NCERT Solutions for Class 10 Maths Chapter 12 – Areas Related to Circles. Furthermore, here we’ve provided you with the latest solution for Class 10 CBSE NCERT Maths Chapter 12 – Areas Related to Circles. As a result here you’ll find solutions to all the exercises. This NCERT Class 10 solution will help you to score good marks in your exam.
Students can refer to our solution for NCERT Class 10 Maths Chapter 12 – Areas Related to Circles. The Chapter 12 Solution of NCERT will help students prepare for the exams and easily crack the exam. Below we’ve provided you with the exercise wise latest solution.
NCERT Solutions for Class 10 Maths Chapter 12 – Areas Related to Circles Exercise Wise Solution
Exercise 12.1 – Page 225 of NCERT
Exercise 12.2 – Page 230 of NCERT
Exercise 12.3 – Page 235 of NCERT
NCERT Solutions for Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.1
Here you’ll find NCERT Chapter 12 – Areas Related to Circles Exercise 12.1 Solution.
Exercise 12.1: Solutions of Questions on Page Number: 225.
Q1: The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
Answer :
Radius (r1) of 1st circle = 19 cm Radius (r2) or 2nd circle = 9 cm Let the radius of 3rd circle be r.
Circumference of 1st circle = 2πr1 = 2π (19) = 38π Circumference of 2nd circle = 2πr2 = 2π (9) = 18π Circumference of 3rd circle = 2πr
Given that,
Circumference of 3rd circle = Circumference of 1st circle + Circumference of 2nd circle 2πr = 38π + 18π = 56π
Therefore, the radius of the circle which has circumference equal to the sum of the circumference of the given two circles is 28 cm.
Q2: The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
Answer :
Radius (r1) of 1st circle = 8 cm Radius (r2) of 2nd circle = 6 cm Let the radius of 3rd circle be r.
Area of 1st circle
Area of 2nd circle
Given that,
Area of 3rd circle = Area of 1st circle + Area of 2nd circle
However, the radius cannot be negative. Therefore, the radius of the circle having area equal to the sum of the areas of the two circles is 10 cm.
Q3: Given figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.
Answer:
Radius (r1) of gold region (i.e., 1st circle)
Given that each circle is 10.5 cm wider than the previous circle. Therefore, radius (r2) of 2nd circle = 10.5 + 10.5
21 cm
Radius (r3) of 3rd circle = 21 + 10.5
= 31.5 cm
Radius (r4) of 4th circle = 31.5 + 10.5
= 42 cm
Radius (r5) of 5th circle = 42 + 10.5
= 52.5 cm
Area of gold region = Area of 1st circle
Area of red region = Area of 2nd circle – Area of 1st circle
Area of blue region = Area of 3rd circle – Area of 2nd circle
Area of black region = Area of 4th circle – Area of 3rd circle
Area of white region = Area of 5th circle – Area of 4th circle
Therefore, areas of gold, red, blue, black, and white regions are 346.5 cm2, 1039.5 cm2, 1732.5 cm2, 2425.5 cm2, and 3118.5 cm2 respectively.
Q4 :
The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is traveling at a speed of 66 km per hour?
Answer:
Diameter of the wheel of the car = 80 cm Radius (r) of the wheel of the car = 40 cm Circumference of wheel = 2πr
= 2π (40) = 80π cm Speed of car = 66 km/hour
Distance travelled by the car in 10 minutes
= 110000 × 10 = 1100000 cm
Let the number of revolutions of the wheel of the car be n. n × Distance travelled in 1 revolution (i.e., circumference)
= Distance travelled in 10 minutes
Therefore, each wheel of the car will make 4375 revolutions.
Q5: Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units (B) π units (C) 4 units (D)7 units
Answer :
Let the radius of the circle be r.
Circumference of circle = 2πr Area of circle = πr2
Given that, the circumference of the circle and the area of the circle are equal. This implies 2πr = πr2
2 = r
Therefore, the radius of the circle is 2 units. Hence, the correct answer is A.
