
                          = 132 × ₹10
                          = ₹1320
Hence, the correct option is (a).

Page No 21.17:

Question 4:

If the diameter of a circle is equal to the diagonal of a square, then the ratio of their areas is

(a) 7 : 1                       (b) 1 : 1                           (c) 11 : 7                           (d) 22 : 7

Answer:

Let r and a be the diameter of the circle and side of the square respectively. Then
Diameter of circle = 2r
Diagonal of square = a2
Now, as per the question
Diameter of circle = Diagonal of square
2r = a2  a=2r
Therefore
Area of circleArea of square=πr2a2=227×r22r2=227×r22r2=117
Hence, the correct option is (c).

Page No 21.17:

Question 5:

A circle is inscribed in a square of side 14 m.  The ratio of the area of the circle and that of
the square is

(a) π : 3                       (b) π : 4                           (c) π : 2                           (d) π : 1

Answer:

Let a and r be the side of the square and radius of the circle respectively.
Here, the diameter of the circle is equal to the side of the square. So
Diameter of circle = 2r = a
Therefore
Area of circleArea of square=πr2a2=π×r22r2=π4
Hence, the correct option is (b).

Page No 21.17:

Question 6:

How many times should a wheel of radius 7 m rotate to go around the perimeter
of a rectangular field of length 60 m and breadth 50 m?

(a) 3                       (b) 4                           (c) 5                           (d) 6

Answer:

Here, Radius (r) = 7 m, Length (l) = 60 m and Breadth (b) = 50 m.
Perimeter of circle = 2πr=2×227×7=44 m
Perimeter of rectangle = 2l+b=260+50=220 m
Therefore
Number of turns = Perimeter of rectanglePerimeter of circle=22044=5
Hence, the correct option is (c).

Page No 21.17:

Question 7:

The minute hand of a clock is 14 cm long. How far does the tip of the minute hand
move in 60 minutes?
(a) 22 cm                    (b) 44 cm                           (c) 33 cm                           (d) 88 cm

Answer:

Length of minute hand = 14 cm
Distance covered by minute hand in one round = 2πr=2×227×14=88 cm
Thus, the minute hand move 88 cm in 60 minutes.
Hence, the correct option is (d).

Page No 21.17:

Question 8:

The cost of fencing a semi-circular garden of radius 14 m at ₹10 per metre is 

(a) ₹1080                   (b) ₹1020                       (c) ₹700                        (d) ₹720

Answer:

Radius of circle (r) = 14 m
Perimeter of semi-circular garden
   =πr+2r=227×14+2×14=44+28=72 m
Cost of fencing = 72 × ₹10 = ₹720
Hence, the correct option is (d).

Page No 21.17:

Question 9:

The area of a square is equal to the area of a circle. The ratio between the side of the square
and the radius of the circle is

(a) π : 1               (b) 1 : π                       (c) 1 : π                  (d) π : 1

Answer:

Let a and r be respectively the side of the square and radius of the circle.
Here, the area of square is equal to the area of the circle. So
                                                      a2=πr2a2r2=πar=π
Hence, the correct option is (a).

Page No 21.17:

Question 10:

If A is the area and C be the circumference of a circle, then its radius is
(a) AC                (b) 2AC                       (c) 3AC                  (d) 4AC

Answer:

Let r be the radius of the circle. Then
A=πr2 and C=2πrAC=πr22πrAC=r2r=2AC
Hence, the correct option is (b).

Page No 21.17:

Question 11:

The area of a circle of circumference C is

(a) C24π                (b) C22π                       (c) C2π                  (d) 4C2π

Answer:

Let r be the radius of the circle. Then
C=2πr r=C2π
Therefore
Area of circle = πr2=πC2π2=C24π
Hence, the correct option is (a).

Page No 21.17:

Question 12:

The circumference of a circle is 44 cm. Its area is

(a) 77 cm2                        (b) 154 cm2                       (c) 208 cm2                  (d) 144 cm2
 

Answer:

Let r be the radius of the circle. Then
44=2πr   r=442×227=7 cm
Therefore
Area of circle = πr2=227×72=154 cm2
Hence, the correct option is (b).

