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Class- X-CBSE-Mathematics Areas Realated to Circles
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CBSE NCERT Solutions for Class 10 Mathematics Chapter 12 Back of Chapter Questions
1. The radii of two circle are 19 cm and 9 cm respectively. Find the radius of thecircle which has circumference equal to the sum of the circumferences of the twocircles. οΏ½use ππ = 22
7 οΏ½
Solution:
Given radius of first circle be ππ1 = 19 cm
Radius of second circle be ππ2 = 9 cm
Let radius of required circle be ππ
Circumference of first circle = 2ππππ1 = 2ππ(19) = 38ππ cm
Circumference of second circle = 2ππππ2 = 2ππ(9) = 18ππ cm
Circumference of required circle = 2ππππ
Given that
Circumference of required circle = Circumference of first circle + Circumferenceof second circle
β 2ππππ = 38ππ + 18ππ = 56ππ cm
β ππ =56ππ2ππ = 28
Hence, the radius of required circle is 28 cm.
2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of thecircle having area equal to the sum of the areas of the two circles.οΏ½use ππ = 22
7. οΏ½
Solution:
Given radius of first circle be ππ1 = 8 cm
Radius of second circle be ππ2 = 6 cm
Let radius of required circle be ππ
Area of first circle = ππππ12 = ππ(8)2 = 64ππ cm2
Area of second circle = ππππ22 = ππ(6)2 = 36ππ cm2
Given that
Area of required circle = area of first circle + area of second circle
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β ππππ2 = 64ππ + 36ππ
β ππππ2 = 100ππ
β ππ2 = 100 cm2
β ππ = Β±10
But radius cannot be negative, so radius of required circle is 10 cm.
3. 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. οΏ½use ππ = 22
7οΏ½
Solution:
Given radius (r1) of gold region = 212
= 10.5 cm
Given that each circle is 10.5 cm wider than previous circle.
So, radius of second circle ππ2 = 10.5 + 10.5 = 21 cm
Radius of third circle r3 = 21 + 10.5 = 31.5 cm
Radius of fourth circle r4 = 31.5 + 10.5 = 42 cm
Radius of fifth circle r5 = 42 + 10.5 = 52.5 cm
Area of golden region = area of 1st circle = ππππ12 = ππ(10.5)2 = 346.5 cm2
Now, area of Red region = area of second circle - area of first circle
= ππππ22 β ππππ12
= ππ[(21)2 β (10.5)2]
= 441ππ β 110.25ππ = 330.75ππ
= 1039.5 cm2
Area of blue region = area of third circle - area of second circle
= ππππ32 β ππππ12
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= ππ[(31.5)2 β (21)2]
= 992.25ππ β 441ππ = 551.25ππ
= 1732.5 cm2
Area of black region = area of fourth circle - area of third circle
= ππππ42 β ππππ32
= ππ[(42)2 β ππ(31.5)2]
= 1764ππ β 992.25ππ
= 771.75ππ = 2425.5 cm2
Area of white region = area of fifth circle β area of fourth circle
= ππππ52 β ππππ42
= ππ[(52.5)2 β (42)2]
= 2756.25ππ β 1764ππ
= 992.25ππ = 3118.5 cm2
Hence, areas of gold, red, blue, black, white regions are 346.5 cm2, 1039.5 cm2, 1732.5 cm2, 2425.5 cm2 and 3118.5 cm2 respectively.
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 traveling at a speed of 66 km per hour? οΏ½use ππ = 22
7οΏ½
Solution:
Given diameter of wheel of car = 80 cm
Therefore, radius (r) of wheel of car = 802
= 40 cm
Now, circumference of wheel = 2ππππ
= 2ππ (40) = 80ππ cm
Given speed of car = 66 km/hour
=66 Γ 100000
60 cm/min
= 110000 cm/min
Distance travelled by car in 10 minutes = speed Γ time
= 110000 Γ 10 = 1100000 cm
Let number of revolutions of wheel of car is ππ.
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ππ Γ distance travelled in 1 revolution (i.e. circumference) = distance traveled in 10 minutes.
