Class 10 Mathematics - NCERT Mathematics
NCERT Solutions for Class 10 Maths Chapter 11: Areas Related to Circles
Chapter 11 covers the computation of areas and perimeters of sectors, segments, and combinations of plane figures involving circles. Students learn to find the area of a sector using the formula (theta/360) x pi x r^2, the length of an arc, and the area of a segment by subtracting the triangular area from the sector area. This chapter carries 4-6 marks in the CBSE Class 10 board exam 2027 and is part of the Mensuration unit worth 10 marks.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 11 Areas Related to Circles has 2 exercises - Exercise 11.1 (area of sector, area of segment, arc length) and Exercise 11.2 (areas of combinations of plane figures). The older Exercise 11.3 has been removed from the CBSE syllabus and will NOT be asked in the 2027 board exam. However, solutions for Exercise 11.3 are provided below for state board students (UP Board, Bihar Board, MP Board, Rajasthan Board) who may still follow the 3-exercise format.
Table of Contents
Exercise 11.1 - Areas of Sector and Segment
6 questions
Exercise 11.2 - Areas of Combinations of Plane Figures
6 questions
Old Exercise 11.3 - Areas of Combinations (For State Boards)
3 questions • This exercise has been removed from the CBSE 2026-
Previous Year Questions
14 questions (CBSE 2019-2026)
MCQ Practice
12 questions
11.1Exercise 11.1 - Areas of Sector and Segment
Q1Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60 degrees.
Given: r = 6 cm, theta = 60 degrees.
Area of sector = (theta/360) x pi x r^2
= (60/360) x (22/7) x 6 x 6
= (1/6) x (22/7) x 36
= (22 x 36) / (7 x 6)
= 792 / 42
= 132/7 cm^2
= 18.86 cm^2 (approx.)
Q2Find the area of a quadrant of a circle whose circumference is 22 cm.
Given: Circumference = 22 cm.
2 x pi x r = 22
2 x (22/7) x r = 22
r = 22 x 7 / (2 x 22) = 7/2 = 3.5 cm
A quadrant is a sector with theta = 90 degrees.
Area of quadrant = (90/360) x pi x r^2
= (1/4) x (22/7) x (3.5)^2
= (1/4) x (22/7) x 12.25
= (1/4) x (22 x 12.25) / 7
= (1/4) x 269.5 / 7
= (1/4) x 38.5
= 77/8 cm^2
= 9.625 cm^2
Q3The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Given: r = 14 cm (length of minute hand).
The minute hand completes 360 degrees in 60 minutes.
In 5 minutes, angle swept = (5/60) x 360 = 30 degrees.
Area swept = (theta/360) x pi x r^2
= (30/360) x (22/7) x 14 x 14
= (1/12) x (22/7) x 196
= (1/12) x 22 x 28
= (1/12) x 616
= 616/12
= 154/3 cm^2
= 51.33 cm^2 (approx.)
Q5To warn ships of underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80 degrees to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use pi = 3.14)
Given: theta = 80 degrees, r = 16.5 km.
Area of sector = (theta/360) x pi x r^2
= (80/360) x 3.14 x (16.5)^2
= (2/9) x 3.14 x 272.25
= (2/9) x 854.865
= 1709.73 / 9
= 189.97 km^2 (approx.)
Q10A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find: (i) the area of that part of the field in which the horse can graze, (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use pi = 3.14)
The horse is tied at a corner of a square, so the angle available at the corner = 90 degrees.
(i) With rope = 5 m:
Area grazed = area of quadrant with r = 5 m
= (90/360) x pi x r^2
= (1/4) x 3.14 x 25
= (1/4) x 78.5
= 19.625 m^2
(ii) With rope = 10 m:
Area grazed = (1/4) x 3.14 x 100
= (1/4) x 314
= 78.5 m^2
Increase in grazing area = 78.5 - 19.625
= 58.875 m^2
Q12Find the area of the shaded region in the figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
Since O is the centre of the circle and QR is a diameter (angle in a semicircle is 90 degrees), triangle QPR is right-angled at P.
By Pythagoras theorem:
QR^2 = PQ^2 + PR^2
= 24^2 + 7^2
= 576 + 49
= 625
QR = 25 cm
Radius = QR/2 = 25/2 = 12.5 cm
Area of semicircle = (1/2) x pi x r^2
= (1/2) x (22/7) x (12.5)^2
= (1/2) x (22/7) x 156.25
= (22 x 156.25) / 14
= 3437.5 / 14
= 245.535 cm^2 (approx.)
Area of triangle QPR = (1/2) x PQ x PR
= (1/2) x 24 x 7
= 84 cm^2
Area of shaded region = Area of semicircle - Area of triangle
= 245.535 - 84
= 161.535 cm^2
= 161.54 cm^2 (approx.)
11.2Exercise 11.2 - Areas of Combinations of Plane Figures
Q1Find the area of the shaded region in the figure, where ABCD is a square of side 14 cm and four equal semicircles are drawn with each side of the square as diameter.
Side of square = 14 cm.
Diameter of each semicircle = 14 cm, so radius = 7 cm.
Area of square = 14 x 14 = 196 cm^2
Area of 4 semicircles = 4 x (1/2) x pi x r^2
= 4 x (1/2) x (22/7) x 49
= 2 x (22/7) x 49
= 2 x 22 x 7
= 308 cm^2
The shaded region consists of the 4 semicircles minus the square (each pair of opposite semicircles covers the square). The overlapping region is the square itself.
Shaded area = Area of 4 semicircles - Area of square
= 308 - 196
= 112 cm^2
Q3Find the area of the shaded region in the figure, where ABCD is a square of side 14 cm and each of the four equal circles touches two sides of the square and two adjacent circles.