NCERT Solutions for Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.2
Here you’ll find NCERT Chapter 12 – Areas Related to Circles Exercise 12.2 Solution.
Exercise 12.2: Solutions of Questions on Page Number: 230.
Q1: Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Answer:
Let OACB be a sector of the circle making 60° angle at centre O of the circle.
Area of sector of angle θ =
Area of sector OACB =
Therefore, the area of the sector of the circle making 60° at the centre of the circle is
Q2: Find the area of a quadrant of a circle whose circumference is 22 cm.
Answer:
Let the radius of the circle be r. Circumference = 22 cm
2πr = 22
Quadrant of circle will subtend 90° angle at the centre of the circle.
Area of such quadrant of the circle
Q3: The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Answer:
We know that in 1 hour (i.e., 60 minutes), the minute hand rotates 360°.
In 5 minutes, minute hand will rotate =
Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of 30° in a circle of 14 cm radius.
Area of sector of angle θ =
Area of sector of 30°
Therefore, the area swept by the minute hand in 5 minutes is
Q4: A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:
- Minor segment
- Major sector [Use π = 3.14]
Answer:
Let AB be the chord of the circle subtending 90° angle at centre O of the circle.
Area of major sector OADB =
Area of minor sector OACB =
Area of ΔOAB =
= 50 cm2
Area of minor segment ACB = Area of minor sector OACB – Area of ΔOAB = 78.5 – 50 = 28.5 cm2
Q5: In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
- The length of the arc
- Area of the sector formed by the arc
- Area of the segment forced by the corresponding chord
Answer :
Radius (r) of circle = 21 cm
Angle subtended by the given arc = 60°
Length of an arc of a sector of angle θ =
Length of arc ACB =
= 22 cm
Area of sector OACB =
In ΔOAB,
∠OAB = ∠OBA (As OA = OB)
∠OAB + ∠AOB + ∠OBA = 180° 2∠OAB + 60° = 180°
∠OAB = 60°
Therefore, ΔOAB is an equilateral triangle.
Area of ΔOAB =
Area of segment ACB = Area of sector OACB – Area of ΔOAB
Q6: A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle.
[Use π = 3.14 and ]
Answer :
Radius (r) of circle = 15 cm
Area of sector OPRQ =
In ΔOPQ,
∠OPQ = ∠OQP (As OP = OQ)
∠OPQ + ∠OQP + ∠POQ = 180° 2 ∠OPQ = 120°
∠OPQ = 60°
ΔOPQ is an equilateral triangle.
Area of ΔOPQ =
Area of segment PRQ = Area of sector OPRQ – Area of ΔOPQ
= 117.75 – 97.3125
= 20.4375 cm2
Area of major segment PSQ = Area of circle – Area of segment PRQ
∙
Q7: A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.
[Use π = 3.14 and ]
Answer:
Let us draw a perpendicular OV on chord ST. It will bisect the chord ST. SV = VT
In ΔOVS,
Area of ΔOST =
Area of sector OSUT =
Area of segment SUT = Area of sector OSUT – Area of ΔOST
= 150.72 – 62.28
= 88.44 cm2
Q8: A 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 (see the given figure). Find
- The area of that part of the field in which the horse can graze.
- The increase in the grazing area of the rope were 10 m long instead of 5 m. [Use À = 3.14]
Answer:
From the figure, it can be observed that the horse can graze a sector of 90° in a circle of 5 m radius. Area that can be grazed by horse = Area of sector OACB
Area that can be grazed by the horse when length of rope is 10 m long
Increase in grazing area = (78.5 – 19.625) m2
= 58.875 m2
Q9: A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure. Find.
- The total length of the silver wire required.
- The area of each sector of the brooch
Answer :
Total length of wire required will be the length of 5 diameters and the circumference of the brooch.
Radius of circle =
Circumference of brooch = 2πr
= 110 mm
Length of wire required = 110 + 5 × 35
= 110 + 175 = 285 mm
It can be observed from the figure that each of 10 sectors of the circle is subtending 36° at the centre of the circle.
