
1.Number System
14
Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

Lecture1.9

Lecture1.10

Lecture1.11

Lecture1.12

Lecture1.13

Lecture1.14


2.Polynomials
10
Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Lecture2.5

Lecture2.6

Lecture2.7

Lecture2.8

Lecture2.9

Lecture2.10


3.Coordinate Geometry
8
Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7

Lecture3.8


4.Linear Equations
8
Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6

Lecture4.7

Lecture4.8


5.Euclid's Geometry
7
Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5

Lecture5.6

Lecture5.7


6.Lines and Angles
10
Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4

Lecture6.5

Lecture6.6

Lecture6.7

Lecture6.8

Lecture6.9

Lecture6.10


7.Triangles
11
Lecture7.1

Lecture7.2

Lecture7.3

Lecture7.4

Lecture7.5

Lecture7.6

Lecture7.7

Lecture7.8

Lecture7.9

Lecture7.10

Lecture7.11


8.Quadrilaterals
13
Lecture8.1

Lecture8.2

Lecture8.3

Lecture8.4

Lecture8.5

Lecture8.6

Lecture8.7

Lecture8.8

Lecture8.9

Lecture8.10

Lecture8.11

Lecture8.12

Lecture8.13


9.Area of Parallelogram
11
Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5

Lecture9.6

Lecture9.7

Lecture9.8

Lecture9.9

Lecture9.10

Lecture9.11


10.Constructions
7
Lecture10.1

Lecture10.2

Lecture10.3

Lecture10.4

Lecture10.5

Lecture10.6

Lecture10.7


11.Circles
11
Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6

Lecture11.7

Lecture11.8

Lecture11.9

Lecture11.10

Lecture11.11


12.Heron's Formula
8
Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4

Lecture12.5

Lecture12.6

Lecture12.7

Lecture12.8


13.Surface Area and Volume
16
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