Page No 21.17:

Question 13:

Each side of an equilateral triangle is equal to the radius of a circle whose area is 154 cm2.
The area of the triangle is

(a) 734 cm2                        (b) 4932 cm2                       (c) 4934 cm2                  (d) 732 cm2

Answer:

Let r be the radius of the circle and a be the side of the equilateral triangle. Then
Area of circle = 154 cm2
154=227×r2   r=154×722=7 cm
Therefore
Area of equilateral triangle = a234=7234=4934        a=r=7 cm
Hence, the correct option is (c).



Page No 21.18:

Question 14:

The area of a circle is 9π cm2. Its circumference is

(a) 6π cm                        (b) 36π cm                       (c) 9π cm                  (d) 36π2 cm

Answer:

Let r be the radius of the circle. Then
Area of circle = 9π cm2
πr2=9π   r=3 cm
Therefore
Circumference of the circle = 2πr=2π×3=6π cm
Hence, the correct option is (a).

Page No 21.18:

Question 15:

The area of a circle is increased by 22 cm2 when its radius is increased by 1 cm.
The original radius of the circle is

(a) 6 cm                        (b) 3 cm                       (c) 4 cm                  (d) 3.5 cm

Answer:

Let r be the radius of the circle. Then
Area of original circle = πr2 cm2
Radius of circle after increment = (r + 1) cm
Thus,as per the question
πr+12-πr2=22r+12-r2=22227=7r2+2r+1-r2=7r=3 cm
Thus, the original radius of the circle is 3 cm.
Hence, the correct option is (b).

Page No 21.18:

Question 16:

The radii of two circles are in the ratio 2 : 3. The ratio of their areas is

(a) 2 : 3                        (b) 4 : 9                       (c) 3 : 2                          (d) 9 : 4

Answer:

Let r1 and r2 be the radius of the two circles. So
r1r2=23
Now
Ratio of areas=πr12πr22=r1r22=232=49
Thus, the required ratio is 4 : 9.
Hence, the correct option is (b).

Page No 21.18:

Question 17:

The areas of two circles are in the ratio 49 : 36. The ratio of their circumferences is

(a) 7 : 6                        (b) 6 : 7                       (c) 3 : 2                          (d) 2 : 3

Answer:

Let r1 and r2 be the radius of the two circles. Then
πr12πr22=4936r1r22=762r1r2=76
Now
Ratio of circumferences=2πr12πr2=r1r2=76
Thus, the required ratio is 7 : 6.
Hence, the correct option is (a).

Page No 21.18:

Question 18:

The circumferences of two circles are in the ratio 3 : 4. The ratio of their areas is

(a) 3 : 4                        (b) 4 : 3                       (c) 9 : 16                          (d) 16 : 9

Answer:

Let r1 and r2 be the radius of the two circles. Then
2πr12πr2=34r1r2=34
Now
Ratio of areas=πr12πr22=r1r22=342=916
Thus, the required ratio is 9 : 16.
Hence, the correct option is (c).

Page No 21.18:

Question 19:

The difference between the circumference and radius of a circle is 37 cm. The area of the circle is

(a) 111 cm2                        (b) 148 cm2                       (c) 154 cm2                          (d) 258 cm2

Answer:

Let r1 and r2 be the radius of the two circles. Then
2πr-r=372×227×r-r=3744r-7r7=37r=37×737=7 cm
Now
Area of circle=πr2=227×7×7=154 cm2
Hence, the correct option is (c).



Page No 21.7:

Question 1:

Find the circumference of a circle whose radius is
(i) 14 cm
(ii) 10 m
(iii) 4 km

Answer:

(i) The circumference C of a circle of radius 'r' is given by:
    C = 2πr
    Here, r = 14 cm
∴ C = 2×227×14 = 88 cm.
(ii) The circumference C of a circle of radius 'r' is given by:
    C = 2πr
   Here, r = 10 m
∴ C = 2×227×10 = 62.86 m.
(iii) ​The circumference C of a circle of radius 'r' is given by:
    C = 2πr
   Here, r = 4 km
∴ C = 2×227×4 = 25.142 km.