Therefore, area of each sector =
Q10: An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
Answer :
There are 8 ribs in an umbrella. The area between two consecutive ribs is subtending at the centre of the assumed flat circle.
Area between two consecutive ribs of circle =
Q11: A car has two wipers which do not overlap. Each wiper has blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Answer:
It can be observed from the figure that each blade of wiper will sweep an area of a sector of 115° in a circle of 25 cm radius.
Area of such sector =
Area swept by 2 blades =
Q12: To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships warned. [Use π = 3.14]
Answer :
It can be observed from the figure that the lighthouse spreads light across a sector of 80° in a circle of 16.5 km radius.
Area of sector OACB =
Q13: A round table cover has six equal designs as shown in 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 cm2. [Use √3= 1.7 ]
Answer:
It can be observed that these designs are segments of the circle.
Consider segment APB. Chord AB is a side of the hexagon. Each chord will substitute 
In ΔOAB,
∠OAB = ∠OBA (As OA = OB)
∠AOB = 60°
∠OAB + ∠OBA + ∠AOB = 180° 2∠OAB = 180° – 60° = 120°
∠OAB = 60°
Therefore, ΔOAB is an equilateral triangle.
Area of ΔOAB =
Area of sector OAPB =
Area of segment APB = Area of sector OAPB – Area of ΔOAB
Cost of making 1 cm2 designs = Rs 0.35
Cost of making 464.76 cm2 designs = 
Therefore, the cost of making such designs is Rs 162.68.
Q14: Tick the correct answer in the following:
Area of a sector of angle p (in degrees) of a circle with radius R is
Answer:
We know that area of sector of angle θ = 
Hence, (D) is the correct answer.
Q15: Tick the correct answer in the following:
Area of a sector of angle p (in degrees) of a circle with radius R is
Answer :
We know that area of sector of angle θ =
Area of sector of angle P =
Hence, (D) is the correct answer.
NCERT Solutions for Class 10 Maths Chapter 12 – Areas Related to Circles Exercise 12.3
Here you’ll find NCERT Chapter 12 – Areas Related to Circles Exercise 12.3 Solution.
Exercise 12.3: Solutions of Questions on Page Number: 235.
Q1: Find the area of the shaded region in the given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°. [ Use π = 22/7]
Answer :
Radius of inner circle = 7 cm Radius of outer circle = 14 cm
Area of shaded region = Area of sector OAFC – Area of sector OBED
=40°360°×π(14)2 – 40°
Q2: Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
Answer :
It can be observed from the figure that the radius of each semi-circle is 7 cm.
Area of each semi-circle =
Area of square ABCD = (Side)2 = (14)2 = 196 cm2 Area of the shaded region
= Area of square ABCD – Area of semi-circle APD – Area of semi-circle BPC
= 196 – 77 – 77 = 196 – 154 = 42 cm2
Q3: Find the area of the shaded region in the given 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.
Answer:
We know that each interior angle of an equilateral triangle is of measure 60°.
Area of sector OCDE
Area of
Area of circle = πr2
Area of shaded region = Area of ΔOAB + Area of circle – Area of sector OCDE
Q4: From 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 given figure. Find the area of the remaining portion of the
square.
Answer :
Each quadrant is a sector of 90° in a circle of 1 cm radius.
Area of each quadrant
Area of square = (Side)2 = (4)2 = 16 cm2
Area of circle = πr2 = π (1)2
Area of the shaded region = Area of square – Area of circle – 4 × Area of quadrant
Q5: In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the given figure. Find the area of the design (Shaded region).
Answer :
Radius (r) of circle = 32 cm
AD is the median of 
AD = 48 cm In ΔABD,
AB2 = AD2 + BD2
Area of equilateral triangle,
Area of circle = πr2
Area of design = Area of circle – Area of ΔABC
Q6: In the given figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded
region.
Answer :
Area of each of the 4 sectors is equal to each other and is a sector of 90° in a circle of 7 cm radius.
