Complete NCERT solutions for class 9 Maths Chapter 13 Surface Areas And Volumes

Solution for Exercise 13.1

Q1 ) A plastic box 1.5 m long, 1.25 m wide and 65 cm 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 1m2 costs Rs.20.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

image
Given :
length (l) of box = 1.5m
Breadth (b) of box = 1.25 m
Depth (h) of box = 0.65 m

i) Box is to be open at top,
So, Area of sheet required
= 2lh + 2bh + lb
=[2x1.5x0.65+2x1.25x0.65+1.5x1.25]
= [1.95 + 1.625 + 1.875] m2
= 5.45 m2
Hence, the area of the sheet required for making the box is 5.45 m2

ii) Cost of sheet per 1 m2 area = Rs.20
Therefore, Cost of sheet of 5.45 m2 area
= Rs. (5.45 x 20)
= Rs. 109

Hence, The cost of sheet for it is Rs. 109

Q2 ) 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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Length (l) of room = 5 m
Breadth (b) of room = 4 m
Height (h) of room = 3 m

It can be observed that four walls and the ceiling of the room are to be white- washed. The floor of the room is not to be white-washed.

So, we get,

Area to be white-washed
= Area of walls + Area of ceiling of room
= 2lh + 2bh + lb
= [2 x 5 x 3 + 2 x 4 x 3 + 5 x 4] m2
= [30 + 24 + 20] m2
= 74 m2.

Now, we have,
Cost of white-washing per m2 area = Rs. 7.50
Cost of white-washing 74 m2 area = Rs. (74 x 7.50) = Rs. 555

Therefore, the cost of white washing the walls of the room and the ceiling is Rs. 555.

Q3 ) 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.
[Hint : Area of the four walls = Lateral surface area.]


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Let the length, breadth, and height of the rectangular hall be l m, b m, and h m respectively.

Now, perimeter of the floor of hall
= 2(l + b) = 250 m ...(Given)

Also, Area of four walls = 2lh + 2bh
? Area of four walls = 2(l + b)h = 250 h m2

Now, Cost of painting per m2 area = Rs. 10

? Cost of painting 250 h area = Rs. (250h x 10) = Rs. 2500h

However, it is given that the cost of painting the walls is Rs 15000.

Hence, 15000 = 2500h
? h = 150002500
? h = 6

Therefore, the height of the hall is 6 m.

Q4 ) The paint in a certain container is suffcient 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?


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

We know that,

Total surface area of one brick
= 2(lb + bh + lh)
= [2(22.5 X 10 + 10 x 7.5 + 22.5 x 7.5)] cm2
= [2(225 + 75 + 168.75)] cm2
= [2 x 468.75] cm2
= 937.5 cm2

Now, Let m bricks can be painted out by the paint of the container.

Thus, we get,

Area of m bricks = (m x 937.5) = 937.5m cm2 ...(i)

Therefore, area that can be painted by the paint of the container = 9.375 m2 ...(ii)
= 93750 cm2

Thus, equating (i) and (ii), we get,

93750 = 937.5m
? m = 93750937.5
? m = 100

Therefore, 100 bricks can be painted out by the paint of the container.

Q5 ) 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?


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given:
Edge of cube = 10 cm
Length (l) of box = 12.5 cm
Breadth (b) of box = 10 cm
Height (h) of box = 8 cm

i) We know that, Lateral surface area of cubical box
= 4×(Edge)2
= 4×(10)2cm2
= 400 cm2

Also, We have,

Lateral surface area of cuboidal box
= 2 [lh + bh]
= [2 (12.5 x 10 x 8)] cm2
= [2 x 180] cm2
= 360 cm2

Clearly, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box.

Now, Lateral surface area of cubical box - Lateral surface area of cuboidal box
= 400 cm2 - 360 cm2 = 40 cm2

Therefore, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box by 40 cm2


ii)Similarly, Total surface area of cubical box
=6×(Edge)2
= 6×(10)2cm2
= 600 cm2

Also, We have,

Total surface area of cuboidal box
= 2(lh + bh + lb)
=[2 (12.5 x 8 + 10 x 8 + 12.5 x 100)] cm2
= 610 cm2

Clearly, the total surface area of the cubical box is smaller than that of the cuboidal box.

Now, Total surface area of cuboidal box - Total surface area of cubical box
= 610 cm2 - 600cm2
= 10 cm2

Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm2.

Q6 ) 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?


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Length (l) of greenhouse = 30 cm
Breadth (b) of greenhouse = 25 cm
Height (h) of greenhouse = 25 cm

We know that,
Total surface area of greenhouse
=2 [lb + lh + bh]
= [2 (30 x 25 + 30 x 25 + 25 x 25)] cm2
= [2(750 + 750 + 625)] cm2
= [2 x 2125] cm2
= 4250 cm2

Therefore, the area of glass is 4250 cm2.
image

It can be observed that tape is required alongside AD, OC, CD, DA, EF, FG, GH, HE, AH, BE, DG, and CF.

We have, Total length of tape
= 4(l+ b + h)
= [4(30 + 25 + 25)] cm
= 320 cm

Therefore, 320 cm tape is required for all the 12 edges.