Page No 21.7:

Question 2:

Find the circumference of a circle whose diameter is
(i) 7 cm
(ii) 4.2 cm
(iii) 11.2 km

Answer:

(i) We have:
    Diameter = 7 cm
∴ Radius = 72 cm
    Let C be the circumference of a circle. Then,
    C = 2πr
∴ C = 2×227×72=22 cm.
(ii)   We have:
        Diameter = 4.2 cm
    ∴ Radius = 4.22 cm= 2.1 cm
       Let C be the circumference of a circle. Then,
       C = 2πr
   ∴ C = 2×227×2.1=13.2 cm.
(iii) ​​We have:
     Diameter = 11.2 km
∴ Radius = 11.22 km= 5.6 km 
    Let C be the circumference of a circle. Then,
    C = 2πr
   ∴ C = 2×227×5.6 =35.2 km.

Page No 21.7:

Question 3:

Find the radius of a circle whose circumference is
(i) 52.8 cm
(ii) 42 cm
(iii) 6.6 km

Answer:

(i) We know that the circumference C of a circle of  radius 'r' is given by:
    C = 2πr
    r = C2πHere, C = 52.8 cmr = 52.8×72×22 cm = 369.644 cm = 8.4 cm..

​(ii) We know that the circumference C of a circle of radius 'r' is given by:
     C = 2πr
      r = C2πHere, C = 42 cmr = 42×72×22 cm = 29444 cm = 6.68 cm.

(iii) We know that the circumference C of a circle of radius 'r' is given by:
      C = 2πr
      r = C2πHere, C = 6.6 kmr = 6.6×72×22 km = 2.12 km = 1.05 km.

Page No 21.7:

Question 4:

Find the diameter of a circle whose circumference is
(i) 12.56 cm
(ii) 88 m
(iii) 11.0 km

Answer:

(i) ​We know that the circumference C of a circle of  diameter 'd' is given by:
 C = πd
d = CπHere, C = 12.56 cmd = 12.56×722 cm = 87.9222 cm = 3.99 cm.

(ii) ​We know that the circumference C of a circle of  diameter 'd' is given by:
        C = πd
   d = CπHere, C = 88 md = 88×722 m = 28 m .

(iii) ​We know that the circumference C of a circle of diameter 'd' is given by:
        C = πd
   d = CπHere, C = 11 kmd = 11×722 km =72km=  3.5 km.

Page No 21.7:

Question 5:

The ratio of the radii of two circles is 3 : 2. What is the ratio of their circumferences?

Answer:

We have, the ratio of the radii = 3:2
So, let the radii of the two circles be 3r and 2r, respectively.
Let C1 and C2 be the circumferences of the two circles of radii 3r and 2r, respectively. Then,
C1= 2πx 3r = 6πr,  and C2 = ​2πx2r = 4πr

   C1C2 = 6πr4πr = 64 = 32, or
 
C1: C2 = 3:2.

Page No 21.7:

Question 6:

A wire in the form of a rectangle 18.7 cm long and 14.3 cm wide is reshaped and bent into the form of a circle. Find the radius of the circle so formed.

Answer:

Length of the wire = Perimeter of the rectangle
                           = 2 ( l +) = 2 x (18.7 + 14.3) = 66 cm

Let the wire be bent in the form of a circle of radius r cm. Then,

circumference = 66 cm

2πr = 66 cm2 ×227×r = 66 cm r= 66×722×2 = 212 cm= 10.5 cm.

Page No 21.7:

Question 7:

A piece of wire is bent in the shape of an equilateral triangle of each side 6.6 cm. it is re-bent to form a circular ring. What is the diameter of the ring?

Answer:

We have:
Length of the wire = The perimeter of  the equilateral triangle   

                       =  3 x side = 3 x 6.6 =   19.8 cm.

