## NCERT Solutions Class 10 Maths Chapter 12: Areas Related To Circles## Exercise 12.1
Let the radius of the required circle be x. Circumference of the first circle = 2πR = 2 × 22/7 × 19 = 836/7 cm Circumference of the second circle = 2πr = 2 × 22/7 × 9 = 396/7 cm Circumference of the third circle = 2πR + 2πr = 836/7 + 396/7 cm = 1232/7 cm 2πx = 1232/7 2 × 22x/7 = 1232/7 44x = 1232 x = 28 cm Hence, the radius of the required circle is 28 cm.
Let the radius of the required circle be x. Area of the first circle = πR = 22/7 × 64 = 1408/7 cm Area of the second circle = πr = 792/7 cm Area of the third circle = πR = 2200/7 cm πx 22(x 22x x x = 10 cm Hence, the radius of the required circle is 10 cm.
Area of Gold region = r = 22/7 × (10.5) = 22(15.75) =
Area of Red region = πr = 22/7 × 21 = 22(63) - 346.5 = 1386 - 346.5 =
Area of Blue region = πr = 22/7 × (31.5) = 22(141.75) - 1386 = 3118.5 - 1386 =
Area of Black region = πr = 22/7 × (42) = 22(252) - 3118.5 = 5544 - 3118.5 =
Area of White region = πr = 22/7 × (52.5) = 22(393.75) - 5544 = 8662.5 - 5544 =
Distance travelled by the car in an hour = 66 km = 66000 m Distance travelled by the car in a minute = 66000/60 = 1100 m Distance travelled by the car in 10 minutes = 1100 × 10 = 11000 m Radius of each wheel = 80/2 = 40 cm = 0.4 m Distance covered by each wheel in one revolution = 2πr = 2 × 22/7 × 0.4 = 17.6/7 m Revolutions required to cover 11000 m = 11000/(17.6/7) = 11000(7)/17.6 = 7000/1.6 = 4375 Hence, each car wheel makes 4375 complete revolutions in 10 minutes when the car is travelling at a speed of 66 km per hour.
Area of the circle = Circumference of the circle πr r r = 2 units Hence, (A) is the correct answer. ## Exercise 12.2
Angle of the sector of given circle = 60° Area of sector = (θ/360°) × πr = (60°/360°) × 22/7 × 6 = 1/6 × 22/7 × 6 = 22/7 × 6 = 132/7 cm
Circumference of the given circle = 22 cm 2πr = 22 2 × 22/7 × r = 22 2r = 7 r = 7/2 cm Area of quadrant of a circle = (1/4) × πr = 1/4 × 22/7 × (7/2) = 11/2 × 7/4 = 77/8 cm
Length of the minute hand of clock = r = 14 cm Area swept by the minute hand in 60 minutes = πr = 22/7 × 14 = 22 × 2 × 14 = 616 cm Angle covered by the minute hand in 60 minutes = 360° Angle covered by the minute hand in 5 minutes = 360°/60° × 5 = 30° Area swept by minute hand of the clock in 5 minutes = (θ/360°) × πr = (30°/360°) × 616 = 1/12 × 616 = 154/3 cm
Radius of the circle = 10 cm Area of the circle = πr = 3.14 × 10
= (90°/360°) × 314 = 1/4 × 314 = 76 cm
= (270°/360°) × 314 = 3/4 × 314 = 228 cm
- the length of the arc
- area of the sector formed by the arc
- area of the segment formed by the corresponding chord
= 2 × 22 × 3 = 132 cm Length of the arc = (θ/360°) × 2πr = (60°/360°) × 132 = 1/6 × 132 = 22 cm
= 22/7 × 21 = 22 × 3 × 21 = 1386 cm Area of the sector formed by the arc = (θ/360°) × πr = 1/6 × 1386 = 231 cm
= 231 - √3/4 × r = 231 - √3/4 × 21 = 231 - 441√3/4 = (924 - 441√3)/4 cm
Area of the circle = πr = 3.14 × 15 = 3.14 × 225 = 706.5 cm Area of the sector = (θ/360°) × πr = (60°/360°) × 706.5 = 1/6 × 706.5 = 706.5/6 cm Area of the minor segment = Area of sector - Area of triangle AOB Since, two sides are equal (both radii of same circle) and their angle is 60° therefore, AOB is an equilateral triangle. Area of the minor segment = 706.5/6 - √3/4 × 15 = 706.5/6 - 1.73/4 × 225 = 706.5/6 - 97.31 = 122.64/6 = Area of major segment = Area of circle - Area of minor segment = 706.5 - 20.44 =
Area of the circle = πr = π × 12 = 144π cm Area of the sector = (θ/360°) × πr = (120°/360°) × 144π = 1/3 × 144π = 48π cm Area of the segment = Area of sector - Area of triangle AOB Draw OD ⊥ AB. AOB has two sides equal (both are radii of same circle) so it is an isosceles triangle. Therefore, the perpendicular to the third side will act as the angle bisector for ∠O. In AOB, we have: sin O = AD/OA sin 60° = AD/12 √3/2 = AD/12 6√3 cm = AD Similarly, BD = 6√3 cm. AB = 12√3 cm. Also, cos O = OD/OA (In AOB) cos 60° = OD/12 1/2 = OD/12 6 cm = OD Area of AOB = 1/2 × 12√3 × 6 = 36√3 cm Area of the segment = 48π - 36√3 = 48(3.14) - 36(1.73) =150.72 - 62.28
Hence, area of the corresponding segment will be 88.44 cm
- the area of that part of the field in which the horse can graze.
