Frank Solutions for Chapter 19 Mensuration I Class 10 ICSE Mathematics


1. Find the area and perimeter of the circles with the following:
(i) Radius = 2.8 cm
(ii) Radius = 10.5 cm
(iii) Diameter = 77 cm
(iv) Diameter = 35 cm
Answer

(i) From the question it is given that, radius of circle, r = 2.8 cm

We know that, Area of the circle = πr2

= (22/7) × (2.8)2

= 24.64 cm2

Then, perimeter of the circle = 2πr

= 2 × (22/7) × (2.8)

= 17.6 cm

(ii) From the question it is given that, radius of circle, r = 10.5 cm

We know that, Area of the circle = πr2

= (22/7) × (10.5)2

= 346.5 cm2

Then, perimeter of the circle = 2πr

= 2 × (22/7) × (10.5)

= 66 cm

(iii) From the question it is given that, diameter of circle, d = 77 cm

We know that, radius of circle = half of the diameter of the circle

r = d/2

⇒ r = 77/2

⇒ r = 38.5 cm

Area of the circle = πr2

= (22/7) × (38.5)2

= 4658.5 cm2

Then, perimeter of the circle = 2πr

= 2 × (22/7) × (38.5)

= 242 cm

(iv) From the question it is given that, diameter of circle, d = 35 cm

We know that, radius of circle = half of the diameter of the circle

r = d/2

⇒ r = 35/2

⇒ r = 17.5 cm

Area of the circle = πr2

= (22/7) × (17.5)2

= 962.5 cm2

Then, perimeter of the circle = 2πr

= 2 × (22/7) × (17.5)

= 110 cm


2. Find the area and perimeter of the following semi-circles:

(i) Radius = 1.4 cm

(ii) Diameter = 7 cm

(iii) Diameter = 5.6 cm

Answer

(i) From the question it is given that, radius of semi-circle, r = 1.4 cm

We know that, Area of the semi-circle = ½πr2

= ½ × (22/7) × (1.4)2

= 3.08 cm2

Then, perimeter of the semi-circle = (πr + 2r)

= ((22/7) × (1.4)) + (2 × 1.4)

= 7.2 cm

(ii) From the question it is given that, diameter of semi-circle, d = 7 cm

We know that, radius of semi-circle = half of the diameter of the semi-circle

r = d/2

⇒ r = 7/2

⇒ r = 3.5 cm

Area of the semi-circle = ½ πr2

= ½ × (22/7) × (3.5)2

= 19.25 cm2

Then, perimeter of the semi-circle = (πr + 2r)

= (22/7 × 3.5) + (2 × 3.5)

= 18 cm

(iii) From the question it is given that, diameter of semi-circle, d = 5.6 cm

We know that, radius of semi-circle = half of the diameter of the semi-circle

r = d/2

⇒ r = 5.6/2

⇒ r = 2.8 cm

Area of the semi-circle = ½ πr2

= ½ × (22/7) × (2.8)2

= 12.32 cm2

Then, perimeter of the semi-circle = (πr + 2r)

= (22/7 × 2.8) + (2 × 2.8)

= 14.4 cm


3. Find the area and perimeter of the following sectors:

(i) Radius = 4.2 cm, angle at the center is 60o

(ii) Radius = 6 cm, angle at the center is 70o

(iii) Diameter = 42 cm, angle at the center is 100o.

Answer

(i) From the question it is given that, radius = 4.2 cm, angle at the center = 60o

We know that, Area of the sector = (πr2 × θ/360o)

= [(22/7) × 4.22 × (60o/360o)]

= 9.24 cm2

Then, perimeter of the sector = 2r + (2πr × θ/360o)

= (2 × 4.2) + [2 × (22/7) × 4.2 × (60o/360o)]

= 8.4 + 4.4

= 12.8 cm

(ii) From the question it is given that, radius = 6 cm, angle at the center = 70o

We know that, Area of the sector = (πr2 × θ/360o)

= [(22/7) × 62 × (70o/360o)]

= 22 cm2

Then, perimeter of the sector = 2r + (2πr × θ/360o)

= (2 × 6) + [2 × (22/7) × 6 × (70o/360o)]

= 12 + 7.33

= 19.33 cm

(iii) From the question it is given that, diameter = 42 cm, angle at the center = 100o

Radius of circle = half of the diameter of the circle

r = d/2

⇒ r = 42/2

⇒ r = 21 cm

We know that, Area of the sector = (πr2 × θ/360o)

= [(22/7) × 212 × (100o/360o)]

= 385 cm2

Then, perimeter of the sector = 2r + (2πr × θ/360o)

= (2 × 21) + [2 × (22/7) × 21 × (100o/360o)]

= 42 + 36.66

= 78.66 cm


4. Find the area of a circular field that has a circumference of 396 m.

Answer

From the question it is given that, circumference of circular field is 396 m

We know that circumference of circle = 2πr

396 = 2πr

⇒ 396 = 2 × (22/7) × r

⇒ r = 396 × (7/22) × ½

⇒ r = 63 m

Then, area of circular field = πr2

= (22/7) × (63)2

= 12,474 cm2

Therefore, area of the circular field is 12,474 cm2.


