Table of contents  
Previous Year Questions 2024  
Previous Year Questions 2023  
Previous Year Questions 2022  
Previous Year Questions 2020  
Previous Year Questions 2019 
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 = πr^{2}
OR
New radius = 10 m
So, area grazed by one horse =
(C) Area of ungrazed portion = Area of square field – Area of circle with radius 7 m
Ans: (c)
Given diameter of semi circle = d
∴ Radius, r = d/2
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 semicircular 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?
(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 = πr^{2} / 2
=
= 19.25 sq. units
Area of two quadrants (I) a n d [II) =1/2 x 1/4 x πr^{2}
=
= 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 cm^{2}
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 cm^{2}
= 17.8 cm^{2}
Area of the major segment
= Area of the circle  Area of the minor segment
= (616  17.8) cm^{2 }= 598.2 cm^{2}
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)
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.
Ans: Here radius (r) = 21 cm
5ector angle (θ) = 120°
∴ Area cleaned by each sweep of the blades
[∵ there are 2 blades]
=
= 22 x 7 x 3 x 2 cm^{2}
= 924 cm^{2}
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 cm^{2}
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^{2 }
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
Area of shaded region =
=
=
= 77 cm^{2}
116 videos420 docs77 tests

1. How to find the area of a circle when only the radius is given? 
2. What is the relationship between the area of a circle and its circumference? 
3. How to find the area of a sector of a circle? 
4. What is the difference between the area of a circle and the area of a sector? 
5. Can the area of a circle be negative? 

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