Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  Previous Year Questions: Areas Related to Circles

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:     (CBSE 2024)
(a) 35 cm
(b) 11 cm
(c) 22 cm
(d) 25 cm

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

Ans: (d)
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

Perimeter of sector

= 2r +πr θ360°

= 2 × 7 + 227 × 7 × 90°180°

= 14 + 11 = 25 cm


Q2: A stable owner has four horses. He usually ties 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 a fence around the area so that each horse could graze.        (CBSE 2024)

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

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 the area of the field that is left ungrazed, if the length of the rope of each horse is 7 m?

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

Ans:
(A) Area of square shaped field
= 20 × 20
= 400 sq. m.

(B) Area of 4 quadrants

= Area of a circle of radius 7m = πr2

=4 × 14 × 227 × 7 × 7

= 154 m2

OR

New radius = 10 m

So, area grazed by one horse

= 14 (Area of circle with radius 10 m)

= 14 × π × (10)2

= 3.14 × 10 × 104

= 78.5 m2

(C) Area of ungrazed portion

= Area of square field - Area of circle with radius 7 m

= 20 × 20 - 227 × 7 × 7

= 400 - 154

= 246 m2

Previous Year Questions 2023

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

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

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

Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
∴ Radius, r = d/2
Area of semi circle
= Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to CirclesClass 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles


Q4: Case Study:  The 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.

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

After the survey, it was decided to build a 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) What is the total area of parking and the two quadrants? (CBSE 2023)

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

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

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

227 × 72 + 2 × 72

= 11 + 7 = 18 Units

(ii) Total area of parking and the two quadrants

= Area of semi-circular region + Area of 2 quadrants

= πR²2 + 2 × 14 × πr²

= π2 [ R² + r² ]

= 227 × 12 × (7/2)² + (2)²

= 117 × 49 + 44

= 117 × (49 + 16)4

= 117 × 654

= 71528

= 25.54 unit² (approx)

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)

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

Ans: Here, radius t(r) = 14 cm and  Sector angle (θ) = 60°
∴ Area of the sector
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

= 102.67 cm2
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Since ∠O = 60° and OA = OB = 14 cm
∴ AOB is an equilateral triangle.
⇒ AS = 14 cm and ∠A = 60°
Draw OM ⊥ AB.
In ΔAMO

OMOA = sin60° = √32 ⇒ OM = OA × √32 = 14√32 cm = 7√3 cm

Now,

Area of △AOB = 12 × AB × OM

= 12 × 14 × 7√3 cm² = 49√3 cm²

= 49 × 1.732 cm² = 84.87 cm²

Now, area of the minor segment

= (Area of minor sector) - (Area of △AOB)

= 102.67 - 84.87 cm2

= 17.8 cm2

Area of the major segment

= Area of the circle - Area of the minor segment

= πr²1 - 17.8

= 227 × 14 × 14 - 17.8 cm2

= (616 - 17.8) cm2 = 598.2 cm2

Previous Year Questions 2022


Q6: The area swept by a 7 cm long minute band of a clock in 10 minutes is        (CBSE 2022)
(a) 77 cm2
(b) Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

(c) Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
(d) Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

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

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


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. The total area of all the dotted regions assuming the thickness of the rings to be negligible is        (2022)Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
(a) 4 (π12) (√34) cm²
(b) (π6) - (√34) cm²
(c) 4 (π6) - (√34) cm²
(d) 8 (π6) - (√34) cm²

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

Ans: (d)
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
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)
= 8 x ( θ360º x πr- √34a2)
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

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)

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

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.
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
∵ Length of arc = Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
= 21 cm
Hence, radius of the circle is 21cm.