Q7 ) 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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given:
Length (l) of bigger box = 25 cm
Breadth (b) of bigger box = 20 cm
Height (h) of bigger box = 5 cm

we know that,

Total surface area of bigger box
= 2(lb + lh + bh)
= [2 (25 x 20 + 25 x 5 + 20 x 5)] cm2
= [2(500 + 125 + 100)] cm2
= 1450 cm2

Therefore, Extra area required for overlapping
= 1450×5100 cm2 = 72.5 cm2

While considering all overlaps, total surface area of 1 bigger box
= (1450 + 72.5) cm2 = 1522.5 cm2

So, Area of cardboard sheet required for 250 such bigger boxes = (1522.5 x 250) cm2 = 380625 cm2

Similarly, total surface area of smaller box = [2 (15 + 15 x 5 + 12 x 5)]
= [2 (180 + 75 + 60)] cm2
= (2 x 315) cm2
= 630 cm2

Therefore, extra area required for overlapping
= 630×5100 cm2 = 31.5 cm2

So, Total surface area of 1 smaller box while considering all overlaps
= (630 + 31.5) cm2
= 661.5 cm2

So, Area of cardboard sheet required for 250 smaller boxes
= (250 x 661.5) cm2 = 165375 cm2

Thus, Total cardboard sheet required
= (380625 + 165375) cm2
= 546000 cm2

Cost of 1000 cm2 cardboard sheet = Rs. 4

Therefore, Cost of 546000 cm2 cardboard sheet will be:
Rs.546000×4100 cm2 = Rs. 2184

Therefore, the cost of cardboard sheet required for 250 such boxes of each kind will be Rs 2184.

Q8 ) 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 4m × 3 m?


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given:
Length (l) of shelter = 4 m
Breadth (b) of shelter = 3 m
Height (h) of shelter = 2.5 m

Tarpaulin will be required for the top and four wall sides of the shelter.

So, Area of Tarpaulin required
= 2(lh + bh) + lb
= [2(4 x 2.5 + 3 x 2.5) + 4 x 3] m2
= [2(10 + 7.5) + 12] m2
= 47 m2

Therefore, 47 m2 tarpaulin will be required.

Solution for Exercise 13.2

Q1 ) 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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Height (h) of cylinder = 14 cm
Curved surface area of cylinder = 88 cm2

Let the diameter of the cylinder be d.
So, (2?rh) cm2= 88 cm2
(where, r is the radius of the base of the cylinder)

??dh = 88 cm2 ...(Since, d = 2r)
? d = 88×722×14 cm ...(?=227)
? d = 42 cm
? d = 2 cm.

Therefore, the diameter of the base of the cylinder is 2 cm.

Q2 ) It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
height(h) of the cylindrical tank = 1m
Base radius(r) of cylindrical tank
=( 1402 )cm = 70 cm = 0.7 m

Now, Area of sheet required = total surface of tank

But, we know that,
Area of sheet required
= 2?r(r+h) m2
= 2×227×0.7(0.7+1) m2
= 4.4×1.7 m2
= 7.48 m2

Therefore , it will required 7.48 m2 area of sheet.

Q3 ) A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. Find its
i) inner curved surface area
ii) outer curved surface area
iii) total surface area.
image


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Inner radius (r) of cylindrical pipe = 42 cm = 2 cm
outer radius (R) of cylindrical pipe = 4.42 cm = 2.2 cm
Height (h) Of cylindrical pipe = Length Of cylindrical pipe = 77 cm

i) CSA of inner surface of pipe
= 2?rh
= 2×227×2×77 cm2
= 2×22×2×11 cm2
= 968 cm2

ii) CSA of inner surface of pipe
= 2?Rh
= 2×227×2.2×77
= 2×22×2.2 cm2
= 1064.8 cm2

iii) Total surface area of pipe
= CSA of inner surface + CSA of outer surface + Area of both circular ends of pipe
= 2?rh + 2?Rh + 2?[(R)2?(r)2]
= [968+1064.8+2?(2.2)2?(2)2]cm2
= 2032.8+2×227×0.84 cm2
= (2032.8 + 5.28) cm2
= 2038.08 cm2

Therefore, the total surface area of the cylindrical pipe is 2038.08 cm2.

Q4 ) 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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Height (h) of cylindrical roller = Length of roller = 120 cm
And radius (r) of the circular end roller = 842 = 42 cm

It can be observed that a roller is cylindrical.

Now, we know that,
CSA of rollar
= 2?rh
= 2×227×42×120 cm2
= 2×22×6×120
= 31680 cm2

Therefore, Area of field
= 500 x CSA of roller
= 31680×500 cm2
= 1584 m2
? the area of the playground in m2 is 1584.

Q5 ) A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Height (h) cylindrical pillar = 3.5 m
Radius (r) of the circular end of pillar = 502 cm = 25 cm = 0.25m

Now, we know that,
CSA of cylindrical pillar = 2?rh
= 2×227×0.25×3.5 m2
= 44×0.125 m2
= 5.5 m2

Now, Cost of painting 1 m2 area = Rs. 12.50
Cost of painting 5.5 m2 area = Rs. (5.5 x 12.50) = Rs. 68.75

Therefore, the cost of painting the CSA of the pillar is Rs. 68.75

Q6 ) Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Radius (r) of the base of cylinder = 0.7 m
CSA Of cylinder = 4.4 m2

Let the height of the circular cylinder be h.

Therefore, we have,

? 2?rh = 4.4 m2
? 2×227×0.7×h m = 4.4 m2
? 2×22×0.1×h m = 4.4 m2
? h = 1 m.

Therefore, the height of the cylinder is 1 m.

Q7 ) 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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Inner radius (r) of circular well = 3.52 m = 1.75 m
Depth (h) of circular well = 10 m

i) We know that,

Inner curved surface area
= 2?rh
= 2×227×1.75×10 m2
= (44 × 0.25 × 10) m2
= 110 m2

Therefore, the inner curved surface area of the circular well is 110 m2

ii) We have, Cost of plastering 1 m2 area = Rs. 40
Cost of plastering 100 m2 area = Rs. (110 x 40) = Rs. 4400

Therefore, the cost of plastering the CSA of this well is Rs. 4400.

Q8 ) 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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Height (h) of cylindrical pipe Length of cylindrical pipe = 28 m
Radius (r) of circular end of pipe
= 52 cm = 2.5 cm = 0.025 m

We know that,

CSA of cylindrical pipe
= 2?rh
= 2×227×0.025×28 m2
= 4.4 m2

Therefore, the area of the radiating surface of the system is 4.4 m2.

