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 1\({m}^2\) costs Rs.20.

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

Answer :

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

\( = [2 x 1.5 x 0.65 + 2 x 1.25 x 0.65 + 1.5 x 1.25] \)

= [1.95 + 1.625 + 1.875] \({m}^2\)

= 5.45 \({m}^2\)

Hence, the area of the sheet required for making the box is 5.45 \({m}^2\)

ii) Cost of sheet per 1 \({m}^2\) area = Rs.20

Therefore, Cost of sheet of 5.45 \({m}^2\) 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 \({m}^2\).

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] \({m}^2\)

= [30 + 24 + 20] \({m}^2\)

= 74 \({m}^2\).

Now, we have,

Cost of white-washing per \({m}^2\) area = Rs. 7.50

Cost of white-washing 74 \({m}^2\) 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 \({m}^2\) is Rs.15000, find the height of the hall.

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

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

\(\Rightarrow \) Area of four walls = 2(l + b)h = 250 h \({m}^2\)

Now, Cost of painting per \({m}^2\) area = Rs. 10

\(\therefore \) 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

\(\Rightarrow \) h = \(\frac{15000}{2500}\)

\(\Rightarrow \) 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 \({m}^2\). How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?

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)] \({cm}^2\)

= [2(225 + 75 + 168.75)] \({cm}^2\)

= [2 x 468.75] \({cm}^2\)

= 937.5 \({cm}^2\)

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 \({cm}^2\) ...(i)

Therefore, area that can be painted by the paint of the container = 9.375 \({m}^2\) ...(ii)

= 93750 \({cm}^2\)

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

93750 = 937.5m

\(\Rightarrow \) m = \(\frac{93750}{937.5}\)

\(\Rightarrow \) 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?

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

= \(4 × (Edge)^2\)

= \(4 × (10)^2 {cm}^2\)

= 400 \({cm}^2\)

Also, We have,

Lateral surface area of cuboidal box

= 2 [lh + bh]

= [2 (12.5 x 10 x 8)] \({cm}^2\)

= [2 x 180] \({cm}^2\)

= 360 \({cm}^2\)

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 \({cm}^2\) - 360 \({cm}^2\) = 40 \({cm}^2\)

Therefore, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box by 40 \({cm}^2\)

ii)Similarly, Total surface area of cubical box

=\(6 × (Edge)^2\)

= \(6 × (10)^2 {cm}^2\)

= 600 \({cm}^2\)

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)] \({cm}^2\)

= 610 \({cm}^2\)

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 \({cm}^2\) - 600\({cm}^2\)

= 10 \({cm}^2\)

Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 \({cm}^2\).

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?

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)] \({cm}^2\)

= [2(750 + 750 + 625)] \({cm}^2\)

= [2 x 2125] \({cm}^2\)

= 4250 \({cm}^2\)

Therefore, the area of glass is 4250 \({cm}^2\).

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 \({cm}^2\), find the cost of cardboard required for supplying 250 boxes of each kind.

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)] \({cm}^2\)

= [2(500 + 125 + 100)] \({cm}^2\)

= 1450 \({cm}^2\)

Therefore, Extra area required for overlapping

= \(\frac{1450 × 5}{100}\) \({cm}^2\) = 72.5 \({cm}^2\)

While considering all overlaps, total surface area of 1 bigger box

= (1450 + 72.5) \({cm}^2\) = 1522.5 \({cm}^2\)

So, Area of cardboard sheet required for 250 such bigger boxes = (1522.5 x 250) \({cm}^2\) = 380625 \({cm}^2\)

Similarly, total surface area of smaller box = [2 (15 + 15 x 5 + 12 x 5)]

= [2 (180 + 75 + 60)] \({cm}^2\)

= (2 x 315) \({cm}^2\)

= 630 \({cm}^2\)

Therefore, extra area required for overlapping

= \(\frac{630 × 5}{100}\) \({cm}^2\) = 31.5 \({cm}^2\)

So, Total surface area of 1 smaller box while considering all overlaps

= (630 + 31.5) \({cm}^2\)

= 661.5 \({cm}^2\)

So, Area of cardboard sheet required for 250 smaller boxes

= (250 x 661.5) \({cm}^2\) = 165375 \({cm}^2\)

Thus, Total cardboard sheet required

= (380625 + 165375) \({cm}^2\)

= 546000 \({cm}^2\)

Cost of 1000 \({cm}^2\) cardboard sheet = Rs. 4

Therefore, Cost of 546000 \({cm}^2\) cardboard sheet will be:

\(Rs. \frac{546000 × 4}{100}\) \({cm}^2\) = 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?

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] \({m}^2\)

= [2(10 + 7.5) + 12] \({m}^2\)

= 47 \({m}^2\)

Therefore, 47 \({m}^2\) tarpaulin will be required.

Q1 ) The curved surface area of a right circular cylinder of height 14 cm is 88 \({cm}^2\). Find the diameter of the base of the cylinder.

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 \({cm}^2\)

Let the diameter of the cylinder be d.

So, (\(2{\pi}rh\)) \({cm}^2\)= 88 \({cm}^2\)

(where, r is the radius of the base of the cylinder)

\(\Rightarrow \)\({\pi}dh\) = 88 \({cm}^2\) ...(Since, d = 2r)

\(\Rightarrow \) d = \(\frac{88 × 7}{22 × 14}\) cm ...(\({\pi} = \frac{22}{7} \))

\(\Rightarrow \) d = \(\frac{4}{2}\) cm

\(\Rightarrow \) 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?

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

=( \(\frac{140}{2}\) )cm = 70 cm = 0.7 m

Now, Area of sheet required = total surface of tank

But, we know that,

Area of sheet required

= \(2{\pi}r(r + h)\) \({m}^2\)

= \(2 × \frac{22}{7} × 0.7(0.7 + 1)\) \({m}^2\)

= \(4.4 × 1.7\) \({m}^2\)

= 7.48 \({m}^2\)

Therefore , it will required 7.48 \({m}^2\) 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.