- the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)
The area in which the horse can graze is a quadrant of a circle with radius equal to the length of the rope.
= (3.14)5 = (3.14)25 = 78.5 m Area of the field in which the horse can graze = 1/4 × πr = 1/4 × 78.5
= (3.14)10 = (3.14)100 = 314 m Area of the field which the horse can graze = 1/4 × πR = 1/4 × 314
Increase in the grazing area when rope length is increased to 10 m = πR = 78.5 - 19.625
- the total length of the silver wire required.
- the area of each sector of the brooch.
Length of wire used for making the circle = 2πr = 2 × 22/7 × 35/20 = 2 × 22/7 × 1.75 = 44 × 0.25 = 11 cm Total length of wire used in the brooch = 11 cm.
= 22/7 × (1.75) = 22 × 0.25 × 1.75 = 9.625 cm Area of each sector = 1/10 × πr = 1/10 × 9.625 = 0.9625 cm Hence, area of the each of the sectors is 96.25 mm
Assuming that the umbrella is a flat circle, the ribs divide it into 8 sectors of equal area. Area of a sector will be the area between two consecutive ribs. Area of the circle = πr = 22/7 × 45 = 22/7 × 2025 = 6364.29 cm Area of each sector = 1/8 × πr = 1/8 × 6364.29 = 795.5 cm Hence, the area between two consecutive ribs of the umbrella is 795.5 cm
Area swept by each wiper can be imaginarily extended to form a circle. Radius of this circle = 25 cm Area swept by each wiper through an angle of 360° = Area of the circle = πr = 22/7 × 25 = 22/7 × 625 = 1964.29 cm Area swept by each wiper through an angle of 115° = (θ/360°) × πr = (115°/360°) × πr = 23/72 × 1964.29 = 627.48 cm Total area swept by the wipers = 2(627.48) = 1254.96 cm Hence, the total area cleaned at each swipe of the blade is 1254.96 cm
Area covered by the light can be imaginarily extended to form a circle. Radius of the circle = 16.5 km Area of circle = πr = (3.14) × (16.5) = 854.87 km Area of the sector = (θ/360°) × πr = (80°/360°) × 854.87 = 2/9 × 854.87 = 189.971 km Hence, the area of sea over which ships are warned = 189.97 km
We need to join the edges of each of the designs to the centre of the circle. Area of the circle = πr = 22/7 × 28 = 22 × 4 × 28 = 2464 cm Area of one sector = 1/6 × πr = 1/6 × 2464 = 1232/3 cm Since, the circle has six sectors which are equal therefore, the angle of each sector = 360°/6 = 60° Triangle AOB has two equal sides (both are radii of same circle) and their angle is 60°, therefore AOB is an equilateral triangle. Area of triangle AOB = √3/4 × r = √3/4 × 28 = √3 × 7 × 28 = 196√3 cm Area of each design = Area of the corresponding minor segment = Area of the sector - Area of AOB = 1232/3 - 196√3 = (1232 - 588√3)/3 = (1232 - 999.6)/3 = 232.4/3 cm Area of six designs = 6 × 232.4/3 = 464.8 cm Cost of making 1 cm Cost of making 464.8 cm = Rs 162.68 Hence, the cost of making the designs on the table cover is Rs 162.68.
