{"id":119275,"date":"2022-05-04T12:56:07","date_gmt":"2022-05-04T07:26:07","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=119275"},"modified":"2022-05-04T12:56:07","modified_gmt":"2022-05-04T07:26:07","slug":"chapter-12-areas-related-to-circles-questions-and-answers-ncert-solutions-for-class-10-maths","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-12-areas-related-to-circles-questions-and-answers-ncert-solutions-for-class-10-maths","title":{"rendered":"Chapter 12 – Areas Related to Circles Questions and Answers: NCERT Solutions for Class 10 Maths"},"content":{"rendered":"

Exercise: 12.1<\/h2>\n

1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.<\/h2>\n

Solution:
\nThe radius of the 1stcircle = 19 cm (given)
\n\u2234 Circumference of the 1stcircle = 2\u03c0\u00d719 = 38\u03c0 cm
\nThe radius of the 2ndcircle = 9 cm (given)
\n\u2234 Circumference of the 2ndcircle = 2\u03c0\u00d79 = 18\u03c0 cm
\nSo,
\nThe sum of the circumference of two circles = 38\u03c0+18\u03c0 = 56\u03c0 cm
\nNow, let the radius of the 3rdcircle = R
\n\u2234 The circumference of the 3rdcircle = 2\u03c0R
\nIt is given that sum of the circumference of two circles = circumference of the 3rdcircle
\nHence, 56\u03c0 = 2\u03c0R
\nOr, R = 28 cm.<\/h3>\n

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.<\/h2>\n

Solution:
\nRadius of 1stcircle = 8 cm (given)
\n\u2234 Area of 1stcircle = \u03c0(8)2= 64\u03c0
\nRadius of 2ndcircle = 6 cm (given)
\n\u2234 Area of 2ndcircle = \u03c0(6)2= 36\u03c0
\nSo,
\nThe sum of 1stand 2ndcircle will be = 64\u03c0+36\u03c0 = 100\u03c0
\nNow, assume that the radius of 3rdcircle = R
\n\u2234 Area of the circle 3rdcircle = \u03c0R2
\nIt is given that the area of the circle 3rdcircle = Area of 1stcircle + Area of 2ndcircle
\nOr, \u03c0R2= 100\u03c0cm2
\nR2= 100cm2
\nSo, R = 10cm<\/h3>\n

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.<\/h2>\n

\"\"<\/p>\n

Solution:
\nThe radius of 1stcircle, r1= 21\/2 cm (as diameter D is given as 21 cm)
\nSo, area of gold region = \u03c0 r12= \u03c0(10.5)2= 346.5 cm2
\nNow, it is given that each of the other bands is 10.5 cm wide,
\nSo, the radius of 2ndcircle, r2= 10.5cm+10.5cm = 21 cm
\nThus,
\n\u2234 Area of red region = Area of 2ndcircle \u2212 Area of gold region = (\u03c0r22\u2212346.5) cm2
\n= (\u03c0(21)2\u2212 346.5) cm2
\n= 1386 \u2212 346.5
\n= 1039.5 cm2
\nSimilarly,
\nThe radius of 3rdcircle, r3= 21 cm+10.5 cm = 31.5 cm
\nThe radius of 4thcircle, r4= 31.5 cm+10.5 cm = 42 cm
\nThe Radius of 5thcircle, r5= 42 cm+10.5 cm = 52.5 cm
\nFor the area of nthregion,
\nA = Area of circle n \u2013 Area of circle (n-1)
\n\u2234 Area of blue region (n=3) = Area of third circle \u2013 Area of second circle
\n= \u03c0(31.5)2\u2013 1386 cm2
\n= 3118.5 \u2013 1386 cm2
\n= 1732.5 cm2
\n\u2234 Area of black region (n=4) = Area of fourth circle \u2013 Area of third circle
\n= \u03c0(42)2\u2013 1386 cm2
\n= 5544 \u2013 3118.5 cm2
\n= 2425.5 cm2
\n\u2234 Area of white region (n=5) = Area of fifth circle \u2013 Area of fourth circle
\n= \u03c0(52.5)2\u2013 5544 cm2
\n= 8662.5 \u2013 5544 cm2
\n= 3118.5 cm2<\/h3>\n

