Important Questions Class 10 Maths Chapter 12 Areas Related to Circles

# Important Questions Class 10 Maths Chapter 12 Areas Related to Circles

It is believed in 5BC the Greek mathematicians discovered that the area enclosed by a circle is proportional to the square of its radius and thus we obtained the formula, Area of a circle =Ï€rÂ². â€˜Ï€â€™ is a Greek letter read as â€˜piâ€™ and in mathematics known as a constant ratio.The legendary Indian Mathematician Shri Aryabhatta gave the nearest value of Ï€ which is 3.1416 in C.E (476-550). As Ï€ is an irrational and non-terminating number so for calculation purposes we take its value as 22/7 or 3.14. Similarly, the circumference of a circle can be known from the formula 2Ï€r(r is the radius of the circle)

The important question for class 10 maths Chapter 12 Areas Related to Circle can be obtained from the following topics:

• Circumference of a circle
• Area of a circle
• Length of an arc of a sector of a circle
• Area of a sector of a circle
• Arc of a segment of a circle

## Important Questions Chapter 12 – Areas Related to Circles

Q1. Calculate the area of a sector of angle 60Â°. Given, the circle has a radius of 6 cm.

Solution:

Given,

The angle of the sector = 60Â°

Using the formula,

The area of sector = (Î¸/360Â°)Ã—Ï€ r^2

= (60Â°/360Â°) Ã— Ï€ r^2 cm^2

Or, area of the sector = 6 Ã— 22/7 cm^2 = 132/7 cm^2

Q2. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

Solution:

The radius of carâ€™s wheel = 80/2 = 40 cm (as D = 80 cm)

So, the circumference of wheels = 2Ï€r = 80 Ï€ cm

Now, in one revolution, the distance covered = circumference of the wheel = 80 Ï€ cm

It is given that the distance covered by the car in 1 hr = 66km

Converting km into cm we get,

Distance covered by the car in 1hr = (66 Ã— 10^5) cm

In 10 minutes, the distance covered will be = (66 Ã— 10^5 Ã— 10)/60 = 1100000 cm/s

âˆ´ Distance covered by car = 11 Ã— 10^5 cm

Now, the no. of revolutions of the wheels = (Distance covered by the car/Circumference of the wheels) = 11 Ã— 105 /80 Ï€ = 4375.

Q3. Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular card board of dimensions 14 cm Ã— 7 cm. Find the area of the remaining card board. [Use Ï€ = 22/7]

Solution:

Here r = 7/2cm, L = 14 cm, B = 7 cm

Area of the remaining card board

= ar(rectangle) â€“ 2(area of circle)

= L x B â€“ 2Ï€r^2)

= 14 Ã— 7 â€“ 2 Ã— 22/7Ã— 7/2Ã— 7/2

= 98 â€“ 77 = 21 cm^2

Q4. Find the perimeter of the D shaded region in Figure, if ABCD is a square of side 14 cm and APB and CPD are semicircles. [Use Ï€ = 22/7]

Solution:Â

= Circumference of circle + AD + BC

= 2Ï€r + 14 + 14

= 2 Ã— 22/7 Ã— 7 + 28 â€¦.[. r = 14/2 = 7 cm

= 44 + 28 = 72 cm

Q5. From a rectangular sheet of paper ABCD with AB = 40 cm and AD = 28 cm, a semi-circular portion with BC as diameter is cut off. Find the area of the remaining paper. (Use Ï€ = 22/7)

Solution:

Length of paper,

AB, l = 40 cm

Width of paper,

Area of paper = l Ã— h

= 40 Ã— 28 = 1120 sq. cm

Diameter of semi-circle = 28 cm â€¦(i)

Radius of semi-circle, r = 14 cm

Area of semi-cirice =Â

1/2Ï€r^2=1/2Ã—22/7Ã—14Ã—14

= 308 sq.cm â€¦(ii)

âˆ´ Area of remaining paper = 1120 â€“ 308

= 812 sq.cm

Q6. Find the area of the shaded region in Figure, where ABCD is a square of side 28 cm.Â

Solution:

Here r = 28/4 = 7 cm

= ar(square) â€“ 4(circle)

= (side)^2â€“ 4 (Ï€r^2)

= (28)^2 â€“ 4 Ã— 22/7 Ã— 7 Ã— 7 = 784 â€“ 616 = 168 cm^2

Q7. The circumference of a circle exceeds its diameter by 180 cm. Then its radius is

1. Â  32 cm
2. Â  36 cm
3. Â  40 cm
4. Â  42 cm

Solution:Â

1. 42 cm

Let the diameter of the circle be D.

â‡’D=2R.

Circumference of a circle is given by, 2Ï€R.

âˆ´2Ï€R=D+180

â‡’2Ï€R=2R+180

â‡’R=42 cm

Q8. The circumference of a circular field is 154 m. Then its radius is

1. Â  7 m
2. Â  14 m
3. Â  7.5 m
4. Â  28 cm

Solution:

Circumference of a circle is given by, 2Ï€R.

âˆ´2Ï€R=154

â‡’R=24.5 m.

Q9. In figure, O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region.

Solution:

Here, BCÂ² = ABÂ²-ACÂ²

= 169-144 = 25 BC = 5

Area of shaded region = Area of semicircle â€“ Area of right triangle ABC

=1/2 x Ï€rÂ²-1/2AC x BC

=1/2 x 3.14(13/2)Â²-1/2 x 12 x 5

=66.33 â€“ 30 = 36.33 cmÂ²

Q10. If the perimeter of a semi-circular protractor is 36 cm, find its diameter.

Solution:

Perimeter of a semicircular protractor = Perimeter of a semicircle

= (2r + Ï€r) cm

â‡’ 2r + Ï€r = 36

â‡’ r(2+22/7) = 36

â‡’ r = 7cm

Diameter 2r = 2 Ã— 7 = 14 cm.