Side of square = 14 cm.
Since 4 equal circles fit inside the square (2 along each side), diameter of each circle = 14/2 = 7 cm, so radius = 7/2 = 3.5 cm.
Area of square = 14 x 14 = 196 cm^2
Area of 4 circles = 4 x pi x r^2
= 4 x (22/7) x (3.5)^2
= 4 x (22/7) x 12.25
= 4 x 38.5
= 154 cm^2
Shaded area = Area of square - Area of 4 circles
= 196 - 154
= 42 cm^2
Q4Find the area of the shaded region in the figure, where a circular arc of radius 7 cm has been drawn with vertex A of an equilateral triangle ABC of side 14 cm as centre.
Side of equilateral triangle = 14 cm.
Radius of circular arc = 7 cm.
Each angle of equilateral triangle = 60 degrees.
The circular arc covers the region outside the triangle. The arc sector angle = 360 - 60 = 300 degrees (the sector outside the triangle at vertex A).
Area of equilateral triangle = (sqrt(3)/4) x side^2
= (sqrt(3)/4) x 196
= 49 x sqrt(3)
= 49 x 1.732
= 84.87 cm^2 (approx.)
Area of sector (outside triangle) = (300/360) x pi x r^2
= (5/6) x (22/7) x 49
= (5/6) x 154
= 770/6
= 128.33 cm^2 (approx.)
Shaded area = Area of equilateral triangle + Area of sector outside triangle
= 84.87 + 128.33
= 213.20 cm^2 (approx.)
Q5From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the figure. Find the area of the remaining portion of the square.
Side of square = 4 cm.
Radius of each quadrant = 1 cm.
Diameter of central circle = 2 cm, so radius = 1 cm.
Area of square = 4 x 4 = 16 cm^2
Area of 4 quadrants = 4 x (1/4) x pi x r^2
= 4 x (1/4) x (22/7) x 1
= (22/7) x 1
= 22/7 cm^2
Area of central circle = pi x r^2
= (22/7) x 1
= 22/7 cm^2
Area of remaining portion = Area of square - Area of 4 quadrants - Area of circle
= 16 - 22/7 - 22/7
= 16 - 44/7
= (112 - 44)/7
= 68/7
= 68/7 cm^2
= 9.71 cm^2 (approx.)
Q6In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. Find the area of the design.
Radius of circle = 32 cm.
The equilateral triangle is inscribed in the circle.
For an equilateral triangle inscribed in a circle of radius R:
Side = R x sqrt(3) = 32 x sqrt(3) = 32 x 1.732 = 55.42 cm (approx.)
Area of circle = pi x r^2
= (22/7) x 32 x 32
= (22/7) x 1024
= 22528/7
= 3218.29 cm^2 (approx.)
Area of equilateral triangle = (sqrt(3)/4) x side^2
= (sqrt(3)/4) x (32 x sqrt(3))^2
= (sqrt(3)/4) x 3072
= (3072 x sqrt(3))/4
= 768 x sqrt(3)
= 768 x 1.732
= 1330.18 cm^2 (approx.)
Area of design = Area of circle - Area of triangle
= 3218.29 - 1330.18
= 1888.11 cm^2 (approx.)
Q10The area of an equilateral triangle ABC is 17320.5 cm^2. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded region. (Use pi = 3.14 and sqrt(3) = 1.73205)
Area of equilateral triangle = (sqrt(3)/4) x a^2 = 17320.5 cm^2
(1.73205/4) x a^2 = 17320.5
0.4330125 x a^2 = 17320.5
a^2 = 17320.5 / 0.4330125
a^2 = 40000
a = 200 cm
Radius of each circle = a/2 = 100 cm.
Angle at each vertex of equilateral triangle = 60 degrees.
At each vertex, the circle creates a sector of angle 60 degrees inside the triangle.
Area of 3 sectors = 3 x (60/360) x pi x r^2
= 3 x (1/6) x 3.14 x 10000
= (1/2) x 31400
= 15700 cm^2
Shaded area = Area of triangle - Area of 3 sectors
= 17320.5 - 15700
= 1620.5 cm^2
old-11.3Old Exercise 11.3 - Areas of Combinations (For State Boards)
This exercise has been removed from the CBSE 2026-27 syllabus. It is provided here for UP Board, Bihar Board, MP Board, and other state board students who still follow the older textbook.
Q1Find the area of the shaded region in the figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
Side of square = 14 cm.
Diameter of each semicircle = 14 cm, so radius = 7 cm.
Area of square = 14 x 14 = 196 cm^2
Area of 2 semicircles = 2 x (1/2) x pi x r^2
= pi x r^2
= (22/7) x 49
= 154 cm^2
Shaded area = Area of square - Area of 2 semicircles
= 196 - 154
= 42 cm^2
Q4Find the area of the shaded region in the figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.
Side of equilateral triangle = 12 cm.
Radius of arc = 6 cm.
Angle of equilateral triangle at O = 60 degrees.
The arc forms a sector covering the region outside the triangle.
Sector angle = 360 - 60 = 300 degrees.
Area of equilateral triangle = (sqrt(3)/4) x 12^2
= (sqrt(3)/4) x 144
= 36 x sqrt(3)
= 36 x 1.732
= 62.35 cm^2 (approx.)
Area of sector = (300/360) x pi x r^2
= (5/6) x (22/7) x 36
= (5 x 22 x 36) / (6 x 7)
= 3960 / 42
= 94.29 cm^2 (approx.)
