Previous Year Questions: Areas Related to Circles

Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

 Table of contents Previous Year Questions 2024 Previous Year Questions 2023 Previous Year Questions 2022 Previous Year Questions 2020 Previous Year Questions 2019

Previous Year Questions 2024

Q1: Perimeter of a sector of a circle whose central angle is 90º and radius 7 cm is:     (2024)
(a) 35 cm
(b) 11 cm
(c) 22 cm
(d) 25 cm

Ans: (d)

Q2: A stable owner has four horses. He usually tie these horses with 7 m long rope to pegs at each corner of a square shaped grass field of 20 m length, to graze in his farm. But tying with rope sometimes results in injuries to his horses, so he decided to build fence around the area so that each horse can graze.        (2024)

Based on the above, answer the following questions:
(A) Find the area of the square shaped grass field.
(B) Find the area of the total field in which these horses can graze.
OR
If the length of the rope of each horse is increased from 7 m to 10 m, find the area grazed by one horse.
(Use π = 3.14)
(C) What is area of the field that is left ungrazed, if the length of the rope of each horse is 7 m?

Ans:
(A) Area of square shaped field
= 20 × 20
= 400 sq. m.
(B) Area of 4 quadrant = area of a. circle of radius 7m = πr2

OR
So, area grazed by one horse =

(C) Area of ungrazed portion = Area of square field – Area of circle with radius 7 m

Previous Year Questions 2023

Q3: What is the area of a semi-circle of diameter 'd' ?        (2023)
(a) 1/16πd2
(b) 1/4πd2
(c) 1/8πd2
(d) 1/2πd2

Ans: (c)
Given diameter of semi circle = d

Area of semi circle
=

Q4: Case Study : Governing council of a local public development authority of Dehradun decided to build an adventurous playground on the top of a bill, which will have adequate space for parking.

After survey, it was decided to build rectangular playground, with a semi-circular area allotted for parking at one end of the playground. The length and breadth of the rectangular playground are 14 units and 7 units, respectively. There are two quadrants of radius 2 units on one side for special seats. Based on the above information, answer the following questions:

(i) What is the total perimeter of the parking area ?
(ii) (a) What is the total area of parking and the two quadrants?

OR

(b) What is the ratio of area of playground to the area of parking area ?
(iii) Find the cost of fencing the playground and parking area at the rate of ? 2 per unit.         (2023)

Ans: (i) Length of play ground . AB = 14 units, Breadth of play ground. AD = 7 units
Radius of semi - circular part is 7/2 units
Total perimeter of parking area = πr + 2r

=
= 11 + 7 = 18 Units
(ii) (a): Area of parking = πr2 / 2
=
= 19.25 sq. units
Area of two quadrants (I) a n d [II) =1/2 x 1/4 x πr2
=
= 6.29 sq. units
Total area of parking and two quadrant
= 19.25 + 6.29
= 25.54 sq. units

Q5: A chord of a circle of radius 14 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor segment of the circle. Also find the area of the major segment of the circle.         (2023)

Ans: Here, radius t(r) = 14 cm and  Sector angle (θ) = 60°
∴ Area of the sector

= 102.67 cm2

Since ∠O = 60° and OA = OB = 14 cm
∴ AOB is an equilateral triangle.
⇒ AS = 14 cm and ∠A = 60°
Draw OM ⊥ AB.
In ΔAMO

Now,

Now, area of the minor segment= (Area of minor sector) - (ar ΔAOB)
= 102.67 - 84.87 cm2
= 17.8 cm2
Area of the major segment
= Area of the circle - Area of the minor segment

= (616 - 17.8) cm= 598.2 cm2

Previous Year Questions 2022

Q6: The area swept by 7 cm long minute band of a clock in 10 minutes is        (2022)
(a) 77 cm2
(b)

(c)
(d)