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?<\/h2>\n

Solution:
\nThe radius of car\u2019s wheel = 80\/2 = 40 cm (as D = 80 cm)
\nSo, the circumference of wheels = 2\u03c0r = 80 \u03c0 cm
\nNow, in one revolution, the distance covered = circumference of the wheel = 80 \u03c0 cm
\nIt is given that the distance covered by the car in 1 hr = 66km
\nConverting km into cm we get,
\nDistance covered by the car in 1hr = (66\u00d7105) cm
\nIn 10 minutes, the distance covered will be = (66\u00d7105\u00d710)\/60 = 1100000 cm\/s
\n\u2234 Distance covered by car = 11\u00d7105cm
\nNow, the no. of revolutions of the wheels = (Distance covered by the car\/Circumference of the wheels)
\n=( 11\u00d7105)\/80 \u03c0 = 4375.<\/h3>\n

5. Tick the correct Solution: 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
\n(A) 2 units
\n(B) \u03c0 units
\n(C) 4 units
\n(D) 7 units<\/h2>\n

Solution:
\nSince the perimeter of the circle = area of the circle,
\n2\u03c0r = \u03c0r2
\nOr, r = 2
\nSo, option (A) is correct i.e. the radius of the circle is 2 units.<\/h3>\n

Exercise: 12.2<\/h2>\n

1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60\u00b0.<\/h2>\n

Solution:
\nIt is given that the angle of the sector is 60\u00b0
\nWe know that the area of sector = (\u03b8\/360\u00b0)\u00d7\u03c0r2
\n\u2234 Area of the sector with angle 60\u00b0 = (60\u00b0\/360\u00b0)\u00d7\u03c0r2cm2
\n= (36\/6)\u03c0 cm2
\n= 6\u00d722\/7 cm2= 132\/7 cm2<\/h3>\n

2. Find the area of a quadrant of a circle whose circumference is 22 cm.<\/h2>\n

Solution:
\nCircumference of the circle, C = 22 cm (given)
\nIt should be noted that a quadrant of a circle is a sector which is making an angle of 90\u00b0.
\nLet the radius of the circle = r
\nAs C = 2\u03c0r = 22,
\nR = 22\/2\u03c0 cm = 7\/2 cm
\n\u2234 Area of the quadrant = (\u03b8\/360\u00b0) \u00d7 \u03c0r2
\nHere, \u03b8 = 90\u00b0
\nSo, A = (90\u00b0\/360\u00b0) \u00d7 \u03c0 r2cm2
\n= (49\/16) \u03c0 cm2
\n= 77\/8 cm2= 9.6 cm2<\/h3>\n

3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.<\/h2>\n

Solution:
\nLength of minute hand = radius of the clock (circle)
\n\u2234 Radius (r) of the circle = 14 cm (given)
\nAngle swept by minute hand in 60 minutes = 360\u00b0
\nSo, the angle swept by the minute hand in 5 minutes = 360\u00b0 \u00d7 5\/60 = 30\u00b0
\nWe know,
\nArea of a sector = (\u03b8\/360\u00b0) \u00d7 \u03c0r2
\nNow, area of the sector making an angle of 30\u00b0 = (30\u00b0\/360\u00b0) \u00d7 \u03c0r2cm2
\n= (1\/12) \u00d7 \u03c0142
\n= (49\/3)\u00d7(22\/7) cm2
\n= 154\/3 cm2<\/h3>\n

4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:
\n(i) minor segment
\n(ii) major sector. (Use \u03c0 = 3.14)<\/h2>\n

Solution:<\/h3>\n

\"\"<\/p>\n

Here AB be the chord which is subtending an angle 90\u00b0 at the center O.
\nIt is given that the radius (r) of the circle = 10 cm
\n(i)Area of minor sector = (90\/360\u00b0)\u00d7\u03c0r2
\n= (\u00bc)\u00d7(22\/7)\u00d7102
\nOr, Area of minor sector = 78.5 cm2
\nAlso, area of \u0394AOB = \u00bd\u00d7OB\u00d7OA
\nHere, OB and OA are the radii of the circle i.e. = 10 cm
\nSo, area of \u0394AOB = \u00bd\u00d710\u00d710
\n= 50 cm2
\nNow, area of minor segment = area of minor sector \u2013 area of \u0394AOB
\n= 78.5 \u2013 50
\n= 28.5 cm2
\n(ii)Area of major sector = Area of circle \u2013 Area of minor sector
\n= (3.14\u00d7102)-78.5
\n= 235.5 cm2<\/h3>\n