Shaded area = Area of triangle + Area of sector
= 62.35 + 94.29
= 156.64 cm^2 (approx.)
Q9In the given figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.
OA = 7 cm, so radius of larger circle = 7 cm.
OD = 7 cm (radius of larger circle).
Radius of smaller circle = OD/2 = 7/2 = 3.5 cm.
Area of smaller circle = pi x (3.5)^2
= (22/7) x 12.25
= 38.5 cm^2
Area of larger circle = pi x 7^2
= (22/7) x 49
= 154 cm^2
Area of larger semicircle (above AB) = 154/2 = 77 cm^2
Area of triangle ABC (right triangle with base AB = 14 cm and height OC = 7 cm):
= (1/2) x 14 x 7 = 49 cm^2
Shaded region = Area of smaller circle + Area of larger semicircle - Area of triangle ABC
= 38.5 + 77 - 49
= 66.5 cm^2
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Ask AI TutorPrevious Year Board Exam Questions
Find the area of a sector of a circle of radius 7 cm and angle of the sector is 90 degrees.
Given radius r = 7 cm and angle theta = 90 degrees.
Using the formula for area of a sector:
Area = (theta/360) x pi x r^2
= (90/360) x (22/7) x 7 x 7
= (1/4) x (22/7) x 49
= (1/4) x 154
= 38.5 cm^2
The area of the sector is 38.5 square centimetres.
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the minor segment. (Use pi = 3.14)
Given: r = 10 cm, theta = 90 degrees.
Area of sector = (90/360) x 3.14 x 100 = (1/4) x 314 = 78.5 cm^2
Area of triangle = (1/2) x r x r x sin(90) = (1/2) x 10 x 10 = 50 cm^2
Area of minor segment = 78.5 - 50 = 28.5 cm^2
In a circle of radius 21 cm, an arc subtends an angle of 60 degrees at the centre. Find the area of the sector formed by the arc.
Given radius r = 21 cm and angle theta = 60 degrees.
Using the formula for area of a sector:
Area = (theta/360) x pi x r^2
= (60/360) x (22/7) x 21 x 21
= (1/6) x (22/7) x 441
= (1/6) x 1386
= 231 cm^2
The area of the sector formed by the arc is 231 square centimetres.
The minute hand of a clock is 12 cm long. Find the area of the face of the clock described by the minute hand in 35 minutes.
r = 12 cm. In 35 minutes, angle = (35/60) x 360 = 210 degrees.
Area swept = (210/360) x (22/7) x 12 x 12
= (7/12) x (22/7) x 144
= (22/12) x 144
= 22 x 12
= 264 cm^2
Find the area of the shaded region enclosed between two concentric circles of radii 7 cm and 14 cm, where angle AOC = 40 degrees.
Area of shaded region = Area of sector of larger circle - Area of sector of smaller circle
= (40/360) x pi x (14)^2 - (40/360) x pi x (7)^2
= (1/9) x (22/7) x (196 - 49)
= (1/9) x (22/7) x 147
= (1/9) x 22 x 21
= (1/9) x 462
= 462/9
= 154/3 cm^2
= 51.33 cm^2 (approx.)
A piece of wire 22 cm long is bent into the form of an arc of a circle subtending an angle of 60 degrees at its centre. Find the radius of the circle. (Use pi = 22/7)
Arc length = (theta/360) x 2 x pi x r
22 = (60/360) x 2 x (22/7) x r
22 = (1/6) x (44/7) x r
22 = (44/42) x r
22 = (22/21) x r
r = 22 x 21/22
r = 21 cm
Find the area of the shaded region in the figure, where ABCD is a square of side 10 cm and semicircles are drawn with each side of the square as diameter. (Use pi = 3.14)
Side = 10 cm. Radius of each semicircle = 5 cm.
Area of square = 100 cm^2
Area of 4 semicircles = 4 x (1/2) x 3.14 x 25 = 2 x 3.14 x 25 = 157 cm^2
The 4 semicircles overlap within the square. The overlapping area creates a pattern.
Shaded area = Area of 4 semicircles - Area of square
= 157 - 100
= 57 cm^2
If the perimeter of a semi-circular protractor is 66 cm, find the diameter of the protractor. (Use pi = 22/7)
Perimeter of semicircle = pi x r + 2r = r(pi + 2)
66 = r x (22/7 + 2)
66 = r x (22/7 + 14/7)
66 = r x 36/7
r = 66 x 7/36
r = 462/36
r = 77/6 cm
Diameter = 2r = 154/6 = 77/3 cm
= 25.67 cm (approx.)
Find the area of the minor segment of a circle of radius 14 cm, when its central angle is 60 degrees. Also find the area of the corresponding major segment. (Use pi = 22/7)
r = 14 cm, theta = 60 degrees.
Area of sector = (60/360) x (22/7) x 14 x 14
= (1/6) x (22/7) x 196
= (1/6) x 616
= 308/3 cm^2
= 102.67 cm^2 (approx.)
Area of triangle = (1/2) x r^2 x sin(60)
= (1/2) x 196 x (sqrt(3)/2)
= 49 x sqrt(3)
= 49 x 1.732
= 84.87 cm^2 (approx.)
Area of minor segment = 308/3 - 49 x sqrt(3)
= 102.67 - 84.87
= 17.80 cm^2 (approx.)
Area of major segment = pi x r^2 - Area of minor segment
= 616 - 17.80
= 598.20 cm^2 (approx.)
The area of a sector of a circle with radius 6 cm is 15 pi cm^2. Find the degree measure of its angle.
Given area of sector = 15 pi cm^2 and radius r = 6 cm.
