
01. Real Numbers
10
Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

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02. Polynomials
12
Lecture2.1

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03. Linear Equation
12
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04. Quadratic Equation
10
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05. Arithmetic Progressions
11
Lecture5.1

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Lecture5.11


06. Some Applications of Trigonometry
7
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Lecture6.6

Lecture6.7


07. Coordinate Geometry
17
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Lecture7.17


08. Triangles
18
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Lecture8.18


09. Circles
9
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Lecture9.9


10. Areas Related to Circles
12
Lecture10.1

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11. Introduction to Trigonometry
9
Lecture11.1

Lecture11.2

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Lecture11.6

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Lecture11.9


12. Surface Areas and Volumes
9
Lecture12.1

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Lecture12.6

Lecture12.7

Lecture12.8

Lecture12.9


13. Statistics
14
Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

Lecture13.5

Lecture13.6

Lecture13.7

Lecture13.8

Lecture13.9

Lecture13.10

Lecture13.11

Lecture13.12

Lecture13.13

Lecture13.14


14. Probability
9
Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8

Lecture14.9


15. Construction
7
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

NCERT Solutions – Areas Related to Circles
Q.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 circumference of the two circles.
Sol. Let r be the radius of the circle whose circumference is equal to the sum of the circumference of two circles of radii 19 cm and 9 cm.
Therefore, 2πr=2π(19)+2π(9)
⇒ r=19+9=28
Hence, the radius of the new circles is 28 cm.
Q.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 area of the two circles.
Sol. Let r be the radius of the circle whose area is equal to the sum of the area of the circles of radii 8 cm and 6 cm.
Therefore, πr2=π(8)2+π(6)2
⇒ r2=82+62
⇒ r2=64+36
⇒ r2=100
⇒ r=100−−−√=10
Hence, the radius of the new circle is 10 cm.
Q.3 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 reach of the other bands is 10.5 cm wide.
Find the area of each of the five scoring regions.
Sol. The area of each of five scoring regions are as under :
For Gold : π(10.5)2cm2=227×110.25cm2
=2425.57cm2=346.5cm2
For Red : π[(21)2−(10.5)2]cm2=227(441−110.25)cm2
=227×330.75cm2
=7276.57cm2=1039.5cm2
For Blue : π[(31.5)2−(21)2]cm2=227(992.25−441)cm2
=227×551.25cm2
=12127.57cm2=1732.5cm2
For Black : π[(42)2−(31.5)2]cm2=227(1764−992.25)cm2
=227×771.75cm2
=16978.57cm2=2425.5cm2
For White : π[(52.5)2−(42)2]cm2=227(2756.25−1764)cm2
=227×992.25cm2
=21829.57cm2=3118.5cm2
Q.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?
Sol. Distance covered by the car in 10 minutes
=(66×1000×100×1060)cm
=66000006cm
Circumference of the wheel of the car
=(2×227×40)cm
Therefore, Number of revolutions in 10 minutes when the car is travelling at a speed of 66 km/hr.
=(66000006×72×22×40)=4375
Hence, the wheel of car makes 4375 revolutions in 10 minutes.
Q.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) πunits
(c) 4 units (d) 7 units
Sol. (a) Because ,
Here, 2πr=πr2, where r is the radius
⇒ r2−2r=0
⇒ r(r – 2) = 0
⇒ r = 0 or r = 2
But, r≠0 Therefore, r = 2 units
Q.1 Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60º
Sol. We know that the area A of a sector of angle θ in a circle of radius r is given by A=θ360×πr2.
Here r = 6 cm, θ=60o
Therefore, A=[60360×227×36]cm2=(1327)cm2
Q.2 Find the area of a quadrant of a circle whose circumference is 22 cm.
Sol. Let r be the radius of the circle. Then,
Circumference = 22 cm
⇒ 2πr=22
⇒ 2×227×r=22
⇒ r=72cm
Area of the quadrant of a circle
=14πr2=(14×227×494)cm2
=53956cm2=778cm2