5. In a circle of radius 21 cm, an arc subtends an angle of 60\u00b0 at the centre. Find:
\n(i) the length of the arc
\n(ii) area of the sector formed by the arc
\n(iii) area of the segment formed by the corresponding chord<\/h2>\n

Solution:<\/h3>\n

\"\"<\/p>\n

Given,
\nRadius = 21 cm
\n\u03b8 = 60\u00b0
\n(i)Length of an arc = \u03b8\/360\u00b0\u00d7Circumference(2\u03c0r)
\n\u2234 Length of an arc AB = (60\u00b0\/360\u00b0)\u00d72\u00d7(22\/7)\u00d721
\n= (1\/6)\u00d72\u00d7(22\/7)\u00d721
\nOr Arc AB Length = 22cm
\n(ii)It is given that the angle subtend by the arc = 60\u00b0
\nSo, area of the sector making an angle of 60\u00b0 = (60\u00b0\/360\u00b0)\u00d7\u03c0 r2cm2
\n= 441\/6\u00d722\/7 cm2
\nOr, the area of the sector formed by the arc APB is 231 cm2
\n(iii)Area of segment APB = Area of sector OAPB \u2013 Area of \u0394OAB
\nSince the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60\u00b0, \u0394OAB is an equilateral triangle. So, its area will be \u221a3\/4\u00d7a2sq. Units.
\nArea of segment APB = 231-(\u221a3\/4)\u00d7(OA)2
\n= 231-(\u221a3\/4)\u00d7212
\nOr, Area of segment APB = [231-(441\u00d7\u221a3)\/4] cm2<\/h3>\n

6. A chord of a circle of radius 15 cm subtends an angle of 60\u00b0 at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use \u03c0 = 3.14 and \u221a3 = 1.73)<\/h2>\n

Solution:<\/h3>\n

\"\"<\/p>\n

Given,
\nRadius = 15 cm
\n\u03b8 = 60\u00b0
\nSo,
\nArea of sector OAPB = (60\u00b0\/360\u00b0)\u00d7\u03c0r2cm2
\n= 225\/6 \u03c0cm2
\nNow, \u0394AOB is equilateral as two sides are the radii of the circle and hence equal and one angle is 60\u00b0
\nSo, Area of \u0394AOB = (\u221a3\/4) \u00d7a2
\nOr, (\u221a3\/4) \u00d7152
\n\u2234 Area of \u0394AOB = 97.31 cm2
\nNow, area of minor segment APB = Area of OAPB \u2013 Area of \u0394AOB
\nOr, area of minor segment APB = ((225\/6)\u03c0 \u2013 97.31) cm2= 20.43 cm2
\nAnd,
\nArea of major segment = Area of circle \u2013 Area of segment APB
\nOr, area of major segment = (\u03c0\u00d7152) \u2013 20.4 = 686.06 cm2<\/h3>\n

7. A chord of a circle of radius 12 cm subtends an angle of 120\u00b0 at the centre. Find the area of the corresponding segment of the circle. (Use \u03c0 = 3.14 and \u221a3 = 1.73)<\/h2>\n

Solution:
\nRadius, r = 12 cm
\nNow, draw a perpendicular OD on chord AB and it will bisect chord AB.
\nSo, AD = DB<\/h3>\n

\"\"<\/p>\n

Now, the area of the minor sector = (\u03b8\/360\u00b0)\u00d7\u03c0r2
\n= (120\/360)\u00d7(22\/7)\u00d7122
\n= 150.72 cm2
\nConsider the \u0394AOB,
\n\u2220 OAB = 180\u00b0-(90\u00b0+60\u00b0) = 30\u00b0
\nNow, cos 30\u00b0 = AD\/OA
\n\u221a3\/2 = AD\/12
\nOr, AD = 6\u221a3 cm
\nWe know OD bisects AB. So,
\nAB = 2\u00d7AD = 12\u221a3 cm
\nNow, sin 30\u00b0 = OD\/OA
\nOr, \u00bd = OD\/12
\n\u2234 OD = 6 cm
\nSo, the area of \u0394AOB = \u00bd \u00d7 base \u00d7 height
\nHere, base = AB = 12\u221a3 and
\nHeight = OD = 6
\nSo, area of \u0394AOB = \u00bd\u00d712\u221a3\u00d76 = 36\u221a3 cm = 62.28 cm2
\n\u2234 Area of the corresponding Minor segment = Area of the Minor sector \u2013 Area of \u0394AOB
\n= 150.72 cm2\u2013 62.28 cm2= 88.44 cm2<\/h3>\n

