NCERT Solutions Class 9 Mathematics Chapter 13 Surface Area and Volume have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 9 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 9 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 9 Mathematics are an important part of exams for Class 9 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 9 Mathematics and also download more latest study material for all subjects. Chapter 13 Surface Area and Volume is an important topic in Class 9, please refer to answers provided below to help you score better in exams
Chapter 13 Surface Area and Volume Class 9 Mathematics NCERT Solutions
Class 9 Mathematics students should refer to the following NCERT questions with answers for Chapter 13 Surface Area and Volume in Class 9. These NCERT Solutions with answers for Class 9 Mathematics will come in exams and help you to score good marks
Chapter 13 Surface Area and Volume NCERT Solutions Class 9 Mathematics
Exercise 13.1
Q.1) A plastic box 1.5 π long, 1.25 π wide and 65 ππ deep is to be made. It is opened 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 1π2 costs π
π 20.
Sol.1) Length of plastic box = 1.5 π
Width of plastic box = 1.25 π
Depth of plastic box = 1.25 π
(i) The area of sheet required to make the box is equal to the surface area of the box excluding the top.
Surface area of the box = Lateral surface area + Area of the base
= 2(π + π) Γ β + (π Γ π)
= 2[(1.5 + 1.25) Γ 1.25] + (1.5 Γ 1.25) π2
= (3.575 + 1.875) π2
= 5.45 π2
The sheet required required to make the box is 5.45 π2
(ii) Cost of 1 π2 of sheet = π
π 20
β΄ Cost of 5.45 π2 of sheet = π
π (20 Γ 5.45) = π
π 109
Q.2) The length, breadth and height of a room are 5π, 4π and 3π respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of π
π . 7.50 πππ π2.
Sol.2) length of the room = 5π
breadth of the room = 4π
height of the room = 3π
Area of four walls including the ceiling = 2(π + π) Γ β + (π Γ π)
= 2(5 + 4) Γ 3 + (5 Γ 4) π2
= (54 + 20) π2
= 74 π2 Cost of white washing = π
π . 7.50 πππ π2
Total cost = π
π . (74 Γ 7.50) = π
π . 555
Q.3) The floor of a rectangular hall has a perimeter 250 π. If the cost of painting the four walls at the rate of π
π . 10 πππ π2 is Rs.15000, find the height of the hall.
[Hint : Area of the four walls = Lateral surface area.]
Sol.3) Perimeter of rectangular hall = 2(π + π) = 250 π
Total cost of painting = π
π . 15000
Rate πππ π2 = π
π . 10
Area of four walls = 2(π + π)β π2 = (250 Γ β) π2
A/q
β β = 15000/2500π
β β = 6 π
Thus the height of the hall is 6 π.
Q.4) The paint in a certain container is sufficient to paint an area equal to 9.375 ππ2. How many bricks of dimensions 22.5 ππ Γ 10 ππ Γ 7.5 ππ can be painted out of this container?
Sol.4) Volume of paint = 9.375 π2 = 93750 ππ2
Dimension of brick = 22.5 ππ Γ 10 ππ Γ 7.5 ππ
Total surface area of a brick = 2(ππ + πβ + πβ) ππ2
= 2(22.5 Γ 10 + 10 Γ 7.5 + 22.5 Γ 7.5) ππ2
= 2(225 + 75 + 168.75) ππ2
= 2 Γ 468.75 ππ2 = 937.5 ππ2
Number of bricks can be painted = 93750/937.5 = 100
Q.5) A cubical box has each edge 10 ππ and another cuboidal box is 12.5 ππ long, 10 ππ wide and 8 ππ 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.5) (i) Lateral surface area of cubical box of edge 10ππ = 4 Γ 102 ππ2 = 400 ππ2
Lateral surface area of cuboid box = 2(π + π) Γ β
= 2 Γ (12.5 + 10) Γ 8 ππ2
= 2 Γ 22.5 Γ 8 ππ2 = 360 ππ2
Thus, lateral surface area of the cubical box is greater by (400 β 360) ππ2 = 40 ππ2
(ii) Total surface area of cubical box of edge 10 ππ = 6 Γ 102ππ2 = 600ππ2
Total surface area of cuboidal box = 2(ππ + πβ + πβ)
= 2(12.5 Γ 10 + 10 Γ 8 + 8 Γ 12.5)ππ2
= 2(125 + 80 + 100) ππ2
= (2 Γ 305) ππ2 = 610 ππ2
Thus, total surface area of cubical box is smaller by 10 ππ2
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.6) i) Dimensions of greenhouse:
π = 30 ππ, π = 25 ππ, β = 25 ππ
Total surface area of greenhouse = 2(ππ + πβ + πβ)
= 2(30 Γ 25 + 25 Γ 25 + 25 Γ 30) ππ2
= 2(750 + 625 + 750) ππ2
= 4250 ππ2
(ii) Length of the tape needed = 4(π + π + β)
= 4(30 + 25 + 25) ππ
= 4 Γ 80 ππ = 320 ππ
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 ππ Γ 20 ππ Γ 5 ππ and the smaller of dimensions 15 ππ Γ 12 ππ Γ 5 ππ. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs.4 for 1000 ππ2 , find the cost of cardboard required for supplying 250 boxes of each kind.