Area of the sector = (θ/360°) × πR = p/360 × πR (A) = p/180 × 2πR = p/90 × πR, so (A) is incorrect. (B) = p/180 × πR (C) = p/360 × 2πR = p/180 × πR, so (C) is incorrect. (D) = p/720 × 2πR Hence, the correct answer is (D). ## Exercise 12.3
∠P is angle in a semi-circle subtended by the diameter. Therefore, it is a right angle. This implies that PQR is a right angled triangle with QR as the hypotenuse. By applying Pythagoras Theorem in PQR, we get QR QR QR QR QR = 25 cm Radius of the circle = 25/2 = 12.5 cm Area of the semi-circle = πr = 22/7 × 12.5 = 11/7 × 156.25 = 245.54 cm Area of PQR = 1/2 × PQ × PR = 1/2 × 24 × 7 = 84 cm Area of the shaded region = Area of semi-circle - Area of PQR = 245.54 - 84 = 161.54 cm Hence, the area of shaded region = 161.54 cm
Area of the smaller circle = πr = 22/7 × 7 = 22 × 7 = 154 cm Area of the bigger circle = πR = 22/7 × 14 = 22 × 2 × 14 = 616 cm Area of the sector BOD = (θ/360°) × πr = (40°/360°) × 154 = 1/9 × 154 = 154/9 cm Area of sector AOC = (θ/360°) × πR = 1/9 × 616 = 616/9 cm Area of the shaded region = Area of AOC - Area of BOD = 616/9 - 154/9 = 462/9 = 51.33 cm Hence, the area of the shaded region is 51.33 cm
AD = BC = 14 cm (Given) Since, APD and BPC are semicircles of same radius, they will also have equal areas. Area of the two semi-circles = 2 × (πr = 22/7 × 14/2 × 14/2 = 22 × 7 = 154 cm Area of the square ABCD = s Area of the shaded region = Area of ABCD - Area of two semi-circles = 196 - 154 = 42 cm Hence, the area of the shaded region is 42 cm
Area of the circle = πr = 22/7 × 6 = 792/7 cm Area of the triangle = √3/4 × a = √3/4 × 12 = 36√3 cm Total area of the figure = Area of circle + Area of triangle = 792/7 + 36√3 cm The triangle and the circle have an overlapping area. Overlapping area = Area of the minor sector = (θ/360°) × πr = (60°/360°) × πr = 1/6 × 792/7 = 132√3 cm Area of the shaded region = Total area - Overlapping area = 792/7 + 36√3 - 132/7 = 660/7 + 36√3 cm Hence, the area of the shaded region is (660/7 + 36√3) cm
Area of the square = s = 4 Area of the circle inside the square = πr = 22/7 × (2/2) = 22/7 × 1 × 1 = 22/7 cm Area of the quadrant of a circle in each corner = 4 × πr = 22/7 cm Area of shaded region = Area of square - Area of 4 quadrants - Area of the circle = 16 - 22/7 - 22/7 = 16 - 44/7 = (112 - 44)/7 = 68/7 cm Hence, the remaining area of the square is 68/7 cm
We need to draw the media AD in triangle ABC. AO = 32 cm Since ABC is an equilateral triangle circumscribed in a circle, so the centre O will coincide with the triangle?s centroid. AO = 2/3 AD as centroid divides the median in 2: 3 ratio. AD = 3/2 AO = 3/2 × 32 = 48 cm We know that the median of an equilateral triangle is also a perpendicular bisector. Therefore, by using Pythagoras Theorem in ABD: AB AB AB 3/4 AB AB AB AB = 32√3 cm Area of the triangle = √3/4 × a = √3/4 × (32√3) = √3/4 × 3072 = 768√3 cm Area of the circle = πr = 22/7 × 32 = 22528/7 cm Area of the design = Area of the circle - Area of the triangle = 22528/7 - 768√3 cm Hence, the area of the design on the cover is (22528/7 - 768√3) cm
Since, all angles are 90° in a square. Therefore, the square has 4 quadrants of equal circle. Area of one quadrant = 1/4 × πr Area of 4 quadrants = 4 × 1/4 × πr = 22/7 × (14/2) = 22 × 7 = 154 cm Area of the square = s = 14 Area of the shaded region = Area of square - Area of the 4 quadrants = 196 - 154 = 42 cm Hence, the area of the shaded region is 42 cm
- the distance around the track along its inner edge
- the area of the track.