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
\n(i) the area of that part of the field in which the horse can graze.
\n(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use \u03c0 = 3.14)<\/h2>\n

\"\"<\/p>\n

Solution:
\nAs the horse is tied at one end of a square field, it will graze only a quarter (i.e. sector with \u03b8 = 90\u00b0) of the field with radius 5 m.
\nHere, the length of rope will be the radius of the circle i.e. r = 5 m
\nIt is also known that the side of square field = 15 m
\n(i)Area of circle = \u03c0r2= 22\/7 \u00d7 52= 78.5 m2
\nNow, the area of the part of the field where the horse can graze = \u00bc (the area of the circle) = 78.5\/4 = 19.625 m2
\n(ii)If the rope is increased to 10 m,
\nArea of circle will be = \u03c0r2=22\/7\u00d7102= 314 m2
\nNow, the area of the part of the field where the horse can graze = \u00bc (the area of the circle)
\n= 314\/4 = 78.5 m2
\n\u2234 Increase in the grazing area = 78.5 m2\u2013 19.625 m2= 58.875 m2<\/h3>\n

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:
\n(i) the total length of the silver wire required.
\n(ii) the area of each sector of the brooch.<\/h2>\n

\"\"<\/p>\n

Solution:
\nDiameter (D) = 35 mm
\nTotal number of diameters to be considered= 5
\nNow, the total length of 5 diameters that would be required = 35\u00d75 = 175
\nCircumference of the circle = 2\u03c0r
\nOr, C = \u03c0D = 22\/7\u00d735 = 110
\nArea of the circle = \u03c0r2
\nOr, A = (22\/7)\u00d7(35\/2)2= 1925\/2 mm2
\n(i)Total length of silver wire required = Circumference of the circle + Length of 5 diameter
\n= 110+175 = 285 mm
\n(ii)Total Number of sectors in the brooch = 10
\nSo, the area of each sector = total area of the circle\/number of sectors
\n\u2234 Area of each sector = (1925\/2)\u00d71\/10 = 385\/4 mm2<\/h3>\n

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.<\/h2>\n

\"\"<\/p>\n

Solution:
\nThe radius (r) of the umbrella when flat = 45 cm
\nSo, the area of the circle (A) = \u03c0r2= (22\/7)\u00d7(45)2=6364.29 cm2
\nTotal number of ribs (n) = 8
\n\u2234 The area between the two consecutive ribs of the umbrella = A\/n
\n6364.29\/8 cm2
\nOr, The area between the two consecutive ribs of the umbrella = 795.5 cm2<\/h3>\n

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\u00b0. Find the total area cleaned at each sweep of the blades.<\/h2>\n

Solution:
\nGiven,
\nRadius (r) = 25 cm
\nSector angle (\u03b8) = 115\u00b0
\nSince there are 2 blades,
\nThe total area of the sector made by wiper = 2\u00d7(\u03b8\/360\u00b0)\u00d7\u03c0 r2
\n= 2\u00d7(115\/360)\u00d7(22\/7)\u00d7252
\n= 2\u00d7158125\/252 cm2
\n= 158125\/126 = 1254.96 cm2<\/h3>\n

12. To warn ships for underwater rocks, a lighthouse spreads a red colored light over a sector of angle 80\u00b0 to a distance of 16.5 km. Find the area of the sea over which the ships are warned.
\n(Use \u03c0 = 3.14)<\/h2>\n

Solution:
\nLet O bet the position of Lighthouse.<\/h3>\n

\"\"<\/p>\n

Here the radius will be the distance over which light spreads.
\nGiven, radius (r) = 16.5 km
\nSector angle (\u03b8) = 80\u00b0
\nNow, the total area of the sea over which the ships are warned = Area made by the sector
\nOr, Area of sector = (\u03b8\/360\u00b0)\u00d7\u03c0r2
\n= (80\u00b0\/360\u00b0)\u00d7\u03c0r2km2
\n= 189.97 km2<\/h3>\n

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 \u20b9 0.35 per cm2. (Use \u221a3 = 1.7)<\/h2>\n