Lecture13.15

Lecture13.16


14.Statistics
15
Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8

Lecture14.9

Lecture14.10

Lecture14.11

Lecture14.12

Lecture14.13

Lecture14.14

Lecture14.15


15.Probability
8
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

Lecture15.8

NCERT Solutions – Surface Area and Volume Exercise 13.1 – 13.9
Q.1 A plastic box 1. 5 m long , 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine :
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1 m2 costs Rs 20.
Sol.
We have, Length l = 1.5 m Breadth b = 1.25 m and depth = Height. h = 65 cm = .65 m
(i) Since the plastic box is open at the top. Therefore , Plastic sheet required for making such a box.
=[2(ℓ+b)×h+ℓb]m2
=[2(1.5+1.25)×.65+1.5×1.25]m2
=[2×2.75×.65+1.875]m2
=(3.575+1.875)m2=5.45m2
(ii) Cost of 1m^{2} of sheet = Rs. 20
Therefore Total cost of 5.45m2ofsheet=Rs(5.45×20)=Rs.109
Q.2 The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs. 7.50 per m2.
Sol.
Here, Length l = 5m , Breadth b = 4 m and Height h = 3m
Area of four walls including ceiling =[2(ℓ+b)×h+ℓb]m2
=[2(5+4)×3+5×4]m2
=[2×9×3+20]m2
=(54+20)m2=74m2
Cost of white washing is Rs 7.50 per square metre.
Therefore Cost of white washing = Rs (74 × 7.50) = Rs 555
Q.3 The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs 10 per m2 is Rs 15000, find the height of the hall.
Sol.
Cost of painting the four walls = Rs 15000
Rate of painting is Rs 10 per m2
Therefore Area of four walls =(1500010)m2=1500m2
⇒ 2(ℓ+b)h=1500
⇒ Perimeter × Height = 1500
⇒ 250× Height = 1500
⇒ Height=1500250=6
Hence , the height of the hall = 6 metres.
Q.4 The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm ×10 cm × 7.5 cm can be painted out of this container?
Sol.
Length l = 22.5 cm , Breadth b = 10 cm and Height h = 7.5 cm
Surface area of one brick =2(ℓb+bh+hℓ)
=2(22.5100×10100+10100×7.5100+7.5100×22.5100)m2
=2×1100×1100(22.5×10+10×7.5+7.5×22.5)m2
=15000×(225+75+168.75)m2
=15000×468.75m2=0.09375m2
Area for which the paint is just sufficient is 9.375 m2
Therefore Number of bricks that can be painted with the available paint =9.3750.09375=100 bricks
Q.5 A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much ?
(ii) Which box has the smaller total surface area and by how much ?
Sol.
(i) Lateral surface area of cubical box of edge 10cm=4×102cm2=400cm2
Lateral surface area of cuboid box =2(ℓ+b)×h
=2×(12.5+10)×8cm2
=2×22.5×8cm2
=360cm2
Thus, lateral surface area of the cubical box is greater and is more by (400 – 360) cm2 i.e., 40cm2
(ii) Total surface area of cubical box of edge 10 cm
=6×102cm2=600cm2
Total surface area of cuboidal box =2(ℓb+bh+hℓ)
=2(12.5×10+10×8+8×12.5)cm2
=2(125+80+100)cm2
=(2×305)cm2=610cm2
Thus, total surface area of cubical box is smaller by 10cm2
Q.6 A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?
Sol.
Here, l = 30 cm , b = 25 cm and h = 25 cm
(i) Area of the glass = Total surface area =2(ℓb+bh+hℓ)
=2(30×25+25×25+25×30)cm2
=2(750+625+750)cm2
=(2×2125)cm2=4250cm2
(ii) Tap needed for all the 12 edges =Thesumofalltheedges
=4(ℓ+b+h)=4(30+25+25)cm
=4×80cm=320cm
Q.7 Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm . For all the overlaps, 5 % of the total surface area is required extra. If the cost of the cardboard is Rs 4 for 1000 cm2, Find the cost of cardboard required for supplying 250 boxes of each kind.
Sol.
In case of bigger box :
l = 25 cm , b = 20 cm and h = 5 cm
Totalsurfacearea=2(ℓb+bh+hℓ)
=2(25×20+20×5+5×25)cm2
=2(500+100+125)cm2
=(2×725)cm2
=1450cm2
In case of smaller box :
l = 15 cm , b = 12 cm and h = 5 cm
Total surface area =2(ℓb+bh+hℓ)
=2(15×12+12×5+5×15)cm2
=2(180+60+75)cm2
=(2×315)cm2=630cm2
Total surface area of 250 boxes of each type
=250(1450+630)cm2
=(250×2080)cm2=520000cm2
Cardboard required (i.e., including 5% extra for overlaps etc.)
=(520000×105100)cm2
=546000cm2
Cost of 1000cm2 of cardboard
=Rs4
Therefore Total cost of cardboard
=Rs(5460001000×4)=Rs2184
Q.8 Parveen wanted to make a temporary shelter for her car, by making a box like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m?
Sol.
Dimensions of the box like structure are l = 4m, b = 3m and h = 2.5 m.
Since there is no tarpaulin for the floor.
Therefore Tarpaulin required =[2(ℓ+b)×h+ℓb]m2
=[2(4+3)×2.5+4×3]m2
=[2×7×2.5+12]m2
=[35×12]m2=47m2
Assume π=227, unless stated otherwise.
Q.1 The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder.
Sol.
Let r be the radius of the base and h = 14 cm be the height of the cylinder. Then,
Curved surface area of cylinder =2πrh