AB = CD = PQ = RS = 106 m CR = DS = 60 m AC = BD = PR = QS = 10 m
Length of track from both semi-circles = 2πr = 2 × 22/7 × 30 = 1320/7 m Total distance of the inner track = Perimeter of the two semi-circles + RS + CD = 1320/7 + 106 + 106 = 1320/7 + 212 = (1320 + 1484)/7
= 106 × 10 = 1060 m AB = PQ and AC = PR, so Area of PQRS = Area of ABCD = 1060 m Area of the 2 outer semi-circles = πR = 22/7 × (60 + 10 + 10) = 22/7 × 1600 = 35200/7 m Area of the 2 inner semi-circles = πr = 22/7 × 30 = 19800/7 m Area of the track = 2 × Area of ABCD + Area of 2 outer semi-circles - Area of 2 inner semi-circles = 2 × 1060 + 35200/7 - 19800/7 = 2120 + 15400/7 = 2120 + 2200
It can be noted that OA, OB, OC and OD are radii of the circle and hence are equal. Area of the bigger circle = πR = 22/7 × 7 = 154 cm OD acts as the diameter for the smaller circle. Area of the smaller circle = πr = 22/7 × 7/2 × 7/2 = 77/2 cm AB = OB + OA = 7 + 7 = 14 cm Area of the triangle ABC = 1/2 × Base × Height = 1/2 × AB × OC = 1/2 × 14 × 7 = 49 cm Area of the shaded region = Area of the smaller circle + 1/2 × Area of bigger circle - Area of triangle ABC = 77/2 + 1/2 × 154 - 49 = 77/2 + 77 - 49 = 77/2 + 28 = (77 + 56)/2 = 133/2 = Hence, the area of the shaded region is 66.5 cm
Area of equilateral triangle = √3/4 × s 17320.5 = √3/4 × s 17320.5 = 1.73205/4 × s 10000 × 4 = s 40000 = s s = 200 cm Radius of each circle = s/2 = 100 cm Area of the minor sector of each circle = (θ/360°) × πr = (60°/360°) × 3.14 × 100 = 1/6 × 31400 = 15700/3 cm Area of shaded region = Area of the triangle - 3 × Area of the minor sector of each circle = 17320.5 - 3 × 15700/3 = 17320.5 - 15700
Hence, area of the shaded region is 1620.5 cm
Length of the side of handkerchief = 3 × Diameter of one circular design = 3 × 7(2) = 3 × 14 = 42 cm Area of the handkerchief = 42 Area of one circular design = πr = 22/7 × 7 = 154 cm Remaining Area of the handkerchief = Area of handkerchief - 9 × Area of one circular design = 1764 - 9 × 154 = 1764 - 1386 = 378 cm
= 1/4 × 22/7 × (3.5) = 77/8 cm = 9.625 cm
= 1/2 × OB × OD = 1/2 × 3.5 × 2 = 3.5 cm Area of the shaded region = Area of the quadrant OACB - Area of the triangle OBD = 9.625 - 3.5
Join O with B. OAB is a right triangle, so by applying Pythagoras Theorem, we get OB OB OB OB OB = 20√2 cm OB is the radius of the given quadrant of a circle. Area of the quadrant = 1/4 × πr = 1/4 × 3.14 × (20√2) = 1/4 × 3.14 × 800 = 628 cm Area of the square OABC = OA = 20 Area of the shaded region = Area of the quadrant - Area of the square = 628 - 400
Hence, the area of the shaded region is 228 cm
Area of the sector of the bigger circle = (θ/360°) × πR = (30°/360°) × 22/7 × 21 = 1/12 × 22/7 × 441 = 231/2 cm Area of the sector of the smaller circle = (θ/360°) × πr = (30°/360°) × 22/7 × 7 = 1/12 × 154 = 77/6 cm Area of the shaded region = Area of sector with radius R - Area of sector with radius r = 231/2 - 77/6 = (693 - 77)/6 = 616/6
Hence, the area of the shaded region is 308/3 cm
Area of the quadrant = 1/4 × πR = 1/4 × 22/7 × 14 = 154 cm Area of the triangle ABC = 1/2 × base × height = 1/2 × 14 × 14 = 98 cm Area of the corresponding segment of the quadrant = Area of quadrant - Area of triangle = 154 - 98 = 56 cm By applying Pythagoras Theorem in ABC, we get BC BC BC BC BC = 14√2 cm Area of the semi-circle = 1/2 × πr = 1/2 × 22/7 × (14√2/2) = 154 cm Area of the shaded region = Area of Semi-circle - Area of the segment = 154 - 56
Hence, the area of the shaded region is 98 cm
Since, both the quadrants have the same radius. Therefore, their area will be equal. This implies that area of their corresponding segments will also be equal. Area of triangle ABC = 1/2 × Base × Height = 1/2 × 8 × 8 = 32 cm Area of each quadrant = 1/4 × πr = 1/4 × 22/7 × 8 = 352/7 cm Area of each segment = Area of quadrant - Area of triangle = 352/7 - 32 = (352 - 224)/7 = 128/7 cm Area of the design = 2 × Area of one segment = 2 × 128/7
Hence, the area of the design is 256/7 cm |