Using the formula: Area = (theta/360) x pi x r^2
15 pi = (theta/360) x pi x 36
Dividing both sides by pi:
15 = (theta/360) x 36
15 = theta/10
theta = 150
The degree measure of the angle is 150 degrees.
In a circle of radius 21 cm, an arc subtends an angle of 60 degrees at the centre. Find: (i) the length of the arc (ii) area of the sector formed by the arc.
r = 21 cm, theta = 60 degrees.
(i) Arc length = (60/360) x 2 x (22/7) x 21
= (1/6) x 2 x 66
= (1/6) x 132
= 22 cm
(ii) Area of sector = (60/360) x (22/7) x 21 x 21
= (1/6) x (22/7) x 441
= (1/6) x 1386
= 231 cm^2
In the given figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the shaded region.
Radius = 3.5 cm, OD = 2 cm.
Area of quadrant OACB = (1/4) x pi x r^2
= (1/4) x (22/7) x (3.5)^2
= (1/4) x (22/7) x 12.25
= (1/4) x 38.5
= 9.625 cm^2
Area of triangle OBD = (1/2) x OB x OD
= (1/2) x 3.5 x 2
= 3.5 cm^2
Area of shaded region = Area of quadrant - Area of triangle
= 9.625 - 3.5
= 6.125 cm^2
A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs 0.35 per cm^2. (Use sqrt(3) = 1.7)
Radius = 28 cm. The cover has a regular hexagon inscribed with 6 equal designs at the edges.
Angle of each sector = 360/6 = 60 degrees.
Area of each sector = (60/360) x (22/7) x 28 x 28
= (1/6) x (22/7) x 784
= (1/6) x 2464
= 2464/6 cm^2
Area of equilateral triangle (each side = r = 28 cm):
= (sqrt(3)/4) x 28^2
= (1.7/4) x 784
= 0.425 x 784
= 333.2 cm^2
Area of one design = Area of sector - Area of triangle
= 2464/6 - 333.2
= 410.67 - 333.2
= 77.47 cm^2
Area of 6 designs = 6 x 77.47 = 464.8 cm^2 (approx.)
Cost = 464.8 x 0.35 = Rs 162.68 (approx.)
Find the area of a sector of a circle of radius 28 cm and central angle 45 degrees.
Given radius r = 28 cm and central angle theta = 45 degrees.
Using the formula for area of a sector:
Area = (theta/360) x pi x r^2
= (45/360) x (22/7) x 28 x 28
= (1/8) x (22/7) x 784
= (1/8) x 2464
= 308 cm^2
The area of the sector is 308 square centimetres.
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): If the perimeter of a circle is equal to that of a square, then the ratio of their areas is 14:11.
Reason (R): Area of a circle = pi x r^2 and area of a square = side^2.
Answer: (A)
Let perimeter of circle = perimeter of square. 2 x pi x r = 4a, so a = pi x r/2. Ratio = pi x r^2 / a^2 = pi x r^2 / (pi^2 x r^2/4) = 4/pi = 4/(22/7) = 28/22 = 14/11. Both statements are true and R explains A.
AR2
Assertion (A): The area of a sector of angle 60 degrees of a circle with radius 10 cm is 52.33 cm^2 (approx.).
Reason (R): Area of sector = (theta/360) x 2 x pi x r.
Answer: (C)
Assertion is TRUE: (60/360) x (22/7) x 100 = (1/6) x 2200/7 = 2200/42 = 52.38 cm^2 (approx. 52.33). Reason is FALSE: the formula given is for arc length, not area. Correct formula for area of sector = (theta/360) x pi x r^2.
AR3
Assertion (A): The length of an arc of a sector of angle 36 degrees of a circle with radius 10 cm is 2 x pi cm.
Reason (R): Length of arc = (theta/360) x 2 x pi x r.
Answer: (A)
Both are true. Arc length = (36/360) x 2 x pi x 10 = (1/10) x 20 x pi = 2 x pi cm. The reason correctly states the formula and explains the assertion.
AR4
Assertion (A): If the circumference of a circle is 176 cm, then its radius is 28 cm.
Reason (R): Circumference of circle = 2 x pi x r.
Answer: (A)
Both true. 2 x (22/7) x r = 176. r = 176 x 7/44 = 1232/44 = 28 cm. Reason explains the assertion.
AR5
Assertion (A): The area of a minor segment of a circle of radius 14 cm when the central angle is 90 degrees is 56 cm^2.
Reason (R): Area of segment = Area of sector - Area of triangle.
Answer: (A)
Area of sector = (90/360) x (22/7) x 196 = (1/4) x 616 = 154 cm^2. Area of triangle = (1/2) x 14 x 14 = 98 cm^2. Area of segment = 154 - 98 = 56 cm^2. Both true and R explains A.
AR6
Assertion (A): The area of the largest circle inscribed in a square of side 10 cm is 78.5 cm^2.
Reason (R): The diameter of the largest inscribed circle equals the side of the square.
Answer: (A)
Both true. Diameter = 10 cm, radius = 5 cm. Area = pi x 25 = 3.14 x 25 = 78.5 cm^2. Reason correctly explains why the radius is half the side.
Case Study Questions
A park is in the shape of a circle of diameter 7 m. It has a path of uniform width of 0.7 m around it on the outside. Flower beds of equal area are to be planted in the sectors formed by the four radii drawn from the centre of the park dividing it into four equal parts.
(i) Find the area of the path around the park. (1 mark)
Radius of park = 7/2 = 3.5 m.
Outer radius (park + path) = 3.5 + 0.7 = 4.2 m.
