## NCERT Solutions of Class 10 Maths Ch 12 Areas Related to Circles

In this page, you will find Chapter 12 Areas Related to Circles NCERT Solutions for Class 10 Maths which are very helpful in finding solutions of difficult problems given in the exercise. Class 10 NCERT Solutions are very beneficial in increasing concertation among students. These NCERT Solutions are updated as per the latest syllabus of CBSE.

Page No: 230

Exercise: 12.1

Unless stated otherwise, use Ï€ =22/7.

1. 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.

Let the radius of the third circle be R.
Circumference of the circle with radius R = 2Ï€R
Circumference of the circle with radius 19 cm = 2Ï€ × 19 = 38Ï€ cm
Circumference of the circle with radius 9 cm = 2Ï€ × 9 = 18Ï€ cm
Sum of the circumference of two circles = 38Ï€ + 18Ï€ = 56Ï€ cm
Circumference of the third circle = 2Ï€R = 56Ï€
⇒ 2Ï€R = 56Ï€ cm
⇒ R = 28 cm
The radius of the circle which has circumference equal to the sum of the circumferences of the two circles is 28 cm.

2.  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.

Let the radius of the third circle be R.
Area of the circle with radius R = Ï€R2
Area of the circle with radius 8 cm = Ï€ × 8= 64Ï€ cm2
Area of the circle with radius 6 cm = Ï€ × 6= 36Ï€ cm2
Sum of the area of two circles = 64Ï€ cm+ 36Ï€ cm= 100Ï€ cm2
Area of the third circle = Ï€R2 = 100Ï€ cm2
⇒ Ï€R2 = 100Ï€ cm2
⇒ R2 = 100 cm2
⇒ R = 10 cm
Thus, the radius of the new circle is 10 cm.

3. Fig. 12.3 depicts an archery target marked with its five scoring regions 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.

Diameter of Gold circle (first circle) = 21 cm
Radius of first circle, r1 = 21/2 cm = 10.5 cm
Each of the other bands is 10.5 cm wide,
Radius of second circle, r2 = 10.5 cm + 10.5 cm = 21 cm
Radius of third circle, r3 = 21 cm + 10.5 cm = 31.5 cm
Radius of fourth circle, r4 = 31.5 cm + 10.5 cm = 42 cm
Radius of fifth circle, r5 = 42 cm + 10.5 cm = 52.5 cm
Area of gold region = Ï€ r1= Ï€ (10.5)= 346.5 cm2
Area of red region = Area of second circle - Area of first circle = Ï€ r22 - 346.5 cm2
= Ï€(21)2 - 346.5 cm2 = 1386 - 346.5 cm= 1039.5 cm2
Area of blue region = Area of third circle - Area of second circle = Ï€ r3- 1386 cm2
=  Ï€(31.5)2 - 1386 cm= 3118.5 - 1386 cm= 1732.5 cm2
Area of black region = Area of fourth circle - Area of third circle = Ï€ r32 - 3118.5 cm2
= Ï€(42)2 - 1386 cm= 5544 - 3118.5 cm= 2425.5 cm2
Area of white region = Area of fifth circle - Area of fourth circle = Ï€ r42 - 5544 cm2
= Ï€(52.5)2 - 5544 cm= 8662.5 - 5544 cm= 3118.5 cm2

4. 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 travelling at a speed of 66 km per hour?

Diameter of the wheels of a car = 80 cm
Circumference of wheels = 2Ï€r = 2r × Ï€ = 80 Ï€ cm
Distance travelled by car in 10 minutes = (66 × 1000 × 100 × 10)/60 = 1100000 cm/s
No. of revolutions = Distance travelled by car/Circumference of wheels
= 1100000/80 Ï€ = (1100000 × 7)/(80×22) = 4375
4375 complete revolutions does each wheel make in 10 minutes.