Q.3 The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Sol. Clearly, minute hand of a clock describes a circle of radius equal to its length i.e., 14 cm. Since the minutes hand rotates through 6º in one minute, therefore, area swept by the minute hand in one minute is the area of sector of angle 6º in a circle of radius 14 cm. Hence, the required area i.e., the area swept in 5 minutes.
=(θ360×πr2×5)cm2
=[6360×227×(14)2×5]cm2
=(160×227×196×5)cm2=1543cm2
Q.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 π = 3.14)
Sol. Here, r = 10 cm, θ=90o
(i) Area of the minor segment
=r2[πθ360−12sinθ]
=(10)2[3.14×90360−12sin90o]cm2
=100(0.785−0.5)cm2
=(100×0.285)cm2=28.5cm2
(ii) Area of the major sector
=θ360×πr2, where r = 10 and θ=270o
=(270360×3.14×102)cm2
=(34×3.14×100)cm2=235.5cm2
Q.5 In a circle of a radius 21 cm, an arc substends an angle of 60º 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.
Sol. Here r = 21 cm θ=60o
(i) Length of the arc, ℓ=θ180×πr
=(60180×227×21)cm
=22cm
(ii) Area of the sector, A=θ360×πr2
=(60360×227×21×21)cm2
=231cm2
(iii) Area of the segment
=r2[πθ360−12sinθ]
=(21)2[227×60360−12sin60o]cm2
=441(1121−12×3√2)cm2
=(21×11−4413√4)cm2
=(231−4413√4)cm2
Q.6 A chord of a circle of radius 15 cm subtends an angle of 60º at the centre. Find the areas of the corresponding minor and major segment of the circle.
(Use π=3.14and3–√=1.73)
Sol. Here r = 15 cm and θ=60o
Therefore, Area of the minor segment
=r2[πθ360−12sinθ]
=(15)2[314×60360−12×sin60]cm2
=225[3146−12×3√2]cm2
=225(3.14×2−33√12)cm2
=225(6.28−3×1.7312)cm2
=225×(6.28−5.19)12cm2
=225×1.0912cm2=245.2512cm2=20.4375cm2
Area of the major segment
= Area of the circle – Area of the minor segment
= (3.14 × 225 – 20.4375) cm2
= (706.5 – 20.4375)cm2 = 686.0625 cm2
Q.7 A chord of a circle of radius 12 cm subtends an angle of 120º at the centre. Find the area of the corresponding segment of the circle.
(Use π=3.14and3–√=1.73)
Sol. Here r = 12 cm θ=120o
Area of the corresponding segment of the circle
= Area of the minor segment
=r2[πθ360−12sinθ]
=(12)2[3.14×120360−12sin120o]cm2
=144(3.143−12×3√2)cm2
[sincesin120o=sin(180o−60o)=sin60o=3√2]
=144×3.14×4−33√12cm2
=12×(12.56−3×1.73)cm2
=12×(12.56−5.19)cm2
=12×7.37cm2=88.44cm2
Q.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 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 π=3.14)
Sol. (i) The horse will graze over a quadrant of a circle with centre at the corner A of the field and radius AF = 5 m.
Then the area of the quadrant of this circle
=π×524m2=3.14×254m2
=78.54m2=19.625m2
(ii) In the 2nd case, radius = 10 m.
The area of the quadrant of this circle
=π×1024m2=3.14×1004m2
=3144m2=78.5m2
Therefore, Increase in the grazing area
=(78.5−19.625)m2=58.875m2
Q.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 sector as shown in figure. Find :
(i) The total length of the silver wire required.
(ii) The area of each sector of the brooch.
Sol. (i) Silver wire used to make a brooch
=2πr,wherer=352mm
=2×227×352mm=110mm
Wire used in 5 diameters = 5 × 35 mm = 175 mm
Therefore, Total wire used = (110 + 175) mm = 285 mm
(ii) The area of each sector of the brooch
=110× Area of the circle
=110×πr2,wherer=352mm
=(110×227×352×352)mm2
=3854mm2
Q.10 An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
Sol. Area between two consecutive ribs
=18×πr2,wherer=45cm
=(18×227×45×45)cm2
=2227528cm2
Q.11 A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115º. Find the total area cleaned at each sweep of the blades.
Sol. Here r = 25 cm, θ=115o
Total area cleaned at each sweep of the blades
= 2 × Area of the sector (having r = 25 and θ=115o
=2×θ360×πr2
=(2×115360×227×25×25)cm2
=(23×11×62518×7)cm2=158125126cm2
Q.12 To warn ships for underwater rocks a lighthouse spreads a red coloured light over a sector of angle 80º to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π=3.14)
Sol. Area of the sea over which the ships are warned
= Area of the sector (having r = 16.5 km and θ=80o)
=θ360×πr2
=(80360×3.14×16.5×16.5)sq.km
=68389.2360sq.km
=189.97km2
Q.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 3–√=1.7)
Sol. Clearly from the figure,
Area of one desgin
= Area of the sector AOB – Area (ΔAOB)
=(θ360×πr2−12×r×r×sin60o)cm2, where r = 28 cm
=(60360×227×28×28−12×28×28×3√2)cm2