Q9: A circular park is surrounded by a road 21 m wide. If the radius of the park is 105 m, find the area of the road. (CBSE 2020)

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

Ans: 

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

Given: Width of road = 21 m 
Radius of park, r= 105 m  
⇒ Radius of the whole circular portion (park + road) 
r2 = 105 + 21 = 126 m 

So, Area of road = Area of park and road - Area of park

= πr22 - πr12

= π (r22 - r12)

= 227 × [(126)2 - (105)2

= 227 × (126 + 105) (126 - 105)   [∵ a2 - b2 = (a + b)(a - b)]

= 227 × 231 × 21 = 15246

Hence, the area of the road is 15246 m2

Previous Year Questions 2019


Q10: 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)

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

Ans: Here radius (r) = 21 cm
5ector angle (θ) = 120°
∴ Area cleaned by each sweep of the blades
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles [∵ there are 2 blades]
= Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
= 22 x 7 x 3 x 2 cm2
= 924 cm2

Q11: Find the area of the segment shown in the given figure, if the radius of the circle is 21 cm and ∠AOB = 120°. (Take π = 22/7)      (2019)Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

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

Ans: 
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Draw OM ⊥ AB.

Area of the minor segment AMBP = Area of sector OAPB - Area of △AOB

Now, area of sector OAPB

= 120°360° × πr2

= 120°360° × 227 × 21 × 21

= 462 cm2

Since, OM ⊥ AB.

∠AOM = ∠BOM = 120°/2 = 60°

[∵ Perpendicular from the centre to the chord bisects the angle subtended by the chord at the centre.]

In △AOM,

sin 60° = AMAO , cos 60° = OMOA

= √32 = AM/21, 12 = OM/21

⇒ AM = 21√32 cm, OM = 212 cm

∴ AB = 2AM = 2 × 21√32 = 21√3 cm

Area of △AOB

= 12 × AB × OM

= 12 × 21√3 × 212

= 441√34 cm2

Hence, required area

= 462 - 441√34

= 462 - 381.92 = 80.08 cm2

Q12: 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)Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

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

Ans: Radius (r) of circle = 7 cm
Area of shaded region =

π(7)2 × 40°360° + π(7)2 × 60°360° + π(7)2 × 80°360°

[∴ Area of sector = θ360° πr2]

= π(7)2 × 19 + π(7)2 × 16 + π(7)2 × 29

= π(7)2 [1/9 + 1/6 + 2/9]

= 227 × 7 × 7 × 918

= 77 cm2

Previous Year Questions 2017

Q13: A chord PQ of a circle of radius 10 cm subtends an angle of 60° at the centre of the circle. Find the area of major and minor segments of the circle.    (Delhi 2017)

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

Ans:  Radius of the circle = 10 cm
Central angle subtended by chord AB = 60°
Area of minor sector OACB
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

Area of equilateral triangle OAB formed by radii and chord

= √34 a² = √34 × (10)² = 1.7324 × 100 = 43.3 cm2

Area of minor segment ACBD

= Area of sector OACB - Area of triangle OAB

= (52.38 - 43.30) cm2

= 9.08 cm2

Area of circle

∴ Area of circle = πr²

= 227 × (10)² = 22 × 1007

= 314.28 cm2

Area of major segment ADBE

= Area of circle - Area of minor segment

= (314.28 - 9.08) cm2

= 305.20 cm2


Q14: In the given figure, ΔABC is a right-angled triangle in which ∠A is 90°. Semi-circles are drawn on AB, AC and BC as diameters. Find the area of the shaded region.     (Al 2017)Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

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

Ans: 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 = { ar(ΔABC) + ar(semicircle on side AB) + ar(semicircle on side AC) } - ar(semicircle on side BC)

Area of shaded region = [(12 × 3 × 4) + (12 π × (322) + (12 π × (2)2)] - (12 π × (522 cm2)

= 6 + 12 π 94 + 4 - 254 cm2

= 6 cm2

Hence, area of shaded region is 6 cm2


Q15: In the following figure, 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)
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

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

Ans: In right angle triangle ABC

Diameter BC = √(242 + 72) = 25 cm

Area ΔCAB = 12 × base × height

= 12 × 24 × 7 = 84 cm2

Area of shaded region = area of semicircle - area of ΔCAB + area of quadrant BOD

= π2 × (252) 2 - 84 + π4 × (252) 2

= (625π8) + (625π16) - 84

= (1875π16) - 84 = (117.18π - 84) = 283.94 cm2


Q16: Two circles touch internally. The sum of their areas is 116π cm2 and the distance between their centres is 6 cm. Find the radii of the circles. (CBSE 2017)

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

Ans: Let ‘r’ and ‘R’ be the radii of the smaller and bigger circles, respectively. 
Then, OO’ = R – r = 6 cm [Given] ...(i) 
Also, sum of their areas = 116π cm2
i.e., πR2 + πr2 = 116π

Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles⇒ R2 + r2 = 116 ...(ii)
We know, (R – r)2 = R2 + r2 – 2Rr 
⇒ (6)2 = 116 – 2Rr 
⇒ 2Rr = 116 – 36 = 80 
⇒ Rr = 40 ...(iii) 
Also, (R + r)2 = R+ r2 + 2Rr = 116 + 2 × 40 [Using (ii) and (iii)] 
⇒ (R + r)2 = 196 
⇒ R + r = 14 ...(iv) 
Now, adding equations (i) and (iv), we get 
2R = 20 
⇒ R = 10 cm 
Putting the value of R in equation (i), we get 
r = 4 cm 
Hence, the radii of the bigger and smaller circles are 10 cm and 4 cm, respectively.

Previous Year Questions 2016

Q17: In Fig. 12.34, 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. (Take k = 3.14)    (CBSE (AI) 2016)
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

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

Ans:  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 ΔABC

= 12 πr2 - 12 × BC × AC

= 12 × (3.14) (132) 2 - 12 × 5 × 12

= 66.33 - 30 = 36.33 cm2


Q18: In the given figure, are shown two arcs PAQ and PBQ. Arc PAQ is a part of the 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 the area of the shaded region is 25 Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles (CBSE (Delhi) 2016)

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

Ans: Since OP = PQ = QO
⇒ APOQ is an equilateral triangle
∴ ∠POQ = 60°
Area of segment PAQM

= θ360° πr2 - √34 a2

= 60°360° π × 102 - √34 × 102

= 100π6 - 100√34 cm2

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

Area of Semicircle with M as Centre

= π2 (5)2 = 25π2 cm2

Area of Shaded Region

= 25π2 - (50π3) - 25√3

= 25π2  - 503 π + 25√3

= -25π6  + 25√3

= 25(√3 - π6 ) cm2


Q19: In the figure, the boundary of the shaded region consists of four semicircular arcs, two smallest being equal. If the diameter of the largest is 14 cm and that of the smallest is 3.5 cm, calculate the area of the shaded region. Use π = 22/7   (Foreign 2016)Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

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

Ans: Given AD = 14 cm, AB = CD = 3.5 cm
∴ BC = 7 cmClass 10 Maths Chapter 11 Previous Year Questions - Areas Related to CirclesClass 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Substituting the Values 
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

Q20: 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 the centre.    (NCERT, CBSE (F) 2016)
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

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

Ans:

Area of Shaded region = Area of equilateral triangle ABO + Area of Major sector

Area of Equilateral Triangle ABO = √34 × a2 = √34 × 12 × 12 cm2

= 62.352 cm2

Area of major sector = θ360 × π × r2

= 300360 × π × 6 × 6 cm2

= 94.2 cm2

Area of Shaded region = 62.352 + 94.2 = 156.55 cm2

Q21: 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 of 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)
Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

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

Ans: In △AOP, cos θ = 510

⇒ cos θ = 12 ⇒ θ = 60°

∴ Reflex ∠AOB = 240°

∴ Length of belt in contact with pulley = θ360° × 2πr

= 2 × 3.14 × 5 × 240360 = 20.93 cm

Now, APOA = tan 60°

PA = 5√3 cm = BP (Tangents from an external point are equal)

Area (△OAP + △OBP) = 2 × 12 × 5 × 5√3 = 25√3 = 43.25 cm2

Area of sector OACB = θ360° πr2 = 240360 × 3.14 × 25 = 26.17 cm2

Shaded area = 43.25 - 26.17 = 17.08 cm2

Q22: In the figure given, a sector OAP of a circle with centre O, containing angle θ. AB is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of the shaded region is Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles (CBSE (AI) 2016)Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

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

Ans: 

Perimeter of shaded region = AB + PB + arc length AP ...(1)

Arc length AP = θ360 × 2πr = πθr180 ...(2)

In right-angled △OAB,

tan θ = ABr ⇒ AB = r tan θ ...(3)

sec θ = OBr ⇒ OB = r sec θ

OB = OP + PB

∴ r sec θ = r + PB

∴ PB = r sec θ - r ...(4)