β ππ Γ 80ππ = 1100000
β ππ =1100000 Γ 7
80 Γ 22
β=35000
8 = 4375
Therefore, each wheel of car will make 4375 revolutions.
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 οΏ½use ππ = 22
7οΏ½
(A) 2 units
(B) ππ units
(C) 4 units
(D) 7 units
Solution:
Let radius of circle be ππ
Circumference of circle= 2ππππ
Area of circle = ππππ2
Given that circumference of circle and area of circle is equal.
β 2ππππ = ππππ2
β ππ = 2
So radius of circle will be 2 units
12.2 Maths
1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60Β°.
Solution:
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Let OACB be a sector of circle making 60Β° angle at centre O of circle.
We know, area of sector of angle ππ = ππ360ππ
Γ ππππ2
So, area of sector OACB = 60ππ
360ππΓ 22
7Γ (6)2
=16 Γ
227 Γ 6 Γ 6 =
1327 cm2
Therefore, area of sector of circle making 60ππ at centre of circle is 18.85 cm2
2. Find the area of a quadrant of circle whose circumference is 22 cm.
Solution:
Given circumference of the circle = 22 cm
Let radius of circle be ππ.
β 2ππππ = 22
β r =222ππ
β ππ =72
We know, quadrant of circle will subtend 90o angle at centre of circle.
So, area of such quadrant of circle = 90o
360oΓ ππ Γ ππ2
=14 Γ ππ Γ οΏ½
72οΏ½
2
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=14 Γ
227 Γ
494
=778 cm2
Therefore, the area of a quadrant of circle whose circumference is 22 cm is 778
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.
Solution:
We know that in one hour, minute hand rotates 360o.
So, in 5 minutes, minute hand will rotate = 360o
600Γ 5 = 30Β°
We know, area of sector of angle ππ = ππ360o
Γ ππππ2
Area of sector of 30o = 30o
360oΓ 22
7Γ 14 Γ 14
=2212 Γ 2 Γ 14
=11 Γ 14
3
=154
3 cm2
So, area swept by minute hand in 5 minutes is 51.3 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 ππ = 3.14)
Solution:
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Given a chord of a circle of radius 10 cm subtends a right angle at the centre
Let AB be the cord of circle subtending 90o angle at centre O of circle.
(i) Area of minor sector
We know, area of sector of angle ππ = ππ360ππ
Γ ππππ2
Therefore, area of minor sector (OAPB) = 90o
360oΓ ππππ2
=14 Γ
227 Γ 10 Γ 10
=1100
14 = 78.5 cm2
Area of βOAB = 12
Γ OA Γ OB = 12
Γ 10 Γ 10
= 50 cm2
Area of minor segment APB = Area of minor sector OAPB β Area of βOAB
= 78.5 β 50 = 28.5 cm2
(ii) Area of major sector = οΏ½360oβ90o
360oοΏ½Γ ππππ2 = οΏ½270
o
360oοΏ½ππππ2
=34 Γ
227 Γ 10 Γ 10
=3300
14 cm2 = 235.7 cm2
5. In a circle of radius 21 cm, an arc subtends an angle of 60o at the centre. Find:
(i) The length of the arc
(ii) Area of the sector formed by the arc
(ii) Area of the segment formed by the corresponding chord
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Solution:
Given radius (ππ) of circle = 21 cm
And angle subtended by given arc = 60o
We know, length of an arc of a sector of angle ππ = ππ360o
Γ 2ππππ
(i) Length of arc ACB = 60ππ
360ππΓ 2 Γ 22
7Γ 21
= 16
Γ 2 Γ 22 Γ 3
= 22 cm
(ii) Area of sector formed by the arc.
We know, area of sector of angle ππ = ππ360ππ
Γ ππππ2
β Area of sector OACB = 60o
360oΓ ππππ2
=16 Γ
227 Γ 21 Γ 21
= 231 cm2
(iii) Area of the segment formed by the corresponding chord
In βOAB
β OAB = β OBA (as OA = OB)
We know, sum of angles in a triangle = 180Β°
β β OAB + β A0B + β OBA = 180Β°
β 2β OAB + 60o = 180o
β β OAB = 60Β°
So, βOAB is an equilateral triangle.