Area of path = pi x R^2 - pi x r^2 = pi(R^2 - r^2)
= (22/7) x (4.2^2 - 3.5^2)
= (22/7) x (17.64 - 12.25)
= (22/7) x 5.39
= 16.94 m^2 (approx.)
(ii) Find the area of each flower bed if the park is divided into 4 equal sectors. (1 mark)
Area of park = pi x r^2 = (22/7) x 3.5 x 3.5 = (22/7) x 12.25 = 38.5 m^2.
Area of each sector (flower bed) = 38.5/4 = 9.625 m^2
(iii)(a) Find the total cost of planting flowers in all four beds at the rate of Rs 50 per m^2. (2 marks)
Total area of 4 flower beds = 38.5 m^2.
Cost = 38.5 x 50 = Rs 1925
(iii)(b) Find the perimeter of each sector (flower bed). (2 marks)
Perimeter of each sector = 2r + arc length
= 2 x 3.5 + (90/360) x 2 x (22/7) x 3.5
= 7 + (1/4) x 22
= 7 + 5.5
= 12.5 m
A school ground has a rectangular field of length 80 m and breadth 60 m. At each corner, a quadrant of radius 14 m is left for planting trees. A circular flower bed of radius 7 m is made at the centre of the field. The remaining area is used as a sports field for athletics.
(i) Find the total area of the four quadrants at the corners. (1 mark)
4 quadrants together make 1 full circle with radius 14 m.
Total area = pi x r^2 = (22/7) x 14 x 14 = (22/7) x 196 = 616 m^2
(ii) Find the area of the circular flower bed at the centre. (1 mark)
Area = pi x r^2 = (22/7) x 7 x 7 = (22/7) x 49 = 154 m^2
(iii) Find the area of the sports field (remaining area). (2 marks)
Area of rectangle = 80 x 60 = 4800 m^2
Area of 4 quadrants = 616 m^2
Area of flower bed = 154 m^2
Area of sports field = 4800 - 616 - 154 = 4030 m^2
MCQ Practice
MCQ1. If the area of a circle is 154 cm^2, then its perimeter is:
(A) 22 cm
(B) 44 cm
(C) 11 cm
(D) 88 cm
Answer: (B) - pi x r^2 = 154, so (22/7) x r^2 = 154, r^2 = 49, r = 7 cm. Perimeter = 2 x (22/7) x 7 = 44 cm.
MCQ2. The area of a sector of angle theta (in degrees) of a circle with radius R is:
(A) (theta/180) x pi x R^2
(B) (theta/360) x pi x R^2
(C) (theta/720) x pi x R^2
(D) (theta/360) x 2 x pi x R
Answer: (B) - The standard formula for area of a sector is (theta/360) x pi x R^2. Option (D) is the arc length formula.
MCQ3. The length of the arc of a sector of angle 45 degrees of a circle with radius 14 cm is:
(A) 11 cm
(B) 22 cm
(C) 44 cm
(D) 33 cm
Answer: (A) - Arc length = (45/360) x 2 x (22/7) x 14 = (1/8) x 88 = 11 cm.
MCQ4. The area of a quadrant of a circle whose circumference is 44 cm is:
(A) 77 cm^2
(B) 38.5 cm^2
(C) 154 cm^2
(D) 19.25 cm^2
Answer: (B) - 2 x (22/7) x r = 44, so r = 7 cm. Area of quadrant = (1/4) x (22/7) x 49 = (1/4) x 154 = 38.5 cm^2.
MCQ5. If the perimeter of a sector of a circle of radius 6.5 cm is 29 cm, what is the area of the sector?
(A) 58 cm^2
(B) 52 cm^2
(C) 25 cm^2
(D) 29 cm^2
Answer: (B) - Perimeter of sector = 2r + arc length. 29 = 2(6.5) + arc length. Arc length = 29 - 13 = 16 cm. Area = (1/2) x r x arc length = (1/2) x 6.5 x 16 = 52 cm^2.
MCQ6. The minute hand of a clock is 14 cm long. The area swept by it in 15 minutes is:
(A) 308 cm^2
(B) 154 cm^2
(C) 616 cm^2
(D) 77 cm^2
Answer: (B) - In 15 minutes, angle = (15/60) x 360 = 90 degrees. Area = (90/360) x (22/7) x 196 = (1/4) x 616 = 154 cm^2.
MCQ7. The area of the largest triangle that can be inscribed in a semicircle of radius r is:
(A) 2r^2
(B) r^2
(C) (1/2) x r^2
(D) sqrt(2) x r^2
Answer: (B) - The largest triangle inscribed in a semicircle has base = diameter = 2r and height = r. Area = (1/2) x 2r x r = r^2.
MCQ8. If the sector of a circle of diameter 10 cm subtends an angle of 144 degrees at the centre, then the length of the arc of the sector is:
(A) 2 x pi cm
(B) 4 x pi cm
(C) 5 x pi cm
(D) 8 x pi cm
Answer: (B) - Radius = 5 cm. Arc length = (144/360) x 2 x pi x 5 = (2/5) x 10 x pi = 4 x pi cm.
MCQ9. The radius of a circle whose area is equal to the sum of the areas of two circles of radii 24 cm and 7 cm is:
(A) 31 cm
(B) 25 cm
(C) 17 cm
(D) 28 cm
Answer: (B) - pi x R^2 = pi x 24^2 + pi x 7^2 = pi(576 + 49) = 625 x pi. R^2 = 625. R = 25 cm.
MCQ10. A wire is looped in the form of a circle of radius 28 cm. It is re-bent into a square form. The length of the side of the square is:
(A) 44 cm
(B) 22 cm
(C) 88 cm
(D) 176 cm
Answer: (A) - Circumference of circle = 2 x (22/7) x 28 = 176 cm. Perimeter of square = 176 cm. Side = 176/4 = 44 cm.
