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 =πr². ‘π’ is a Greek letter read as ‘pi’ and in mathematics known as a constant ratio.The legendary Indian Mathematician Shri Aryabhatta gave the nearest value of π which is 3.1416 in C.E (476-550). As π 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 2πr(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 60°. Given, the circle has a radius of 6 cm.

Solution:

Given,

The angle of the sector = 60°

Using the formula,

The area of sector = (θ/360°)×π r^2

= (60°/360°) × π r^2 cm^2

Or, area of the sector = 6 × 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 car’s wheel = 80/2 = 40 cm (as D = 80 cm)

So, the circumference of wheels = 2πr = 80 π cm

Now, in one revolution, the distance covered = circumference of the wheel = 80 π 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 × 10^5) cm

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

∴ Distance covered by car = 11 × 10^5 cm

Now, the no. of revolutions of the wheels = (Distance covered by the car/Circumference of the wheels) = 11 × 105 /80 π = 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 × 7 cm. Find the area of the remaining card board. [Use π = 22/7]

Solution:

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

Area of the remaining card board

= ar(rectangle) – 2(area of circle)

= L x B – 2πr^2)

= 14 × 7 – 2 × 22/7× 7/2× 7/2

= 98 – 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 π = 22/7]

Solution: 

Perimeter of the shaded region

= Circumference of circle + AD + BC

= 2πr + 14 + 14

= 2 × 22/7 × 7 + 28 ….[. 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 π = 22/7)

Solution:

Length of paper,

AB, l = 40 cm

Width of paper,

AD, b = 28 cm

Area of paper = l × h

= 40 × 28 = 1120 sq. cm

Diameter of semi-circle = 28 cm …(i)

Radius of semi-circle, r = 14 cm

Area of semi-cirice = 

1/2πr^2=1/2×22/7×14×14

= 308 sq.cm …(ii)

∴ Area of remaining paper = 1120 – 308

= 812 sq.cm

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

Solution:

Here r = 28/4 = 7 cm

Area of the shaded region

= ar(square) – 4(circle)

= (side)^2– 4 (πr^2)

= (28)^2 – 4 × 22/7 × 7 × 7 = 784 – 616 = 168 cm^2

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

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

Solution: 

1. 42 cm

Let the diameter of the circle be D.

⇒D=2R.

Circumference of a circle is given by, 2πR.

∴2πR=D+180

⇒2πR=2R+180

⇒R=42 cm

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

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

Solution:

Circumference of a circle is given by, 2πR.

∴2πR=154

⇒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, BC² = AB²-AC²

= 169-144 = 25 BC = 5

Area of shaded region = Area of semicircle – Area of right triangle ABC

=1/2 x πr²-1/2AC x BC

=1/2 x 3.14(13/2)²-1/2 x 12 x 5

=66.33 – 30 = 36.33 cm²

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 + πr) cm

⇒ 2r + πr = 36

⇒ r(2+22/7) = 36

⇒ r = 7cm

Diameter 2r = 2 × 7 = 14 cm.