Q9 ) Find
i) the lateral or curved surface area of a closed 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 th of the steel actually used was wasted in making the tank.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Height (h) of the cylindrical tank = 4.5 m
radius(r) of the circular end of cylindrical tank = 4.22 m = 2.1 m

i) lateral or curved surface area of tank
= 2?rh
= 2×227×2.1×4.5 m2
= (44 × 0.3 × 4.5) m2
= 59.4 m2

Therefore , CSA of tank is 59.4 m2.

ii) Now, so as to find steel used,

total surface area
= 2?r(r+h)m2
= 2×227×2.1(2.1+4.5)m2
= 4.4×0.3×6.6m2
= 87.12 m2

Now, Let actual area of x m2 sheet be used in making the tank.

Since, 112th of the actual steel wasted, the area of steel which has gone into the tank = 1112th of x.

This means that the actual area used
= 1211th×87.12 m2 = 95.04 m2.

Therefore, 95.04 m2 steel was used in actual while making such a tank.

Q10 ) In figure below, 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.
image


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

image

Given :
Height (h) of the frame of lampshade = (2.5 + 30 + 2.5) cm = 35cm
Radius (r) of the circular end of the frame of lampshade = 202 m = 10 m

Now, Cloth required for covering the lampshade
= 2?rh
= 2×227×10×35 cm2
= (44 × 10 × 5) cm2
= 2200 cm2

Hence, for covering the lampshade, 2200 cm2 cloth will be required.

Q11 ) 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 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?


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Radius (r) of the circular end of cylindrical penholder = 3 cm
Height (h) Of penholder = 10.5 cm

We have,

Surface area of 1 penholder
= CSA of penholder + Area of base of penholder
= 2?rh + ?r2
= [2×227×3×10.5+227×32cm2]
= [132×1.5+1987cm2]
= 198+1987cm2
= 15847cm2

So, we get,

Area Of cardboard sheet used by 1 competitor = 15847cm2

Thus, Area of cardboard sheet used by 35 competitors
= (15847×35)cm2 = 7920 cm2

? 7920 cm2 cardboard sheet will be bought for the competition.

Solution for Exercise 13.3

Q1 ) Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Radius (r) of the base of cone = 10.52cm=5.25 cm
Slant height (l) of cone = 10 cm

We know that, CSA of cone
= ?rl
= [227×5.25×10 cm2]
= [22×0.75×10 cm2]
= 165 cm2

Therefore, the curved surface area of the cone is 165 cm2.

Q2 ) Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Radius (r) of the base of cone = 242cm=12cm
Slant height (l) of cone = 21 cm

We know that, TSA of cone
= ?r(r+l)
= [227×12×(12+21) m2]
= [22×1.71×33 m2
= 1244.57 m2

Therefore, the total surface area of the cone is 1244.57 m2.

Q3 ) 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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Slant height (l) of cone = 14 cm
Curved surface area of a cone is 308 cm2

i) Let the radius Of the circular end of the cone be r.

We know that,

CSA of cone
= ?rl
= [227×r×14cm]
= [22×2×rcm
= 44r cm

? 308 cm2 = 44r cm ...(Given)
? r = 30844cm = 7cm

Therefore, the radius Of the circular end Of the cone is = 7 cm.

ii) Total surface area of cone
= CSA of cone + Area of base
= [?rl + ?r2]
= [308+227×(7)2 cm2]
= [308 + 154] cm2
= 462 cm2

Therefore, the total surface area of the cone is 462 cm2.

Q4 ) 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 1 m2 canvas is Rs. 70.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

image

i) Let ABC be a conical tent.
Given :
Height (h) of conical tent = 10 m
Radius (r) of conical tent = 24 m

Let the slant height of the tent be l.
In ?ABO,
? AB2=AO2+BO2
? l2=h2+r2
? l2=102+242m2
? l2=676m2
? l2=262m
? l = 26 m

Therefore, the slant height Of the tent is 26 m.

ii) We know, CSA of tent
= ?rl
= [227×24×26m]
= 137287m2

Cost of 1 m2 canvas = Rs. 70
Cost of [137287m2] canvas
= Rs. [137287×70 m2]
= Rs. 137280.

Therefore, the cost of the canvas required to make such a tent is Rs. 137280.

Q5 ) What length of tarpaulin 3m 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 ? = 3.14).


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Height (h) of conical tent = 8 cm
Radius (r) of base of tent = 6 m
Slant height (l) of tent =
? l2=h2+r2
? l2=82+62m2
? l2=100m2
? l2=102m
? l = 10 m

Therefore, the slant height Of the tent is 10 m.

We know, CSA of tent
= ?rl
= [3.14×6×10m]
= 188.4m2

Now, let the length of tarpaulin sheet required be l.

As, 20 cm will be wasted, therefore, the effective length will be (l- 0.2) m.

Breadth of tarpaulin = 3m ...(Given)

We have, Area of sheet = CSA of tent
? [(l - 0.2) × 3]m = 188.4m2
? (l - 0.2) m = 62.8 m2
? l = 73 m.

? the length of the required tarpaulin sheet will be 63 m.

Q6 ) 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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Slant height (l) of conical tomb = 25 m
Base radius (r) of tomb = 142=7m

We know, CSA of tent = ?rl
= [227×7×25m]
= [22×1×25m]
= 550m2

Now, we know that,
Cost of white-washing 100 m2 area = Rs. 210
So, Cost of white-washing 550 m2 area
= Rs. [210×550100]
= Rs. 1155

Therefore, it will cost of Rs. 1155 while white-washing such a conical tomb.

Q7 ) 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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Radius (r) of conical cap = 7 cm
Height (h) of conical cap = 24 cm

Slant height (l) of the conical cap
? l2=h2+r2
? l2=72+242 cm2
? l2=625 cm2
? l2=252cm
? l = 25 cm

Therefore, the slant height Of the tent is 25 cm.

We know, CSA of tent
= ?rl
= [227×7×25 cm2]
= [22×1×25 cm2]
= 550 cm2

Thus, CSA of 10 such caps
= (10 x 550) cm2
= 5500 cm2

? 5500 cm2 sheet will be required.