Therefore, area of βOAB = β34
Γ (side)2
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=β34 Γ (212) =
441β34 cm2
Now, area of segment ACB = Area of sector OACB β Area of βOAB
= οΏ½231β441β3
4 οΏ½ cm2
6. A chord of a circle of radius 15 cm subtends an angle of 60Β° at the centre. Find the areas of the corresponding minor and major segments of the circle.
(Use Ο = 3.14 and β3 = 1.73)
Solution:
Given radius (ππ) of circle = 15 cm
We know, area of sector of angle ππ = ππ360ππ
Γ ππππ2
Area of sector OPRQ = 60o
360oΓ ππππ2
=16 Γ
227 Γ (15)2
=11 Γ 75
7 =825
7
= 117.85 cm2
In βOPQ
β OPQ = β OQP (as OP = OQ)
β OPQ + β OQP + β POQ = 180Β°
2β OPQ = 120Β°
β OPQ = 60Β°
βOPQ is an equilateral triangle
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Area of equilateral βOPQ = β34
(side)2
=β34 Γ (15)2 =
225β34 cm2
= 56.25β3
= 97.3125 cm2
Area of segment PRQ = Area of sector OPRQ β Area of βOPQ
= 117.85β 97.3125
= 20.537 cm2
Area of major segment PSQ = Area of circle β Area of segment PRQ
= ππ(15)2 β 20.537
=225 Γ 22
7 β 20.537
=4950
7 β 20.537 = 707.14β 20.537
= 686.605 cm2
7. A chord of a circle of radius 15 cm subtends an angle of 120o at the centre. Find the areas of the corresponding segment of the circle. (Use ππ = 3.14 and β3 =1.73)
Solution:
Given radius of circle = 15 cm
Draw a perpendicular OV on chord ST. It will bisect the chord ST.
SV = VT
In βOVS
cos 60o =OVOS
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β12 =
OV12
β OV = 6
And sin 60o = β32
= SVSO
βSV12 =
β32
β SV = 6β3
β΄ ST = 2SV = 2 Γ 6β3 = 12β3
Now, area of βOST = 12
Γ ST Γ OV
=12 Γ 12β3 Γ 6
= 36β3 = 36 Γ 1.73 = 62.28
We know, area of sector of angle ππ = ππ360ππ
Γ ππππ2
Here ππ = 120o
Area of sector OSUT = 120o
360oΓ ππ(12)2
=13 Γ
227 Γ 144 = 150.72
Area of segment SUT = Area of sector OSUT β Area of βOST
= 150.72β 62.28
= 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 as given in figure. 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 ππ = 3.14)
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Solution:
Given horse is tired with a rope of length 5 m.
Horse can graze a sector of 90o in a circle of 5 m radius.
(i) The area of that part of the field in which the horse can graze
The area that can be grazed by horse = area of sector
=90o
360o ππππ2 =
14 Γ ππ(5)2
=25ππ
4 = 19.62 m2
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(iii) Area that can be grazed by horse when length of rope is 10 m long
Therefore, radius of sector becomes 10 m
Area that can be grazed by horse = 90o
360oΓ ππ Γ (10)2
=14 Γ ππ Γ 100
= 25ππ
Hence increase in grazing area = 25ππ β 25ππ4
= 25ππ οΏ½1β14οΏ½
= 25 Γ 3.14 Γ34 = 58.87 cm2
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 figure. Find.
(i) The total length of the silver wire required.
(ii) The area of each sector of the brooch
Solution:
(i) The total length of the silver wire required.
Here, total length of wire required will be = length of 5 diameters + circumference of brooch.
Given diameter of circle = 35 mm
β Radius of circle = 352
mm
We know, circumference of circle = 2ππππ
Therefore, circumference of brooch = 2ππ οΏ½352οΏ½
= 2 Γ227 Γ οΏ½
352 οΏ½
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= 110 mm
Total length of silver wire required = 110 + 5 Γ 35
= 110 + 175 = 285 mm
(ii) Each of 10 sectors of circle subtend 360o
10= 36o at centre of circle.