Sol.7) Dimension of bigger box = 25 ππ Γ 20 ππ Γ 5 ππ
Total surface area of bigger box = 2(ππ + πβ + πβ)
= 2(25 Γ 20 + 20 Γ 5 + 25 Γ 5) ππ2
= 2(500 + 100 + 125) ππ2
= 1450 ππ2
Dimension of smaller box = 15 ππ Γ 12 ππ Γ 5 ππ
Total surface area of smaller box = 2(ππ + πβ + πβ)
= 2(15 Γ 12 + 12 Γ 5 + 15 Γ 5) ππ2
= 2(180 + 60 + 75) ππ2
= 630 ππ2
Total surface area of 250 boxes of each type = 250(1450 + 630)ππ2
= 250 Γ 2080 ππ2 = 520000 ππ2
Extra area required = 5/100(1450 + 630) Γ 250 ππ2 = 26000 ππ2
Total Cardboard required = 520000 + 26000 ππ2 = 546000 ππ2
Total cost of cardboard sheet = π
π .
(546000 Γ 4) / 1000 = π
π . 2184
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 4π Γ 3π?
Sol.8) Dimensions of the box- like structure = 4π Γ 3π Γ 2.5
Tarpaulin only required for all the four sides and top.
Thus, Tarpaulin required = 2(π + π) Γ β + ππ
= [2(4 + 3) Γ 2.5 + 4 Γ 3] π2
= (35 Γ 12) π2
= 47 π2
Exercise 13.2
Q.1) The curved surface area of a right circular cylinder of height 14 cm is 88 ππ2. Find the diameter of the base of the cylinder.
Sol.1) Let π be the radius of the base and β = 14 ππ be the height of the cylinder.
Curved surface area of cylinder = 2ππβ = 88 ππ2
β π = 1 ππ
Thus, the diameter of the base = 2π = 2 Γ 1 = 2ππ
Q.2) It is required to make a closed cylindrical tank of height 1 π and base diameter 140 ππ from a metal sheet. How many square metres of the sheet are required for the same?
Sol.2) Let r be the radius of the base and h be the height of the cylinder.
Base diameter = 140 ππ and Height (β) = 1π
Radius of base (π) = 140/2 = 70 ππ = 0.7 π
Metal sheet required to make a closed cylindrical tank = 2ππ(β + π)
= 7.48 π2
Q.3) A metal pipe is 77 ππ long. The inner diameter of a cross section is 4 ππ, the outer diameter being 4.4 ππ (see Fig.). Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.
Sol.3) Let π be external radius and π be the internal radius β be the length of the pipe.
Q.4) The diameter of a roller is 84 ππ and its length is 120 ππ. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in π2.
Sol.4) Length of the roller (β) = 120 ππ = 1.2 π
Radius of the cylinder = 84/2 ππ = 42 ππ = 0.42 π
Total no. of revolutions = 500
Distance covered by roller in one revolution = Curved surface area = 2ππβ
= (2 Γ 22/7 Γ 0.42 Γ 1.2) π2
= 3.168 π2
Area of the playground = (500 Γ 3.168) π2 = 1584 π2
Q.5) A cylindrical pillar is 50 ππ in diameter and 3.5 π in height. Find the cost of painting the curved surface of the pillar at the rate of π
π . 12.50 πππ π2.
Sol.5) Radius of the pillar (π) = 50/2 ππ = 25 ππ = 0.25 π
Height of the pillar (h) = 3.5 m.
Rate of painting = π
π . 12.50 πππ π2
Curved surface = 2ππβ
= (2 Γ 22 / 7 Γ 0.25 Γ 3.5) π2
= 5.5 π2
Total cost of painting = (5.5 Γ 12.5) = π
π . 68.75
Q.6) Curved surface area of a right circular cylinder is 4.4 π2. If the radius of the base of the cylinder is 0.7 π, find its height.
Sol.6) Let r be the radius of the base and h be the height of the cylinder.