Let the wire be bent to form a circular ring of radius r cm.Then,

Circumference =  19.8 cm

2πr = 19.8 cm2 ×227×r = 19.8 cm r= 19.8×722×2 = 138.644 cm= 3.15 cm.

So, the diameter of the ring = 2 x 3.15 = 6.30 cm.
 

Page No 21.7:

Question 8:

The diameter of a wheel of a car is 63 cm. Find the distance travelled by the car during the period, the wheel makes 1000 revolutions.

Answer:

   It may be noted that in one revolution, the cycle covers a distance equal to the circumference of the wheel.
   Now, the diameter of the wheel = 63 cm
∴ Circumference of the wheel = πd = 227×63 cm= 198 cm.
   Thus, the cycle covers 198 cm in one revolution.
∴ The distance covered by the cycle in 1000 revolutions = (198 x 1000 ) cm = 198000 cm = 1980 m.

Page No 21.7:

Question 9:

The diameter of a wheel of a car is 98 cm. How many revolutions will it make to travel 6160 metres.

Answer:

We have:
    Diameter of the wheel of the car = 98 cm
∴ Circumference of the wheel of the car = πd = 227×98 cm= 308 cm.
     Note that, in one revolution of the wheel, the car travels a distance equal to the circumference of the wheel.

∴ The distance travelled by the car in one revolution of the wheel = 308 cm.
    Total distance travelled by the car = 6160 m = 616000 cm.
​​∴ Number of revolutions = 616000308 = 2000.

Page No 21.7:

Question 10:

The moon is about 384400 km from the earth and its path around the earth is nearly circular. Find the circumference of the path described by the moon in lunar month.

Answer:

    We have:
    The radius of the path described by the moon around the earth = 384400 km
∴ The circumference of the path described by the moon C = 2πr =2× 227×384400 km= 2416228.57 km.
    

Page No 21.7:

Question 11:

How long will John take to make a round of a circular field of radius 21 m cycling at the speed of 8 km/hr?

Answer:

 We have:
    The radius of the circular field = 21 m
∴ Circumference of the circular field = 2πr =2× 227×21 m= 132 m.
     If John cycles at the speed of 8 km/hr (In 1 hour John covers 8 km = 8000 m ), then,
     John covers 8000 m in 1 hour.

​​∴ Time required to cover 132 m  = 1328000 = 0.0165 hours
    1 hour = 3600 seconds
∴ 0.0615 hours = 0.0615 x 3600 = 59.4 seconds.

Page No 21.7:

Question 12:

The hour and minute hands of a clock are 4 cm and 6 cm long respectively. Find the sum of the distances travelled by their tips in 2 days.

Answer:

The radius of the path inscribed by the hour hand = Length of the hour hand = 4 cm
The radius of the path inscribed by the minute hand = Length of the minute hand = 6 cm

The circumference of the path inscribed by the hour hand = 2πr = 2×227x 4 = 1767 cm.
The hour hand makes 2 revolutions in one day.
∴ The distance covered by the hour hand in 2 days = 1767×2×2 = 100.57 cm.
    The distance covered by the minute hand in 1 revolution = 2πr = 2×227x 6 = 2647 cm
  The minute hand makes 1 revolution in one hour. 
∴ In  1 day, it makes 24 revolutions.
    In 2 days, it makes  2 x 24 revolutions.
∴ ​The distance covered by the minute hand in ​2 days = 2 ×24×2647cm =126727  cm = 1810.28 cm
The sum of the distances travelled by the hour and minute hands in 2 days = 1810.28 + 100.57 = 1910.85 cm.

Page No 21.7:

Question 13:

A rhombus has the same perimeter as the circumference of a circle. If the side of the rhombus is 2.2 m, find the radius of the circle.

Answer:

We have:

The side of a rhombus  = 2.2 m
Let C be the circumference of a circle having a radius r cm. Then,
the perimeter of the rhombus = 4×side = 4 x 2.2 = 8.8 m.