\"\"<\/p>\n

Solution:<\/h3>\n

\"\"<\/p>\n

Total number of equal designs = 6
\nAOB= 360\u00b0\/6 = 60\u00b0
\nRadius of the cover = 28 cm
\nCost of making design = \u20b9 0.35 per cm2
\nSince the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60\u00b0, \u0394AOB is an equilateral triangle. So, its area will be (\u221a3\/4)\u00d7a2sq. units
\nHere, a = OA
\n\u2234 Area of equilateral \u0394AOB = (\u221a3\/4)\u00d7282= 333.2 cm2
\nArea of sector ACB = (60\u00b0\/360\u00b0)\u00d7\u03c0r2cm2
\n= 410.66 cm2
\nSo, area of a single design = area of sector ACB \u2013 area of \u0394AOB
\n= 410.66 cm2\u2013 333.2 cm2= 77.46 cm2
\n\u2234 Area of 6 designs = 6\u00d777.46 cm2= 464.76 cm2
\nSo, total cost of making design = 464.76 cm2\u00d7Rs.0.35 per cm2
\n= Rs. 162.66<\/h3>\n

14. Tick the correct solution in the following:
\nArea of a sector of angle p (in degrees) of a circle with radius R is
\n(A) p\/180 \u00d7 2\u03c0R
\n(B) p\/180 \u00d7 \u03c0 R2
\n(C) p\/360 \u00d7 2\u03c0R
\n(D) p\/720 \u00d7 2\u03c0R2<\/h2>\n

Solution:
\nThe area of a sector = (\u03b8\/360\u00b0)\u00d7\u03c0r2
\nGiven, \u03b8 = p
\nSo, area of sector = p\/360\u00d7\u03c0R2
\nMultiplying and dividing by 2 simultaneously,
\n= (p\/360)\u00d72\/2\u00d7\u03c0R2
\n= (2p\/720)\u00d72\u03c0R2
\nSo, option (D) is correct.<\/h3>\n

Exercise: 12.3<\/h2>\n

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.<\/h2>\n

\"\"<\/p>\n

Solution:
\nHere, P is in the semi-circle and so,
\nP = 90\u00b0
\nSo, it can be concluded that QR is hypotenuse of the circle and is equal to the diameter of the circle.
\n\u2234 QR = D
\nUsing Pythagorean theorem,
\nQR2= PR2+PQ2
\nOr, QR2= 72+242
\nQR= 25 cm = Diameter
\nHence, the radius of the circle = 25\/2 cm
\nNow, the area of the semicircle = (\u03c0R2)\/2
\n= (22\/7)\u00d7(25\/2)\u00d7(25\/2)\/2 cm2
\n= 13750\/56 cm2= 245.54 cm2
\nAlso, area of the \u0394PQR = \u00bd\u00d7PR\u00d7PQ
\n=(\u00bd)\u00d77\u00d724 cm2
\n= 84 cm2
\nHence, the area of the shaded region = 245.54 cm2-84 cm2
\n= 161.54 cm2<\/h3>\n

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\u00b0.<\/h2>\n

Solution:<\/h3>\n

\"\"<\/p>\n

Given,
\nAngle made by sector = 40\u00b0,
\nRadius the inner circle = r = 7 cm, and
\nRadius of the outer circle = R = 14 cm
\nWe know,
\nArea of the sector = (\u03b8\/360\u00b0)\u00d7\u03c0r2
\nSo, Area of OAC = (40\u00b0\/360\u00b0)\u00d7\u03c0r2cm2
\n= 68.44 cm2
\nArea of the sector OBD = (40\u00b0\/360\u00b0)\u00d7\u03c0r2cm2
\n= (1\/9)\u00d7(22\/7)\u00d772= 17.11 cm2
\nNow, area of the shaded region ABDC = Area of OAC \u2013 Area of the OBD
\n= 68.44 cm2\u2013 17.11 cm2= 51.33 cm2<\/h3>\n

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.<\/h2>\n

\"\"<\/p>\n

Solution:
\nSide of the square ABCD (as given) = 14 cm
\nSo, Area of ABCD = a2
\n= 14\u00d714 cm2= 196 cm2
\nWe know that the side of the square = diameter of the circle = 14 cm
\nSo, side of the square = diameter of the semicircle = 14 cm
\n\u2234 Radius of the semicircle = 7 cm
\nNow, area of the semicircle = (\u03c0R2)\/2
\n= (22\/7\u00d77\u00d77)\/2 cm2
\n= 77 cm2
\n\u2234 Area of two semicircles = 2\u00d777 cm2= 154 cm2
\nHence, area of the shaded region = Area of the Square \u2013 Area of two semicircles
\n= 196 cm2-154 cm2
\n= 42 cm2<\/h3>\n