Curved surface area = 2ππβ = 4.4 π2
β 2 Γ 22 / 7 Γ 0.7 Γ β = 4.4
β β = 4.4/(2 Γ 22 / 7 Γ 0.7) = 1π
β β = 1π
Q.7) The inner diameter of a circular well is 3.5 π. It is 10 π deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of π
π . 40 πππ π2.
Sol.7) Radius of circular well (π) = 3.5/2 π = 1.75 π
Depth of the well (β) = 10 π
Rate of plastering = π
π . 40 πππ π2
(i) Curved surface = 2ππβ
= (2 Γ 22 / 7 Γ 1.75 Γ 10) π2 = 110 π2
(ii) Cost of plastering = π
π . (110 Γ 40) = π
π . 4400
Q.8) In a hot water heating system, there is a cylindrical pipe of length 28 π and diameter 5 ππ. Find the total radiating surface in the system.
Sol.8) Radius of the pipe (π) = 5/2 ππ = 2.5 ππ = 0.025 π
Length of the pipe (β) = 28/2 π = 14 π
Total radiating surface = Curved surface area of the pipe = 2ππβ
= (2 Γ 22 / 7 Γ 0.025 Γ 28) π2
= 4.4 π2
Q.9) Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 π in diameter and 4.5 π high.
(ii) how much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank.
Sol.9) (i) Radius of the tank (π) = 4.2/2 π = 2.1 π
Height of the tank (β) = 4.5 π
Curved surface area = 2ππβ π2
Q.10) In Fig., you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 ππ and height of 30 ππ. A margin of 2.5 ππ 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.10) Radius of the frame (π) = 20/2 ππ = 10 ππ
Height of the frame (β) = 30 ππ + 2 Γ 2.5 ππ = 35 ππ
2.5 ππ of margin will be added both side in the height.
Cloth required for covering the lampshade = curved surface area = 2ππβ
= (2 Γ 22 / 7 Γ 10 Γ 35) ππ2
= 2200 ππ2
Q.11) The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 ππ and height 10.5 ππ. 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.11) Radius of the penholder (π) = 3ππ
Height of the penholder (β) = 10.5ππ
Cardboard required by 1 competitor = CSA of one penholder + area of the base
= 2ππβ + ππ2
Exercise 13.3
Q.1) Diameter of the base of a cone is 10.5 ππ and its slant height is 10 ππ. Find its curved surface area.
Sol.1) Radius (π) = 10.5/2 ππ = 5.25 ππ
Slant height (π) = 10 ππ
Curved surface area of the cone = (πππ)ππ2
= (22 / 7 Γ 5.25 Γ 10) ππ2
= 165 ππ2
Q.2) Find the total surface area of a cone, if its slant height is 21 π and diameter of its base is 24 π.
Sol.2) Radius (π) = 24/2 π = 12 π
Slant height (π) = 21 π
Total surface area of the cone = ππ (π + π) π2
= 22/7 Γ 12 Γ (21 + 12) π2
= (22/7 Γ 12 Γ 33) π2
= 1244.57 π2
Q.3) Curved surface area of a cone is 308 ππ2 and its slant height is 14 ππ. Find
(i) radius of the base and (ii) total surface area of the cone.
Sol.3) (i) Curved surface of a cone = 308 ππ2
Slant height (π) = 14ππ
Let r be the radius of the base
β΄ πππ = 308
Q.4) A conical tent is 10 π high and the radius of its base is 24 π. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 π2 canvas is Rs.70
Sol.4) Radius of the base (π) = 24 π
Height of the conical tent (β) = 10 π
Let l be the slant height of the cone.
β΄ π2 = β2 + π2
β π = ββ2 + βπ2
β π = β102 + 242
β π = β100 + 576
β π = 26 π
(ii) Canvas required to make the conical tent = Curved surface of the cone Cost of 1 π2 canvas = βΉ70
Q.5) What length of tarpaulin 3π wide will be required to make conical tent of height 8m and base radius 6π? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20ππ (ππ π π = 3.14).
Sol.5) Radius of the base (π) = 6 π
Height of the conical tent (β) = 8 π
Let l be the slant height of the cone.
β π = β100
β π = 10 π
CSA of conical tent = πππ
= (3.14 Γ 6 Γ 10) π2 = 188.4 π2
Breadth of tarpaulin = 3 m
Let length of tarpaulin sheet required be π₯.
20 cm will be wasted in cutting.
So, the length will be (π₯ β 0.2 π)
Breadth of tarpaulin = 3 π
Area of sheet = CSA of tent
[(π₯ β 0.2 π) Γ 3] π = 188.4 π2
β π₯ β 0.2 π = 62.8 π
β π₯ = 63 π
Q.6) The slant height and base diameter of a conical tomb are 25 π and 14 π respectively.
Find the cost of white-washing its curved surface at the rate of βΉ210 per 100 π2.