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

Answer :

Given :

Inner radius (r) of cylindrical pipe = \(\frac{4}{2}\) cm = 2 cm

outer radius (R) of cylindrical pipe = \(\frac{4.4}{2}\) cm = 2.2 cm

Height (h) Of cylindrical pipe = Length Of cylindrical pipe = 77 cm

i) CSA of inner surface of pipe

= \(2{\pi}rh\)

= \(2 × \frac{22}{7} × 2 × 77\) \({cm}^2\)

= \(2 × 22 × 2 × 11\) \({cm}^2\)

= 968 \({cm}^2\)

ii) CSA of inner surface of pipe

= \(2{\pi}{R}h\)

= \(2 × \frac{22}{7} × 2.2 × 77\)

= \(2 × 22 × 2.2\) \({cm}^2\)

= 1064.8 \({cm}^2\)

iii) Total surface area of pipe

= CSA of inner surface + CSA of outer surface + Area of both circular ends of pipe

= \(2{\pi}{r}h\) + \(2{\pi}{R}h\) + \(2{\pi} [(R)^2 - (r)^2]\)

= \([968 + 1064.8 + 2{\pi}{(2.2)^2 - (2)^2}] {cm}^2\)

= \(2032.8 + 2 × \frac{22}{7} × 0.84\) \({cm}^2\)

= (2032.8 + 5.28) \({cm}^2\)

= 2038.08 \({cm}^2\)

Therefore, the total surface area of the cylindrical pipe is 2038.08 \({cm}^2\).

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 \({m}^2\).

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 = \(\frac{84}{2}\) = 42 cm

It can be observed that a roller is cylindrical.

Now, we know that,

CSA of rollar

= \(2{\pi}rh\)

= \(2 × \frac{22}{7} × 42 × 120\) \({cm}^2\)

= \(2 × 22 × 6 × 120\)

= 31680 \({cm}^2\)

Therefore, Area of field

= 500 x CSA of roller

= \(31680 × 500\) \({cm}^2\)

= 1584 \({m}^2\)

\(\therefore\) the area of the playground in \({m}^2\) 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 \({m}^{2}\).

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 = \(\frac{50}{2}\) cm = 25 cm = 0.25m

Now, we know that,

CSA of cylindrical pillar = \(2{\pi}rh\)

= \(2 × \frac{22}{7} × 0.25 × 3.5\) \({m}^2\)

= \(44 × 0.125\) \({m}^2\)

= 5.5 \({m}^2\)

Now, Cost of painting 1 \({m}^2\) area = Rs. 12.50

Cost of painting 5.5 \({m}^2\) 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 \({m}^2\). If the radius of the base of the cylinder is 0.7 m, find its height.

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 \({m}^2\)

Let the height of the circular cylinder be h.

Therefore, we have,

\(\Rightarrow \) \(2{\pi}rh\) = 4.4 \({m}^2\)

\(\Rightarrow \) \(2 × \frac{22}{7} × 0.7 × h\) m = 4.4 \({m}^2\)

\(\Rightarrow \) \(2 × 22 × 0.1 × h\) m = 4.4 \({m}^2\)

\(\Rightarrow \) 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 \({m}^2\).

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

Answer :

Given :

Inner radius (r) of circular well = \(\frac{3.5}{2}\) m = 1.75 m

Depth (h) of circular well = 10 m

i) We know that,

Inner curved surface area

= \(2{\pi}rh\)

= \(2 × \frac{22}{7} × 1.75 × 10\) \({m}^2\)

= (44 × 0.25 × 10) \({m}^2\)

= 110 \({m}^2\)

Therefore, the inner curved surface area of the circular well is 110 \({m}^2\)

ii) We have, Cost of plastering 1 \({m}^2\) area = Rs. 40

Cost of plastering 100 \({m}^2\) 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.

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

= \(\frac{5}{2}\) cm = 2.5 cm = 0.025 m

We know that,

CSA of cylindrical pipe

= \(2{\pi}rh\)

= \(2 × \frac{22}{7} × 0.025 × 28\) \({m}^2\)

= 4.4 \({m}^2\)

Therefore, the area of the radiating surface of the system is 4.4 \({m}^2\).

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 \(\frac{1}{12} \) th of the steel actually used was wasted in making the tank.

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 = \(\frac{4.2}{2}\) m = 2.1 m

i) lateral or curved surface area of tank

= \(2{\pi}rh\)

= \(2 × \frac{22}{7} × 2.1 × 4.5\) \({m}^2\)

= (44 × 0.3 × 4.5) \({m}^2\)

= 59.4 \({m}^2\)

Therefore , CSA of tank is 59.4 \({m}^2\).

ii) Now, so as to find steel used,

total surface area

= \(2{\pi}r(r + h) {m}^2\)

= \(2 × \frac{22}{7} × 2.1(2.1 + 4.5) {m}^2\)

= \(4.4 × 0.3 × 6.6 {m}^2\)

= 87.12 \({m}^2\)

Now, Let actual area of x \({m}^2\) sheet be used in making the tank.

Since, \(\frac{1}{12}th\) of the actual steel wasted, the area of steel which has gone into the tank = \(\frac{11}{12}th\) of x.

This means that the actual area used

= \(\frac{12}{11}th × 87.12\) \({m}^2\) = 95.04 \({m}^2\).

Therefore, 95.04 \({m}^2\) 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.

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

Answer :

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 = \(\frac{20}{2}\) m = 10 m

Now, Cloth required for covering the lampshade

= \(2{\pi}rh\)

= \(2 × \frac{22}{7} × 10 × 35\) \({cm}^2\)

= (44 × 10 × 5) \({cm}^2\)

= 2200 \({cm}^2\)

Hence, for covering the lampshade, 2200 \({cm}^2\) 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?

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

= \(2{\pi}rh\) + \({\pi}{r}^2\)

= [\(2 × \frac{22}{7} × 3 × 10.5 + \frac{22}{7} × {3}^2 {cm}^2\)]

= [\(132 × 1.5 + \frac{198}{7} {cm}^2\)]

= \(198 + \frac{198}{7} {cm}^2\)

= \(\frac{1584}{7} {cm}^2\)

So, we get,

Area Of cardboard sheet used by 1 competitor = \(\frac{1584}{7} {cm}^2\)

Thus, Area of cardboard sheet used by 35 competitors

= \((\frac{1584}{7} × 35) {cm}^2\) = 7920 \({cm}^2\)

\(\therefore \) 7920 \({cm}^2\) cardboard sheet will be bought for the competition.