We know:
 Perimeter of the rhombus = Circumference of the circle
                    8.8 m = 2πrr = 8.82πm = 8.8×72×22m  =  88×72×22×10mr = 2×710m = 1.4 m.
The radius of the circle is 1.4 m.

Page No 21.7:

Question 14:

A wire is looped in the form of a circle of radius 28 cm. It is re-bent into a square form. Determine the length of the side of the square.

Answer:

We have:

The radius of the circle  = 28 cm
∴ Circumference of the circle  = 2πr = 2×227×28 = 176 cm.
Let a cm be the side of the square. Then,
the circumference of the circle  = the perimeter of the square

                    176 = 4×aa = 1764 cm = 44 cm.
The side of the square is 44 cm.

Page No 21.7:

Question 15:

A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.

Answer:

We have:

    Total distance covered in 5000 revolutions =  11 km = 11,000 m
∴​ Distance covered in 1 revolution =  11,0005000 = 115 m.
    Distance covered in 1 revolution =  Circumference of the wheel  
                                   ​115  = πdd = 115×π = 11×75×22 md =710 = 0.7 m.
Thus, the diameter of the wheel is 0.7 m = 70 cm.

Page No 21.7:

Question 16:

A boy is cycling such that the wheels of the cycle are making 140 revolutions per minute. If the diameter of the wheel is 60 cm, calculate the speed per hour with which the boy is cycling.

Answer:

​​We have:
The diameter of the wheel =  60 cm
Distance covered by the wheel in 1 revolution = Circumference of the wheel
∴ Distance covered by the wheel in 1 revolution = πd = 227×60 cm.
∴ Distance covered in 140 revolutions =  227×60×140 cm   = 1848007cm = 26400 cm.

Thus, the wheel covers 26400 cm in 1 minute. Then,
    Speed = 26400100×60m/hr = 264×60m/hr Speed= 264×601000km/hr = 15.84 km/hr.
The speed with which the boy is cycling is 15.84 km/hr.                                 ​

Page No 21.7:

Question 17:

The diameter of the driving wheel of a bus is 140 cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 km per hour?

Answer:

We have:
   Diameter of the wheel = 140 cm, desired speed of the bus = 66 km/hr
∴ Distance covered in 1 revolution = The circumference of the wheel = πd = 227×140cm = 440 cm.
   Now, the desired speed of the bus = 66 km/hr  =66×1000×10060 = 1,10,000 cm/min.
∴ Number of revolutions per minute = 110000440 = 10004 = 250.

Thus, the bus must make 250 revolutions per minute to keep the speed at 66 km/hr.



Page No 21.8:

Question 18:

A water sprinkler in a lawn sprays water as far as 7 m in all directions. Find the length of the outer edge of wet grass.

Answer:

The wet grass forms a circular region of radius 7 m.
∴ The length of the outer edge of the wet grass is 2πr = 2×227×7m = 44 m.

Page No 21.8:

Question 19:

A well of diameter 150 cm has a stone parapet around it. If the length of the outer edge of the parapet is 660 cm, then find the width of the parapet.

Answer:

We have :
Diameter of the well  = 150 cm
Length of the outer edge of the parapet = 660 cm
Width of the parapet = ?
 Radius of well = 150 ÷ 2 = 75 cm.
Let the width of  the stone parapet be x cm.Clearly, the outer edge of the parapet forms a circular region of radius (x +75 cm). Therefore,

  660 cm= 2×227×x+75 660×72×22 = x+75 15×7 =  x+75 x = 105 - 75x = 30 cm.
Thus, the width of the parapet is 30 cm.

Page No 21.8:

Question 20:

An ox in a kolhu (an oil processing apparatus) is tethered to a rope 3 m long. How much distance does it cover in 14 rounds?

Answer:

We have :
Radius of the circular path traced by the ox in a kolhu = 3 m
​Distance covered by the ox in 1 round = Circumference of the circular path = 2πr = 2×227×3m.
∴ Distance covered in 14 rounds = ​ 2×227×3m ×14 = 22×12 = 264 m.



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