⇒ 88=2×227×r×14
⇒ r=88×72×22×14=1
Therefore Diameter of the base = 2r =2×1=2cm
Q.2 It is required to make a closed cylindrical tank of height 1m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same?
Sol.
Let r be the radius of the base and h be the height of the cylinder.
Base diameter = 140 cm, radius of base = 1402 = 70 cm = .70 m
Height = 1m
Metal sheet required to make a closed cylindrical tank
= Its total surface area
=2πr(h+r) =2×227×0.7(1+0.70)m2
[Because h = 1 m, r = 1402 cm = 70 cm = 0.70 cm]
=2×22×0.1×1.70m2
=7.48m2
Hence, the sheet required =7.48m2
Q.3 A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see figure). Find its.
(i) Inner curved surface area,
(ii) Outer curved surface area,
(iii) Total surface area.
Sol.
We have, R = external radius =4.42cm=2.2cm
r = internal radius =42cm=2cm
h = length of the pipe = 77 cm
(i) Inner curved surface =2πrhcm2
=2×227×2×77cm2
=968cm2
(ii) Outer curved surface =2πRhcm2
=2×227×2.2×77cm2
=1064.8cm2
(iii) Total surface area of a pipe
= Inner curved surface area + outer curved surface area + areas of two bases
= 2πrh+2πRh+2π(R2−r2)
= [968+1064.8+2×227(4.84−4)] cm^{2}
= (2032.8+447×0.84)cm2
=(2032.8+5.28)cm2=2038.08cm2
Q.4 The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2.
Sol.
The length of the roller is 120 cm i.e., h = 1.2 m and ,
radius of the cylinder (i.e., roller) =842cm=42cm=0.42m.
Distance covered by roller in one revolution
= Its curved surface area = 2πrh
= (2×227×0.42×1.2)m2
= 3.168m2
Area of the playground = Distance covered by roller in 500 revolution.
= (500×3.168)m2=1584m2
Hence , the area of playground is 1584m2
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Q.5 A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curve surface of the pillar at the rate of Rs 12.50 per m2.
Sol.
Let r be the radius of the base and h be the height of the pillar.
Therefore r=502 cm = 25 cm = .25 m and h = 3.5 m.
Curved surface = 2πrh
=(2×227×0.25×3.5)m2=5.5m2
Cost of painting the curved surface @ Rs. 12. 50 per m2
=(5.5×12.5) = Rs. 68.75
Q.6 Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder 0.7 m, Find its height.
Sol.
Let r be the radius of the base and h be the height of the cylinder. Then,
Curved surface area = 4.4 m2
⇒ 2πrh=4.4
⇒ 2×227×0.7×h=4.4 [since r = 0.7]
⇒ h=(4.4×72×22×0.7)m=1m
Thus, the height of the cylinder = 1 metre.
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Q.7 The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find.
(i) Its inner curved surface area.
(ii) The cost of plastering this curved surface at the rate of Rs 40 per m2
Sol.
(i) Let r be the radius of the face and h be depth of the well. Then,
Curved surface = 2πrh
= (2×227×3.52×10)m2=110m2
(ii) Cost of plastering is Rs 40 per m2
Therefore , cost of plastering the curved surface = Rs (110 × 40) = Rs 4400.
Q.8 In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Sol.
Total radiating surface in the system
= Curved surface area of the pipe = 2πrh
[where r=52cm=2.5cm=2.5100m=0.025mandh=28m]
=(2×227×0.025×28)m2=4.4m2
Q.9 Find
(i) the lateral or curved surface area of cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) How much steel was actually used, if 112 of the steel actually used was wasted in making the closed tank.
Sol.
(i) Here, r=(4.22)m=2.1mandh=4.5m
Lateral surface area =2πrhm2
=(2×227×2.1×4.5)m2
=59.4m2
(ii) Since 112 of the actual steel used was wasted,
the area of the steel which has gone into the tank = (1−112) of x=1112ofx.
Steel used = Letarl surface area + 2 x Area of base
= (2πrh+2πr2)cm2 = (59.4+2×227×2.1×2.1)cm2
= (59.4+27.72)cm=87.12cm2
Therefore, actual steel used =1112×x=87.12
⇒ x=87.12×1211=95.04m2
Hence, the actual area of the steel used = 95.04m2
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Q.10 In figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.Sol.
Here , r=(202) cm = 10 cm and h = 30 cm + 2 × 2.5 cm (i.e., margin) = 35 cm
Cloth required for covering the lampshade
= Its curved surface area = 2πrh
= (2×227×10×35)cm2
= 2200cm2
Q.11 The students of a Vidyalaya were asked to participate in a competition for making and decorating pen holders in the shape of a cylinder with a base, using cardboard. Each pen holder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors , how much cardboard was required to be bought for the competition?
Sol.
Cardboard required by each competitor
= Curved surface area of one penholder + base area = 2πrh+πr2
[ wherer=3cm,h=10.5cm]
= [(2×227×3×10.5)+227×9]cm2