5. Find the circumference of the circle whose area is 81π cm2.

Answer

From the question it is given that, area of circle 81π cm2.

We know that area of circle = πr2

81π = πr2

⇒ 81π/π = r2

⇒ r2= 81

⇒ r = √81

⇒ r = 9 cm

Then, circumference of circle = 2πr

= 2 × π × 9

= 18π cm


6. The diameter of a wheel is 1.4 m. How many revolutions does it make in moving a distance of 2.2 kms ?

Answer

From the question it is given that,

Diameter of the wheel = 1.4 m

r = 1.4/2

⇒ r = 0.7 m

We know that, circumference = 2πr

= 2 × 22/7 × 0.7

= 4.4 m

Distance travelled is given = 2.2 km

We know that, 1 km = 1000 m

So, 2.2 km = 2.2 × 1000

=2,200 m

Then, number of revolutions = 2,200/4.4

= 500

Therefore, wheel makes 500 revolutions in travelling 2.2 km.


7. The wheel of the car makes 10 revolutions per second. If its diameter is 70 cm, find the speed of the car in km per hour.

Answer

From the question it is given that,

Diameter of the wheel = 70 cm

r = 70/2

⇒ r = 35 cm

We know that, circumference = 2πr

= 2 × 22/7 × 35

= 220 cm

= 2.2 m

Number of revolutions = 10 per second

Then, number of revolutions per hour = 10 × 3600

= 36000 per hour

Distance covered in one hour = 36000 × 2.2 = 79200 m/hour

Therefore, speed of the car in km/hour = 79200/1000 = 79.2 km/h


8. The speed of the car is 66 km/h. If each wheel of the car is 140 cm in diameter, find the number of revolutions made by each wheel per minute.

Answer

From the question it is given that,

The speed of the car = 66 km/h = 66000 m/h

Diameter of the wheel = 140 cm

Radius of wheel, r = 140/2 = 70 cm

We know that, circumference of circle = 2πr

= 2 × 22/7 × 70

= 440 cm

= 4.4 m

Distance travelled by car in one minute = 66000/60

= 1100 m

Number of revolutions in one minute = distance /circumference

= 1100/4.4

= 250

Therefore, the number of revolutions made by each wheel = 250 rpm.



9. A cart wheel makes 9 revolutions per second. If the diameter of the wheel is 42 cm, find its speed in km/hr.
Answer

From the question it is given that,
Diameter of the wheel = 42 cm
Radius of the wheel, r = 42/2 = 21 cm
Number of revolutions made by a cart wheel = 9 revolutions per second
Number of revolutions made by a cart wheel per hour = 9 × 3600 = 32,400 per hour
We know that, circumference of circle = 2πr
= 2 × 22/7 × 21
= 132 cm
= 1.32 m

Distance covered in one hour = 32,400 × 1.32
= 42,768m/hour
Then, speed of the wheel in km/hour = 42,768/1000
= 42.768 km/h
Therefore, the speed of the wheel is approximately equal to 43km/h.


10. A bucket is raised from a well by means of a rope wound round a wheel of diameter 35 cm. If the bucket ascends in 2 minutes with a uniform speed of 1.1 m/sec, calculate the number of complete revolutions the wheel makes in raising the bucket.

Answer

From the question it is given that,

Diameter of wheel = 35 cm

Radius of wheel = 35/2 = 17.5 cm

We know that, circumference of circle = 2πr

= 2 × 22/7 × 17.5

= 110 cm

Then, time taken by the bucket to ascend = 2 minutes = 120 seconds … [given]

Speed of the rope while ascending = 1.1 m/s

So, the length of rope = 120 × 1.1 = 132 m

Therefore, number of revolutions made by rope = (132 × 100)/110 = 120


11. The circumference of a garden roller is 280 cm. How many revolutions make in moving 490 meters?

Answer

From the question it is given that,

The circumference of a garden roller = 280 cm

Total distance travelled = 490 m

Therefore, number of revolutions = 490/2.8 = 175


12. The diameter of a cycle wheel is 4.5/11 cm. How many revolutions will it make in moving 6.3 km?

Answer

From the question it is given that,

Diameter of a cycle wheel = 4.5/11 cm = 49/11 cm

We know that, circumference of circle = πd

= 22/7 × 49/11

= 14 cm

= 1.4 m

Then, 

distance travelled = 6.3 km = 6,300 m

Therefore, number of revolutions = 6,300/0.14 = 45,000

 

13. The area of a circle ring enclosed between two concentric circles is 88 cm2. Find the radii of the two circles, if their difference is 1 cm.