5. 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) Ï€ units                  (C) 4 units              (D) 7 units

Let the radius of the circle be r.
∴ Perimeter of the circle = Circumference of the circle = 2Ï€r
∴ Area of the circle = Ï€ r2
A/q,
2Ï€r = Ï€ r2
⇒ 2 = r
Thus, the radius of the circle is 2 units. (A) is correct.

Page No: 230

Exercise: 12.2

Unless stated otherwise, use Ï€ =22/7.
1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Area of the sector making angle Î¸ = (Î¸/360°)×Ï€ r
Area of the sector making angle 60° = (60°/360°)×Ï€ rcm2
= (1/6)×62Ï€  = 36/6 Ï€ cm2 = 6 × 22/7 cm= 132/7 cm2

2. Find the area of a quadrant of a circle whose circumference is 22 cm.

Quadrant of a circle means sector is making angle 90°.
Circumference of the circle = 2Ï€r = 22 cm
Radius of the circle = r = 22/2Ï€ cm = 7/2 cm
Area of the sector making angle 90° = (90°/360°)×Ï€ rcm2
= (1/4)×(7/2)2Ï€  = (49/16) Ï€ cm2 = (49/16) × (22/7) cm= 77/8 cm2

3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Here, Minute hand of clock acts as radius of the circle.
∴ Radius of the circle (r) = 14 cm
Angle rotated by minute hand in 1 hour = 360°
∴ Angle rotated by minute hand in 5 minutes = 360° × 5/60 = 30°
Area of the sector making angle 30° = (30°/360°)×Ï€ rcm2
= (1/12) × 142Ï€  = 196/12 Ï€ cm2 = (49/3) × (22/7) cm= 154/3 cm2
Area swept by the minute hand in 5 minutes = 154/3 cm2

4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use Ï€ = 3.14)

Radius of the circle = 10 cm
Major segment is making 360° - 90° = 270°
Area of the sector making angle 270°
= (270°/360°) × Ï€ rcm2
= (3/4) × 102Ï€  = 75 Ï€ cm2
= 75 × 3.14 cm= 235.5 cm2
∴ Area of the major segment = 235.5 cm2
Height of Î”AOB = OA = 10 cm
Base of Î”AOB = OB = 10 cm
Area of Î”AOB = 1/2 × OA × OB
= 1/2 ×10 × 10 = 50 cm2
Major segment is making  90°
Area of the sector making angle 90°                                                                = (90°/360°) × Ï€ rcm2
= (1/4) × 102Ï€  = 25 Ï€ cm2
= 25 × 3.14 cm= 78.5 cm2
Area of the minor segment = Area of the sector making angle 90° - Area of Î”AOB
= 78.5 cm-  50 cm= 28.5 cm2

5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) the length of the arc
(ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord

Radius of the circle = 21 cm

(i) Length of the arc AB = Î¸/360° × 2Ï€r
= 60°/360° × 2 × 22/7 × 21
= 1/6 × 2 × 22/7 × 21 = 22
The length of the arc is  22 cm.

(ii) Angle subtend by the arc = 60°
Area of the sector making angle 60° = (60°/360°) × Ï€ rcm2
= (1/6) × 212Ï€  = 441/6 Ï€ cm2
= 441/6 × 22/7 cm= 231 cm2
∴ Area of the sector formed by the arc is 231 cm2

(iii) Area of equilateral Î”AOB = √3/4 × (OA)= √3/4 × 21= (441√3)/4 cm2
Area of the segment formed by the corresponding chord
= Area of the sector formed by the arc - Area of equilateral Î”AOB
= 231 cm- (441√3)/4 cm

6. 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 Ï€ = 3.14 and √3 = 1.73)

Radius of the circle = 15 cm
Î”AOB is isosceles as two sides are equal.
∴ ∠A = ∠B
Sum of all angles of triangle = 180°
∠A + ∠B + ∠C = 180°
⇒ 2 ∠A = 180° - 60°
⇒ ∠A = 120°/2
⇒ ∠A = 60°
Triangle is equilateral as ∠A = ∠B = ∠C = 60°
∴ OA = OB = AB = 15 cm
Area of equilateral Î”AOB = √3/4 × (OA)= √3/4 × 15
= (225√3)/4 cm= 97.3 cm2