We know, area of sector of angle ππ = ΞΈ360o
Γ ππππ2
So, area of each sector = 36o
360oΓ ππππ2
=1
10 Γ227 Γ οΏ½
352 οΏ½Γ οΏ½
352 οΏ½
=385
4 mm2
= 96.25 mm2
10. An umbrella has 8 ribs which are equally spaced as shown in figure. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
Solution:
Given there are 8 ribs in umbrella.
Each of 8 sectors of circle subtend 360o
8= 45o at centre of circle.
We know, area of sector of angle ππ = ΞΈ360o
Γ ππππ2
So, area between two consecutive ribs of circle = 45o
360oΓ ππππ2
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=18 Γ
227 Γ (45)2
=1128 Γ 2025 =
2227528 cm2
= 795.8 cm2
11. A car has two wipers which do not overlap. Each wiper has blade of length 25 cm sweeping through an angle of 115o. Find the total area cleaned at each sweep of the blades.
Solution:
Given that each blade of wiper will sweep an area of a sector of 115o in a circle of 25 cm radius.
We know, area of sector of angle ππ = ΞΈ360o
Γ ππππ2
Here ππ = 115o and ππ = 25
Area of such sector = 115o
360oΓ ππ Γ (25)2
=2372 Γ
227 Γ 25 Γ 25
=158125
252 cm2
Hence area swept by 2 blades = 2 Γ 158125252
=158125
126 cm2
= 1254.96 cm2
12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80Β° to a distance of 16.5 km. Find the area of the sea over which the ships warned. (Use ππ = 3.14)
Solution:
Given lighthouse spreads light over a sector of angle 800 in a circle of 16.5 km radius
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We know, area of sector of angle ππ = ππ360o
Γ ππππ2
Here ππ = 80o and ππ = 16.5
Area of sector OACB = 80o
360oΓ ππππ2
=29 Γ 3.14 Γ 16.5 Γ 16.5
= 189.97 km2
Therefore, area of the sea over ships warned is 189.97 km2
13. 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 β3 = 1.7 )
Solution:
Designs are segments of circle.
Consider segment APB.
Chord AB is a side of hexagon. Each chord will subtend 360o
6= 60o at centre of
circle.
In βOAB
β OAB = β OBA (as OA = OB)
And β AOB = 60o
We know, sum of interior angles of triangle = 180o
β β OAB + β OBA + β AOB = 180o
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β 2β OAB = 180o β 60o = 120o
β β OAB = 60o
So, β OAB is an equilateral triangle
Hence area of βOAB = β34
Γ (side)2
=β34 Γ (28)2 = 196β3 cm2 = 333.2 cm2
We know, area of sector of angle ππ = ΞΈ360o
Γ ππππ2
Here ππ = 60o and ππ = 28
Area of sector OAPB = 60o
360oΓ ππππ2
=16 Γ
227 Γ 28 Γ 28
=1232
3 = 410.66 cm2
Area of segment APB = Area of sector OAPB β Area of βOAB
= 410.66β 333.2
= 77.46 cm2
Hence, total area of designs = 6 Γ 77.46 = 464.76 cm2
Cost occurred in making 1 cm2 designs = Rs. 0.35
Cost occurred in making 464.76 cm2 designs = 464.76 Γ 0.35 = 162.66
So, cost of making such designs is Rs. 162.66.
14. Tick the correct answer in the following: Area of a sector of angle ππ (in degrees) of a circle with radius R is
(A) ππ180
Γ 2ππ R
(B) ππ180
Γ ππ R2
(C) ππ360
Γ 2ππ R
(D) ππ720
Γ 2ππ R2
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We know that area of sector of angle ππ = ππ360o
ππ r2
Here ππ = ππ and r = R
Hence, Area of sector of angle ππ = ππ360o
(ππ R2)
= οΏ½ππ
720oοΏ½(2ππ R2)
12.3 Maths
1. Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle
Solution:
Given PQ = 24 cm, and PR = 7 cm
RQ is the diameter of circle, so β RPQ will be 90o.