Q8 ) 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 perm2 , what will be the cost of painting all these cones? (Use ? = 3.14 and take 1.04 = 1.02)


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Radius (r) of conical cap = 402cm=20cm=0.2m
Height (h) of conical cap = 1 m

Slant height (l) of the conical cap
? l2=h2+r2
? l2=12+0.22m2
? l2=1.04m2
? l2=1.022m
? l = 1.02 m

? the slant height Of the tent is 1.02 m.

We know, CSA of tent
= ?rl
= [3.14×0.2×1.02 m2]
= 0.64056 m2

So, CSA of 50 such cones
= 50×0.64056 m2
= 32.028 m2.

Now, we have,

Cost of painting 1 m2 area = Rs. 12
Cost of painting 32.02 m2 area = Rs. (32.028 x 12)
= Rs. 384.336
= Rs 384.34 (approximately)

Therefore, it Will cost Rs. 384.34 in painting 50 such hollow cones.

Solution for Exercise 13.4

Q1 ) Find the surface area of a sphere of radius:
i) 10.5 cm
ii) 5.6 cm
iii) 14 cm


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

i) Radius (r) of sphere = 10.5 cm
Surface area of sphere
= 4?r2
= [4×227×(10.5)2 cm2]
= [4×227×10.5×10.5 cm2]
= (88 × 1.5 × 10.5) cm2
= 1386 cm2

Therefore, the surface area of a sphere having radius 10.5cm is 1386 cm2.


ii) Radius (r) of sphere = 5.6 cm
Surface area of sphere
= 4?r2
= [4×227×(5.6)2 cm2]
= [4×227×5.6×5.6 cm2]
= (88 × 0.8 × 5.6) cm2
= 394.24 cm2

Therefore, the surface area of a sphere having radius 5.6 cm is 394.24 cm2.


iii) Radius (r) of sphere = 14 cm
Surface area of sphere
= 4?r2
= [4×227×(14)2 cm2]
= [4×227×14×14 cm2]
= (4 × 44 × 14) cm2
= 2464 cm2

Therefore, the surface area of a sphere having radius 14 cm is 2464 cm2.

Q2 ) Find the surface area of a sphere of diameter:
i) 14 cm
ii) 21 cm
iii) 3.5 m


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Answer :

i) Radius (r) of sphere = Diameter2=142=7cm

Surface area of sphere
= 4?r2
= [4×227×(7)2 cm2]
= [4×227×7×7 cm2]
= (88 × 7) cm2
= 616 cm2

Therefore, the surface area of a sphere having radius 14 cm is 616 cm2.


ii) Radius (r) of sphere = Diameter2=212=10.5cm

Surface area of sphere
= 4?r2
= [4×227×(10.5)2 cm2]
= [4×227×10.5×10.5 cm2]
= (88 × 1.5 × 10.5) cm2
= 1386 cm2

Therefore, the surface area of a sphere having radius 10.5cm is 1386 cm2.


iii) Radius (r) of sphere = Diameter2=3.52=1.75cm

Surface area of sphere
= 4?r2
= [4×227×(1.75)2 m2]
= [4×227×1.75×1.75 m2]
= ((88 × 1.75) m2
= 38.5 m2

Therefore, the surface area of a sphere having radius 3.5 m is 38.5 m2.

Q3 ) Find the total surface area of a hemisphere of radius 10 cm. (Use ? = 3.14)


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Answer :

image

We know that,

total surface area Of hemisphere
= CSA of hemisphere + Area of circular end of hemisphere
= 2?r2+?r2
= 3?r2
= [3×3.14×(10)2cm2]
= 942 cm2

Therefore, the total surface area of such a hemisphere is 942 cm2.

Q4 ) 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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Radius (r) of spherical balloon = 7 cm
Radius (R) of spherical balloon, when air is pumped into it = 14 cm

= InitialsurfaceareaSurfaceareaafterpumpingairintoballoon
= [4×?×r24×?×R2] =(rR)2
= (714)2=14

Therefore, the ratio between the surface areas in these two cases is 1 : 4.

Q5 ) 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.


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Answer :

Inner Radius (r) of sphere = Diameter2=10.52=5.25cm

Surface area of sphere
= 4?r2
= [4×227×(5.25)2 cm2]
= [4×227×5.25×5.25 cm2]
= ((88 × 5.25) cm2
= 173.25 cm2

Therefore, the surface area of a sphere having radius 5.25 cm is 173.25 cm2.

Cost of tin-plating 100 cm2 area = Rs. 16
Cost of tin-plating 173.25 cm2 area = Rs.16×173.25100 = Rs. 27.72

Therefore, the cost of tin-plating the inner side of the hemispherical bowl is Rs. 27.72.

Q6 ) Find the radius of a sphere whose surface area is 154 cm2.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Let the radius of the sphere be r.
Given :
Surface area of sphere = 154 cm2
? 4?r2=154 cm2
? r2=154×74×22 cm2
? r2=7×72×2 cm2
? r=72cm=3.5cm

? the radius of the sphere whose surface area is 154 cm2 is 3.5 cm.

Q7 ) The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Let the diameter of earth be d.
Therefore, the diameter of moon will be d4.
Thus, Radius of earth = d2
So, Radius of moon = 12 × d4 = d8
Hence, Surface area of moon = 4?(d8)2
And Surface area of earth = 4?(d8)24?(d2)2
So, Required ratio = 464=116

Therefore, the ratio between their surface areas will be 1 : 16.

Q8 ) 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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Inner radius Of hemispherical bowl = 5cm
Thickness of the bowl = 0.25 cm
Outer radius (r) Of hemispherical bowl = (5 + 0.25) cm = 5.25 cm

We also know that,
Outer CSA of hemispherical bowl
= 2?r2
= 2×227×5.252
= 173.25cm2

Therefore, the outer curved surface area Of the bowl is 173.25 cm2

Q9 ) 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).