Angle subtend at the centre by minor segment = 60°
Area of Minor sector making angle 60° = (60°/360°) × Ï€ rcm2
= (1/6) × 15Ï€  cm2 =  225/6 Ï€  cm2
=  (225/6) × 3.14 cm= 117.75  cm2

Area of the minor segment = Area of Minor sector - Area of equilateral Î”AOB
= 117.75  cm2 - 97.3 cm= 20.4 cm2

Angle made by Major sector = 360° - 60° = 300°
Area of the sector making angle 300° = (300°/360°) × Ï€ rcm2
= (5/6) × 15Ï€  cm2 =  1125/6 Ï€  cm2
=  (1125/6) × 3.14 cm= 588.75  cm2

Area of major segment = Area of Minor sector + Area of equilateral Î”AOB
= 588.75  cm2 + 97.3 cm= 686.05 cm2

7. 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 Ï€ = 3.14 and √3 = 1.73)

Radius of the circle, r = 12 cm
Draw a perpendicular OD to chord AB. It will bisect AB.
∠A = 180° - (90° + 60°) = 30°
⇒ AB = 2 × AD = 12√3 cm
sin 30° = OD/OA
⇒ 1/2 = OD/12
⇒ OD = 6 cm
Area of Î”AOB = 1/2 × base × height
= 1/2 × 12√3 × 6 = 36√3 cm
= 36 × 1.73 = 62.28 cm2

Angle made by Minor sector = 120°
Area of the sector making angle 120° = (120°/360°) × Ï€ rcm2
= (1/3) × 12Ï€  cm2 =  144/3 Ï€  cm2
=  48 × 3.14 cm= 150.72  cm2
∴ Area of the corresponding Minor segment = Area of the Minor sector - Area of Î”AOB
= 150.72  cm- 62.28 cm2
= 88.44 cm2

8. 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 Fig. 12.11). 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 Ï€ = 3.14)

Side of square field = 15 m
Length of rope is the radius of the circle, r = 5 m
Since, the horse is tied at one end of square field, it will graze only quarter of the field with radius 5 m.

(i) Area of circle = Ï€ r= 3.14 × 52 = 78.5 m2
Area of that part of the field in which the horse can graze = 1/4 of area of the circle = 78.5/4 = 19.625 m2

(ii) Area of circle if the length of rope is increased to 10 m = Ï€ r2 = 3.14 × 102 = 314 m2
Area of that part of the field in which the horse can graze = 1/4 of area of the circle
= 314/4 = 78.5 m2
Increase in grazing area = 78.5 m2 - 19.625 m2 = 58.875 m2

9. 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 Fig. 12.12. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.

Number of diameters = 5
Length of diameter = 35 mm

(i) Total length of silver wire required = Circumference of the circle +  Length of 5 diameter
= 2Ï€ r + (5×35) mm = (2 × 22/7 × 35/2) + 175 mm
= 110 + 175 mm = 185 mm

(ii) Number of sectors = 10
Area of each sector = Total area/Number of sectors
Total Area = Ï€ r2 = 22/7 × (35/2)= 1925/2 mm2
∴ Area of each sector = (1925/2) × 1/10 = 385/4 mm2

10. An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Number of ribs in umbrella = 8
Radius of umbrella while flat = 45 cm
Area between the two consecutive ribs of the umbrella =
Total area/Number of ribs
Total Area = Ï€ r2 = 22/7 × (45)= 6364.29 cm2
∴ Area between the two consecutive ribs = 6364.29/8 cm2
= 795.5 cm2