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

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

Answer :

Given :

Radius (r) of the base of cone = \(\frac{10.5}{2} cm = 5.25 \) cm

Slant height (l) of cone = 10 cm

We know that, CSA of cone

= \({\pi}rl\)

= [\(\frac{22}{7} × 5.25 × 10\) \({cm}^2\)]

= [\(22 × 0.75 × 10\) \({cm}^2\)]

= 165 \({cm}^2\)

Therefore, the curved surface area of the cone is 165 \({cm}^2\).

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

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

Answer :

Given :

Radius (r) of the base of cone = \(\frac{24}{2} cm = 12 cm\)

Slant height (l) of cone = 21 cm

We know that, TSA of cone

= \({\pi}r(r + l)\)

= [\(\frac{22}{7} × 12 × (12 + 21)\) \({m}^2\)]

= [\(22 × 1.71 × 33\) \({m}^2\)

= 1244.57 \({m}^2\)

Therefore, the total surface area of the cone is 1244.57 \({m}^2\).

Q3 )
Curved surface area of a cone is 308 \({cm}^2\) and its slant height is 14 cm. Find

i) radius of the base and

ii) total surface area of the cone.

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 \({cm}^2\)

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

We know that,

CSA of cone

= \({\pi}rl\)

= [\(\frac{22}{7} × r × 14 cm\)]

= [\(22 × 2 × r cm\)

= 44r cm

\(\Rightarrow \) 308 \({cm}^2\) = 44r cm ...(Given)

\(\Rightarrow \) r = \(\frac{308}{44} cm\) = 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

= [\({\pi}rl\) + \({\pi}{r}^2\)]

= [\(308 + \frac{22}{7} × (7)^2\) \({cm}^2\)]

= [308 + 154] \({cm}^2\)

= 462 \({cm}^2\)

Therefore, the total surface area of the cone is 462 \({cm}^2\).

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 \({m}^2\) canvas is Rs. 70.

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

Answer :

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 \(\triangle{ABO}\),

\(\Rightarrow \) \({AB}^2 = {AO}^2 + {BO}^2\)

\(\Rightarrow \) \({l}^2 = {h}^2 + {r}^2\)

\(\Rightarrow \) \(l^2= {10}^2 + {24}^2 {m}^2\)

\(\Rightarrow \) \(l^2= 676 {m}^2\)

\(\Rightarrow \) \({l}^2 = {26}^2 m\)

\(\Rightarrow \) l = 26 m

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

ii) We know, CSA of tent

= \({\pi}rl\)

= [\(\frac{22}{7} × 24 × 26 m\)]

= \(\frac{13728}{7} {m}^2\)

Cost of 1 \({m}^2\) canvas = Rs. 70

Cost of [\(\frac{13728}{7} {m}^2\)] canvas

= Rs. [\(\frac{13728}{7} × 70\) \({m}^2\)]

= 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 \({\pi}\) = 3.14).

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 =

\(\Rightarrow \) \({l}^2 = {h}^2 + {r}^2\)

\(\Rightarrow \) \(l^2= {8}^2 + {6}^2 {m}^2\)

\(\Rightarrow \) \(l^2= 100 {m}^2\)

\(\Rightarrow \) \({l}^2 = {10}^2 m\)

\(\Rightarrow \) l = 10 m

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

We know, CSA of tent

= \({\pi}rl\)

= [\(3.14 × 6 × 10 m\)]

= \(188.4 {m}^2\)

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

\(\Rightarrow \) [(l - 0.2) × 3]m = \(188.4 {m}^2\)

\(\Rightarrow \) (l - 0.2) m = 62.8 \({m}^2\)

\(\Rightarrow \) l = 73 m.

\(\therefore \) 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 \({m}^2\).

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 = \(\frac{14}{2} = 7 m\)

We know, CSA of tent = \({\pi}rl\)

= [\(\frac{22}{7} × 7 × 25 m\)]

= [\(22 × 1 × 25 m\)]

= \(550 {m}^2\)

Now, we know that,

Cost of white-washing 100 \({m}^2\) area = Rs. 210

So, Cost of white-washing 550 \({m}^2\) area

= Rs. [\(\frac{210 × 550}{100}\)]

= 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.

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

\(\Rightarrow \) \({l}^2 = {h}^2 + {r}^2\)

\(\Rightarrow \) \(l^2= {7}^2 + {24}^2\) \({cm}^2\)

\(\Rightarrow \) \(l^2= 625\) \({cm}^2\)

\(\Rightarrow \) \({l}^2 = {25}^2 cm\)

\(\Rightarrow \) l = 25 cm

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

We know, CSA of tent

= \({\pi}rl\)

= [\(\frac{22}{7} × 7 × 25\) \({cm}^2\)]

= [\(22 × 1 × 25\) \({cm}^2\)]

= \(550\) \({cm}^2\)

Thus, CSA of 10 such caps

= (10 x 550) \({cm}^2\)

= 5500 \({cm}^2\)

\(\therefore \) 5500 \({cm}^2\) 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 per\( m^2 \) , what will be the cost of painting all these cones? (Use \({\pi}\) = 3.14 and take 1.04 = 1.02)

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

Answer :

Given :

Radius (r) of conical cap = \(\frac{40}{2} cm = 20 cm = 0.2 m\)

Height (h) of conical cap = 1 m

Slant height (l) of the conical cap

\(\Rightarrow \) \({l}^2 = {h}^2 + {r}^2\)

\(\Rightarrow \) \(l^2= {1}^2 + {0.2}^2 {m}^2\)

\(\Rightarrow \) \(l^2= 1.04 {m}^2\)

\(\Rightarrow \) \({l}^2 = {1.02}^2 m\)

\(\Rightarrow \) l = 1.02 m

\(\therefore \) the slant height Of the tent is 1.02 m.