Sol.6) Radius (r) = 14/2 m = 7 m
Slant height tomb (l) = 25 m
Curved surface area = Οrl π2
= (227Γ25Γ7) π2
= 550 π2
Rate of white- washing = βΉ210 per 100 π2
Total cost of white-washing the tomb = βΉ(550 Γ 210/100) = βΉ1155
Q.7) A jokerβs cap is in the form of a right circular cone of base radius 7 ππ and height 24 ππ.
Find the area of the sheet required to make 10 such caps.
Sol.7) Radius of the cone (π) = 7 ππ
Height of the cone (β) = 24 ππ
Let l be the slant height
β΄ π = ββ2 + π2
β π = β242 + 72
β π = β625
β π = 25 π
Sheet required for one cap = Curved surface of the cone
= πππ ππ2
= (22 / 7 Γ 7 Γ 25) ππ2 = 550 ππ2
Sheet required for 10 ππππ = 550 Γ 10 ππ2 = 5500 ππ2
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 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs.12 per π2, what will be the cost of painting all these cones? (ππ π π = 3.14 and take β1.04 = 1.02)
Sol.8) Radius of the cone (π) = 40/2 ππ = 20 ππ = 0.2 π
Height of the cone (β) = 1 π
Let l be the slant height of a cone.
β΄ π = ββ2 + π2
β π = β12 + 0.22
β π = β1.04
β π = 1.02 π
Rate of painting = π
π . 12 πππ π2
Curved surface of 1 ππππ = πππ π2
= (3.14 Γ 0.2 Γ 1.02) π2
= 0.64056 π2
Curved surface of such 50 cones = (50 Γ 0.64056) π2
= 32.028 π2
Cost of painting all these cones = π
π . (32.028 Γ 12)
= π
π . 384.34
Exercise 13.4
Q.1) Find the surface area of a sphere of radius:
(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm
Sol.1) (i) Radius of the sphere (π) = 10.5 ππ
Surface area = 4ππ2
Q.3) Find the total surface area of a hemisphere of radius 10 ππ. (ππ π π = 3.14)
Sol.3) π = 10 ππ
Total surface area of hemisphere = 3ππ2
= (3 Γ 3.14 Γ 10 Γ 10)ππ2
= 942 ππ2
Q.4) The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Sol.4) Let r be the initial radius and R be the increased radius of balloons.
π = 7ππ and π
= 14ππ
Ratio of the surface area = 4ππ2/4ππ
2
= π2/π
2
= 7Γ7/14Γ14 = 1/4
Thus, the ratio of surface areas = 1 : 4
Q.5) A hemispherical bowl made of brass has inner diameter 10.5 ππ. Find the cost of tinplating it on the inside at the rate of Rs.16 per 100 ππ2.
Sol.5) Radius of the bowl (π) = 10.5/2 ππ = 5.25 ππ
Curved surface area of the hemispherical bowl = 2ππ2
= (2 Γ 22/7 Γ 5.25 Γ 5.25) ππ2
= 173.25 ππ2
Rate of tin - plating is = π
π . 16 πππ 100 ππ2
Therefore, cost of 1 ππ2 = π
π . 16/100
Total cost of tin-plating the hemisphere bowl = 173.25 Γ 16 / 100
= π
π . 27.72
Q.6) Find the radius of a sphere whose surface area is 154 ππ2.
Sol.6) Let r be the radius of the sphere.
Surface area = 154 ππ2
β 4ππ2 = 154
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.7) Let the diameter of earth be r and that of the moon will be r/4
Radius of the earth = π/2
Radius of the moon = π/8
Q.8) A hemispherical bowl is made of steel, 0.25 ππ thick. The inner radius of the bowl is 5ππ. Find the outer curved surface area of the bowl.
Sol.8) Inner radius of the bowl (π) = 5 ππ
Thickness of the steel = 0.25 ππ
β΄ outer radius (π
) = (π + 0.25) ππ
= (5 + 0.25) ππ = 5.25 ππ
Outer curved surface = 2ππ2
= (2 Γ 22/7 Γ 5.25 Γ 5.25) ππ2
= 173.25 ππ2
Q.9) A right circular cylinder just encloses a sphere of radius r (see Fig.). 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.9) (i) The surface area of the sphere with raius π = 4ππ2
(ii) The right circular cylinder just encloses a sphere of radius r.
β΄ the radius of the cylinder = π and its height = 2π
β΄ Curved surface of cylinder = 2ππβ
= 2π Γ π Γ 2π = 4ππ2
(iii) Ratio of the areas = 4ππ2: 4ππ2 = 1: 1
Exercise 13.5
Q.1) A matchbox measures 4ππ Γ 2.5ππ Γ 1.5ππ. What will be the volume of a packet containing 12 such boxes?