= (198+28.28)cm2 = 226.28cm2(approx)
Cardboard required for 35 competitors = (35×226.28)cm2
= 7920cm2(approx)
Assume π=227, unless stated otherwise.
Q.1 Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Sol.
Here, radiusr=(10.52) cm = 5.25 cm and slant height( l) = 10 cm
Curved surface area of the cone = (πrℓ)cm2
=(227×5.25×10)cm2
=165cm2
Q.2 Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Sol.
Here, radius r=(242) m = 12 m and slant height( l)= 21 m.
Total surface area of the cone =(πrℓ+πr2)m2
=πr(ℓ+r)m2
=227×12×(21+12)m2
=(227×12×33)m2
=1244.57m2(approx)
Q.3 Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find
(i) radius of the base and
(ii) total surface area of the cone.
Sol.
(i) Curved surface of a cone = 308 cm2
Slant height ℓ=14cm
Let r be the radius of the base .
Therefore πrℓ=308
⇒ 227×r×14=308
⇒ r=308×722×14=7 cm
Thus, the radius of the base = 7 cm
(ii) Total surface area of the cone =πr(ℓ+r)cm2
=227×7×(14+7)cm2
=(22×21)cm2
=462cm2
Q.4 A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) Slant height of the tent.
(ii) Cost of the canvas required to make the tent, if the cost of 1m2 canvas is Rs 70.
Sol.
(i) Here r = 24 m , h = 10 m
Let l be the slant height of the cone. Then,
ℓ2=h2+r2
⇒ ℓ=h2+r2−−−−−−√
⇒ ℓ=242+102−−−−−−−−√
=576+100−−−−−−−−√
=676−−−√=26m
(ii) Canvas required to make the conical tent = Curved surface of the cone
πrℓ=(227×24×26)m2
Rate of canvas per 1m2 is Rs 70
Therefore cost of canvas =(227×24×26×70) = Rs 137280
Q.5 What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use π=3.14)
Sol.
Let r m be the radius , h m be the height and l m be the slant height of the tent. Then ,
r = 6 m , h = 8m
⇒ ℓ=r2+h2−−−−−−√=62+82−−−−−−√=36+64−−−−−−√
=100−−−√=10m
Area of the canvas used for the tent = curved surface area of the tent
=πrℓ=(3.14×6×10)m2=188.4m2
Now, this area is bought in the form of a rectangle of width 3m.
Therefore length of tarpaulin required =AreaoftarpaulinrequiredWidthoftarpaulin
=(188.43)m=62.8m
The extra material required for stitching margins and cutting =20cm=0.2m
So, the total length of tarpaulin required =(62.8+0.2)m=63m
Q.6 The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white – washing its curved surface at the rate of Rs 210 per 100 m2.
Sol.
Here , slant height (l) = 25 m and radius r=(142)m=7m
Curved surface area = πrℓm2
=(227×25×7)m2
=550m2
Rate of white washing is Rs 210 per 100m2
Therefore , cost of white – washing the tomb =(550×210100)=Rs1155
Q.7 A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Sol.
Let r cm be the radius, h cm be the height and l cm be the slant height of the joker’s cap. Then ,
r = 7 cm , h = 24 cm
ℓ=h2+r2−−−−−−√=242+72−−−−−−−√
=576+49−−−−−−−√
=625−−−√=25cm
Sheet required for one cap = Curved surface of the cone
=πrℓcm2
=(227×7×25)cm2
=550cm2
Sheet required for 10 such caps = 5500cm2
Q.8 A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1m. If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per m2, what will be the cost of painting all these cones? (Use π=3.14 and take 1.04−−−−√=1.02).
Sol.
Let r be the radius , h be the height and l be the slant height of a cone. Then ,
r=(402)cm = 20 cm = .2 m,
and h = 1m.
Therefore ℓ=r2+h2−−−−−−√=0.04+1−−−−−−−√=1.04−−−−√=1.02
Curved surface of 1 cone =πrℓm2
=(3.14×.2×1.02)m2
Curved surface of such 50 cones =(50×3.14×.2×1.02)m2
Cost of painting @ Rs. 12 per m2 =(50×3.14×.2×1.02×12)
=384.68(approx)
Q.1 Find the surface area of a sphere of radius :
(i) 10. 5 cm (ii) 5.6 cm (iii) 14 cm
Sol.
(i) We have :
r = radius of the sphere = 10.5 cm
Surface area =4πr2
=(4×227×10.5×10.5)cm2
=1386cm2
(ii) We have :
r = radius of the sphere = 5.6 cm
Surface area =4πr2 =(4×227×5.6×5.6)cm2
=394.24cm2
(iii) We have :
r = radius of the sphere = 14 cm
Surface area =(4×227×14×14)cm2
=2464cm2
Q.2 Find the surface area of a sphere of diameter :
(i) 14 cm (ii) 21 cm (iii) 3.5 m
Sol.
(i) Here , r=(142)cm=7cm
Surface area = 4πr2 =(4×227×7×7)cm2
=616cm2
(ii) Here , r=(212)cm=10.5cm
Therefore Surface area = 4πr2 =(4×227×10.5×10.5)cm2
=1386cm2
(iii) Here , r=(3.52)cm=1.75m
=(4×227×1.75×1.75)m2
=38.5m2
Q.3 Find the total surface area of a hemisphere of radius 10 cm (Use π=3.14)
Sol.
Here , r = 10 cm
Total surface area of hemisphere =3πr2 =(3×3.14×10×10)cm2
= 942cm2
Q.4 The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Sol.
Let r1andr2 be the radius of balloons in the two cases.
Here , r1=7cmandr2=14cm
Therefore , ratio of their surface area =4πr124πr22=r21r22
=7×714×14=14