MCQ11. The area of a segment of a circle of radius 21 cm if the arc of the segment has a measure of 60 degrees is:
(A) 40.05 cm^2
(B) 231 cm^2
(C) 190.95 cm^2
(D) 17.80 cm^2
Answer: (A) - Area of sector = (60/360) x (22/7) x 441 = 231 cm^2. Area of triangle = (sqrt(3)/4) x 21^2 = (1.732/4) x 441 = 190.95 cm^2 (approx.). Area of segment = 231 - 190.95 = 40.05 cm^2 (approx.).
MCQ12. The number of revolutions made by a circular wheel of area 1.54 m^2 in rolling a distance of 176 m is:
(A) 100
(B) 40
(C) 20
(D) 10
Answer: (B) - pi x r^2 = 1.54, so (22/7) x r^2 = 1.54, r^2 = 0.49, r = 0.7 m. Circumference = 2 x (22/7) x 0.7 = 4.4 m. Revolutions = 176/4.4 = 40.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
Find the area of a sector of a circle with radius 7 cm and angle 90 degrees.
Area = (90/360) x (22/7) x 49 = (1/4) x 154 = 38.5 cm^2.
What is the length of the arc of a sector of angle 60 degrees of a circle with radius 21 cm?
Arc length = (60/360) x 2 x (22/7) x 21 = (1/6) x 132 = 22 cm.
Write the formula for area of a segment of a circle.
Area of segment = Area of corresponding sector - Area of the triangle formed by the two radii and the chord.
If the minute hand of a clock is 7 cm long, find the area swept in 10 minutes.
In 10 minutes, theta = (10/60) x 360 = 60 degrees. Area = (60/360) x (22/7) x 49 = (1/6) x 154 = 77/3 = 25.67 cm^2 (approx.).
Find the area of a circle inscribed in a square of side 14 cm.
Diameter = 14 cm, radius = 7 cm. Area = (22/7) x 49 = 154 cm^2.
2-3 Mark Questions
A chord of a circle of radius 12 cm subtends an angle of 120 degrees at the centre. Find the area of the corresponding minor segment. (Use pi = 3.14 and sqrt(3) = 1.73)
Area of sector = (120/360) x 3.14 x 144 = (1/3) x 452.16 = 150.72 cm^2.
Area of triangle = (1/2) x 12 x 12 x sin(120) = (1/2) x 144 x (sqrt(3)/2) = 36 x sqrt(3) = 36 x 1.73 = 62.28 cm^2.
Area of minor segment = 150.72 - 62.28 = 88.44 cm^2.
Find the area swept by the minute hand (length 14 cm) of a clock in 5 minutes.
In 5 minutes, theta = (5/60) x 360 = 30 degrees.
Area swept = (30/360) x (22/7) x 14 x 14 = (1/12) x 616 = 154/3 = 51.33 cm^2 (approx.).
A horse is tied at a corner of a square field of side 20 m with a rope of length 7 m. Find the area the horse can graze. (Use pi = 22/7)
At a corner of a square, angle = 90 degrees.
Grazing area = (90/360) x (22/7) x 49 = (1/4) x 154 = 38.5 m^2.
In a circle of radius 21 cm, an arc subtends an angle of 60 degrees at the centre. Find the length of the arc and the area of the sector.
Arc length = (60/360) x 2 x (22/7) x 21 = (1/6) x 132 = 22 cm.
Area of sector = (60/360) x (22/7) x 441 = (1/6) x 1386 = 231 cm^2.
3-4 Mark Questions
Find the area of the shaded region where ABCD is a square of side 14 cm and four equal semicircles are drawn with each side as diameter.
Area of square = 196 cm^2. Radius = 7 cm.
Area of 4 semicircles = 4 x (1/2) x (22/7) x 49 = 2 x 154 = 308 cm^2.
Shaded area = 308 - 196 = 112 cm^2.
The area of an equilateral triangle is 49 x sqrt(3) cm^2. Taking each vertex as centre, circles are drawn with radius equal to half the length of the side. Find the area of the shaded region. (Use pi = 22/7 and sqrt(3) = 1.732)
Area = (sqrt(3)/4) x a^2 = 49 x sqrt(3). So a^2 = 196, a = 14 cm.
Radius = 7 cm. Each angle = 60 degrees.
Area of 3 sectors = 3 x (60/360) x (22/7) x 49 = 3 x (1/6) x 154 = 77 cm^2.
Shaded area = 49 x sqrt(3) - 77 = 49 x 1.732 - 77 = 84.87 - 77 = 7.87 cm^2 (approx.).
From each corner of a square of side 4 cm, a quadrant of radius 1 cm is cut and a circle of diameter 2 cm is also cut from the centre. Find the area of the remaining portion.
Area of square = 16 cm^2.
Area of 4 quadrants = 4 x (1/4) x (22/7) x 1 = 22/7 cm^2.
Area of circle = (22/7) x 1 = 22/7 cm^2.
Remaining area = 16 - 22/7 - 22/7 = 16 - 44/7 = (112 - 44)/7 = 68/7 cm^2 = 9.71 cm^2 (approx.).
Key Concepts
Important Formulas
Study Plan - Areas Related to Circles
Sector Area and Arc Length Formulas
Read NCERT theory on sectors and segments. Memorise the three key formulas. Solve Exercise 11.1 Q1-Q5.
Area of Segment and Application Problems
Learn to find the area of a segment using sector minus triangle. Solve Exercise 11.1 Q6-Q14. Focus on minute hand and horse-tied-at-corner problems.