Now in βPQR by applying Pythagoras theorem
RP2 + PQ2 = RQ2
β (7)2 + (24)2 = RQ2
β RQ = β625 = 25
β Radius of circle OR = RQ2
= 252
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Area of semicircle RPQOR = 12Οr2
=12Ο οΏ½
252 οΏ½
2
=625
8 Ο cm2
= 245.53 cm2
Area of βPQR =12 Γ PQ Γ PR
=12 Γ 24 Γ 7
= 84 cm2
Area of shaded region = area of semicircle RPQOR β area of βPQR
= 245.53β 84
= 161.53 cm2
2. 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 β AOC =40o.
Solution:
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Given radius of inner circle = 7 cm
And radius of outer circle = 14 cm
Area of shaded region
= area of sector OAFCOβ area of sector OBEDO
= 40o
360oΓ ππ(14)2 = 40o
360oΓ ππ(7)2 οΏ½since area of sector of angle ππ =
ππ360o
Γ ππππ2οΏ½
=19 Γ
227 Γ 14 Γ 14 β
19 Γ
227 Γ 7 Γ 7
=616
9 β154
9 =462
9
=154
3 = 51.33 cm2
Therefore, area of shaded region is 51.33 cm2.
3. 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.
Solution:
Given length of side of square = 14 cm = length of diameter of semicircle.
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β radius of semicircle = 7 cm
We know, area of semicircle = 12ππππ2
=12 Γ
227 Γ (7)2
= 77 cm2
Now, area of square ABCD = (side)2 = (14)2 = 196 cm2
Area of shaded region = area of square ABCD β area of semicircle APD β area of semicircle BPC
= 196β 77β 77
= 196β 154
= 42 cm2
4. 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.
Solution:
Given length of side of equilateral triangle = 12 cm and radius of circle = 6 cm
β COE = 60o (β΅ interior angle of equilateral triangle)
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We know, area of sector of angle ππ = ππ360o
Γ ππππ2
Hence. area of sector OCDE = 60o
360o ππππ2
=16 Γ
227 Γ 6 Γ 6
=132
7 cm2
Now, area of βOAB = β34
(12)2 (β΅ area of equilateral triangle = β34
(side)2)
=β3 Γ 12 Γ 12
4
= 36β3 cm2
And area of circle = ππππ2
=227 Γ 6 Γ 6 =
7927 cm2
Area of shaded region = area of AOAB + area of circle β area of sector OCDE
= 36β3 +792
7 β132
7
= οΏ½36β3 +660
7 οΏ½ cm2
= 156.64 cm2
Therefore, area of shaded region is 156.64 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 the given figure. Find the area of the remaining portion of the square.
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Solution:
Given length of side of square is 4 cm and radius of quadrant circle = 1 cm.
And diameter of inner circle = 2 cm β radius = 1 cm.
We know, area of quadrant of a circle = 14 (area of a circle)
=14 Γ
227 Γ (1)2 =
2228 cm2
Hence, area of 4 quadrant circles = 4 Γ area of one quadrant circle
= 4 Γ 2228 =
227 cm2
Area of square = (side)2 = (4)2 = 16 cm2
Area of inner circle = ππππ2 = ππ(1)2
=227 cm2
Area of shaded region = area of square β area of circle β area of 4 quadrant circles
= 16 β227 β
227
= 16 β447
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=112β 44
7 =687 cm2
= 9.71 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 the given figure. Find the area of the design (Shaded region).
Solution:
Given radius (ππ) of circle = 32 cm
Let AD be the median of βABC
β AO =23 AD = 32
β AD = 48 cm
In βABD, using Pythagoras theorem
AB2 = AD2 + BD2
β AB2 = (48)2 + οΏ½AB2 οΏ½
2
β3AB2
4 = (48)2
β AB = 48Γ2β3
= 96β3
= 32β3 cm
Now, area of equilateral triangle βABC = β34οΏ½32β3οΏ½
2
=β34 Γ 32 Γ 32 Γ 3
= 96 Γ 8 Γ β3
= 768β3 cm2
Area of circle = Οr2
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=227 Γ (32)2
=227 Γ 1024
=22528
7 cm2
Area of design = area of circle β area of βABC
= οΏ½22528
7 β 768β3οΏ½
= (3218.28β 1330.18)
= 1888.10 cm2
Therefore, area of given design is 1888.10 cm2
7. 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.