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

image

i) Surface area of sphere = 4?r2

ii) Height of cylinder = r + r = 2r
Radius of cylinder = r
We know that,

CSA of cylinder
= 2?rh
= 2?r(2h)
= 4?r2


iii) ratio of the areas obtained
= SurfaceareaofsphereCSAofcylinder
= 4?r24?r2
= 11

Therefore, the ratio between these two surface areas is 1:1.

Solution for Exercise 13.5

1. A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes?


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Answer :

Given :
Matchbox is a cuboid having its length (l), breadth (b), height (h) as 4 cm, 2.5cm,and 1.5 cm.

We know that,
Volume of 1 match box = l x b x h
= (4 x 2.5 x 1.5) cm2
= 15 cm2

So, the volume Of 12 match boxes is = Volume of 1 match box x 12
i.e., 15 cm2 x 12
= 180 cm2
Therefore, the volume Of 12 match boxes is 180 cm2

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? (1 (\{m}^{2}\) = 1000 L)


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Answer :

Given :
The cuboidal water tank has its length (l) as 6 m, breadth (b) as 5 m, and height (h) as 4.5 m.

We know that,
Volume of tank = l x b x h
= (6 x 5 x 4.5) m2
= 135 m2
But, Amount of water that 1 m2 volume can hold = 1000 litres
Thus, Amount of water that 135 volume can hold = (135 x 1000) litres = 135000 litres
Therefore, such tank can hold up to 135000 litres of Water.

3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 m3 of a liquid?


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Answer :

Given :
Length (l) of vessel = 10 m
Width (b) Of vessel = 8 m

We know that,
Volume of tank = l x b x h
But, Volume of vessel = 380 m3
Thus, l x b x h = 380 m3
(10 x 8 x h) m2 = 380 m3
(80 x h) m2 = 380 m3
h = 38080 m
h = 4.75 m
Therefore, the height Of the vessel should be 4.75 m.

4. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ` 30 per m3.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given: The cuboidal pit has its length (l) as 8 m, width (b) as 6 m, and depth (h)as 3 m.

We know that,
Volume Of pit = (8 x 6 x 3) m3 = 144 m3
Now, Cost of digging per m3 volume = Rs. 30
Therfore, Cost of digging 144 m3 volume = Rs. (144 x 30) = Rs. 4320

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.


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Answer :

Given :
Length (l) and depth (h) of tank is 2.5 m and 10 m respectively.

Let the breadth of the tank be b m.
Volume of tank = l x b x h
= (2.5 x b x 10) m3
= 25b m3

But, Capacity of tank = 50000 litres of water
Thus, 25b m3= 25000 b litres
i.e., 25000 b = 50000
i.e., b = 20 m.
Therefore, the breadth of the tank is 2 m.

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?


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
The tank is cuboidal in shape having its length (l) as 20 m, breadth (b) as 15 m, and height (h) as 6 m.

Capacity of tank = l x b x h
= (20 x 15 x 6) m3
= 1800 m3 = 1800000 litres

Thus, Water consumed by the people Of the village in 1 day = (4000 x 150) litres
i.e., = 600000 litres
Now, Let water in this tank last for n days.
So, we get that,
Water consumed by all people of village in n days = Capacity of tank
i.e., n x 600000 = 1800000
i.e., n = 3
Therefore, the water of this tank will last for 3 days.

7. A godown measures 40 m × 25 m × 15 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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

The godown has its length (l1) as 40 m, breadth (b1) as 25 m, height (h1) as 10 m,
while the wooden crate has its length (l2) as 1.5 m, breadth (b2) as 1.25 m, and height (h2) as 0.5 m.

We get that, volume of godown = (40 x 25 x 10) m3
= 10000 m3

Also, Volume of 1 wooden crate
= (1.5 x 1.25 x 0.5) m3
= 0.9375 m3

Now, Let n wooden crates can be stored in the godown.
Therefore, volume of n wooden crates = Volume of godown
= 0.9375 x n = 10000
i.e., n = 10666.66
Therefore, 10666 Wooden crates can be stored in the godown.

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, and the ratio between their surface areas.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Side (a) of cube = 12 cm

We know that,
Volume of 8 cubes = a3
= 123 cm3
= 1728 cm3.

Now, Let the side of the smaller cube be (a1).
Thus, volume of 1 smaller cube = 17288 cm3
= 216 cm3
i.e., (a1)3 = 216 cm3
i.e., (a1)=6cm
Therefore, the side of the smaller cubes will be 6 cm.

Ratio between surface areas of cube = SurfaceareaofbiggercubeSurfaceareaofsmallercube
= [6×(a2)26×(a1)2] = (126)2
= (21)2 = 41
Therefore, the ratio between the surface areas of these cubes is 4:1.

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?


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Rate of water flow = 2 km per hour
= 200060 m/min
= 1003 m/min
Depth (h) of river = 3 m
Width (b) of = 40 m

So, we get,
Volume of water flowed in 1 min = (1003×40×3) m3 = 4000 m3.
Therefore, in 1 minute, 4000 m3 water will fall in the sea.

Solution for Exercise 13.6

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 cm3 = 1 L)


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Height (h) of vessel = 25 cm
Circumference of vessel = 132 cm

Let the radius Of the cylindrical vessel be r.
i.e., 2?r=132cm
i.e., 2×227×r=132cm
r=132×72×22cm
i.e., r=21cm

We know that,
Volume of cylindrical vessel =?r2h
=227×212×25 cm3
= 34650 cm3
i.e., 346501000L=34.65L
Therefore, such vessel can hold 34.65 litres Of water.

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 cm3 of wood has a mass of 0.6 g.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Inner radius of cylindrical pipe (r) = 242cm = 12 cm
Outer radius of cylindrical pipe (R) = 282cm = 14 cm
Height (h) of pipe = Length of pipe = 35 cm

We know that,
Volume of pipe = ?(R2?r2)h
= 227×(142?122)×35 cm3
= 110 x 52 cm3
= 5720 cm3

Now, Mass of 1 cm3 wood = 0.6 g
So, Mass of 5720 cm3 wood = (5720 x 0.6) g
= 3432 g
= 3.432 kg
Therefore, mass of the pipe is 3.432 kg.