Sol.1) Dimension of matchbox = 4ππ Γ 2.5ππ Γ 1.5ππ
π = 4 ππ, π = 2.5 ππ and β = 1.5 ππ
Volume of one matchbox = (π Γ π Γ β)
= (4 Γ 2.5 Γ 1.5) ππ3 = 15 ππ3
Volume of a packet containing 12 such boxes = (12 Γ 15) ππ3 = 180 ππ33
Q.2) A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1π3 = 1000 π)
Sol.2) Dimensions of water tank = 6π Γ 5π Γ 4.5π
π = 6π , π = 5π and β = 4.5π
Therefore Volume of the tank = ππβ π3
= (6 Γ 5 Γ 4.5)π3 = 135 π3
[πππππ 1π3 = 1000πππ‘πππ ] = 135000 πππ‘πππ of water.
Therefore , the tank can hold = 135 Γ 1000 πππ‘πππ
Q.3) A cuboidal vessel is 10 π long and 8 π wide. How high must it be made to hold 380 cubic metres of a liquid?
Sol.3) Length = 10 π , Breadth = 8 π and Volume = 380 π3
Volume of cuboid = πΏππππ‘β Γ π΅πππππ‘β Γ π»πππβπ‘
β Height = ππππ’ππ ππ ππ’ππππ/πΏππππ‘β Γ π΅πππππ‘β
= 380/10 Γ 8 π
= 4.75π
Q.4) Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of π
π . 30 πππ π3.
Sol.4) π = 8 π, π = 6 π and β = 3 π
Volume of the pit = ππβ π3
= (8 Γ 6 Γ 3)π3
= 144 π3
Rate of digging = π
π . 30 πππ π3
Total cost of digging the pit = π
π . (144 Γ 30) = π
π . 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 π and 10 π.
Sol.5) length = 2.5 π, depth = 10 π and volume = 50000 πππ‘πππ
1π3 = 1000 πππ‘πππ
β΄ 50000 πππ‘πππ = (50000/1000)π3 = 50 π3
Breadth = ππππ’ππ ππ ππ’ππππ / πΏππππ‘β Γ π΅πππππ‘β
= (50/2.5 Γ 10) π = 2 π
Q.6) A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20π Γ 15π Γ 6π. For how many days will the water of this tank last?
Sol.6) Dimension of tank = 20π Γ 15π Γ 6π
π = 20 π , π = 15 π and β = 6 π
Capacity of the tank = ππβ π3
= (20 Γ 15 Γ 6)π3
= 1800 π3
Water requirement per person per day = 150 πππ‘πππ
Water required for 4000 person per day = (4000 Γ 150) π
= 4000 Γ 150/1000
= 600 π3
Number of days the water will last = Capacity of tank Total water required per day
= (1800/600) = 3
The water will last for 3 days.
Q.7) A godown measures 40π Γ 25π Γ 15π. Find the maximum number of wooden crates each measuring 1.5π Γ 1.25π Γ 0.5π that can be stored in the godown.
Sol.7) Dimension of godown = 40 π Γ 25 π Γ 15 π
Volume of the godown = (40 Γ 25 Γ 15)π3 = 10000 π3
Dimension of crates = 1.5π Γ 1.25π Γ 0.5π
Volume of 1 crates = (1.5 Γ 1.25 Γ 0.5)π3 = 0.9375 π3
Number of crates that can be stored =Volume of the godown/Volume of 1 crate
= 10000/0.9375
= 10666.66 = 10666
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.8) Edge of the cube = 12 ππ.
Volume of the cube = (ππππ)3 ππ3
= (12 Γ 12 Γ 12)ππ3
= 1728 ππ3
Number of smaller cube = 8
Volume of the 1 smaller cube = (1728/8) ππ3 = 216 ππ3
Side of the smaller cube = π
π3 = 216
β π = 6 ππ
Surface area of the cube = 6 (π πππ)2
Ratio of their surface area = 6 Γ 12 Γ 12/6 Γ 6 Γ 6
= 4/1 = 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.9) Depth of river (β) = 3 π
Width of river (π) = 40 π
Rate of flow of water (π) = 2 ππ πππ βππ’π = (2000/60) π πππ ππππ’π‘π
= (100/3)π πππ ππππ’π‘π
Volume of water flowing into the sea in a minute = ππβ π3 = (100/3 Γ 40 Γ 3)π3
= 4000 π3
Exercise 13.6
Q.1) The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm.
How many litres of water can it hold? (1000 ππ3 = 1π)
Sol.1) Here, β = 25 ππ, 2ππ = 132 ππ.