11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Angle of the sector of circle made by wiper = 115°
Radius of wiper = 25 cm
Area of the sector made by wiper = (115°/360°) × Ï€ rcm2
= 23/72 × 22/7 × 25=  23/72 × 22/7 × 625 cm2
= 158125/252 cm2

Total area cleaned at each sweep of the blades = 2 ×158125/252 cm2 =  158125/126 = 1254.96 cm2

12. 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 are warned.
(Use Ï€ = 3.14)

Let O bet the position of Lighthouse.
Angle made by the sector = 80°
Area of the sea over which the ships are warned = Area made by the sector.
Area of sector = (80°/360°) × Ï€ rkm2
= 2/9 × 3.14 × (16.5)km2
=  189.97 km2

13. A round table cover has six equal designs as shown in Fig. 12.14. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm2 . (Use √3 = 1.7)

Number of equal designs = 6
Radius of round table cover = 28 cm
Cost of making design = ₹ 0.35 per cm2
∠O = 360°/6 = 60°

Î”AOB is isosceles as two sides are equal. (Radius of the circle)
∴ ∠A = ∠B
Sum of all angles of triangle = 180°
∠A + ∠B + ∠O = 180°
⇒ 2 ∠A = 180° - 60°
⇒ ∠A = 120°/2
⇒ ∠A = 60°
Triangle is equilateral as ∠A = ∠B = ∠C = 60°
Area of equilateral Î”AOB = √3/4 × (OA)= √3/4 × 28= 333.2 cm2

Area of sector ACB = (60°/360°) × Ï€ rcm2
= 1/6 × 22/7 × 28 × 28 = 410.66 cm2
Area of design = Area of sector ACB - Area of equilateral Î”AOB
= 410.66 cm2 - 333.2 cm= 77.46 cm2

Area of 6 design = 6 × 77.46 cm= 464.76 cm2

Total cost of making design = 464.76 cm× ₹ 0.35 per cm2 = ₹ 162.66

14. Tick the correct answer in the following :
Area of a sector of angle p (in degrees) of a circle with radius R is
(A) p/180 × 2Ï€R             (B) p/180 × Ï€ R2                   (C) p/360 × 2Ï€R           (D) p/720 × 2Ï€R

Area of a sector of angle p = p/360 × Ï€ R2
= p/360 × 2/2 × Ï€ R
=  2p/720 × 2Ï€R
Hence, Option (D) is correct.

Page No: 234

Exercise: 12.3

Unless stated otherwise, use Ï€ =22/7
1. Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

PQ = 24 cm and PR = 7 cm
∠P = 90° (Angle in the semi-circle)
∴ QR is hypotenuse of the circle = Diameter of the circle.
By Pythagoras theorem,
QR= PR2  + PQ2
⇒ QR= 72  + 242
⇒ QR= 49  + 576
⇒ QR= 625
⇒ QR = 25 cm
∴ Radius of the circle = 25/2 cm
Area of the semicircle = (Ï€ R2)/2
= (22/7 × 25/2 × 25/2)/2 cm2
= 13750/56 cm= 245.54 cm2
Area of the Î”PQR = 1/2 × PR × PQ
= 1/2 × 7 × 24 cm2
= 84 cm2
Area of the shaded region = 245.54 cm2 - 84 cm= 161.54 cm2

2. Find the area of the shaded region in Fig. 12.20, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Radius inner circle = 7 cm
Radius of outer circle = 14 cm
Angle made by sector = 40°
Area of the sector OAC = (40°/360°) × Ï€ rcm2
= 1/9 × 22/7 × 142 = 68.44 cm2
Area of the sector OBD = (40°/360°) × Ï€ rcm2
= 1/9 × 22/7 × 72  = 17.11 cm2

Area of the shaded region ABDC = Area of the sector OAC - Area of the sector circle OBD
= 68.44 cm2 - 17.11 cm= 51.33 cm2

3. Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

There are two semicircles in the figure.
Side of the square = 14 cm
Diameter of the semicircle = 14 cm
∴ Radius of the semicircle = 7 cm
Area of the square = 14 × 14 = 196 cm2
Area of the semicircle = (Ï€ R2)/2
= (22/7 × 7 × 7)/2 cm= 77 cm
Area of two semicircles = 2 × 77 cm = 154 cm2

Area of the shaded region = 196 cm- 154 cm2 = 42 cm2

4. Find the area of the shaded region in Fig. 12.22, 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.

OAB is an equilateral triangle with each angle equal to 60°.
Area of the sector is common in both.
Radius of the circle = 6 cm.
Side of the triangle = 12 cm.
Area of the equilateral triangle = √3/4 × (OA)= √3/4 × 12= 36√3 cm2
Area of the circle = Ï€ R2 = 22/7 × 6= 792/7 cm2
Area of the sector making angle 60° = (60°/360°) × Ï€ rcm2
= 1/6 × 22/7 × 6cm2  = 132/7 cm2
Area of the shaded region = Area of the equilateral triangle + Area of the circle - Area of the sector
= 36√3 cm2 + 792/7 cm2 - 132/7 cm2
= (36√3 + 660/7) cm2

5. 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 Fig. 12.23. Find the area of the remaining portion of the square.

Side of the square = 4 cm
Radius of the circle = 1 cm
Four quadrant of a circle are cut from corner and one circle of radius are cut from middle.
Area of square = (side)= 4= 16 cm2
Area of the quadrant = (Ï€ R2)/4 cm2 = (22/7 × 12)/4 = 11/14 cm2
∴ Total area of the 4 quadrants = 4 × (11/14) cm2 = 22/7 cm2
Area of the circle = Ï€ Rcm2 = (22/7 × 12) = 22/7 cm2
Area of the shaded region = Area of square - (Area of the 4 quadrants + Area of the circle)
= 16 cm- (22/7 + 22/7) cm2
= 68/7 cm2

6. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 12.24. Find the area of the design.

Radius of the circle = 32 cm
Draw a median AD of the triangle passing through the centre of the circle.
⇒ BD = AB/2
Since, AD is the median of the triangle
⇒ 2/3 AD = 32 cm
By Pythagoras theorem,
⇒ AB= 482  + (AB/2)2
⇒ AB= 2304  + AB2/4
⇒ 3/4 (AB2) = 2304
⇒ AB= 3072
⇒ AB = 32√3 cm
Area of Î”ADB = √3/4 × (32√3)cm= 768√3 cm2
Area of circle = Ï€ R2 = 22/7 × 32 × 32 = 22528/7 cm2

Area of the design = Area of circle - Area of Î”ADB
= (22528/7 - 768√3) cm2

Page No: 236

7. In Fig. 12.25, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Side of square = 14 cm
Four quadrants are included in the four sides of the square.
∴ Radius of the circles = 14/2 cm = 7 cm
Area of the square ABCD = 14= 196 cm2
Area of the quadrant = (Ï€ R2)/4 cm2 = (22/7 × 72)/4 cm2
= 77/2 cm2
Total area of the quadrant = 4 × 77/2 cm= 154 cm

Area of the shaded region = Area of the square ABCD - Area of the quadrant
= 196 cm- 154 cm2
= 42 cm

8. Fig. 12.26 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 :
(i) the distance around the track along its inner edge
(ii) the area of the track.

Width of track = 10 m
Distance between two parallel lines = 60 m
Length of parallel tracks = 106 m
DE = CF = 60 m
Radius of inner semicircle, r = OD = O'C
= 60/2 m = 30 m
Radius of outer semicircle, R = OA = O'B
= 30 + 10 m = 40 m
Also, AB = CD = EF = GH = 106 m

Distance around the track along its inner edge = CD + EF + 2 × (Circumference of inner semicircle)
= 106 + 106 + (2 × Ï€r) m = 212 + (2 × 22/7 × 30) m
= 212 + 1320/7 m = 2804/7 m