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.<\/h2>\n

\"\"<\/p>\n

Solution:
\nIt is given that OAB is an equilateral triangle having each angle as 60\u00b0
\nArea of the sector is common in both.
\nRadius of the circle = 6 cm.
\nSide of the triangle = 12 cm.
\nArea of the equilateral triangle = (\u221a3\/4) (OA)2= (\u221a3\/4)\u00d7122= 36\u221a3 cm2
\nArea of the circle = \u03c0R2= (22\/7)\u00d762= 792\/7 cm2
\nArea of the sector making angle 60\u00b0 = (60\u00b0\/360\u00b0) \u00d7\u03c0r2cm2
\n= (1\/6)\u00d7(22\/7)\u00d7 62cm2= 132\/7 cm2
\nArea of the shaded region = Area of the equilateral triangle + Area of the circle \u2013 Area of the sector
\n= 36\u221a3 cm2+792\/7 cm2-132\/7 cm2
\n= (36\u221a3+660\/7) cm2<\/h3>\n

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.<\/h2>\n

\"\"<\/p>\n

Solution:
\nSide of the square = 4 cm
\nRadius of the circle = 1 cm
\nFour quadrant of a circle are cut from corner and one circle of radius are cut from middle.
\nArea of square = (side)2= 42= 16 cm2
\nArea of the quadrant = (\u03c0R2)\/4 cm2= (22\/7)\u00d7(12)\/4 = 11\/14 cm2
\n\u2234 Total area of the 4 quadrants = 4 \u00d7(11\/14) cm2= 22\/7 cm2
\nArea of the circle = \u03c0R2cm2= (22\/7\u00d712) = 22\/7 cm2
\nArea of the shaded region = Area of square \u2013 (Area of the 4 quadrants + Area of the circle)
\n= 16 cm2-(22\/7) cm2\u2013 (22\/7) cm2
\n= 68\/7 cm2<\/h3>\n

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.<\/h2>\n

\"\"<\/p>\n

Solution:
\nRadius of the circle = 32 cm
\nDraw a median AD of the triangle passing through the centre of the circle.
\n\u21d2 BD = AB\/2
\nSince, AD is the median of the triangle
\n\u2234 AO = Radius of the circle = (2\/3) AD
\n\u21d2 (2\/3)AD = 32 cm
\n\u21d2 AD = 48 cm
\nIn \u0394ADB,<\/h3>\n

\"\"<\/p>\n

By Pythagoras theorem,
\nAB2= AD2+BD2
\n\u21d2 AB2= 482+(AB\/2)2
\n\u21d2 AB2= 2304+AB2\/4
\n\u21d2 3\/4 (AB2)= 2304
\n\u21d2 AB2= 3072
\n\u21d2 AB= 32\u221a3 cm
\nArea of \u0394ADB = \u221a3\/4 \u00d7(32\u221a3)2cm2= 768\u221a3 cm2
\nArea of circle = \u03c0R2= (22\/7)\u00d732\u00d732 = 22528\/7 cm2
\nArea of the design = Area of circle \u2013 Area of \u0394ADB
\n= (22528\/7 \u2013 768\u221a3) cm2<\/h3>\n

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.<\/h2>\n

\"\"<\/p>\n

Solution:
\nSide of square = 14 cm
\nFour quadrants are included in the four sides of the square.
\n\u2234 Radius of the circles = 14\/2 cm = 7 cm
\nArea of the square ABCD = 142= 196 cm2
\nArea of the quadrant = (\u03c0R2)\/4 cm2= (22\/7) \u00d772\/4 cm2
\n= 77\/2 cm2
\nTotal area of the quadrant = 4\u00d777\/2 cm2= 154cm2
\nArea of the shaded region = Area of the square ABCD \u2013 Area of the quadrant
\n= 196 cm2\u2013 154 cm2
\n= 42 cm2<\/h3>\n

8. Fig. 12.26 depicts a racing track whose left and right ends are semicircular.
\nThe 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:
\n(i) the distance around the track along its inner edge
\n(ii) the area of the track.<\/h2>\n

\"\"<\/p>\n

Solution:
\nWidth of the track = 10 m
\nDistance between two parallel lines = 60 m
\nLength of parallel tracks = 106 m<\/h3>\n

