Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
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
View AnswerAns: (d)
Perimeter of sector 
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?
View AnswerAns:
(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
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
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
View AnswerAns: (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 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?
(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)
View AnswerAns: (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)
View AnswerAns: 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) cm2 = 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)
View AnswerAns: (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) 
View AnswerAns: (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)
View AnswerAns: 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)
View AnswerAns: 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 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)
View AnswerAns: 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 cm2
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)
View AnswerAns: Radius (r) of circle = 7 cm
Area of shaded region =
= 
= 
= 77 cm2
Previous Year Questions 2017
Q10: A chord PQ of a circle of radius 10 cm subtends an angle of 60° at the centre of circle. Find the area of major and minor segments of the circle. (Delhi 2017)
View AnswerAns: Radius of the circle = 10 cm
Central angle subtended by chord AB = 60°
Area of minor sector OACB
Area of equilateral triangle OAB formed by radii and chord
∴ Area of minor segment ACBD
= Area of sector OACB - Area of triangle OAB
= (52.38 - 43.30) cm2 = 9.08 cm2
Area of circle = πr2 
∴ Area of major segment ADBE
= Area circle - Area of minor segment
= (314.28 - 9.08) cm2 = 305.20 cm2
Q11: In the given figure, ΔABC is a right-angled triangle in which ∠A is 90°. Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region. (Al 2017)
View AnswerAns: In right triangle ABC.
AB2 + AC2 = BC2
⇒ (3)2 + (4)2 = BC2 ⇒ 9 + 16 = BC2 ⇒ 25 = BC2
∴ BC = 5 cm
Now, Area of shaded region = Area of semicircle on side AB + area of semicircle on side AC - area of semicircle on side BC + area of ΔABC
Now, Area o f semicircle on side AB 
Hence area of the shaded region = 6 cm2
Q12: In Fig. 12.51, O is the centre of the circle with AC = 24 cm , AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region. (CBSE (AI) 2017)
View AnswerAns: In right angle triangle ABC
Area of shaded region = area of semicircle - area of ΔCAB + area of quadrant BOD
Previous Year Questions 2016
Q13: In Fig. 12.34, O is the centre of a circle such that diameter AB =13 cm and AC = 12 cm. ISC is joined. Find the area of the shaded region. (Take k = 3.14) (CBSE (AI) 2016)
View AnswerAns: In ΔABC, ∠ACB = 90° (Angle in the semicircle)
∴ BC2 + AC2 = AB2
∴ BC2 = AB2 -AC2
= 169 - 144 = 25
∴ BC = 5 cm
Area of the shaded region = area of semicircle - area of right AABC
Q14: In Fig. 12.35, are shown two arcs PAQ and PBQ. Arc PAQ is a part of circle with centre 0 and radius OP while arc PBQ is a semi-circle drawn on PQ as diameter with centre M.
If OP = PQ = 10 cm show that area of shaded region is 25 
(CBSE (Delhi) 2016)
View AnswerAns: Since OP = PQ = QO
⇒ APOQ is an equilateral triangle
∴ ∠POQ = 60°
Area of segment PAQM
Area of semicircle with M as centre 
Area of shaded region 
Q15: In figure, the boundary of shaded region consists of four semicircular arcs, two smallest being equal. If diameter of the largest is 14 cm and that of the smallest is 3.5 cm, calculate the area of the shaded region. 
(Foreign 2016)
View AnswerAns: Given AD = 14 cm, AB = CD = 3.5 cm
∴ BC = 7 cm
Area of shaded region
Q16: Find the area of the shaded region in Fig. 12.23, where a circular arc of radius 6 cm has been drawn with vertex 0 of an equilateral triangle ΔOAB of side 12 cm as centre. (NCERT, CBSE (F) 2016)
View AnswerAns: We have, radius of circular region = 6 cm and each side of ΔOAB = 12 cm.
∴ Area of the circular portion
= area of circle - area of the sector
Now, area of the equilateral triangle OAB
∴ Area of shaded region = area of circular portion + area of equilateral triangle OAB
Q17: An elastic belt is placed around the rim of a pulley of radius 5 cm. (Fig. 12.46). From one point C on the belt, the elastic belt is pulled directly away from the centre O of the pulley until it is at P, 10 cm from the point O. Find the length o f the belt that is still in contact with the pulley. Also find the shaded area. (Use π = 3.14 and √3 = 1.73) (CBSE Delhi 2016)
View AnswerAns: 
PA = 5√S cm = BP (Tangents from an external point are equal)
Shaded area = 43 .25 - 26 .17 = 17.08 cm2
Q18: In Fig. 12.47, a sector OAP of a circle with centre O, containing angle 0. AB is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of shaded region is 
(CBSE (AI) 2016)
View AnswerAns: 
In right ΔAOB
Perimeter of shaded region 
Q19: Find the area of the shaded region in Fig. 12.48, where 
are semicircles of diameter 14 cm, 3.5 cm, 7 cm and 3.5 cm respectively. 
(CBSE (E) 2016)
View AnswerAns: Area of shaded region = 
Previous Year Questions 2015
Q20: In figure, ABCD is a trapezium with AB || DC, AB = 18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If arcs of equal radii 7 cm with centres A, B, C and D have been drawn, then find the area of the shaded region. (Foreign 2015)
View AnswerAns: Area of shaded region = area of trapezium - (area of 4 sectors)
Total area of all sector
[Sum of angles of a quadrilateral is 360°]
From (i) and (ii),
Area of shaded region = 350 - 154 = 196 cm2
Q21: Find the area of the shaded region given in Fig. (Delhi 2015)
View AnswerAns: Side of the square = 14 cm
Area of the square = a2
= 142 - 196 cm2
Let radius of a semicircle=x
radius of two semicircles = 2x
side of inner square = diameter of semicircle = 2x
According to figure 2x + 2x = 8
4x = 8 ⇒ x = 2 cm
⇒ Side of inner square = 4 cm
Area of unshaded region = area of inner square + 4 (Area of a semicircle)
∴ Area of shaded region = area of square - area of unshaded region
= (196-16-871) cm2 = (180-8π) cm2
= 180-8 x 3.14 = 180-25.12 = 154.88 cm2
|
124 videos|457 docs|77 tests
|
FAQs on Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
| 1. What are the key formulas related to the area and circumference of a circle that Class 10 students should know? |  |
| 2. How can we calculate the area of a sector of a circle? | |
| 3. What is the relationship between the radius and diameter of a circle, and how does it affect area calculations? | |
| 4. How do you find the circumference of a circle if you only know the area? | |
| 5. What are some common mistakes students make when solving problems related to circles in Class 10? | |
|
3.1K Views |
|
4.73/5 Rating |
|
Oct 24, 2024 Last updated |
|
Explore Courses for Class 10 exam
|
|
ppt
,Objective type Questions
,Exam
,Sample Paper
,Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
,practice quizzes
,MCQs
,video lectures
,Viva Questions
,Extra Questions
,past year papers
,study material
,Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
,Summary
,Semester Notes
,shortcuts and tricks
,Important questions
,Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
,Free
,Previous Year Questions with Solutions
,mock tests for examination
;
Previous Year Questions: Areas Related to Circles Free PDF Download
Importance of Previous Year Questions: Areas Related to Circles
Previous Year Questions: Areas Related to Circles Notes
Previous Year Questions: Areas Related to Circles Class 10
Study Previous Year Questions: Areas Related to Circles on the App
|
© EduRev
|
Education Revolution
|
Follow Us
|