2ππ = 132
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 1 ππ3 of wood has a mass of 0.6 g.
Sol.2)
Mass of 1 ππ3 of wood = 0.6 π
β΄ Mass of 5720 ππ3 of wood = 0.6 Γ 5720 π = 3432 π = 3.432 ππ
Q.3) A soft drink is available in two packs β
(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.3) For tin can with rectangular 6 base.
π = 5 ππ, π = 4 ππ, β = 15 ππ
Volume of the tin can = ππβ = 5 Γ 4 Γ 15 ππ3 = 300 ππ3
For plastic cylinder with circular base.
Difference in the capacities of the two containers= (385 β 300)ππ3 = 85 ππ3
Hence, the plastic cylinder with circular base has greater capacity by 85 ππ3
Q.4) If the lateral surface of a cylinder is 94.2 ππ2 and its height is 5 cm, then find
(i) radius of its base (ii) its volume (ππ π π = 3.14)
Sol.4) Here, β = 5 ππ, 2ππβ = 94.2 ππ2.
(i) 2ππβ = 94.2
β 2 Γ 3.14 Γ π Γ 5 = 94.2
Hence, base radius of the cylinder = 3 ππ
(ii) Volume of the cylinder = ππ2β
= 3.14 Γ 3 Γ 3 Γ 5 ππ3 = 141.3 ππ3
Q.5) It costs Rs. 2200 to paint the inner curved surface of a cylindrical vessel 10π deep. If the cost of painting is at the rate of π
π 20 πππ π2, find
(i) Inner curved surface area of the vessel, (ii) radius of the base,
(iii) capacity of the vessel.
Sol.5) Here, β = 10 π
(i) Inner curved surface area = Total cost / Cost of painting per sq. metre
Q.6) The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?
Sol.6) Here, β = 1 π, volume = 15.4 πππ‘πππ
= 2 Γ 22/7 Γ 0.07 (1 + 0.07)π2
= 44 Γ 0.01 Γ 1.07 π2 = 0.4708 π2
Hence, 0.4708 π2 of metal sheet would be needed
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 dimeter 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.7) Here, β = 14 ππ.
Radius of the pencil (π
) = 7/2 ππ = 0.35 ππ.
Radius of the graphite (π) = 1/2 ππ = 0.05 ππ.
Volume of the the graphite = ππ2β
= 22/7 Γ 0.05 Γ 0.05 Γ 14 ππ3 = 0.11 ππ3
Volume of the the wood = π (π
2 β π2)β
= 22/7 Γ [(0.35)2 β (0.05)2] Γ 14 ππ3
= 22/7 Γ 0.4 Γ 0.3 Γ 14 ππ3 = 5.28 ππ3
Hence, volume of the wood = 5.28 ππ3 and
volume of the graphite = 0.11 ππ3
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.8) Here, π = 7/2 ππ = 3.5 ππ, β = 4 ππ
Capacity of 1 cylindrical bowl = ππ2β
= 22/7 Γ 3.5 Γ 3.5 Γ 4 ππ3 = 154 ππ3
Hence, soup consumed by 250 patients per day = 250 Γ 154 ππ3 = 38500 ππ3
Exercise 13.7
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.1) (i) Radius (π) = 6 ππ Height (β) = 7 ππ
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.2) (i) Radius (π) = 7 ππ
Slant height (π) = 25 ππ
Let h be the height of the conical vessel.
β΄ β = βπ2 β π2
Q.3) . The height of a cone is 15 ππ. If its volume is 1570 ππ3, find the radius of the base.
(Use π = 3.14)
Sol.3) Height (β) = 15 ππ
Volume = 1570 ππ3
Let the radius of the base of cone be π ππ
β΄ Volume = 1570 ππ3
β (1/3)ππ2β = 1570
β 13 Γ 3.14 Γ π2 Γ 15 = 1570
β π2 = 1570/3.14 Γ 5 = 100
β π = 10
Q.4) If the volume of a right circular cone of height 9 cm is 48π ππ3 , find the diameter of its base
Sol.4) Height (β) = 9 ππ
Volume = 48π ππ3
Let the radius of the base of the cone be π ππ
β΄ Volume = 48π ππ3
β (1/3)ππ2β = 48π
β 13 Γ π2 Γ 9 = 48
β 3π2 = 48
β π2 = 48/3 = 16
β π = 4
Q.5) A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Sol.5) Diameter of the top of the conical pit = 3.5 π
Radius (π) = (3.5/2) π = 1.75 π
Depth of the pit (β) = 12 π
Volume = (1/3)ππ2β
= (13 Γ 22/7 Γ 1.75 Γ 1.75 Γ 12) π3
= 38.5 π3 ( 1 π3 = 1 πππππππ‘ππ )
Capacity of pit = 38.5 πππππππ‘πππ
Q.6) The volume of a right circular cone is 9856 ππ3 . 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.6) (i) Diameter of the base of the cone = 28 ππ
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.7) On revolving the Ξπ΄π΅πΆ along the side 12 ππ,
a right circular cone of height(β) 12 ππ,
radius(r) 5 cm and slant height(π) 13 ππ will be formed.