Answer

Let us assume r be the radius of the inner circle,

And (r + 1) be the radius of the outer circle

Then,

Area of circular ring = Area of outer circle – area of inner circle

π(r + 1)2 – πr2 = 88

⇒ π(r2 + 2r + 1) – πr2 = 88

⇒ πr2 + 2πr + π – πr2 = 88

On simplification we get,

π(2r + 1) = 88

⇒ 22/7(2r + 1) = 88

⇒ 2r + 1 = 88 × 7/22

⇒ 2r + 1 = 28

⇒ 2r = 28 – 1

⇒ 2r = 27

⇒ r = 27/2

⇒ r = 13.5 cm

⇒ (r + 1) = 13.5 + 1 = 14.5 cm

Therefore, radii of the two circles are 13.5 cm and 14.5 cm.


14. Find the area enclosed between two concentric circles, if their radii are 6 cm and 13 cm respectively.

Answer

From the question it is given that,

Radii of the two concentric circles are 6cm and 13 cm respectively.

r1 = 6 cm

r2 = 13 cm

We know that,

Area between two concentric circles = Areas of larger circle – area of smaller circle

Area of circle = πr2

= πr12 – πr22

= [(22/7) × 62] – [(22/7) – 132]

= 531.1429 – 113.1429

= 418 cm2

Therefore, area enclosed between two concentric circles is 418 cm2.


15. The area between the circumferences of two concentric circles is 2464 cm2. If the inner circle has circumference of 132 cm, calculate the radius of outer circle.

Answer

From the question it is given that,

The area between the circumferences of two concentric circles is 2464 cm2.

Circumference of inner circle = 132 cm

We know that, circumference = 2πr

132 = 2πr

⇒ r = 132/2π

⇒ r = (132 × 7)/(2 × 22)

⇒ r = 21 cm

So, radius of inner circle is 21 cm.

Then, area of inner circle = πr2

= 22/7 × 212

= 1,386 cm2

Area of outer circle = area of inner circle + area of concentric circles

= (1386 + 2464) cm2

= 3,850 cm2

Let R be the radius of outer circle,

Then, we know that area of outer circle = πR2

3,850 = πR2

⇒ R2 = (3,850 × 7)/22

⇒ R2 = 1225

⇒ R = √1225

⇒ R = 35 cm

Therefore, radius of outer circle is 35 cm.


16. A wire when bent in the form of a square encloses an area of 484cm2. If the same wire is bent into the form of a circle, find the area of the circle.

Answer


17. A wire bent in the form of an equilateral triangle has an area of 121√3 cm2. If the same wire is bent into the form of a circle, find the area enclosed by the wire.

Answer


18. Find the circumference of the circle whose area is 25 times the area of the circle with radius 7 cm.

Answer


19. The circumference of a circle is equal to the perimeter of a square. The area of the square is 484 sq. m. Find the area of the circle.

Answer


20. A wire is in form of a circle of radius 42 cm. It is bent into a square. Determine the side of the square and compare the area of the regions enclosed in two cases.

Answer


21. A 7 m road surrounds a circular garden whose area is 5544 m2. Find the area of the road and the cost of tarring it at the rate of Rs. 150 per sq. m.

Answer


22. A 4.2 m wide road surrounds a circular plot whose circumference is 176 m. Find the cost of paying the road at Rs 75 m2.

Answer


23. Two circles touch each other externally. The sum of their areas is 58 cm2 and the distance between their centres is 10 cm. Find the radii of the two circles.

Answer


24. The sum of diameter of two circles is 112 cm and the sum of their areas is 5236 cm. Find the radii of the two circles.

Answer


25. The sum of the radii of two circles is 10.5 cm and the difference of their circumferences is 13.2 cm. find the radii of the two circles.

Answer


26. A lawn is in the shape of a semi-circle of diameter 42 m. The lawn is surrounded by a flower bed of width 7 m all around. Find the area of the flower bed in m2.

Answer


27. Find the area enclosed between two concentric circles of radii 6.3 cm and 8.4 cm. A third concentric circle is drawn outside the 8.4 cm circle, so that the area enclosed between it and 8.4 cm circle is the same as that between two inner circles. Find the radii of the third circle correct to two decimal places.

Answer


28. Find the area of the biggest circle that can be cut from a rectangular piece of 44 cm by 28 cm. Also, find the area of the paper left after cutting cut the circle.

Answer


29. Find the area grazed by a horse tied with a 11.2 m rope to a corner of a field measuring 25 m by 15 m.

Answer


30. A horse is tied with a 21 m long rope to the corner of a field which is in the shape of an equilateral triangle. Find the area of the field over which it can graze.

Answer


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