Thus, the required ratio of their surface areas = 1 : 4
Q.5 A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin plating it on the inside at the rate of Rs 16 per 100 cm2
Sol.
Here r=(10.52)cm=5.25cm
Curved surface area of the hemisphere =2πr2=(2×227×5.25×5.25)cm2
=173.25cm2
Rate of tin – plating is Rs 16 per 100 cm2.
Therefore Cost of tin – plating the hemisphere =(173.25×16100)
=Rs.27.72
Q.6 Find the radius of a sphere whose surface area is 154cm2
Sol.
Let r be the radius of the sphere.
Surface area = 154cm2
⇒ 4πr2=154
⇒ 4×227×r2=154
⇒ r2=154×74×22=12.25
⇒ r=12.25−−−−√=3.5
Thus, the radius of the sphere is 3.5 cm
Q.7 The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Sol.
Let the diameter of earth be R and that of the moon will be R4
The radii of moon and earth are R8andR2 respectively.
Ratio of their surface area =4π(R8)24π(R2)2=16414
=164×41=116i.e.,1:16
Q.8 A hemispherical bowl is made of steel , 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Sol.
Inner radius r = 5 cm
Thickness of steel = 0.25 cm
Therefore , outer radius R = (r + 0.25) cm
= (5 + 0.25) cm = 5.25
Therefore Outer curved surface = 2πR2
=(2×227×5.25×5.25)cm2
=173.25cm2
Q.9 A right circular cylinder just encloses a sphere of radius r (see figure). Find
(i) Surface area of the sphere,
(ii) Curved surface area of the cylinder,
(iii) Ratio of the areas obtained in (i) and (ii).
Sol.
(i) The radius of the sphere r, so its surface area = 4πr2
(ii) Since the right circular cylinder just encloses a sphere of radius r. So the radius of cylinder = r and its height = 2r
Curved surface of cylinder =2πrh
2πr(2r) [Since r = r , h = r]
= 4πr2
(iii) Ratio of area = 4πr2:4πr2=1:1
Q.1 A matchbox measures 4 cm × 2.5 cm × 1.5 cm . What will be the volume of a packet containing 12 such boxes?
Sol.
Here , l = 4 cm, b = 2.5 cm and h = 1.5 cm
Therefore Volume of one matchbox =ℓ×b×hcm3
=(4×2.5×1.5)cm3=15cm3
Therefore Volume of a packet containing 12 such boxes =(12×15)cm3=180cm3
Q.2 A cuboid water tank is 6 m long, 5 m wide and 4.5 m deep. How many liters of water can it hold? (1m3=1000ℓ).
Sol.
Here , l = 6 m , b = 5 m and h = 4.5 m
Therefore Volume of the tank =ℓbhm3
=(6×5×4.5)m3=135m3
Therefore , the tank can hold = 135 × 1000 litres [Since 1m3=1000litres]
= 135000 litres of water.
Q.3 A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic meters of a liquid?
Sol.
Here, Length = 10 m , Breadth = 8 m and Volume = 380m3
Volume of cuboid = Length x Breadth x Height
Height=VolumeofcuboidLength×Breadth
=38010×8m
=4.75m
Q.4 Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3m deep at the rate of Rs 30 per m3.
Sol.
Here, l = 8 m, b = 6 m and h = 3 m
Volume of the pit =ℓbhm3
= (8×6×3)m3=144m3
Rate of digging is Rs 30 per m3
Therefore Cost of digging the pit = Rs (144 × 30) = Rs 4320.
Q.5 The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Sol.
Here, length = 2.5 m, depth = 10 m and volume = 50000 litres
=(50000×11000)m3[1m3=1000litres]
=50m3
Breadth=VolumeofcuboidLength×Depth=(5025×10)m=2m
Q.6 A village having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m ×6 m . For how many days will the water of this tank last?
Sol.
Here , l = 20 m , b = 15 m and h = 6 m
Therefore Capacity of the tank =ℓbhm3
=(20×15×6)m3=1800m3
Water requirement per person per day =150litres
Water required for 4000 person per day =(4000×150)ℓ
=(4000×1501000)m3=600m3
Number of days the water will last =CapacityoftankTotalwaterrequiredperday
=(1800600)=3
Thus, the water will last for 30 days.
Q.7 A godown measures 60 m × 25 m × 10 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.
Sol.
Volume of the godown =(60×25×10)m3 =15000m3
Volume of 1 crates =(1.5×1.25×0.5)m3 =0.9375m3
Number of crates that can be stored in the godown
=VolumeofthegodownVolumof1crate
=150000.9375=16000
Q.8 A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
Sol.
Let V1 be volume of the cube of edge 12 cm . So , length = breadth = height = 12 cm
Volume of the cube =(12×12×12)cm3
and V2 = Volume of the cube cut out of the first one
=18×V1=(18×12×12×12)cm3
=(6×6×6)cm3
Therefore , side of the new cube = 6 cm
Ratio of their surface areas =6(side)26(side)2=6×12×126×6×6
=41i.e.,4:1
Q.9 A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Sol.
Since the water flows at the rate of 2 km per hour, the water from 2 km of river flows into the sea in one hour.
Therefore The volume of water flowing into the sea in one hour = Volume of the cuboid
=ℓ×b×hm3
=(2000×40×3)m3
Therefore , the volume of water flowing into the sea in one minute