Volume of solid so obtained = 1/3 ππ 2β
= (1/3 Γ π Γ 5 Γ 5 Γ 12) ππ3 = 100π ππ3
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 Questions 7 and 8.
Sol.8) On revolving the Ξπ΄π΅πΆ along the side 12 ππ,
a cone of radius(π) 12 ππ,
height(β) 5 ππ, and
slant height(π) 13 ππ will be formed.
Volume of solid so obtained = (1/3)ππ2β
= (1/3 Γ π Γ 12 Γ 12 Γ 5) ππ3 = 240π ππ3
Ratio of the volumes = 100π/240π = 5/12 = 5: 12
Q.9) A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m.
Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required
Sol.9) Diameter of the base of the cone = 10.5 π
Exercise 13.8
Q.1) Find the volume of a sphere whose radius is (i) 7 cm (ii) 0.63 m
Sol.1) (i) Radius of the sphere(π) = 7 ππ
Therefore, Volume of the sphere = 4/3 ππ2
Q.2) Find the amount of water displaced by a solid spherical ball of diameter.
(i) 28 cm (ii) 0.21 m
Sol.2) (i) Diameter of the spherical ball = 28 ππ
Q.3) The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 π πππ ππ3?
Sol.3) Diameter of the ball = 4.2 ππ
Radius = (4.2/2) ππ = 2.1 ππ
Volume of the ball = 4/3 ππ3
= (4/3 Γ 22/7 Γ 2.1 Γ 2.1 Γ 2.1) ππ3
= 38.808 ππ3
Density of the metal is 8.9π πππ ππ3
Mass of the ball = (38.808 Γ 8.9) π = 345.3912 π
Q.4) The diameter of the moon is approximately one-fourth of the diameter of the earth.
What fraction of the volume of the earth is the volume of the moon?
Sol.4) Let the diameter of the moon be r Radius of the moon = π/2
A/q,
Diameter of the earth = 4π
Radius(π) = 4π/2 = 2π
Q.5) How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Sol.5) Diameter of a hemispherical bowl = 10.5 ππ
Radius(π) = (10.5/2) ππ = 5.25ππ
= 0.3031875 πππ‘πππ (approx.)
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.6) Internal radius = π = 1π
External radius = π
= (1 + 0.1) ππ = 1.01 ππ
Volume of iron used = External volume β Internal volume
Q.7) Find the volume of a sphere whose surface area is 154 ππ2.
Sol.7) Let π ππ be the radius of the sphere So,
surface area = 154ππ2
β 4ππ2 = 154
β 4 Γ 22/7 Γ π2 = 154
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.8) (i) Inside surface area of the dome =Total cost of white washing/Rate of white washing
= ( 498.96/2.00) π2 = 249.48 π2
(ii) Let π be the radius of the dome.
Surface area = 2ππ2
Q.9) Twenty -seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area Sβ². Find the
(i) radius rβ² of the new sphere, (ii) ratio of S and S
Sol.9)
(ii) Required ratio = π/πβ² = 4ππ2/4ππβ²2
= π 2 /(3π) 2 = π 2 /9π 2
= 1/9 = 1: 9
Q.10) A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (ππ ππ3 ) is needed to fill this capsule?
Sol.10) Diameter of the spherical capsule = 3.5 ππ
Radius(π) = 3.52ππ = 1.75ππ
Medicine needed for its filling = Volume of spherical capsule
= 4/3 ππ3
= (4/3 Γ 22 7 Γ 1.75 Γ 1.75 Γ 1.75) ππ3
= 22.46 ππ3(ππππππ₯. )
Exercise 13.9
Q.1) A wooden bookshelf has external dimensions as follows:
Height = 110 cm, Depth = 25 cm, Breadth = 85 cm [See fig.].
The thickness of the planks 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 ππππ π πππ ππ2 and the rate of painting is 10 ππππ π πππ ππ2, find the total expenses required for polishing and painting the surface of the bookshelf.