=(12323−333.2)cm2
=(1232−999.63)cm2
=232.43cm2
Area of 6 such design =(6×232.43)cm2=464.8cm2
Cost of making such designs @ Rs. 0.35 per cm2
= Rs (0.35 × 464.8)
= Rs . 162.68
Q.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) p180×2πR (b) p180×πR2
(c) p360×2πR (d) p720×2πR2
Sol. We know that area A of a sector of angle θ in a circle of radius r is given by
A=θ360×πr2
But here , r = R and θ=p
Therefore, A=p360×πR2=p720×2πR2
Therefore, (D) is the correct answer.
Q.1 Find the area of the shaded region in figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
Sol. Since ROQ is a diameter, therefore, ∠RPQ=90o
In rt ∠dΔPRQ RQ2=RP2+PQ2
⇒ RQ2=72+242=49+576=625
⇒ RQ=625−−−√=25cm
Therefore, Radius r=12RQ=252cm
Area of the semi circle =12πr2=12×227×252×252cm2
=687528cm2
and area of ΔRPQ=12×RP×PQ
=(12×7×24)cm2=84cm2
Area of the shaded region
= Area of the semi circle – Area (Δ RPQ)
=(687528−84)cm2=(6875−235228)cm2
=452328cm2
Q.2 Find the area of the shaded region in figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC=40o
Sol. Area of the shaded region
= Area of sector AOC – Area of sector OBD
=(40360×π×142−40360×π×72)cm2
=19×227(196−49)cm2
=(2263×147)cm2=1543cm2
Q.3 Find the area of the shaded region in figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
Sol. Area of the square ABCD = (14)2cm2=196cm2
Diameter of the semicircles = AD or BC = 14 cm
Therefore, Radius of each semicircle = 7 cm
Area of the semicircular regions =2×12πr2=πr2
=(227×49)cm2=154cm2
Therefore, Area of the shaded portion
= Area of the square ABCD – Area of the semicircular regions
= (196−154)cm2=42cm2
Q.4 Find the area of the shaded region in 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.
Sol. Area of the circular portion
= Area of the circle – Area of the sector
=πr2−60360πr2=πr2(1−16)
=56πr2, where r = 6
=(56×227×36)cm2=6607cm2
Area of the equilateral ΔOAB
=3√4(side)2=(3√4×144)cm2
=363–√cm2
Therefore, Area of the shaded region
=(6607+363–√)cm2
Q.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 figure. Find the area of the remaining portion of the square.
Sol. The area of the whole square ABCD =42cm2=16cm2
The sum of the area of the four quadrants at the four corners of the square
= The area of a circle of radius 1 cm
=227×12cm2=227cm2
The area of the circle of diameter 2 cm,
i.e., radius 1cm2
=227×12cm2=227cm2
Therefore, Area of the remaining portion
= The area of the square ABCD
– The sum of the area of 4 quadrants at the four corners of the square
– The area of the circle of diameter 2 cm
=(16−227−227)cm2=(112−22−227)cm2
=687cm2
Q.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 figure. Find the area of the design (shaded region).
Sol. Let ABC be an equilateral triangle and let O be the circumcentre of the circumcircle of radius 32 cm.
Area of the circle =πr2
=(227×32×32)cm2
=225287cm2
Area of Δ ABC = 3 × Area of Δ BOC
=3×12×OB×OC×sinBOC
(32×32×32×sin120o)
=(3×16×32×3√2)
=7683–√cm2
Therefore, Area of the design (i.e., shaded region)
= Area of the circle – Area of Δ ABC
=(225287−7683–√)cm2
Q.7 In figure , 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.
Sol. The area of the whole square ABCD=142cm2=196cm2
The sum of the area of the four quadrants at the four corners of the square
= The area of a circle of radius 7cm2
=π(7)2cm2=(227×49)cm2=154cm2
Area of the shaded portion = The area of the square ABCD
– The sum of the area of four quadrants at the four corners of the square
=(196−154)cm2=42cm2
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Q.8 Figure depicts a racking track whose left and right ends are semicircular.
The distance between the two inner parallel line segment 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.
Sol. We have, OB = O’C = 30 m
and AB = CD = 10 m
OA = O’D = (30 + 10) m = 40 m
(i) The distance around the track around along its inner edge.
= BC + EH + 2 × circumference of the semicrircle of radius OB = 30 m
=(106+106+2×12×2π(30))m
=(212+2×227×30)m
=(212+13207)m=(1484+13207)m=28047m
(ii) Area of the track
= Area of the shaded region
= Area of rectangle ABCD + Area of rectangle EFGH
+ 2 (Area of the semicircle of radius 40 m