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?


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

The tin can will be cuboidal in shape while the plastic cylinder Hill be cylindrical in shape.
i)image
Given :
Length (l) of tin can = 5 cm
Breadth (b) of tin can = 4 cm
Height (h) Of tin can = 15 cm
Therefore, Capacity of tin can = l x b x h
= (5 x 4 x 15) cm3
= 300 cm3

ii)image
Given :
Radius (r) of circular end of plastic cylinder = 72cm=3.5cm
Height (h) of plastic cylinder = 10 cm
Therefore, Capacity of plastic cylinder = ?r2h
=227×3.52×10 cm3
= 11×35 cm3
= 385 cm3

Therefore, plastic cylinder has the greater capacity.
So, we get,
Difference in capacity = (385 - 300) cm3 = 85 cm3.

4. If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then and
i) radius of its base
ii) its volume.
(Use ? = 3.14)


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

i) Given :
Height (h) of cylinder = 5 cm

Let radius of cylinder be r.
Also, CSA of cylinder = 94.2 cm2
Thus, we get that,
2?rh = 94.2 cm2
(2 × 3.14 × r × 5) cm = 94.2 cm2
i.e., r = 3 cm

ii) We know that,
Volume of cylinder = ?r2h
= 3.14×32×5 cm3
= 141.3 cm3.

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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

i)Given :
Rs. 20 is the cost Of painting 1 m2 area.
So, when Rs. 2200 is the cost of painting, area is = 120x2200 m2
= 110 m2 area
Therefore, the inner curved surface area of the vessel is 110 m2.

ii)Height (h) of vessel = 10 m
Surface area = 110 m2
Let the radius of the base of the vessel be r.
We know that,
Surface area = 2?rh
But, CSA = 110 m2 ...(from i))
i.e., 110 m2 = 2x227xrx10m
Therefore, r = 74m=1.75m

iii) Now, volume of vessel = ?r2h
= 22×(1.75)2×10m3
= 96.25 m3
Therefore, the capacity of the vessel is 96.25 m3 or 96250 litres.

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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

i)Given :
Rs. 20 is the cost Of painting 1 m2 area.
So, when Rs. 2200 is the cost of painting, area is = 120×2200 m2
= 110 m2 area
Therefore, the inner curved surface area of the vessel is 110 m2.

ii)Height (h) of vessel = 10 m
Surface area = 110 m2
Let the radius of the base of the vessel be r.
We know that,
Surface area = 2?rh
But, CSA = 110 m2 ...(from i))
i.e., 110 m2 = 2×227×r×10m
Therefore, r = 74m=1.75m

iii) Now, volume of vessel = ?r2h
= 22×(1.75)2×10m3
= 96.25 m3
Therefore, the capacity of the vessel is 96.25 m3 or 96250 litres.

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?


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Height (h) Of cylindrical vessel = 1 m
Volume of cylindrical vessel = 15.4 litres = 0.0154 m3

Let the radius of the circular end be r.
But we know that,
Volume of cylinder = ?r2h
Thus, we get that,
?r2h=0.0154 m3
227×r2×1m = 0.0154 m3
i.e., r = 0.07 m.

Also, we have,
Total surface area of vessel = 2?r(r+h)
= 2×227×0.07(1+0.07) m2
= 0.44 × 1.07 m2
= 0.4708 m2
Therefore, 0.4708 m2 of the metal sheet would be required to make the cylindrical vessel.

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, and the volume of the wood and that of the graphite.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

image
Given :
Radius (R) of pencil = 72mm=3.5mm=0.35cm
Radius (r) of graphite = 12mm=0.5mm=0.05cm
Height (h) of pencil = 14 cm

We know that,
Volume of wood in pencil = ?(R2?r2)h
= 227×(0.352?0.052)×14 cm3
= 227×(0.1225?0.0025)×14 cm3
= 44 × 0.12 cm3
= 5.28 cm3

Similarly, Volume of graphite in pencil = ?r2h
= 227×0.052×14 cm3
= 44 × 0.0025 cm3
= 0.11 cm3

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?


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

image
Given :
Radius (r) of cylindrical bowl = 72cm=3.5cm
Height (h) of bowl, up to which bowl is filled with soup = 4 cm

We know that,
Volume Of soup in bowl = ?r2h
= 227×3.52×4 cm3
= 11 × 3.5 × 4 cm3
= 154 cm3
Therefore, Volume of soup given to 250 patients = (250 × 154) cm3 = 38500 cm3

Solution for Exercise 13.7

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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

i)Given :
Radius (r) of cone = 6 cm
Height (h) of cone = 7 cm

We know that,
Volume of cone = 13?r2h
= 13×227×62×7 cm3
= 12 × 22 cm3
= 264 cm3
Therefore, the volume of the cone is 264 cm3

ii)Given :
Radius (r) of cone = 3.5 cm
Height (h) of cone = 12 cm

Similarly,
Volume of cone = 13?r2h
= 13×227×3.52×12 cm3
= 12.833 × 12 cm3
= 154 cm3
Therefore, the volume of the cone is 154 cm3

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 .


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

i)Given :
Radius (r) of cone = 7 cm
Slant height (l) of cone = 25 cm
Height of cone = (l2?r2) cm
= (252?72) cm
= 24 cm

We know that,
Volume of cone = 13?r2h
= 13×227×72×24 cm3
= 154 × 8 cm3
= 1232 cm3
Therefore, capacity of the conical vessel = 1232 cm3 = 1.232 litres

ii) Height (h) of cone = 12 cm
Slant height (l) of cone = 13 cm
Radius of cone = (l2?h2) cm
= (132?122) cm
= 5 cm

We know that,
Volume of cone = 13?r2h
= 13×227×52×12 cm3
= 300 × 1.047 cm3
= 314.28 cm3
Therefore, capacity of the conical vessel = 314.28 cm3 = 0.31428 litres

3. The height of a cone is 15 cm. If its volume is 1570 cm3 , find the radius of the base. (Use ? = 3.14)


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Height (h) of cone = 9 cm
Also, Volume of cone = 1570 cm3

Let the radius of the cone be r.
But, Volume of cone = 13?r2h
i.e., 1570 cm3 = 13×227×r2×15 cm
i.e., r2=100cm2
Thus, r = 10 cm
Therefore, the radius of the base of cone is 10 cm.