=(2000×40×360)m3
=4000m3
Assume π=227, unless stated otherwise.
Q.1 The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many liters of water can it hold? (1000cm3=1ℓ)
Sol.
Let r cm be the radius of the base and h cm be the height of the cylinder.
Circumference of the base = 132 cm
⇒ 2πr=132
⇒ 2×227×r=132
⇒ r=(132×72×22)cm=21cm
Volume of the cylinder = πr2hcm3
=(227×21×21×25)cm3
=34650cm3
Therefore Vessel can hold =(346501000)litres
i.e., 34.65 litres of water.
Q.2 The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1cm3 of wood has a mass of 0.6 g.
Sol.
We have h = Height of the cylindrical pipe = 35 cm
R = External radius =(282)cm=14cm
r = Internal radius =(242)cm=12cm
Volume of the wood used in making the pipe = Volume the external cylinder – Volume of the internal cylinder
πR2h−πr2h=π(R2−r2)h
=227×(142−122)×35cm3
=227×26×2×35cm3
=5720cm3
Weight of 1cm3=0.6g
Therefore Weight of 5720 cm3=(5720×0.6)g
=(5720×0.61000)kg [1 kg = 1000 g]
=3.432kg
Q.3 A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much ?
Sol.
(i) Capacity of tin can =ℓbhcm3
=(5×4×15)cm3
=300cm3
(ii) Capacity of plastic cylinder =πr2hcm3
=(227×72×72×10)cm
=385cm3
Thus , the plastic cylinder has greater capacity by (385 – 300) = 85 cm^{3}
Q.4 If the lateral surface of a cylinder is 94.2cm2 and its height is 5 cm , then find (i) radius of its base (ii) its volume (Use π=3.14)
Sol.
(i) Let r be the radius of the base and h be the height of the cylinder. Then ,
Lateral surface =94.2cm2
⇒ 2πrh=94.2
⇒ 2 × 3.14 × r × 5 = 94.2
⇒ r=94.22×3.14×5=3
Thus , the radius of its base = 3 cm
(ii) Volume of the cylinder =πr2h
=(3.14×32×5)cm3
=141.3cm3
Q.5 It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per m2, Find.
(i) Inner curved surface area of the vessel,
(ii) Radius of the base,
(iii) Capacity of the vessel.
Sol.
(i) Inner curved surface area of the vessel
=TotalcostofpaintingRateofpainting
=(220020)m2=110m2
(ii) Let r be the radius of the base and h be the height of the cylindrical vessel.
Therefore 2πrh=110
⇒ 2×227×r×10=110
⇒ r×110×72×22×10=74=1.75
Thus, the radius of the base = 1.75 m
(iii) Capacity of the vessel = πr2h
=(227×74×74×10)m3
=96.25m3
Q.6 The capacity of a closed cylindrical vessel height 1 m is 15.4 litres. How many square metres metal sheet would be needed to make it?
Sol.
Capacity of a closed cylindrical vessel = 15. 4 litres
=(15.4×11000)m3=0.0154m3 [1 m^{3}= 1000 liters]
Let r be the radius of the base and h be the height of the vessel. Then ,
Volume=πr2h=πr2×1=πr2 [Since h = 1m]
Therefore πr2=0.0154
⇒ 227×r2=0.0154
⇒ r2=0.0154×722=0.0049
⇒ r=0.0049−−−−−√=0.07
Thus, the radius of the base of vessel = 0.07 m
Metal sheet needed to make the vessel = Total surface area of the vessel
=2πrh+2πr2=2πr(h+r)
=2×227×0.07×(1+0.07)m2
=44×0.01×1.07m2=0.4708m2
Q.7 A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
Sol.
Diameter of the graphite cylinder = 1 mm = 110 cm
Therefore Radius=120cm
Length of the graphite = 14 cm
Volume of the graphite cylinder =πr2h
=(227×120×120×14)cm3=0.11cm3
Diameter of the pencil = 7mm = 710cm=0.11cm3
Therefore Radius of the pencil = 720cm
and , length of the pencil = 14 cm
Therefore Volume of the pencil =πr2h
=(227×720×720×14)cm3
=5.39cm3
Volume of wood = Volume of the pencil – Volume of the graphite
=(5.39−0.11)cm3
=5.28cm3
Q.8 A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
Sol.
Diameter of the cylindrical bowl = 7 cm
Therefore Radius = 72cm
Height of serving bowl = 4 cm
Therefore Soup saved in on serving = Volume of the bowl
=πr2h
=(227×72×72×4)cm3
=1.54cm3
Soup served to 250 patients = (250×1.54)cm3
=38500cm3i.e.,38.5ℓ
Hence , the hospital has to prepare 38.5 l soup daily to serve 250 patients.
Assume π=227, unless stated otherwise.
Q.1 Find the volume of the right circular cone with
(i) Radius 6 cm , height 7 cm
(ii) Radius 3.5 cm , height 12 cm
Sol.
(i) Here , r = 6 cm and h = 7 cm
Volume of the cone =13πr2h
=(13×227×6×6×7)cm3
=264cm3
(ii) Here , r = 3.5 cm and h = 12 cm
Volume of the cone =13πr2h
=(13×227×3.5×3.5×12)cm3
=154cm3.
Q.2 Find the capacity in litres of a conical vessel with
(i) Radius 7 cm , slant height 25 cm
(ii) Height 12 cm, slant height 13 cm
Sol.
(i) Here , r = 7 cm and l = 25 cm
Let the height of the cone be h cm. Then ,
h2=ℓ2−r2=252−72
=625−49=576
⇒ h=576−−−√=24cm
Volume of the conical vessel =13πr2h
=(13×227×7×7×24)cm3=1232cm3
Therefore , capacity of the vessel in litres =(12321000)ℓ=1.232ℓ
(ii) Here , h = 12 cm and l = 13 cm