We know, CSA of tent

= \({\pi}rl\)

= [\(3.14 × 0.2 × 1.02\) \({m}^2\)]

= \(0.64056\) \({m}^2\)

So, CSA of 50 such cones

= \(50 × 0.64056\) \({m}^2\)

= 32.028 \({m}^2\).

Now, we have,

Cost of painting 1 \({m}^2\) area = Rs. 12

Cost of painting 32.02 \({m}^2\) 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.

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

= \(4 {\pi} {r}^2\)

= [\(4 × \frac{22}{7} × (10.5)^2\) \({cm}^2\)]

= [\(4 × \frac{22}{7} × 10.5 × 10.5\) \({cm}^2\)]

= (88 × 1.5 × 10.5) \({cm}^2\)

= 1386 \({cm}^2\)

Therefore, the surface area of a sphere having radius 10.5cm is 1386 \({cm}^2\).

ii) Radius (r) of sphere = 5.6 cm

Surface area of sphere

= \(4 {\pi} {r}^2\)

= [\(4 × \frac{22}{7} × (5.6)^2\) \({cm}^2\)]

= [\(4 × \frac{22}{7} × 5.6 × 5.6\) \({cm}^2\)]

= (88 × 0.8 × 5.6) \({cm}^2\)

= 394.24 \({cm}^2\)

Therefore, the surface area of a sphere having radius 5.6 cm is 394.24 \({cm}^2\).

iii) Radius (r) of sphere = 14 cm

Surface area of sphere

= \(4 {\pi} {r}^2\)

= [\(4 × \frac{22}{7} × (14)^2\) \({cm}^2\)]

= [\(4 × \frac{22}{7} × 14 × 14\) \({cm}^2\)]

= (4 × 44 × 14) \({cm}^2\)

= 2464 \({cm}^2\)

Therefore, the surface area of a sphere having radius 14 cm is 2464 \({cm}^2\).

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

Answer :

i) Radius (r) of sphere = \(\frac{Diameter}{2} = \frac{14}{2} = 7 cm\)

Surface area of sphere

= \(4 {\pi} {r}^2\)

= [\(4 × \frac{22}{7} × (7)^2\) \({cm}^2\)]

= [\(4 × \frac{22}{7} × 7 × 7\) \({cm}^2\)]

= (88 × 7) \({cm}^2\)

= 616 \({cm}^2\)

Therefore, the surface area of a sphere having radius 14 cm is 616 \({cm}^2\).

ii) Radius (r) of sphere = \(\frac{Diameter}{2} = \frac{21}{2} = 10.5 cm\)

Surface area of sphere

= \(4 {\pi} {r}^2\)

= [\(4 × \frac{22}{7} × (10.5)^2\) \({cm}^2\)]

= [\(4 × \frac{22}{7} × 10.5 × 10.5\) \({cm}^2\)]

= (88 × 1.5 × 10.5) \({cm}^2\)

= 1386 \({cm}^2\)

Therefore, the surface area of a sphere having radius 10.5cm is 1386 \({cm}^2\).

iii) Radius (r) of sphere = \(\frac{Diameter}{2} = \frac{3.5}{2} = 1.75 cm\)

Surface area of sphere

= \(4 {\pi} {r}^2\)

= [\(4 × \frac{22}{7} × (1.75)^2\) \({m}^2\)]

= [\(4 × \frac{22}{7} × 1.75 × 1.75\) \({m}^2\)]

= ((88 × 1.75) \({m}^2\)

= 38.5 \({m}^2\)

Therefore, the surface area of a sphere having radius 3.5 m is 38.5 \({m}^2\).

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

Answer :

We know that,

total surface area Of hemisphere

= CSA of hemisphere + Area of circular end of hemisphere

= \(2 {\pi} {r}^2 + {\pi} {r}^2\)

= \(3 {\pi} {r}^2 \)

= [\(3 × 3.14 × (10)^2 {cm}^2\)]

= 942 \({cm}^2\)

Therefore, the total surface area of such a hemisphere is 942 \({cm}^2\).

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.

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

= \(\frac{Initial surface area}{Surface area after pumping air into balloon}\)

= \([\frac{4 × {\pi} × {r}^2}{4 × {\pi} × {R}^2}]\) \(= (\frac{r}{R})^2\)

= \((\frac{7}{14})^2 = \frac{1}{4}\)

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 \({cm}^2\).

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

Answer :

Inner Radius (r) of sphere = \(\frac{Diameter}{2} = \frac{10.5}{2} = 5.25 cm\)

Surface area of sphere

= \(4 {\pi} {r}^2\)

= [\(4 × \frac{22}{7} × (5.25)^2\) \({cm}^2\)]

= [\(4 × \frac{22}{7} × 5.25 × 5.25\) \({cm}^2\)]

= ((88 × 5.25) \({cm}^2\)

= 173.25 \({cm}^2\)

Therefore, the surface area of a sphere having radius 5.25 cm is 173.25 \({cm}^2\).

Cost of tin-plating 100 \({cm}^{2}\) area = Rs. 16

Cost of tin-plating 173.25 \({cm}^{2}\) area = Rs.\(\frac{16 × 173.25}{100}\) = Rs. 27.72

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

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 \( cm^2\)

\(\Rightarrow \) \(4 {\pi} {r}^2 = 154\) \({cm}^{2}\)

\(\Rightarrow \) \({r}^2 = \frac{154 × 7}{4 × 22}\) \({cm}^2\)

\(\Rightarrow \) \({r}^2 = \frac{7 × 7}{2 × 2} \) \({cm}^2\)

\(\Rightarrow \) \(r = \frac{7}{2} cm = 3.5 cm\)

\(\therefore \) the radius of the sphere whose surface area is 154 \({cm}^{2}\) 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.

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 \(\frac{d}{4}\).

Thus, Radius of earth = \(\frac{d}{2}\)

So, Radius of moon = \(\frac{1}{2}\) × \(\frac{d}{4}\) = \(\frac{d}{8}\)

Hence, Surface area of moon = \(4{\pi}(\frac{d}{8})^2\)

And Surface area of earth = \(\frac{4{\pi}(\frac{d}{8})^2}{4{\pi}(\frac{d}{2})^2}\)

So, Required ratio = \(\frac{4}{64} = \frac{1}{16}\)

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.