– Area of the semicircle with radius 30 m)
=[(10×106)+(10+106)+2{12×227×(40)2−12×227×(30)2}]m2
=[1060+1060+227(402−302)]m2
=[2120+227(40+30)(40−30)]m2
=(2120+227×70×10)m2
=(2120+2200)m2
=4320m2
Q.9 In 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.
Sol. Area of the sector =(90360×227×7×7)cm2
=772cm2
Area of Δ OCB =12×OC×OB
=12×7×7cm2
=492cm2
Therefore, The area of the segment BPC
=(772−492)cm2=282cm2=14cm2
Similarly the area of the segment AQC = 14cm2
Also, the area of the circle with DO as diameter
=(227×72×72)cm2=772cm2
Hence, the total area of the shaded region
=(14+14+772)cm2
=(28+28+772)cm2=1332cm2
=66.5cm2
Q.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 figure). Find the area of the shaded region.
(Use π=3.14and3–√=1.73205)
Sol. Let each side of the triangle be a cm. Then,
Area = 17320.5cm2
⇒ 3√4a2=17320.5 [sinceArea=3√4(side)2]
⇒ a2=17320.5×43√=17320.5×41.73205=40000
⇒ a = 200
Thus, radius of each circle is 100 cm.
Now, required area
= Area of Δ ABC – 3 ×(Area of a sector of angle 60º in a circle of 100 cm)
=[17320.5−3(60360×3.14×100×100)]cm2
=(17320.5−15700)cm2=1620.5cm2
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Q.11 On a square hand kerchief, nine circular designs each of radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.
Sol. Side of the square ABCD = AB
= 3 × diameter of circular design
= 3 × (2 × 7) cm = 42 cm
Therefore, Area of the square ABCD
=(42×42)cm2=1764cm2
Area of one circular design
=πr2=(227×7×7)cm2=154cm2
Therefore, Area of 9 such designs
=(9×154)cm2=1386cm2
Therefore, Area of the remaining portion of the handkerchief
= Area of the square ABCD – Area of 9 circular designs
=(1764−1386)cm2=378cm2
Q.12 In figure 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) quadrant OACB, (ii) Shaded region,
Sol. (i) Area of quadrant =14πr2
=14×227×(3.5)2cm2
=14×227×72×72cm2
=778cm2
(ii) Area of ΔAOD=12Base×Height
=12(OB×OD)
=12(3.5×2)cm2=72cm2
Hence, area of the shaded region
= Area of quadrant – Area of Δ AOD
=(778−72)cm2=(77−288)cm2
=498cm2
Sol. Radius of the quadrant = OB = OA2+AB2−−−−−−−−−−√
=202+202−−−−−−−−√cm
=201+1−−−−√cm
=202–√cm
Therefore, Area of quadrant OPBQ
=14πr2=14×3.14×(202–√)2cm2
=(14×3.14×800)cm2=628cm2
Area of the square OABC
=(20)2cm2=400cm2
Hence, area of the shaded region
= Area of quadrant – Area of square OABC
=(628−400)cm2=228cm2
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Q.14 AB and CD are respectively across of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If ∠AOB=30o, find the area of the shaded region.
Sol. Let A1andA2 be the areas of sectors OAB and OCD respectively. Then,
A1 = Area of a sector of angle 30º in a circle of radius 21 cm.
=(30360×227×21×21)cm2[UsingA=θ360×πr2]
=(2312)cm2
A2 = Area of a sector of angle 30º in a circle of radius 7 cm
=(30360×227×7×7)cm2
=776cm2
Area of the shaded region
=A1−A2=(2312−776)cm2
=693−776cm2
=6166cm2=3083cm2
Q.15 In 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.
Sol. Let 14 cm be the radius of the quadrant with A as the centre.
Then the area of the quadrant ABMC
=14πr2
=(14×227×196)cm2
=154cm2
Area of Δ BAC
=12×AC×AB
=(12×14×14)cm2=98cm2
Therefore, Area of the segment of the circle, BMC
= Area of the quadrant ABMC – Area of Δ BAC
= (154 – 98) cm2=56cm2
Now, since AC = AB = 14 cm and ∠BAC=90o
Therefore, By Pyrthagoras Theorem,
BC=AC2+AB2−−−−−−−−−−√
=142+142−−−−−−−−√
=142–√
Therefore, Radius of the semicircle BNC = 72–√cm
Area of the semicircle BNC = π2(72–√)2cm2
=(12×227×98)cm2=154cm2
Hence, the area of the region between two arcs BMC and BNC
= The area of the shaded region
= The area of semicircle BNC – The area of the segment of the circle BMC
=(154−56)cm2=98cm2
Q.16 Calculate the area of the the designed region in figure common between the two quadrants of circle of radius 8 cm each.
Sol. Here, 8 cm is the radius of the quadrants ABMD and BNDC.
Sum of their areas
=2×14πr2=12πr2
=(12×227×64)cm2
=7047cm2
Area of the square ABCD
=(8×8)cm2
=64cm2
Area of the designed region
= Area of the shaded region
= Sum of the area of quadrants
– Area of the square ABCD
=(7047−64)cm2=(704−4487)cm2
=2567cm2
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