Let the radius of the base of the cone be r cm . Then,
r2=ℓ2−h2=132−122
=169−144=25
⇒ r=25−−√=5cm
Volume of the conical vessel =13πr2h
=(13×227×5×5×12)cm3=22007cm3
Therefore Capacity of the vessel in litres =(22007×11000)ℓ=1135ℓ
Q.3 The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base. (Use π=3.14)
Sol.
Here , h = 15 cm and volume = 1570 cm3
Let the radius of the base of cone be r cm.
Therefore Volume = 1570cm3
⇒ 13πr2h=1570
⇒ 13×3.14×r2×15=1570
⇒ r2=15703.14×5=100
⇒ r=100−−−√=10
Thus , the radius of the base of cone is 10 cm.
Q.4 If the volume of a right circular cone of height 9 cm is 48πcm3, find the diameter of its base.
Sol.
Here , h = 9 cm and volume = 48πcm3,
Let the radius of the base of the cone be r cm
Therefore Volume =48πcm3,
⇒ 13πr2h=48π
⇒ 13×r2×9=48
⇒ 3r2=48
⇒ r2=483=16
⇒ r=16−−√=4
Thus, the diameter of the base of the cone = 2X4=8 cm
Q.5 A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Sol.
Diameter of the top of the conical pit = 3.5 m
Therefore , radius =(3.52)m=1.75m
Depth of the pit i.e., height = 12 m
Volume =13πr2h
=(13×227×1.75×1.75×12)m3
=38.5m3 [1m3 = 1 kilolitres]
Therefore Capacity of pit = 38.5 kilolitres.
Q.6 The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find
(i) Height of the cone.
(ii) Slant height of the cone,
(iii) Curved surface area of the cone.
Sol.
(i) Diameter of the base of the cone = 28 cm
Therefore , radius = r = (282)cm=14cm
Volume of the cone = 9856cm3
Let the height of the cone be h cm.
Now, volume =9856cm3
⇒ 13πr2h=9856
⇒ 13×227×14×14×h=9856
⇒ h=9856×3×722×14×14=48 cm
Thus , the height of the cone = 48 cm
(ii) Here , r = 14 m and h = 48 cm
Let l be the slant height of the cone. Then,
ℓ2=h2+r2=482+142
=2304+196=2500
⇒ ℓ=2500−−−−√=50
Thus, the slant height of the cone = 50 cm.
(iii) Here , r = 14 m and l = 50 cm.
Curved surface area = πrℓ
=(227×14×50)cm2
=2200cm2
Q.7 A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Sol. On revolving the right Δ ABC about the side AB ( = 12 cm) , we get a cone as shown in the figure.
Volume of solid so obtained =13πr2h
=(13×π×25×12)cm3
=100πcm3
Q.8 If the triangle ABC in the question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in question 7 and 8.
Sol.
On revolving the right Δ ABC about the side BC( = 5 cm), we get a cone as shown in the figure,
Volume of solid so obtained =13πr2h
=13×π×12×12×5cm3 [h = 5 cm , r = 12 cm]
= 240πcm3
Therefore Ratio of their volumes = 100π:240π (i.e., of Q. 7 and Q. 8) = 5 : 12
Q. 9 A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Sol.
Diameter of the base of the cone = 10.5 m
Therefore , radius = r = (10.52)m=5.25m
Height of the cone = 3m
Therefore Volume of the cone (heap) =13πr2h
=(13×227×5.25×5.25×3)m3
=86.625m3
To find the slant height l :
We have, ℓ2=h2+r2=32+(5.25)2
=9+27.5625=36.5625
⇒ ℓ=36.5625−−−−−−√=6.0467(approx)
Canvas required to protect wheat from rain = Curve surface area
=πrℓ=(227×5.25×6.0467)m2
=99.77m2(approx)
( Assume π=227, unless stated otherwise. )
Q.1 Find the volume of a sphere whose radius is
(i) 7 cm (ii) 0.63 cm
Sol.
(i) We have : r = radius of the sphere = 7 cm
Therefore, Volume of the sphere =43πr3
=(43×227×7×7×7)cm3
=43123cm3=143713cm3
(ii) We have : r = radius of the sphere = 0.63 m
Therefore, Volume of the sphere =43πr3
=(43×227×0.63×0.63×0.63)m3
=1.05m3(approx)
Q.2 Find the amount of water displaced by a solid spherical ball of diameter.
(i) 28 cm (ii) 0.21 m
Sol.
(i) Diameter of the spherical ball = 28 cm
Therefore, Radius =(282)cm=14cm
Amount of water displaced by the spherical ball
=Itsvolume=43πr3
=(43×227×14×14×14)cm3
=344963cm3=1149823cm3
(ii) Diameter of the spherical ball = 0.21 m
Therefore, Radius =(0.212)m=0.105m
Amount of water displaced by the spherical ball
=Itsvolume=43πr3
=(43×227×0.105×0.105×0.105)m3
=0.004851m3
Q.3 The diameter of a metallic ball is 4.2 cm. What is mass of the ball, if the density of the metal is 8.9 g per cm3?
Sol.
Diameter of the ball = 4.2 cm
Therefore , Radius =(4.22)cm=2.1cm
Volume of the ball =43πr3
=(43×227×2.1×2.1×2.1)cm3
=38.808cm3
Density of the metal is 8.9g per cm3
Therefore Mass of the ball = (38.808 × 8.9) g = 345.3912 g
Q.4 The diameter of the moon is approximately one – fourth the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Sol.
Let the diameter of the moon be r. Then, the radius of the moon =r2
According to the question, diameter of the earth is 4r, so its radius =4r2=2r.
V1 = The volume of the moon =43π(r2)3
=43πr3×18
⇒ 8V1=43πr3 … (1)
and, V2 = The volume of the earth =43π(2r)3
=43πr3×8