Solution:
Given length of side of square = 14 cm
From the figure, radius of circle = 7cm
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Area of quadrant circle = 14 (area of circle)
Area of each quadrant cirlce = 14
Γ ππ(7)2
=14 Γ
227 Γ 7 Γ 7
=772 cm2
Now, area of square ABCD = (side)2 = (14)2 = 196 cm2
Therefore, area of shaded portion = area of square ABCD β 4 Γ area of each quadrant circle
= 196β 4 Γ772 = 196β 154
= 42 cm2
Hence, area of shaded region is 42 cm2
8. 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:
(i) The distance around the track along its inner edge
(ii) The area of the track
Solution:
Given distance between inner parallel lines = 60 m and length of line segment =106 m
(i) Distance around the track along its inner edge = length of AB +length of arc BEC + length of CD + length of arc DFA
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= 106 +12 Γ 2ππππ + 106 +
12 Γ 2ππππ
= 212 +12 Γ 2 Γ
227 Γ 30 +
12 Γ 2 Γ
227 Γ 30
= 212 + 2 Γ227 Γ 30
= 212 +1320
7
=1484 + 1320
7 =2804
7
= 400.57 m
Therefore, distance around the track along its inner edge is = 400.57 m
(ii) Area of track = area of GHIJ β area of ABCD + area of semicircle HKI β Area of semicircle BEC + area of Semicircle GLJ β area of semicircle AFD
= 106 Γ 80 β 106 Γ 60 +12 Γ
227 Γ (40)2 β
12 Γ
227 Γ (30)2
+12 Γ
227 Γ (40)2 β
12 Γ
227 Γ (30)2
= 106 (80β 60) +227 Γ (40)2 β
227 Γ (30)2
= 106(20) +227
[(40)2 β (30)2]
= 2120 +227
(40β 30)(40 + 30)
= 2120 + οΏ½227 οΏ½ (10)(70)
= 2120 + 2200
= 4320 m2
Hence, area of track is 4320 m2
9. 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.
Class- X-CBSE-Mathematics Areas Realated to Circles
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Solution:
Given radius (ππ1) of larger circle = 7 cm
Hence, radius (ππ2) of smaller circle = 72
cm
Area of smaller circle = ππππ22
=227 Γ
72 Γ
72
=772 cm2
Area of semicircle AECFB of larger circle = 12ππππ12
=12 Γ
227 Γ (7)2
= 77 cm2
Area of βABC = 1 Γ AB Γ OC
= 12
Γ 14 Γ 7 = 49 cm2
Therefore, area of shaded region = area of smaller circle + area of semicircle AECFB - area of βABC
=772 + 77 β 49
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= 28 +772 = 28 + 38.5
= 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 the given figure). Find the area of shaded region. (Use ππ =3.14 and β3 = 1.73205)
Solution:
Given area of equilateral triangle = 17320.5 m2
Let side of equilateral triangle be ππ
Area of equilateral triangle = β34
(ππ2)
ββ34
(ππ2) = 17320.5
β ππ2 = 4 Γ 10000
β ππ = 200 cm
We know, area of sector of angle ππ = ΞΈ360o
Γ ππππ2
Here ππ = 60o (since equilateral triangle) ππ = 100
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So, area of sector ADEF = 60o
360oΓ ππ Γ ππ2
=16 Γ ππ Γ (100)2
=3.14 Γ 10000
6 =15700
3
Area of shaded region = area of equilateral triangle β 3 Γ area of each sector
= 17320.5β 3 Γ15700
3
= 17320.5β 15700 = 1620.5 cm2
11. On a square handkerchief, nine circular designs each of radius 7 cm are made as shown in figure. Find the area of the remaining portion of the handkerchief.
Solution:
Side of square = 6 Γ 7 = 42 cm,
Area of square = (side)2
= (42)2
= 1764 cm2
Area of each circle = Οr2
=227 Γ (7)2 = 154 cm2
Therefore, area of 9 circles = 9 Γ 154
= 1386 cm2
Area of remaining portion of handkerchief = 1764β 1386
= 378 cm2
Hence, area of the remaining portion of the handkerchief = 378 cm2
12. 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
Class- X-CBSE-Mathematics Areas Realated to Circles
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(i) Quadrant OACB
(ii) Shaded region
Solution:
Since OACB is quadrant so it will subtend 90Β° angle at O.