Substitute (2), (3), and (4) in (1), we get:

Perimeter of shaded region = AB + PB + arc length AP

= r tan θ + r sec θ - r + πθr180

= r [tan θ + sec θ + πθ180 - 1]

Q23: In the figure, AB is a chord of a circle with a centre O and a radius of 10 cm, that subtends a right angle at the centre of the circle. Find the area of the minor segment AQBP. Also, find the area of the major segment ALBQA. (Use π = 3.14) (CBSE 2016)

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

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

Ans: Given, A circle of radius (r) = 10 cm in which ∠AOB = 90º. 
Area of the minor segment AQBP = Area of sector OAPB – Area of ΔAOB 
= θ / 360° × πr2 – 1 /2 × OA × OB 
= 90 / 360 °  × 3.14 × 10 × 10 – 1/ 2 × 10 × 10 
= 3.14 × 5 × 5 – 5 × 10 
= 78.5 – 50 
= 28.5 cm2 

So, Area of the minor segment AQBP = 28.5 cm2
Area of the major segment ALBQA = Area of circle – Area of minor segment AQBP 
= 3.14 × (10)2 – 28.5 
= 314 – 28.5 
= 285.5 cm2
Area of major segment ALBQA = 285.5 cm2. Hence, the area of the minor segment AQBP is 28.5 cm2 and the area of the major segment ALBQA is 285.5 cm2.

Previous Year Questions 2015

Q24:  In the 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)

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

Ans: Area of shaded region = area of trapezium - (area of 4 sectors)Class 10 Maths Chapter 11 Previous Year Questions - Areas Related to Circles

Area of trapezium = 12 (sum of parallel sides) × distance between parallel sides

= 12 (18 + 32) × 14 = 12 × 50 × 14 = 350 cm2 ...(i)

Area of sector on A = ∠A360° × π(7)2

Area of sector on B = ∠B360° × π(7)2

Area of sector on C = ∠C360° × π(7)2

Area of sector on D = ∠D360° × π(7)2

Total area of all sectors

= ∠A360° × π(7)2 + ∠B360° × π(7)2 + ∠C360° × π(7)2 + ∠D360° × π(7)2

= π360 (7)2(∠A + ∠B + ∠C + ∠D)

= 227 × 360 × 49 × 360° = 154 cm2 ...(ii)   [Sum of angles of a quadrilateral is 360°]

From (i) and (ii),
Area of shaded region = 350 - 154 = 196 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.
All you need of Class 10 at this link: Class 10
126 videos|457 docs|75 tests

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

1. What is the formula for the area of a circle?
Ans. The formula for the area of a circle is \( A = \pi r^2 \), where \( A \) is the area and \( r \) is the radius of the circle.
2. How do you calculate the circumference of a circle?
Ans. The circumference of a circle can be calculated using the formula \( C = 2\pi r \) or \( C = \pi d \), where \( C \) is the circumference, \( r \) is the radius, and \( d \) is the diameter of the circle.
3. What is the relationship between the radius and diameter of a circle?
Ans. The diameter of a circle is twice the length of the radius. This relationship can be expressed as \( d = 2r \), where \( d \) is the diameter and \( r \) is the radius.
4. How do you find the area of a sector of a circle?
Ans. The area of a sector of a circle can be found using the formula \( A = \frac{\theta}{360} \times \pi r^2 \), where \( A \) is the area of the sector, \( \theta \) is the angle in degrees, and \( r \) is the radius of the circle.
5. What are the properties of a circle that are important for solving problems in geometry?
Ans. Important properties of a circle include: all points on the circle are equidistant from the center, the radius is constant, the diameter is the longest chord, and the relationship between the area and circumference is defined by \( A = \pi r^2 \) and \( C = 2\pi r \).
Related Searches

Important questions

,

video lectures

,

Summary

,

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

,

Exam

,

practice quizzes

,

study material

,

Sample Paper

,

Extra Questions

,

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

,

Viva Questions

,

mock tests for examination

,

shortcuts and tricks

,

Semester Notes

,

pdf

,

Free

,

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

,

ppt

,

Previous Year Questions with Solutions

,

past year papers

,

Objective type Questions

,

MCQs

;