Sol.1) External faces to be polished = Area of six faces of cuboidal bookshelf β 3 (Area of open portion ABCD)
= 2 (110 Γ 25 + 25 Γ 85 + 85 Γ 110) β 3 (75 Γ 30)
[π΄π΅ = 85 β 5 β 5 = 75 ππ and π΄π·
= 1/3 Γ 110 β 5 β 5 β 5 β 5 = 30 ππ]
= 2 (2750 + 2125 + 9350) β 3 Γ 2250
= 2 Γ 14225 β 6750
= 28450 β 6750
= 21700 ππ2
Now, cost of painting outer faces of wodden bookshelf at the rate of 20 paise.
= π
π . 0.20 πππ ππ2 = π
π . 0.20 Γ 21700 = π
π . 4340
Here, three equal five sides inner faces.
Therefore, total surface area
= 3 [ 2 (30 + 75) 20 + 30 Γ 75] [ π·πππ‘β = 25 β 5 = 20 ππ]
= 3 [ 2 Γ 105 Γ 20 + 2250] = 3
[ 4200 + 2250]
= 3 Γ 6450 = 19350 ππ2
Now, cost of painting inner faces at the rate of 10 ππππ π i.e. π
π . 0.10 πππ ππ2.
= π
π . 0.10 Γ 19350 = π
π . 1935
Total expenses required = π
π . 4340 + π
π . 1935 = π
π . 6275
Q.2) The front compound wall of a house is decorated by wooden spheres of diameter 21 ππ, 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 ππ and height 7 ππ and is to be painted black. Find he cost of paint required if silver paint costs 25 ππππ π πππ ππ2 and black paint costs 5 ππππ π πππ ππ2.
Sol.2) Diameter of a wooden sphere = 21 ππ.
β΄ Radius of wooden sphere(π
) = 21/2 ππ
And Radius of the cylinder (π) = 1.5 ππ
Surface area of silver painted part = Surface area of sphere β Upper part of cylinder for support
Surface area of such type of 8 spherical part = 8 Γ 1378.928
= 11031.424 ππ2
Cost of silver paint over 1 ππ2 = π
π . 0.25
Cost of silver paint over 11031.928 ππ2 = 0.25 Γ 11031.928
= π
π . 2757.85
Now, curved surface area of a cylindrical support = 2ππβ
= 2 Γ 22 / 7 Γ 15/10 Γ 7 = 66ππ2
Curved surface area of 8 such cylindrical supports = 66 Γ 8 = 528 ππ2
Cost of black paint over 1 cπ2 of cylindrical support = π
π . 0.50
Cost of black paint over 528 cπ2 of cylindrical support = 0.50 Γ 528
= π
π . 26.40
Total cost of paint required = π
π . 2757.85 + π
π . 26.4
= π
π . 2784.25
Q.3) If diameter of a sphere is decreased by 25% then what percent does its curved surface area decrease?
Sol.3) Diameter of original sphere = π· = 2π
π
= π·/2
Curved surface area of original sphere = 4ππ
2 = 4π (π·/2)2 = ππ·2
According to the question,
Decreased diameter = 25% ππ π·
Q.4) Sameera wants to celebrate the fifth birthday of her daughter with a party. She bought thick paper to make the conical party caps. Each cap is to have a base diameter of 10 cm and height 12 cm. A sheet of the paper is 25 cm by 40 cm and approximately 82% of the sheet can be effectively used for making the caps after cutting. What is the minimum number of sheets of paper that Sameera would need to buy, if there are to be 15 children at the party? (Use Ο = 3.14)
NCERT Solutions Class 9 Mathematics Chapter 1 Number Systems |
NCERT Solutions Class 9 Mathematics Chapter 2 Polynomials |
NCERT Solutions Class 9 Mathematics Chapter 3 Coordinate Geometry |
NCERT Solutions Class 9 Mathematics Chapter 4 Linear Equations In Two Variables |
NCERT Solutions Class 9 Mathematics Chapter 5 Introduction to Euclid's Geometry |
NCERT Solutions Class 9 Mathematics Chapter 6 Lines and Angles Variables |
NCERT Solutions Class 9 Mathematics Chapter 7 Triangles |
NCERT Solutions Class 9 Mathematics Chapter 8 Quadrilaterals |
NCERT Solutions Class 9 Mathematics Chapter 9 Areas of Parallelograms and Triangles |
NCERT Solutions Class 9 Mathematics Chapter 10 Circles |
NCERT Solutions Class 9 Mathematics Chapter 11 Construction |
NCERT Solutions Class 9 Mathematics Chapter 12 Herons Formula |
NCERT Solutions Class 9 Mathematics Chapter 13 Surface Area and Volume |
NCERT Solutions Class 9 Mathematics Chapter 14 Statistics |
NCERT Solutions Class 9 Mathematics Chapter 15 Probability |
NCERT Solutions Class 9 Mathematics Chapter 13 Surface Area and Volume
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