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

= \(2 {\pi} {r}^2\)

= \(2 × \frac{22}{7} × {5.25}^2\)

= \(173.25 {cm}^2\)

Therefore, the outer curved surface area Of the bowl is 173.25 \({cm}^{2}\)

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).

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

Answer :

i) Surface area of sphere = \(4 {\pi} {r}^2\)

ii) Height of cylinder = r + r = 2r

Radius of cylinder = r

We know that,

CSA of cylinder

= \(2{\pi}rh\)

= \(2{\pi}r(2h)\)

= \(4{\pi} {r}^2\)

iii) ratio of the areas obtained

= \(\frac{Surface area of sphere}{CSA of cylinder}\)

= \(\frac{4 {\pi} {r}^2}{4 {\pi} {r}^2}\)

= \(\frac{1}{1}\)

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

Q1 ) A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes?

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

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) \({cm}^{2}\)

= 15 \({cm}^{2}\)

So, the volume Of 12 match boxes is

= Volume of 1 match box x 12

= \( 15 cm^2 x 12 \)

= 180 \({cm}^{2}\)

Therefore, the volume Of 12 match boxes is 180 \({cm}^{2}\)

Q2 ) 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)

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

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) \(m^2\)

= 135 \({m}^{2}\)

But, Amount of water that 1 \({m}^{2}\) 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.

Q3 ) A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 \(m^3\) of a liquid?

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

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 \({m}^{3}\)

\(\therefore \) l x b x h = 380 \({m}^{3}\)

\(\Rightarrow\) (10 x 8 x h) \({m}^{2}\) = 380 \({m}^{3}\)

\(\Rightarrow\) (80 x h) \({m}^{2}\) = 380 \({m}^{3}\)

\(\Rightarrow \) h = \(\frac{380}{80}\) m

\(\Rightarrow \) h = 4.75 m

Therefore, the height Of the vessel should be 4.75 m.

Q4 ) Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ` 30 per \({m}^{3}\).

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) \({m}^{3}\) = 144 \({m}^{3}\)

Now, Cost of digging per \({m}^{3}\) volume = Rs. 30

\(\therefore \) Cost of digging 144 \({m}^{3}\) volume

= Rs. (144 x 30) = Rs. 4320

Q5 ) 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.

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

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) \({m}^{3}\)

= 25b \({m}^{3}\)

But, Capacity of tank = 50000 litres of water

\(\therefore \) 25b \({m}^{3}\)= 25000 b litres

\(\Rightarrow \) 25000 b = 50000

\(\Rightarrow \) b = 20 m.

Therefore, the breadth of the tank is 2 m.

Q6 ) 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?

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) \({m}^{3}\)

= 1800 \({m}^{3}\) = 1800000 litres

\(\therefore \) Water consumed by the people Of the village in 1 day = (4000 x 150) litres

= 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

\(\Rightarrow \) = n x 600000 = 1800000

\(\Rightarrow \) n = 3

Therefore, the water of this tank will last for 3 days.

Q7 ) 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.

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

Answer :

The godown has its length \(\left(l_{1}\right)\) as 40 m, breadth \(\left(b_{1}\right)\) as 25 m, height
\(\left(h_{1}\right)\) as 10 m,

while the wooden crate has its length \(\left(l_{2}\right)\) as 1.5 m, breadth \(\left(b_{2}\right)\) as 1.25 m, and height \(\left(h_{2}\right)\) as 0.5 m.

We get that, volume of godown

= (40 x 25 x 10) \({m}^{3}\)

= 10000 \({m}^{3}\)

Also, Volume of 1 wooden crate

= (1.5 x 1.25 x 0.5) \({m}^{3}\)

= 0.9375 \({m}^{3}\)

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

\(\Rightarrow \) n = 10666.66

Therefore, 10666 Wooden crates can be stored in the godown.

Q8 ) 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.

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

= \({a}^3\)

= \({12}^3\) \({cm}^3\)

= 1728 \({cm}^3\).

Now, Let the side of the smaller cube be \(\left(a_{1}\right)\).

Thus, volume of 1 smaller cube

= \(\frac{1728}{8}\) \({cm}^3\)

\(\Rightarrow \) \({\left(a_{1}\right)}^3\) = 216 \({cm}^3\)

\(\Rightarrow \) \(\left(a_{1}\right)= 6 cm \)

Therefore, the side of the smaller cubes will be 6 cm.

Ratio between surface areas of cube

= \(\frac{Surface area of bigger cube}{Surface area of smaller cube}\)

= \([\frac{6 × \left(a_{2}\right)^2}{6 × \left(a_{1}\right)^2}]\)

= \((\frac{12}{6})^2\)

= \((\frac{2}{1})^2\)

= \(\frac{4}{1}\)

Therefore, the ratio between the surface areas of these cubes is 4:1.

Q9 ) 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?

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

Answer :

Given :

Rate of water flow = 2 km per hour

= \(\frac{2000}{60}\) m/min

= \(\frac{100}{3}\) m/min

Depth (h) of river = 3 m

Width (b) of = 40 m

So, we get,

Volume of water flowed in 1 min

= \((\frac{100}{3} × 40 × 3)\) \({m}^{3}\)

= 4000 \({m}^{3}\).

Therefore, in 1 minute, 4000 \({m}^{3}\) water will fall in the sea.

Q1 ) 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 \({cm}^{3}\) = 1 L)

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.

\(\Rightarrow \) \(2{\pi}r = 132 cm\)

\(\Rightarrow \) \(2 × \frac{22}{7} × r = 132 cm\)

\(\Rightarrow \) \(r = \frac{132 × 7}{2 × 22} cm\)

\(\Rightarrow \) \(r = 21 cm\)

We know that,

Volume of cylindrical vessel

\(= {\pi} {r}^2 h\)

\(= \frac{22}{7} × {21}^2 × 25\) \({cm}^{3}\)

= 34650 \({cm}^{3}\)

\(\therefore \) \(\frac{34650}{1000} L = 34.65 L\)

Therefore, such vessel can hold 34.65 litres Of water.