⇒ V28=43πr3 … (2)
From (1) and (2) , we have
8V1=V28 ⇒ V1=164V2
Hence, the volume of the moon is 164 of the volume of the earth.
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Q.5 How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Sol.
Diameter of a hemispherical bowl = 10.5 cm
Therefore , its radius =(10.52)cm=5.25cm
Volume of the bowl =23πr3
=(23×227×5.25×5.25×5.25)cm3
=303.1875cm3
Hence, the hemispherical bowl can hold 303 l (approx.) of milk.
Q.6 A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Sol.
Let R cm and r cm be respectively the external and internal radii of the hemispherical vessel. Then,
R = 1.01 (as thickness = 1 cm = .01 m)
and r = 1 m.
Volume of iron used = External volume – Internal volume
=23πR3−23πr3
=23π(R3−r3)
=23×227×[(1.01)3−(1)3]m3
=4421×(1.030301−1)m3
=(4421×0.030301)m3
=0.06348m3(approx)
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Q.7 Find the volume of a sphere whose surface area is 154 cm2.
Sol.
Let r cm be the radius of the sphere.
So, surface area = 154cm2
⇒ 4πr2=154
⇒ 4×227×r2=154
⇒ r2=154×74×22=12.25
⇒ r=12.25−−−−√=3.5cm
Now , Volume =43πr3
=(43×227×3.5×3.5×3.5)cm3
=5393cm3=17923cm3
Q.8 A dome of a building is in the form of a hemisphere. From inside it was white – washed at the cost of Rs 498.96. If the cost of white washing is Rs.2.00 per square metre, find the
(i) Inside surface area of the dome,
(ii) Volume of the air inside the dome.
Sol.
(i) Inside surface area of the dome =TotalcostofwhitewashingRateofwhitewashing
= (498.962.00)m2=249.48m2
(ii) Let r be the radius of the dome.
Therefore Surface area =2πr2
⇒ 2×227×r2=249.48
⇒ r2=249.48×72×22=39.69
⇒ r=39.69−−−−√=6.3m
Volume of the air inside the dome = Volume of the dome
=23πr3=23×227×6.3×6.3×6.3m3
=523.9m3(approx)
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Q.9 Twenty seven solid iron spheres , each of radius r and surface area S are melted to form a sphere with surface area S’. Find the
(i) Radius r’ of the new sphere,
(ii) Ratio of S and S’.
Sol.
(i) Volume of 27 solid sphere of radius r=27×43πr3 … (1)
Volume of the new sphere of radius r′=43πr3 … (2)
According to the problem, we have
43πr′3=27×43πr3
⇒ r′3=27r3=(3r)3
Therefore r′=3r
(ii) Required ratio =SS′=4πr24πr′2=r2(3r)2
=r29r2=19=1:9
Q.10 A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (inmm3) is needed to fill this capsule?
Sol.
Diameter of the spherical capsule = 3.5 mm
Radius =3.52mm
=1.75mm
Medicine needed for its filling = Volume of spherical capsule
=43πr3
=(43×227×1.75×1.75×1.75)mm3
=22.46mm3(approx)
Q.1 A wooden bookshelf has external dimensions as follows : Height = 110 cm , Depth = 25 cm, Breadth = 85 cm (see figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted . If the rate of polishing is 20 paise per cm2 and the rate of painting is 10 paise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf.
Area to be polished = (110 × 85 + 2 × 85 × 25 + 2 × 25 × 110 + 4 × 75 × 5 + 2 × 110 × 5)cm2
= (9350 + 4250 + 5500 + 1500 + 1100)cm2
= 21700 cm2
Cost of polishing @ 20 paise per cm2
=(21700×20100)=4340
Area to be painted = (6 × 75 × 20 + 2 × 90 × 20 + 75 × 90)cm2
= (9000 + 3600 + 6750) cm2 = 19350 cm2
Cost of painting @ 10 paise per cm2 =(19350×10100)=Rs1935
Therefore total expenses =Rs (4340 + 1935) =Rs 6275
Q.2 The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2.
Clearly, we have to subtract the part of the sphere that is resting on the sphere while calculating the cost of silver paint.
Surface area to be silver painted = 8 (Curved surface area of the sphere – area of circle on which sphere is resting).
=8(4πR2−πr2)cm2 [where R=212cm,r=1.5cm]
=8π(4×4414−2.25)cm2
=8π(441−2.25)cm2
=8π(438.75)cm2
Therefore Cost of silver paint @ 25 paise per cm2
=Rs(8×227×438.75×25100)
=Rs(193057)
=Rs2757.86(approx)
Surface area to be black painted = 8 × curved area of cylinder.
=8×2πrh
=8×2×227×1.5×7cm2
=528cm2
Cost of black paint @ 5 paise per cm2
Total cost of painting =Rs (2757.86 + 26.40) =Rs 2784.26 (approx)
Q.3 The diameter of a sphere is decreased by 25%, By what percent does its curved surface area decrease?
Sol.
Let d be the diameter of the sphere . Then it’s surface area.
=4π(d2)2=πd2
On decreasing its diameter by 25% ,the new diameter,
d1=(75100×d)=3d4
Therefore New surface area =4π(d12)2
=4π(12×3d4)2
=4π9d264
=πd2916
Decrease in surface area = πd2−πd2.916
=πd2(1−916)
=πd2(716)
Therefore Percentage decrease in surface area = (DecreaseinsurfaceareaInitialsurfacearea×100)%
=(πd2×716×1πd2×100)%
=(70016)%=43.75%