\"\"<\/p>\n

DE = CF = 60 m
\nRadius of inner semicircle, r = OD = O\u2019C
\n= 60\/2 m = 30 m
\nRadius of outer semicircle, R = OA = O\u2019B
\n= 30+10 m = 40 m
\nAlso, AB = CD = EF = GH = 106 m
\nDistance around the track along its inner edge = CD+EF+2\u00d7(Circumference of inner semicircle)
\n= 106+106+(2\u00d7\u03c0r) m = 212+(2\u00d722\/7\u00d730) m
\n= 212+1320\/7 m = 2804\/7 m
\nArea of the track = Area of ABCD + Area EFGH + 2 \u00d7 (area of outer semicircle) \u2013 2 \u00d7 (area of inner semicircle)
\n= (AB\u00d7CD)+(EF\u00d7GH)+2\u00d7(\u03c0r2\/2) -2\u00d7(\u03c0R2\/2) m2
\n= (106\u00d710)+(106\u00d710)+2\u00d7\u03c0\/2(r2-R2) m2
\n= 2120+22\/7\u00d770\u00d710 m2
\n= 4320 m2<\/h3>\n

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.<\/h2>\n

\"\"<\/p>\n

Solution:
\nRadius of larger circle, R = 7 cm
\nRadius of smaller circle, r = 7\/2 cm
\nHeight of \u0394BCA = OC = 7 cm
\nBase of \u0394BCA = AB = 14 cm
\nArea of \u0394BCA = 1\/2 \u00d7 AB \u00d7 OC = (\u00bd)\u00d77\u00d714 = 49 cm2
\nArea of larger circle = \u03c0R2= (22\/7)\u00d772= 154 cm2
\nArea of larger semicircle = 154\/2 cm2= 77 cm2
\nArea of smaller circle = \u03c0r2= (22\/7)\u00d7(7\/2)\u00d7(7\/2) = 77\/2 cm2
\nArea of the shaded region = Area of larger circle \u2013 Area of triangle \u2013 Area of larger semicircle + Area of smaller circle
\nArea of the shaded region = (154-49-77+77\/2) cm2
\n= 133\/2 cm2= 66.5 cm2<\/h3>\n

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 \u03c0 = 3.14 and \u221a3 = 1.73205)<\/h2>\n

\"\"<\/p>\n

Solution:
\nABC is an equilateral triangle.
\n\u2234 \u2220 A = \u2220 B = \u2220 C = 60\u00b0
\nThere are three sectors each making 60\u00b0.
\nArea of \u0394ABC = 17320.5 cm2
\n\u21d2 \u221a3\/4 \u00d7(side)2= 17320.5
\n\u21d2 (side)2=17320.5\u00d74\/1.73205
\n\u21d2 (side)2= 4\u00d7104
\n\u21d2 side = 200 cm
\nRadius of the circles = 200\/2 cm = 100 cm
\nArea of the sector = (60\u00b0\/360\u00b0)\u00d7\u03c0 r2cm2
\n= 1\/6\u00d73.14\u00d7(100)2cm2
\n= 15700\/3cm2
\nArea of 3 sectors = 3\u00d715700\/3 = 15700 cm2
\nThus, area of the shaded region = Area of equilateral triangle ABC \u2013 Area of 3 sectors
\n= 17320.5-15700 cm2= 1620.5 cm2<\/h3>\n

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.<\/h2>\n

\"\"<\/p>\n

Solution:
\nNumber of circular designs = 9
\nRadius of the circular design = 7 cm
\nThere are three circles in one side of square handkerchief.
\n\u2234 Side of the square = 3\u00d7diameter of circle = 3\u00d714 = 42 cm
\nArea of the square = 42\u00d742 cm2= 1764 cm2
\nArea of the circle = \u03c0 r2= (22\/7)\u00d77\u00d77 = 154 cm2
\nTotal area of the design = 9\u00d7154 = 1386 cm2
\nArea of the remaining portion of the handkerchief = Area of the square \u2013 Total area of the design = 1764 \u2013 1386 = 378 cm2<\/h3>\n

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
\n(i) quadrant OACB,
\n(ii) shaded region.<\/h2>\n