Area of quadrant OACB = 14 (area of circle)
=14 Γ
227 Γ (3.5)2 =
14 Γ
227 Γ οΏ½
72οΏ½
2
=(11 Γ 7 Γ 7)2 Γ 7 Γ 2 Γ 2 =
778 cm2
Area of βOBD = 12
Γ OB Γ OD
=12 Γ 3.5 Γ 2
= 3.5 cm2
Area of shaded region = area of quadrant OACB β area of βOBD
=778 β 3.5
=498 cm2
= 6.125 cm2
Therefore, area of shaded region = 6.125 cm2
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13. In the given figure, a square OABC in inscribed in a quadrant OPBQ. If OA =20 cm, find the area of the shaded region. (Use ππ = 3.14)
Solution:
Given OA = 20 cm
In βOAB, using Pythagoras therom
OB2 = OA2 + AB2
β OB2 = (20)2 + (20)2
β OB = 20β2
Therefore, radius (ππ) of circle = 20β2 cm
Area of quadrant OPBQ = 90o
360oΓ 3.14 Γ οΏ½20β2οΏ½
2
=14 Γ 3.14 Γ 800
= 628 cm2
Area of square = (side)2 = (20)2 = 400 cm2
Therefore, area of shaded region = area of quadrant OPBQβ area of square
= 628β 400 = 228 cm2
14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O as shown in the figure. If β AOB = 30Β°, find the area of the shaded region.
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Solution:
Given radius of larger circle = 21 and radius of smaller circle = 7
Area of shaded region = area of sector OAEB β area of sector OCFD
=30o
360o Γ ππ Γ (21)2 β30o
360o Γ ππ Γ (7)2
οΏ½ since area of sector of angle ππ =ππ
360o Γ ππππ2οΏ½
=1
12 Γ ππ[(21)2 β (7)2]
=1
12 Γ227 Γ [(21β 7)(21 + 7)]
=22 Γ 14 Γ 28
12 Γ 7
=308
3 cm2
= 102.6 cm2
Therefore, area of shaded potion = 102.6 cm2
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15. 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.
Solution:
Given radius of circle = 14 cm
As ABDC is a quadrant of circle, β BAC will be of 90o
In βABC, using Pythagoras theorem,
BC2 = AC2 + AB2
β BC2 = (14)2 + (14)2
β BC = 14β2
From the figure, radius (ππ1) of semicircle drawn on BC = 14β22
= 7β2 cm Now,
area of βABC = 12
Γ AB Γ AC
=12 Γ 14 Γ 14
= 98 cm2
And area of sector ABDC = 14
Γ ππππ2
=14 Γ
227 Γ 14 Γ 14
= 154 cm2
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Area of semicircle drawn on BC = 12
Γ ππ Γ ππ12
=12 Γ
227 Γ οΏ½7β2οΏ½
2
=12 Γ
227 Γ 98
= 154 cm2
Area of shaded region = area of semicircle β (area of sector ABDC β area of βABC
= 154β (154β 98)
= 98 cm2
Therefore, area of shade region = 98 cm2
16. Calculate the area of the designed region in the given figure common between the two quadrants of circles of radius 8 cm each.
Solution:
Given radius of each circle = 8 cm
The designed area is common region between two sectors BAEC and DAFC
Area of sector BAEC = 14
Γ 227
Γ (8)2
=14 Γ
227 Γ 64
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=22 Γ 16
7
=352
7 cm2
Area of βBAC = 12
Γ BA Γ BC
=12 Γ 8 Γ 8 = 32 cm2
So, area of segment AEC = area of sector BAEC β area of βBAC
= 3527β 32
Area of designed portion = 2 Γ (area of segment AEC)
= 2 Γ οΏ½352
7 β 32οΏ½
= 2 οΏ½352β 224
7 οΏ½
=2 Γ 128
7
=256
7 cm2
= 36.5 cm2
Therefore, area of designed region is 36.5 cm2.
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