Area of the track = Area of ABCD + Area EFGH + 2 × (area of  outer semicircle) - 2 × (area of  inner                                                                                                                                               semicircle)
= (AB × CD) + (EF × GH) + 2 × (Ï€r2/2) - 2 × (Ï€R2/2) m2
= (106 × 10) + (106 × 10) + 2 × Ï€/2 (r2 -R2) m2
= 2120 + 22/7 × 70 × 10 m2
= 4320 m

9. In Fig. 12.27, 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.

Radius of larger circle, R = 7 cm
Radius of smaller circle, r = 7/2 cm
Height of Î”BCA = OC = 7 cm
Base of Î”BCA = AB = 14 cm
Area of Î”BCA = 1/2 × AB × OC = 1/2 × 7 × 14 = 49 cm
Area of larger circle = Ï€R2 = 22/7 × 72 = 154 cm
Area of larger semicircle = 154/2 cm= 77 cm
Area of smaller circle = Ï€r2 = 22/7 × 7/2 × 7/2 = 77/2 cm2

Area of the shaded region = Area of larger circle - Area of triangle - Area of larger semicircle + Area                                               of smaller circle
Area of the shaded region = (154 - 49 - 77 + 77/2) cm2
= 133/2 cm2 = 66.5 cm2

10. 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 Fig. 12.28). Find the area of the shaded region. (Use Ï€ = 3.14 and √3 = 1.73205)

ABC is an equilateral triangle.
∴ ∠A = ∠B = ∠C = 60°
There are three sectors each making 60°.
Area of Î”ABC = 17320.5 cm2
⇒ √3/4 × (side) = 17320.5
⇒ (side) = 17320.5 × 4/1.73205
⇒ (side) = 4 × 10
⇒ side  = 200 cm
Radius of the circles = 200/2 cm = 100 cm
Area of the sector = (60°/360°) × Ï€ rcm2
= 1/6 × 3.14 × (100)cm2
= 15700/3 cm2
Area of 3 sectors = 3 × 15700/3 = 15700 cm=

Area of the shaded region = Area of equilateral triangle ABC - Area of 3 sectors
= 17320.5 - 15700 cm= 1620.5 cm2

11. On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief.

Number of circular design = 9
Radius of the circular design = 7 cm
There are three circles in one side of square handkerchief.
∴ Side of the square = 3 × diameter of circle = 3 × 14 = 42 cm
Area of the square = 42 × 42 cm2 = 1764 cm2

Area of the circle = Ï€ r= 22/7 × 7 × 7 = 154 cm2
Total area of the design = 9 × 154 = 1386 cm2

Area of the remaining portion of the handkerchief = Area of the square - Total area of the design
= 1764 - 1386 = 378 cm2

12. In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the

(i) Area of quadrant OACB = (Ï€R2)/4 cm2
= (22/7 × 7/2 × 7/2)/4 cm2
= 77/8 cm2
(ii) Area of triangle BOD = 1/2 × 7/2 × 2 cm2
= 7/2 cm2
Area of shaded region = Area of quadrant - Area of triangle BOD
= (77/8 - 7/2) cm= 49/8 cm
= 6.125 cm2

13. In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use Ï€ = 3.14)

Side of square = OA = AB = 20 cm
OAB is right angled triangle
By Pythagoras theorem in Î”OAB ,
OB= AB2  + OA2
⇒ OB= 202  + 202
⇒ OB= 400  + 400
⇒ OB2 = 800
⇒ OB = 20√2 cm
Area of the quadrant = (Ï€R2)/4 cm= 3.14/4 × (20√2)cm= 628 cm2
Area of the square = 20 × 20 = 400  cm2

Area of the shaded region = Area of the quadrant - Area of the square
= 628 - 400 cm= 228 cm2

14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig. 12.32). If ∠AOB = 30°, find the area of the shaded region.