Area of each sector
Area of square ABCD = (Side)2 = (14)2 = 196 cm2
Area of shaded portion = Area of square ABCD – 4 × Area of each sector
Therefore, the area of shaded portion is 42 cm2.
Q7: The given figure depicts a racing track whose left and right ends are semicircular.
The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:
- The distance around the track along its inner edge
- The area of the track
Answer :
Distance around the track along its inner edge = AB + arc BEC + CD + arc DFA
Area of the track = (Area of GHIJ – Area of ABCD) + (Area of semi-circle HKI – Area of semi-circle BEC) + (Area of semi-circle GLJ – Area of semi-circle AFD)
Therefore, the area of the track is 4320 m2.
Q8: In 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.
Answer:
Radius (r1) of larger circle = 7 cm
Radius (r2) of smaller circle 
Area of smaller circle
Area of semi-circle AECFB of larger circle
Area of
Area of the shaded region
= Area of smaller circle + Area of semi-circle AECFB – Area of ΔABC
Q9: The area of an equilateral triangle ABC is 17320.5 cm2. 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 (See the given figure). Find the area of shaded region. [Use π = 3.14 and ]
Answer :
Let the side of the equilateral triangle be a. Area of equilateral triangle = 17320.5 cm2
Each sector is of measure 60°.
Area of sector ADEF
Area of shaded region = Area of equilateral triangle – 3 × Area of each sector
Q10: On a square handkerchief, nine circular designs each of radius 7 cm are made (see the given figure). Find the area of the remaining portion of the handkerchief.
Answer :
From the figure, it can be observed that the side of the square is 42 cm. Area of square = (Side)2 = (42)2 = 1764 cm2
Area of each circle = πr2
Area of 9 circles = 9 × 154 = 1386 cm2
Area of the remaining portion of the handkerchief = 1764 – 1386 = 378 cm2
Q11: In the given figure, OACB is a quadrant of circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
- Quadrant OACB
- Shaded region
Answer :
- Since OACB is a quadrant, it will subtend 90° angle at O.
Area of quadrant OACB
- Area of ΔOBD
Area of the shaded region = Area of quadrant OACB – Area of ΔOBD
Q12: In the given figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. [Use π = 3.14]
Answer :
In ΔOAB,
OB2 = OA2 + AB2
= (20)2 + (20)2
Radius (r) of circle
Area of quadrant OPBQ
Area of OABC = (Side)2 = (20)2 = 400 cm2
Area of shaded region = Area of quadrant OPBQ – Area of OABC
= (628 – 400) cm2
= 228 cm2
Q13: AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see the given figure). If ∠AOB = 30°, find the area of the shaded region.
Answer :
Area of the shaded region = Area of sector OAEB – Area of sector OCFD
Q14: In the given figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.
Answer :
As ABC is a quadrant of the circle, ∠BAC will be of measure 90 º. In ΔABC,
BC2 = AC2 + AB2
= (14)2 + (14)2
Radius (r1) of semi-circle drawn on
Area of
Area of sector
= 154 – (154- 98)
= 98 cm2
Q15: Calculate the area of the designed region in the given figure common between the two quadrants of circles of radius 8 cm each.
Answer :
The designed area is the common region between two sectors BAEC and DAFC.
Area of sector
Area of ΔBAC
Area of the designed portion = 2 × (Area of segment AEC)
= 2 × (Area of sector BAEC – Area of ΔBAC)
NCERT Class 10 Maths All Chapters Solution
Chapter 1: Real Numbers
Chapter 2: Polynomials
Chapter 3: Pair of Linear Equation in Two Variables
Chapter 4: Quadratic Equations
Chapter 5: Arithmetic Progressions
Chapter 6: Triangles
Chapter 7: Coordinate Geometry
Chapter 8: Introduction to Trigonometry
Chapter 9: Some Applications of Trigonometry
Chapter 10: Circles
Chapter 11: Constructions
Chapter 12: Areas Related to Circles
Chapter 13: Surface Areas and Volumes
Chapter 14: Statistics
Chapter 15: Probability