\"\"<\/p>\n

Solution:
\nRadius of the quadrant = 3.5 cm = 7\/2 cm
\n(i)Area of quadrant OACB = (\u03c0R2)\/4 cm2
\n= (22\/7)\u00d7(7\/2)\u00d7(7\/2)\/4 cm2
\n= 77\/8 cm2
\n(ii)Area of triangle BOD = (\u00bd)\u00d7(7\/2)\u00d72 cm2
\n= 7\/2 cm2
\nArea of shaded region = Area of quadrant \u2013 Area of triangle BOD
\n= (77\/8)-(7\/2) cm2= 49\/8 cm2
\n= 6.125 cm2<\/h3>\n

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 \u03c0 = 3.14)<\/h2>\n

\"\"<\/p>\n

Solution:
\nSide of square = OA = AB = 20 cm
\nRadius of the quadrant = OB
\nOAB is right angled triangle
\nBy Pythagoras theorem in \u0394OAB,
\nOB2= AB2+OA2
\n\u21d2 OB2= 202+202
\n\u21d2 OB2= 400+400
\n\u21d2 OB2= 800
\n\u21d2 OB= 20\u221a2 cm
\nArea of the quadrant = (\u03c0R2)\/4 cm2= (3.14\/4)\u00d7(20\u221a2)2cm2= 628cm2
\nArea of the square = 20\u00d720 = 400 cm2
\nArea of the shaded region = Area of the quadrant \u2013 Area of the square
\n= 628-400 cm2= 228cm2<\/h3>\n

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 \u2220AOB = 30\u00b0, find the area of the shaded region.<\/h2>\n

\"\"<\/p>\n

Solution:
\nRadius of the larger circle, R = 21 cm
\nRadius of the smaller circle, r = 7 cm
\nAngle made by sectors of both concentric circles = 30\u00b0
\nArea of the larger sector = (30\u00b0\/360\u00b0)\u00d7\u03c0R2cm2
\n= (1\/12)\u00d7(22\/7)\u00d7212cm2
\n= 231\/2cm2
\nArea of the smaller circle = (30\u00b0\/360\u00b0)\u00d7\u03c0r2cm2
\n= 1\/12\u00d722\/7\u00d772cm2
\n=77\/6 cm2
\nArea of the shaded region = (231\/2) \u2013 (77\/6) cm2
\n= 616\/6 cm2= 308\/3cm2<\/h3>\n

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.<\/h2>\n

\"\"<\/p>\n

Solution:
\nRadius of the quadrant ABC of circle = 14 cm
\nAB = AC = 14 cm
\nBC is diameter of semicircle.
\nABC is right angled triangle.
\nBy Pythagoras theorem in \u0394ABC,
\nBC2= AB2+AC2
\n\u21d2 BC2= 142+142
\n\u21d2 BC = 14\u221a2 cm
\nRadius of semicircle = 14\u221a2\/2 cm = 7\u221a2 cm
\nArea of \u0394ABC =( \u00bd)\u00d714\u00d714 = 98 cm2
\nArea of quadrant = (\u00bc)\u00d7(22\/7)\u00d7(14\u00d714) = 154 cm2
\nArea of the semicircle = (\u00bd)\u00d7(22\/7)\u00d77\u221a2\u00d77\u221a2 = 154 cm2
\nArea of the shaded region =Area of the semicircle + Area of \u0394ABC \u2013 Area of quadrant
\n= 154 +98-154 cm2= 98cm2<\/h3>\n

16. Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each.<\/h2>\n

Solution:<\/h3>\n

\"\"<\/p>\n

AB = BC = CD = AD = 8 cm
\nArea of \u0394ABC = Area of \u0394ADC = (\u00bd)\u00d78\u00d78 = 32 cm2
\nArea of quadrant AECB = Area of quadrant AFCD = (\u00bc)\u00d722\/7\u00d782
\n= 352\/7 cm2
\nArea of shaded region = (Area of quadrant AECB \u2013 Area of \u0394ABC) = (Area of quadrant AFCD \u2013 Area of \u0394ADC)
\n= (352\/7 -32)+(352\/7- 32) cm2
\n= 2\u00d7(352\/7-32) cm2
\n= 256\/7 cm2<\/h3>\n","protected":false},"excerpt":{"rendered":"

Class 10 Maths NCERT book solutions for Chapter 12 – Areas Related to Circles Questions and Answers.<\/p>\n","protected":false},"author":21830,"featured_media":117421,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":{"0":"post-119275","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-education"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119275","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=119275"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119275\/revisions"}],"predecessor-version":[{"id":119310,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/119275\/revisions\/119310"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/117421"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=119275"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=119275"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=119275"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}