Radius of the larger circle, R = 21 cm
Radius of the smaller circle, r = 7 cm
Angle made by sectors of both concentric circles =  30°
Area of the larger sector = (30°/360°) × Ï€ Rcm2
= 1/12 × 22/7 × 21cm2
= 231/2 cm2
Area of the smaller circle = (30°/360°) × Ï€ rcm2
= 1/12 × 22/7 × 7cm2
= 77/6 cm2

Area of the shaded region = 231/2 - 77/6 cm2
= 616/6 cm2 = 308/3 cm2

15. In Fig. 12.33, 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.

AB = AC = 14 cm
BC is diameter of semicircle.
ABC is right angled triangle.
By Pythagoras theorem in Î”ABC,
BC= AB2  + AC2
⇒ BC= 142  + 142
⇒ BC = 14√2 cm
Radius of semicircle = 14√2/2 cm = 7√2 cm

Area of Î”ABC = 1/2 × 14 × 14 = 98 cm2
Area of quadrant = 1/4 × 22/7 × 14 × 14 = 154 cm2
Area of the semicircle = 1/2 × 22/7 × 7√2 × 7√2 = 154 cm2

Area of the shaded region = Area of the semicircle + Area of Î”ABC - Area of quadrant
= 154 + 98 - 154 cm= 98 cm2

Page No: 238

16. Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each.

AB = BC = CD = AD = 8 cm
Area of Î”ABC = Area of Î”ADC = 1/2 × 8 × 8 = 32 cm2
Area of quadrant AECB = Area of quadrant AFCD = 1/4 × 22/7 × 82
= 352/7 cm2

= (352/7 - 32) + (352/7 -32) cm2
= 2 × (352/7 -32) cm2
=  256/7 cm2

Go Back to NCERT Solutions for Class 10 Maths

### NCERT Solutions of Class 10 Maths Ch 12 Areas Related to Circles

Chapter 12 Areas related to Circles is an important chapter that will help you in fetching more marks in the examinations. There are total 4 topics in the chapter which are important for improving your knowledge and for board exam purpose.

• Introduction: In our daily lives, we come across things that are circular in shape such as wheel of bicycles, cake and various others. Through this chapter, we will get to know about perimeter and area of circle and and will also learn terms like segment and sector.

• Perimeter and Area of a Circle — A Review: The distance covered by travelling once around a circle is its perimeter which is usually called its circumference. In the exercise, we are focused on finding radius and area of circles.

• Areas of Sector and Segment of a Circle: The portion (or part) of the circular region enclosed by two radii and the corresponding arc is called a sector of the circle and the portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle. We will be finding the areas of sector of circle.

• Areas of Combinations of Plane Figures: We are focused on finding areas of sector of circles with plane figures which is based on daily lives things like window designs, designs on table covers etc.

Other than NCERT Solutions, you can also find our study materials for Class 10:

There are total 3 exercises in the chapter 12 Class 10 Maths which will help you in knowing application of formulas. Also, you can find exercisewise NCERT Solutions for Class 10 Maths Chapter 12 Areas related to Circles.

NCERT Solutions for Class 10 Maths Chapters:

 Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations 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 13 Surface Areas And Volumes Chapter 14 Statistics Chapter 15 Probability

FAQ on Chapter 12 Areas related to Circles

#### How many exercises in Chapter 12 Areas related to Circles

There are 3 exercise in the Chapter 12 Areas related to Circles which will encourage students to learn topics. These NCERT Solutions are prepared by Studyrankers experts who have vast experience in teaching students.

#### What is Area of a circle?

Area of a circle = Ï€r2

#### What is Perimeter of semicircle?

Perimeter of a semicircle or protractor = Ï€r+2r

#### What is Circumference of a circle or Perimeter of a circle?

The distance around the circle or the length of a circle is called its circumference or perimeter.
Circumference (perimeter) of a circle = Ï€d or 2Ï€r, where r is the radius of the circle and Ï€=22/7 or 3.14.