Ans: (d)
Angle formed by minute hand of a clock in 60 minutes = 360°
∴ Angle formed by minute hand of a clock in 10 minutes = 10/60 x 360° = 60°
Length of minute hand of a dock = radius = 7 cm
∴ Required area
=
=

Q7: Given below is the picture of the Olympic rings made by taking five congruent circles of radius 1 cm each, intersecting in such a way that the chord formed by joining the point of intersection of two circles is also of length 1 cm. Total area of all the dotted regions assuming the thickness of the rings to be negligible is        (2022)(a)
(b)
(c)
(d)

Ans: (d)

Let O be the centre of the circle. So. OA = OB = AB = 1 cm
So ΔOAB is an equilateral triangle.
∴ ∠AOB = 60°

∴  Required area = 8 x area of one segment with r = 1 cm,θ = 60°

= 8 x  (area of sector - area of ΔAOB)

Previous Year Questions 2020

Q8: A piece of wire 22 cm long is bent into the form of an arc of a circle subtending an angle of 60° at its centre. Find the radius of the circle. [Use π = 22/7]        (2020)

Ans: Let AB be the wire of length 22 cm in the form of an art of a circle so blending an ∠AOB - 60° at centre O.

∵ Length of arc =

= 21 cm
Hence, radius of the circle is 21cm.

Previous Year Questions 2019

Q9: A car has two wipers which do not overlap. Each wiper has a blade of length 21 cm sweeping through an angle 120°. Find the total area cleaned at each sweep of the blades. (Take π = 22/7)       (2019)

Ans: Here radius (r) = 21 cm
5ector angle (θ) = 120°
∴ Area cleaned by each sweep of the blades
=
= 22 x  7 x 3 x 2 cm2
= 924 cm2

Q10: Find the area of the segment shown in the given figure, if radius of the circle is 21 cm and ∠AOB = 120°. (Take π = 22/7)      (2019)

Ans: Given. O is the centre of the circle of radius 21cm and AB is the chord that subtends an angle of 120° at the centre.

Draw OM ⊥ AB,
Area of the minor segment AMBP = Area of sector OAPB - Area of ΔAOB
Now, area of sector OAPB
=
= 462 cm2
Since, OM ⊥ AB.
[∵ Perpendicular from the centre to the chord bisects the angle subtended by the chord at the centre.]
In ΔAOM, sin60° = AM/AO, cos60° = OM/OA

Area of ΔAOB =

Hence, Required Area =
= 462 - 381.92= 80.08 cm

Q11: In the given figure, three sectors of a circle of radius 7 cm, making angles of 60°, 80° and 40° at the centre are shaded. Find the area of the shaded region.      (2019)

Ans: Radius (r) of circle = 7 cm

=
=
= 77 cm2

The document Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles is a part of the Class 10 Course Mathematics (Maths) Class 10.
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FAQs on Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

 1. How to find the area of a circle when only the radius is given?
Ans. To find the area of a circle when only the radius is given, you can use the formula A = πr^2, where A is the area and r is the radius of the circle.
 2. What is the relationship between the area of a circle and its circumference?
Ans. The area of a circle is related to its circumference through the formula A = πr^2, where A is the area, r is the radius, and π is a constant value. The circumference of a circle can be calculated using the formula C = 2πr, where C is the circumference and r is the radius.
 3. How to find the area of a sector of a circle?
Ans. To find the area of a sector of a circle, you can use the formula A = (θ/360) x πr^2, where A is the area of the sector, θ is the central angle in degrees, and r is the radius of the circle.
 4. What is the difference between the area of a circle and the area of a sector?
Ans. The area of a circle refers to the total space enclosed by the circle, calculated using the formula A = πr^2. On the other hand, the area of a sector of a circle is the portion of the circle enclosed by a central angle, calculated using the formula A = (θ/360) x πr^2.
 5. Can the area of a circle be negative?
Ans. No, the area of a circle cannot be negative as it represents a physical measurement of the space enclosed by the circle, which is always a positive value.

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