Q2 ) 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 \({cm}^{3}\) of wood has a mass of 0.6 g.

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

Answer :

Given :

Inner radius of cylindrical pipe (r) = \(\frac{24}{2} cm\) = 12 cm

Outer radius of cylindrical pipe (R) = \(\frac{28}{2} cm\) = 14 cm

Height (h) of pipe = Length of pipe = 35 cm

We know that,

Volume of pipe

= \({\pi}({R}^2 - {r}^2)h\)

= \(\frac{22}{7} × ({14}^2 - {12}^2) × 35\) \({cm}^{3}\)

= 110 x 52 \({cm}^{3}\)

= 5720 \({cm}^{3}\)

Now, Mass of 1 \({cm}^{3}\) wood = 0.6 g

So, Mass of 5720 \({cm}^{3}\) wood br> = (5720 x 0.6) g

= 3432 g

= 3.432 kg

Therefore, mass of the pipe is 3.432 kg.

Q3 )
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?

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)

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) \({cm}^{3}\)

= 300 \({cm}^{3}\)

ii)

Given :

Radius (r) of circular end of plastic cylinder = \(\frac{7}{2} cm = 3.5 cm\)

Height (h) of plastic cylinder = 10 cm

Therefore, Capacity of plastic cylinder

= \({\pi} {r}^2 h\)

\( = \frac{22}{7} × {3.5}^2 × 10\) \({cm}^{3}\)

= \(11 × 35\) \({cm}^{3}\)

= \(385\) \({cm}^{3}\)

Therefore, plastic cylinder has the greater capacity.

So, we get,

Difference in capacity

= (385 - 300) \({cm}^{3}\)

= 85 \({cm}^{3}\).

Q4 )
If the lateral surface of a cylinder is 94.2 \({cm}^{2}\) and its height is 5 cm, then and

i) radius of its base

ii) its volume.

(Use \({\pi}\) = 3.14)

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 \({cm}^{2}\)

Thus, we get that,

\(\Rightarrow \) \(2 {\pi} r h\) = 94.2 \({cm}^{2}\)

\(\Rightarrow \)
(2 × 3.14 × r × 5) cm = 94.2 \({cm}^{2}\)

\(\Rightarrow \) r = 3 cm

ii) We know that,

Volume of cylinder

= \({\pi} {r}^2 h\)

= \(3.14 × {3}^2 × 5\) \({cm}^{3}\)

= 141.3 \({cm}^{3}\).

Q5 )
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 \({m}^2\), find

i) inner curved surface area of the vessel,

ii) radius of the base,

iii) capacity of the vessel.

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

Answer :

i)Given :

Rs. 20 is the cost Of painting 1 \({m}^2\) area.

\(\therefore\) when Rs. 2200 is the cost of painting, area is

= \(\frac{1}{20} × 2200\) \({m}^2\)

= 110 \({m}^2\) area

Therefore, the inner curved surface area of the vessel is 110 \({m}^2\).

ii)Height (h) of vessel = 10 m

Surface area = 110 \({m}^2\)

Let the radius of the base of the vessel be r.

We know that,

Surface area = \(2 {\pi} r h\)

But, CSA = 110 \({m}^2\) ...(from i))

\(\Rightarrow \) 110 \({m}^2\) = \(2 x \frac{22}{7} x r x 10 m\)

\(\Rightarrow\) r = \(\frac{7}{4} m = 1.75 m\)

iii) Now, volume of vessel

= \({\pi} {r}^2 h\)

= \(22 × (1.75)^2 × 10 {m}^3\)

= 96.25 \({m}^3\)

Therefore, the capacity of the vessel is 96.25 \({m}^3\) or 96250 litres.

i) inner curved surface area of the vessel,

ii) radius of the base,

iii) capacity of the vessel.

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

Answer :

i)Given :

Rs. 20 is the cost Of painting 1 \({m}^2\) area.

\(\therefore \) when Rs. 2200 is the cost of painting, area is

= \(\frac{1}{20} × 2200\) \({m}^2\)

= 110 \({m}^2\) area

Therefore, the inner curved surface area of the vessel is 110 \({m}^2\).

ii)Height (h) of vessel = 10 m

Surface area = 110 \({m}^2\)

Let the radius of the base of the vessel be r.

We know that,

Surface area = \(2 {\pi} r h\)

But, CSA = 110 \({m}^2\) ...(from i))

\(\Rightarrow \) 110 \({m}^2\) = \(2 × \frac{22}{7} × r × 10 m\)

\(\Rightarrow \) r = \(\frac{7}{4} m = 1.75 m\)

iii) Now, volume of vessel

= \({\pi} {r}^2 h\)

= \(22 × (1.75)^2 × 10 {m}^3\)

= 96.25 \({m}^3\)

Therefore, the capacity of the vessel is 96.25 \({m}^3\) or 96250 litres.

Q6 ) 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?

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 \({m}^3\)

Let the radius of the circular end be r.

But we know that,

Volume of cylinder = \({\pi} {r}^2 h\)

Thus, we get that,

\(\Rightarrow \) \({\pi} {r}^2 h = 0.0154\) \({m}^3\)

\(\Rightarrow \) \(\frac{22}{7} × {r}^2 × 1 m \) = 0.0154 \({m}^3\)

\(\Rightarrow \) r = 0.07 m.

Also, we have,

Total surface area of vessel

= \(2 {\pi}r(r + h)\)

= \(2 × \frac{22}{7} × 0.07 (1 + 0.07)\) \({m}^2\)

= 0.44 × 1.07 \({m}^2\)

= 0.4708 \({m}^2\)

Therefore, 0.4708 \({m}^2\) of the metal sheet would be required to make the cylindrical vessel.

Q7 ) 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.