4. If the volume of a right circular cone of height 9 cm is 48?cm3 , find the diameter of its base.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Height (h) of cone = 9 cm
Also, Volume of cone = 48?cm3

Let the radius of the cone be r.
But, Volume of cone = 13?r2h
i.e., 48?cm3 = 13×?×r2×9 cm
i.e., r2=48?×39×?cm2
i.e., r2=42cm2
i.e., r = 4cm
Thus, Diameter of base = 2r = 8 cm
Therefore, the diameter of the base of cone is 8 cm.

5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Radius (r) of pit = 3.52m=1.75m
Height (h) of pit = depth of pit = 12 m

We know that,
Volume of pit = 13?r2h
= 13×227×3.52×12 m3
= 38.5 m3
Thus, capacity of the pit = (38.5 × 1) kilolitres = 38.5 kilolitres

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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Radius of the base = 282cm=14cm
Also, Volume of a right circular cone = 9856 cm3

i) Let the height of the cone be h.
But, we know that,
Volume of right circular cone = 13?r2h
i.e., 9856 cm3 = 13×227×142×h cm2
i.e., h = 48 cm.
Therefore, height of the cone is 48 cm.

ii)We know, slant height of cone = (h2+r2) cm
= (482+142) cm
l = 50cm
Therefore, slant height of the cone is 50 cm.

iii) We also know that,
Curved surface area of the cone = ?rl
= 227×14×50cm2
= 2200 cm2
Therefore, the curved surface area of the cone is 2200 cm2.

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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

image
When right-angled ?ABC is revolved about its side 12 cm, a cone With height (h) as 12 cm, radius (r) as 5 cm, and slant height (l) 13 cm will be formed.

Now, thus, we know that,
Volume of cone = 13?r2h
= 13×?×52×12 cm3
= 13×300×? cm3
=100? cm3
Therefore, the volume of the cone so formed is 100? cm3

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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

image
When right-angled ?ABC is revolved about its side 5 cm, a cone will be formed having radius (r) as 12 cm, height (h) as 5 cm, and slant height (l) as 13 cm.

Now, thus, we know that,
Volume of cone = 13?r2h
= 13×?×122×5 cm3
= 240×? cm3
= 240? cm3
Therefore, the volume of the cone so formed is 240? cm3

Thus, the ratio of the volumes of the two solids obtained in Questions 7 and 8
= 100240
= 512
i.e., 5 : 12.

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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Radius (r) of heap = 10.52m=5.25m
Height (h) of heap = 3 m

As we know that,
Volume of heap = 13?r2h
= 13×227×5.252×3 m3
= 86.625 m3
Therefore, the volume of the heap of wheat is 86.625 m3.

But, Area of canvas required = CSA of cone
So, we get,
= ?rl
= ?×r×((r)2+(h)2)
= 227×5.25×((5.25)2+(3)2) m2
227××5.25×6.05 m2
= 99.825 m2
Therefore, 99.825 m2 canvas will be required to protect the heap from rain.

Solution for Exercise 13.8

1. Find the volume of a sphere whose radius is
i) 7 cm
ii) 0.63 m


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

i)Given :
Radius of sphere = 7 cm

We know that,
Volume of sphere = 43?r3
= 43×227×73 cm3
= 43123 cm3
= 143713 cm3
Therefore, the volume of the sphere is 143713 cm3.

ii) Given :
Radius of sphere = 0.63 m

We know that,
Volume of sphere = 43?r3
= 43×227×0.633 m3
= 1.0478 m3
Therefore, the volume of the sphere is 1.05 m3 (approx.)

2. Find the amount of water displaced by a solid spherical ball of diameter
i) 28 cm
ii) 0.21 m.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

i)Given :
Radius of sphere = 282cm=14cm

We know that,
Volume of sphere = 43?r3
= 43×227×143 cm3
= 1149823 cm3
Therefore, the volume of the sphere is 1149823 cm3.

ii)Given :
Radius of sphere = 0.212m=0.105m

We know that,
Volume of sphere = 43?r3
= 43×227×0.1053 m3
= 0.004851 m3
Therefore, the volume of the sphere is 0.004851 m3.

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 g per cm3 ?


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Radius of sphere = 4.22cm=2.1cm

We know that,
Volume of sphere = 43?r3
= 43×227×2.13 cm3
= 38.808 cm3

Now, we know that,
Density=MassVolume
i.e., Mass = Density × Volume
i.e, = 8.9 × 38.808 g
i.e., = 345.3912 g
Hence, the mass of the ball is 345.39 g (approximately).

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?


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Let the diameter of earth be d.
Therefore, the diameter of moon will be d4.
Thus, Radius of earth = d2
So, Radius of moon = 12 × d4 = d8

Now, Volume of moon = 43?r3=43?×d83=1512×43?×d3
Similarly,
Volume of earth = 43?r3=43?×d23=18×43?×d3
So, VolumeofmoonVolumeofearth=1512×43?×d318×43?×d3=164
Therefore, the volume of moon = 164 is of the volume of earth.

5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Radius (r) of hemisphere bowl = 10.52cm=5.25cm

We know that,
Volume of hemisphere = 23?r3
= 23×227×5.253 cm3
= 303.1875 cm3
= 303.18751000l
= 0.303 litre (approximately)
Therefore ,the volume hemisphere bowl is 0.303 litres.