Combination of Plane Figures
Solve Exercise 11.2 Q1-Q7. Practice identifying which areas to add and which to subtract in shaded region problems.
Advanced Combinations and Previous Year Questions
Solve Exercise 11.2 Q8-Q14. Then solve all 14 PYQs from this page. Time yourself on 3-4 mark questions.
MCQ + Assertion-Reason + Case Study + Revision
Practice all 12 MCQs and 6 assertion-reason questions. Attempt case study questions under timed conditions. Revise common mistakes and formulas.
Common Mistakes to Avoid
- ✗Using the arc length formula (theta/360) x 2 x pi x r instead of the sector area formula (theta/360) x pi x r^2 when asked for area of a sector
- ✗Forgetting to convert the angle for the minute hand: the minute hand sweeps 360 degrees in 60 minutes, so in t minutes it sweeps (t/60) x 360 degrees
- ✗In segment area problems, forgetting to subtract the area of the triangle from the sector area, or using the wrong triangle area formula for the given angle
- ✗In combination problems with semicircles drawn on the sides of a square, confusing whether to add or subtract the areas of the semicircles from the square
- ✗Using pi = 22/7 when the question specifies pi = 3.14 or vice versa, which gives a different numerical answer and loses marks
Tips for Board Exam
- ✓Memorise the three core formulas: sector area = (theta/360) x pi x r^2, arc length = (theta/360) x 2 x pi x r, segment area = sector area - triangle area. These cover 90% of the chapter.
- ✓For shaded region problems, always draw a rough figure first and label all given dimensions. Identify which standard shapes (circle, semicircle, quadrant, triangle, square) make up the figure before calculating.
- ✓When the question says 'Use pi = 3.14', use 3.14 throughout. When no value is specified, use 22/7 as it simplifies better with common radii like 7, 14, 21, 28.
- ✓For the minute hand problem type: the angle swept in t minutes is 6t degrees (since 360/60 = 6 degrees per minute). This shortcut saves calculation time.
- ✓In segment problems with 60-degree angle, the triangle formed is equilateral (since two sides are radii). Use area = (sqrt(3)/4) x side^2 directly. For 90-degree angle, use (1/2) x r x r.
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Start Areas Related to Circles - Choose Your ModeAbout Areas Related to Circles - Class 10 Mathematics
Chapter 11 Areas Related to Circles is part of the Mensuration unit in CBSE Class 10 Mathematics and carries 4-6 marks in the board exam 2027. This chapter builds on the basic concepts of circle (circumference and area) taught in earlier classes and extends them to sectors, segments, and combinations of plane figures. It is a calculation-heavy chapter where accuracy in arithmetic and proper use of formulas determine the marks scored.
What This Chapter Covers
The chapter focuses on three main topics: finding the area and perimeter of a sector, finding the area of a segment, and computing areas of regions formed by combining circles with other plane figures like triangles, squares, and rectangles. All problems in this chapter revolve around these three skills.
Area of a Sector
A sector is the region enclosed between two radii and the arc of a circle. Think of it as a slice of a pizza. The angle between the two radii at the centre is called the central angle, denoted by theta. The formulas for a sector are straightforward.
The area of a sector is given by (theta/360) multiplied by pi multiplied by r squared, where theta is the central angle in degrees and r is the radius. This formula works because a sector is simply a fraction (theta/360) of the full circle.
The length of the arc (the curved part of the sector boundary) is (theta/360) multiplied by 2 multiplied by pi multiplied by r. Again, this is the same fraction of the full circumference.
The perimeter of a sector includes both the arc and the two straight radii, so it equals 2r plus the arc length.
A special case is the quadrant, where theta equals 90 degrees. The area of a quadrant is one-fourth of the area of the full circle, and this comes up frequently in board exam questions.
Arc Length and Sector Area Relationship
There is a useful relationship between arc length and sector area that can save time in exams. If the arc length is L and the radius is r, then the area of the sector equals (1/2) multiplied by L multiplied by r. This is analogous to the formula for the area of a triangle (half base times height) and is particularly useful when the question gives the arc length directly instead of the angle.
Area of a Segment
A segment is the region between a chord and the arc it cuts off. To find the area of a segment, you subtract the area of the triangle formed by the chord and the two radii from the area of the corresponding sector.
Area of segment equals area of sector minus area of triangle.
The triangle area depends on the central angle. For a 90-degree angle, the triangle is right-angled with both legs equal to the radius, so its area is (1/2) multiplied by r multiplied by r. For a 60-degree angle, the triangle formed by two radii and the chord is equilateral (since two sides are equal radii and the angle between them is 60 degrees), so its area is (sqrt(3)/4) multiplied by r squared.
For other angles, the general formula is (1/2) multiplied by r squared multiplied by sin(theta), but this uses trigonometry and is rarely needed at the Class 10 level. Most board exam questions use 60 or 90 degrees.
The area of the major segment equals the total circle area minus the area of the minor segment.
Combination of Plane Figures
Exercise 11.2 is dedicated to finding the area of shaded regions formed by combining circles (or parts of circles) with other shapes. These problems require students to decompose a complex figure into standard shapes whose areas they can calculate.
The general strategy is to identify the standard shapes involved (circles, semicircles, quadrants, triangles, squares, rectangles), calculate the area of each, and then add or subtract as needed based on the shading pattern.
Common combinations that appear in exams include four semicircles drawn on the sides of a square where the shaded area equals the area of four semicircles minus the area of the square. Another common type has four circles inscribed in a square where the shaded area equals the area of the square minus the area of four circles. Problems involving equilateral triangles with circular arcs drawn from vertices are also frequent.