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

Answer :

Given :

Radius (R) of pencil = \(\frac{7}{2} mm = 3.5 mm = 0.35 cm\)

Radius (r) of graphite = \(\frac{1}{2} mm = 0.5 mm = 0.05 cm\)

Height (h) of pencil = 14 cm

We know that,

Volume of wood in pencil

= \({\pi} ({R}^2 - {r}^2) h\)

= \(\frac{22}{7} × ({0.35}^2 - {0.05}^2) × 14\) \({cm}^3\)

= \(\frac{22}{7} × (0.1225 - 0.0025) × 14\) \({cm}^3\)

= 44 × 0.12 \({cm}^3\)

= 5.28 \({cm}^3\)

Similarly, Volume of graphite in pencil

= \({\pi} {r}^2 h\)

= \(\frac{22}{7} × {0.05}^2 × 14\) \({cm}^3\)

= 44 × 0.0025 \({cm}^3\)

= 0.11 \({cm}^3\)

Q8 ) 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?

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

Answer :

Given :

Radius (r) of cylindrical bowl = \(\frac{7}{2} cm = 3.5 cm\)

Height (h) of bowl, up to which bowl is filled with soup = 4 cm

We know that,

Volume Of soup in bowl

= \({\pi} {r}^2 h\)

= \(\frac{22}{7} × {3.5}^2 × 4\) \({cm}^3\)

= 11 × 3.5 × 4 \({cm}^3\)

= 154 \({cm}^3\)

Therefore, Volume of soup given to 250 patients

= (250 × 154) \({cm}^3\)

= 38500 \({cm}^3\)

Q1 )
Find the volume of the right circular cone with

i) radius 6 cm, height 7 cm

ii) radius 3.5 cm, height 12 cm.

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

= \(\frac{1}{3} {\pi} {r}^2 h\)

= \(\frac{1}{3} × \frac{22}{7} × {6}^2 × 7\) \({cm}^3\)

= 12 × 22 \({cm}^3\)

= 264 \({cm}^3\)

Therefore, the volume of the cone is 264 \({cm}^3\)

ii)Given :

Radius (r) of cone = 3.5 cm

Height (h) of cone = 12 cm

Similarly,

Volume of cone

= \(\frac{1}{3} {\pi} {r}^2 h\)

= \(\frac{1}{3} × \frac{22}{7} × {3.5}^2 × 12\) \({cm}^3\)

= 12.833 × 12 \({cm}^3\)

= 154 \({cm}^3\)

Therefore, the volume of the cone is 154 \({cm}^3\)

Q2 )
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 .

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

= \(\sqrt({l}^2 - {r}^2)\) cm

= \(\sqrt({25}^2 - {7}^2)\) cm

= \(24\) cm

We know that,

Volume of cone

= \(\frac{1}{3} {\pi} {r}^2 h\)

= \(\frac{1}{3} × \frac{22}{7} × {7}^2 × 24\) \({cm}^3\)

= 154 × 8 \({cm}^3\)

= 1232 \({cm}^3\)

Therefore, capacity of the conical vessel

= 1232 \({cm}^3\) = 1.232 litres

ii) Height (h) of cone = 12 cm

Slant height (l) of cone = 13 cm

Radius of cone

= \(\sqrt({l}^2 - {h}^2)\) cm

= \(\sqrt({13}^2 - {12}^2)\) cm

= \(5\) cm

We know that,

Volume of cone

= \(\frac{1}{3} {\pi} {r}^2 h\)

= \(\frac{1}{3} × \frac{22}{7} × {5}^2 × 12\) \({cm}^3\)

= 300 × 1.047 \({cm}^3\)

= 314.28 \({cm}^3\)

Therefore, capacity of the conical vessel = 314.28 \({cm}^3\) = 0.31428 litres

Q3 ) The height of a cone is 15 cm. If its volume is 1570 \({cm}^3\) , find the radius of the base. (Use \({\pi}\) = 3.14)

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 \({cm}^3\)

Let the radius of the cone be r.

But, Volume of cone= \(\frac{1}{3} {\pi} {r}^2 h\)

\(\Rightarrow \) 1570 \({cm}^3\) = \(\frac{1}{3} × \frac{22}{7} × {r}^2 × 15\) \(cm\)

\(\Rightarrow \) \({r}^2 = 100 {cm}^2\)

\(\Rightarrow \) r = 10 cm

Therefore, the radius of the base of cone is 10 cm.

Q4 ) If the volume of a right circular cone of height 9 cm is \(48 {\pi} {cm}^3\) , find the diameter of its base.

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

Answer :

Given :

Height (h) of cone = 9 cm

Also, Volume of cone = \(48 {\pi} {cm}^3\)

Let the radius of the cone be r.

But, Volume of cone = \(\frac{1}{3} {\pi} {r}^2 h\)

\(\Rightarrow \) \(48 {\pi} {cm}^3\) = \(\frac{1}{3} × {\pi} × {r}^2 × 9\) \(cm\)

\(\Rightarrow \)\({r}^2 = \frac{48 {\pi} × 3}{9 × {\pi}} {cm}^2\)

\(\Rightarrow \) \({r}^2 = {4}^2 {cm}^2\)

\(\Rightarrow \) r = 4cm

\(\therefore \) Diameter of base = 2r = 8 cm

Therefore, the diameter of the base of cone is 8 cm.

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

Answer :

Given :

Radius (r) of pit = \(\frac{3.5}{2} m = 1.75 m\)

Height (h) of pit = depth of pit = 12 m

We know that,

Volume of pit

= \(\frac{1}{3} {\pi} {r}^2 h\)

= \(\frac{1}{3} × \frac{22}{7} × {3.5}^2 × 12\) \({m}^3\)

= 38.5 \({m}^3\)

Thus, capacity of the pit

= (38.5 × 1) kilolitres

= 38.5 kilolitres

Q6 )
The volume of a right circular cone is 9856 \({cm}^3\) . 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.

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

Answer :

Given :

Radius of the base = \(\frac{28}{2} cm = 14 cm\)

Also, Volume of a right circular cone = 9856 \({cm}^3\)

i) Let the height of the cone be h.

But, we know that,

Volume of right circular cone = \(\frac{1}{3} {\pi} {r}^2 h\)

\(\Rightarrow \) 9856 \({cm}^3\) = \(\frac{1}{3} × \frac{22}{7} × {14}^2 × h\) \({cm}^2\)

\(\Rightarrow \) h = 48 cm.