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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Inner radius of hemispherical tank (r) = 1 m
Thickness of hemispherical tank 1 cm = 0.01 m
Outer radius of hemispherical tank (R) = (1 + 0.01) m = 1.01 m

We know that,
Volume of iron is used to make such a tank = 23?(R3?r3)
= 23×227×(1.013?13) m3
= 4421×(1.030301?1) m3
= 0.06348 m3 (approx.)
Therefore, the volume of the iron used to make the tank is 0.06348 m3.

7. Find the volume of a sphere whose surface area is 154 cm2 .


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Surface area of sphere = 154 cm2.
Let radius of sphere be r.

But, we also know that,
Surface area of sphere = 4?r2
i.e., 154 cm2= 4×227×r2
i.e., r2=(154×74×22) cm2
i.e., r = 72cm
i.e., r = 3.5 cm

Now, Volume of sphere = 43?r3
= 43×227×3.53 cm3
= 17923 cm3
Therefore , the volume of the sphere is = 17923 cm3.

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 2.00 per square metre, find the
i) inside surface area of the dome,
ii) volume of the air inside the dome.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

i)Given :
Cost of white washing the dome from inside = Rs. 498.96
Also, Cost of white washing 1 m2 area = Rs. 2.00
Therefore, CSA of the inner side of dome = Rs. 498.962m2=249.48m2

ii) Let the inner radius of the hemispherical dome be r.
Now, CSA of inner side of dome = 249.48 m2
But, CSA = 2?r2
i.e., 249.48 m2 = 2×227×r2
i.e., r2=(249.48×72×22) m2
Therefore, r = 6.3 m.

Also, we have,
Volume of air inside the dome = Volume of hemispherical dome
= 23?r3
= 23×227×6.33 m3
= 523.908 m3 (approx.)
Therefore, volume of the air inside the dome is 523.908 m3.

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'.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

i) Radius of 1 solid iron sphere = r
Volume of 1 solid iron sphere = 43?r3
Thus, Volume of 27 solid iron spheres = 27 × 43?r3
27 solid iron spheres are melted to form 1 iron sphere.
Therefore, the volume of this iron sphere will be equal to the volume of 27 solid iron spheres.

Now, Let the radius of this new sphere be r'.
Therefore, Volume of new solid iron sphere = 43?r?3
Thus,43?r?3 = 27 × 43?r3
i.e., r?3=27×r3
i.e., r' = 3r.

ii) Now, we have,
surface area of 1 solid iron sphere of radius r = 4?r2
So, Surface area of iron sphere of radius r' = 4?r?2
= 4?3r2
= 36?r2
Therefore, ratio of S and S' = SS?=4?r236?r2
= 19 = 1 : 9

10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule?


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Given :
Radius (r) of capsule = 3.52mm=1.75mm

We know that,
Volume of spherical capsule = 43?r3
= 43×227×1.753 mm3
= 22.458 mm3
= 22.46 mm3(appro×.)
Therefore ,the volume of the spherical capsule is 22.46 mm3.

Solution for Exercise 13.9

Q.1 Height = 110cm, Depth = 25cm, Breadth = 85cm (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 cm2and the rate of pointing is 10 paise per cm2 find the total expenses required for palishing and painting the surface of the bookshelf.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

External dimensions of book self,
Length, l = 85cm
Breadth, b = 25 cm
Height, h = 110 cm
External surface area of shelf while leaving out the front face of the shelf
= lh+2(lb+bh)
= [85×110+2(85×25+25×110)] = (9350+9750) = 19100
External surface area of shelf is 19100 cm2
Area of front face = [85×110-75×100+2(75×5)] = 1850+750 = 2600 cm2
So, area is 2600 cm2
Area to be polished = (19100+2600) cm2 = 21700 cm2.
Cost of polishing 1 cm2 area = Rs 0.20
Cost of polishing 21700 cm2 area Rs. (21700×0.20) = Rs 4340
Dimensions of row of the book shelf
Length(l) = 75 cm
Breadth (b) = 20 cm and
Height(h) = 30 cm
Area to be painted in one row= 2(l+h)b+lh = [2(75+30)× 20+75×30] = (4200+2250) = 6450
So, area is 6450 cm2.
Area to be painted in 3 rows = (3×6450)cm2 = 19350 cm2.
Cost of painting 1 cm2 area = Rs. 0.10
Cost of painting 19350 cm2 area = Rs (19350 x 0.1) = Rs 1935
Total expense required for polishing and painting = 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.


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Diameter of wooden sphere = 21 cm
Radius of wooden sphere, r = diameter/ 2 = (212) cm = 10.5 cm
Surface area of a wooden sphere =4?r2
=4×(227)×(10.5)^2 = 1386 \)cm2
Radius of the circular end of cylindrical support = 1.5 cm
Height of cylindrical support = 7 cm
Curved surface area of one cylindrical support = 2?rh
= 2×(227)×1.5×7 = 66 cm2
Now,
Area of the circular end of cylindrical support = ?r2
= 227×(1.5)2=7.07
Again,
Area to be painted silver = [8 ×(1386-7.07)] = 8×1378.93 = 11031.44
Area to be painted is 11031.44 cm2
Cost for painting with silver colour = Rs(11031.44×0.25) =Rs 2757.86
Area to be painted black = (8×66) cm2 = 528 cm2
Cost for painting with black colour =Rs (528×0.05) = Rs26.40
Therefore, the total painting cost is:
= Rs(2757.86 +26.40)
= Rs 2784.26

Q.3 The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?


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Best NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes


Answer :

Let the diameter of a sphere be d.
After decreasing, diameter of the sphere
= d – 25100 x d
= d –14d=34d
Since, surface area of a sphere = 4?r2or?(2r)2or?d2
Surface area of a sphere, when diameter of the sphere is
?(34d)2 = ?916d2
Now, decrease percentage in curved surface area
?d2??916d2?d2×100
= 16?916×100=716×100
= 43.75%



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