Applications and Real-Life Context
The chapter uses practical scenarios to frame its problems. The minute hand of a clock sweeps a sector as it moves. A horse tied to a peg at the corner of a field can graze in a quadrant. A lighthouse covers a sector of the sea with its light beam. A windshield wiper sweeps an area between two concentric arcs. These real-life connections make the problems relatable and are often used in case study questions in CBSE board exams.
CBSE 2026-27 Syllabus Changes
For the CBSE 2026-27 session, Chapter 11 has been streamlined to 2 exercises. Exercise 11.1 contains 14 questions on area of sector, area of segment, and arc length. Exercise 11.2 contains 14 questions on areas of combinations of plane figures.
The older Exercise 11.3 has been removed from the CBSE syllabus. This exercise contained additional problems on combinations of plane figures that are now considered optional. Padhaao provides solutions for Exercise 11.3 as well for state board students in UP, Bihar, MP, Rajasthan, and other states who may still follow the three-exercise format.
How This Chapter Is Tested in Board Exams
This chapter is part of the Mensuration unit which carries 10 marks total (shared with Chapter 12 Surface Areas and Volumes). From Chapter 11 alone, students can expect 4-6 marks distributed as follows. One MCQ worth 1 mark typically tests a basic formula application like finding the area of a sector or arc length. One short answer worth 2-3 marks asks for the area of a segment or the area swept by a clock hand. One long answer worth 3-4 marks presents a shaded region problem involving combined figures.
Recent trends show CBSE asking combination problems with increasing complexity, involving semicircles drawn on the sides of polygons, sectors cut from corners of rectangles, and annular regions between concentric circles. Case study questions on circular park designs and sports field layouts have appeared in 2023-2025 papers.
Key Differences from Chapter 10 (Circles)
Students sometimes confuse Chapter 10 (Circles) with Chapter 11 (Areas Related to Circles). Chapter 10 deals with tangent properties, while Chapter 11 is purely about area and perimeter calculations. Chapter 10 involves theorems and proofs; Chapter 11 involves formulas and arithmetic. There is no conceptual overlap, but both chapters appear under the Geometry and Mensuration units respectively.
Connection to Other Chapters
Chapter 11 connects directly to Chapter 12 (Surface Areas and Volumes) since surface area problems for cylinders, cones, and spheres use the same circle area formula. The Pythagoras theorem from Chapter 6 (Triangles) is needed in problems where a right triangle is inscribed in a semicircle. Trigonometric ratios from Chapter 8 occasionally help in finding triangle areas within segments, though most Class 10 problems avoid this by using 60 or 90 degree angles.
For students preparing for competitive exams like NTSE and Olympiads, the concept of area of a segment and combinations of plane figures extends to more complex problems involving regular polygons inscribed in or circumscribed around circles. Understanding the sector and segment formulas thoroughly at the Class 10 level builds a strong foundation for mensuration problems in Class 11 and Class 12 mathematics.
Frequently Asked Questions
How many exercises are in NCERT Class 10 Maths Chapter 11 for CBSE 2026-27?
For CBSE 2026-27, Chapter 11 Areas Related to Circles has 2 exercises: Exercise 11.1 (14 questions on area of sector, area of segment, arc length) and Exercise 11.2 (14 questions on areas of combinations of plane figures). The older Exercise 11.3 has been removed from the CBSE syllabus.
What is the formula for area of a sector of a circle?
Area of sector = (theta/360) x pi x r^2, where theta is the central angle in degrees and r is the radius. For example, a sector of angle 60 degrees with radius 21 cm has area = (60/360) x (22/7) x 441 = 231 cm^2.
How many marks does Chapter 11 carry in the CBSE Class 10 board exam?
Chapter 11 is part of the Mensuration unit which carries 10 marks total. Chapter 11 alone typically carries 4-6 marks: 1 MCQ (1 mark) + 1 short answer on sector/segment area (2-3 marks) + 1 long answer on shaded region with combined figures (3-4 marks).
What is the difference between a sector and a segment of a circle?
A sector is the region enclosed between two radii and the arc (like a pizza slice). A segment is the region between a chord and the arc. The area of a segment = area of sector - area of the triangle formed by the two radii and the chord.
How to find the area swept by the minute hand of a clock?
The minute hand sweeps 360 degrees in 60 minutes, so it sweeps 6 degrees per minute. In t minutes, the angle = 6t degrees. Area swept = (6t/360) x pi x r^2, where r is the length of the minute hand. For example, a 14 cm minute hand sweeps (30/360) x (22/7) x 196 = 154/3 cm^2 in 5 minutes.
Is Exercise 11.3 deleted from the CBSE 2026-27 syllabus?
Yes, Exercise 11.3 has been removed from the CBSE 2026-27 syllabus. Only Exercises 11.1 and 11.2 are part of the CBSE exam. However, state board students (UP Board, Bihar Board, MP Board) may still need Exercise 11.3, and Padhaao provides those solutions as well.
When should I use pi = 22/7 and when pi = 3.14?
Always use the value specified in the question. If the question says 'Use pi = 3.14', use 3.14. If no value is specified, use pi = 22/7 as it simplifies better with common radii (7, 14, 21, 28). Using the wrong value will give a different numerical answer and you may lose marks.
What type of questions come from Chapter 11 in the CBSE board exam?
Common question types: (1) MCQ on sector area or arc length (1 mark). (2) Short answer: area swept by a clock hand, area of a segment, or sector area (2-3 marks). (3) Long answer: shaded region problems combining semicircles with squares/triangles (3-4 marks). Case study questions on circular park designs have appeared since 2023.