Therefore, height of the cone is 48 cm.

ii)We know, slant height of cone

= \(\sqrt({h}^2 + {r}^2)\) cm

= \(\sqrt({48}^2 + {14}^2)\) cm

l = \(50 cm\)

Therefore, slant height of the cone is 50 cm.

iii) We also know that,

Curved surface area of the cone

= \({\pi}rl\)

= \(\frac{22}{7} × 14 × 50 {cm}^2\)

= 2200 \({cm}^2\)

Therefore, the curved surface area of the cone is 2200 \({cm}^2\).

Q7 ) 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.

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

Answer :

When right-angled \(\triangle{ABC}\) 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

= \(\frac{1}{3} {\pi} {r}^2 h\)

= \(\frac{1}{3} × {\pi} × {5}^2 × 12\) \({cm}^3\)

= \(\frac{1}{3} × 300 × {\pi}\) \({cm}^3\)

\(= 100 {\pi}\) \({cm}^3\)

Therefore, the volume of the cone so formed is \(100 {\pi}\) \({cm}^3\)

Q8 ) 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.

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

Answer :

When right-angled \(\triangle{ABC}\) 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

= \(\frac{1}{3} {\pi} {r}^2 h\)

= \(\frac{1}{3} × {\pi} × {12}^2 × 5\) \({cm}^3\)

= \(240 × {\pi}\) \({cm}^3\)

= \(240 {\pi}\) \({cm}^3\)

Therefore, the volume of the cone so formed is \(240 {\pi}\) \({cm}^3\)

Thus, the ratio of the volumes of the two solids obtained in Questions 7 and 8

= \(\frac{100}{240}\)

= \(\frac{5}{12}\)

\(\therefore \) 5 : 12.

Q9 ) 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.

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

Answer :

Given :

Radius (r) of heap = \(\frac{10.5}{2} m = 5.25 m\)

Height (h) of heap = 3 m

As we know that,

Volume of heap

= \(\frac{1}{3} {\pi} {r}^2 h\)

= \(\frac{1}{3} × \frac{22}{7} × {5.25}^2 × 3\) \({m}^3\)

= 86.625 \({m}^3\)

Therefore, the volume of the heap of wheat is 86.625 \({m}^3\).

But, Area of canvas required = CSA of cone

So, we get,

= \({\pi} r l\)

= \({\pi} × r × (\sqrt{(r)^2 + (h)^2})\)

= \(\frac{22}{7} × 5.25 × (\sqrt{(5.25)^2 + (3)^2})\) \({m}^2\)

= \(\frac{22}{7} × × 5.25 × 6.05\) \({m}^2\)

= 99.825 \({m}^2\)

Therefore, 99.825 \({m}^2\) canvas will be required to protect the heap from rain.

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

= \(\frac{4}{3}{\pi}{r}^3\)

= \(\frac{4}{3} × \frac{22}{7} × {7}^3\) \({cm}^3\)

= \(\frac{4312}{3}\) \({cm}^3\)

= \(1437\frac{1}{3}\) \({cm}^3\)

Therefore, the volume of the sphere is \(1437\frac{1}{3}\) \({cm}^3\).

ii) Given :

Radius of sphere = 0.63 m

We know that,

Volume of sphere

= \(\frac{4}{3}{\pi}{r}^3\)

= \(\frac{4}{3} × \frac{22}{7} × {0.63}^3\) \({m}^3\)

= \(1.0478\) \({m}^3\)

Therefore, the volume of the sphere is \(1.05\) \({m}^3\) (approx.)

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

Answer :

i)Given :

Radius of sphere = \(\frac{28}{2} cm = 14 cm\)

We know that,

Volume of sphere

= \(\frac{4}{3}{\pi}{r}^3\)

= \(\frac{4}{3} × \frac{22}{7} × {14}^3\) \({cm}^3\)

= \(11498\frac{2}{3}\) \({cm}^3\)

Therefore, the volume of the sphere is \(11498\frac{2}{3}\) \({cm}^3\).

ii)Given :

Radius of sphere = \(\frac{0.21}{2} m = 0.105 m\)

We know that,

Volume of sphere

= \(\frac{4}{3}{\pi}{r}^3\)

= \(\frac{4}{3} × \frac{22}{7} × {0.105}^3\) \({m}^3\)

= \(0.004851\) \({m}^3\)

Therefore, the volume of the sphere is \(0.004851\) \({m}^3\).

Q3 ) 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 \({cm}^3\) ?

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

Answer :

Given :

Radius of sphere = \(\frac{4.2}{2} cm = 2.1 cm\)

We know that,

Volume of sphere

= \(\frac{4}{3}{\pi}{r}^3\)

= \(\frac{4}{3} × \frac{22}{7} × {2.1}^3\) \({cm}^3\)

= \(38.808\) \({cm}^3\)

Now, we know that,

\(Density = \frac{Mass}{Volume}\)

\(\Rightarrow \) Mass = Density × Volume

\(\Rightarrow \) = 8.9 × 38.808 g

\(\Rightarrow \) = 345.3912 g

Hence, the mass of the ball is 345.39 g (approximately).

Q4 ) 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?

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 \(\frac{d}{4}\).

Thus, Radius of earth = \(\frac{d}{2}\)

\(\therefore \) Radius of moon

= \(\frac{1}{2}\) × \(\frac{d}{4}\)

= \(\frac{d}{8}\)

Now, Volume of moon

= \(\frac{4}{3}{\pi}{r}^3 \)

= \(\frac{4}{3}{\pi} × {\frac{d}{8}}^3 \)

=\( \frac{1}{512} × \frac{4}{3}{\pi} × {d}^3\)

Similarly,

Volume of earth

= \(\frac{4}{3}{\pi}{r}^3 \)

=\( \frac{4}{3}{\pi} × {\frac{d}{2}}^3 \)

= \(\frac{1}{8} × \frac{4}{3}{\pi} × {d}^3\)

\(\therefore \) \(\frac{Volume of moon}{Volume of earth} \)

=\( \frac{\frac{1}{512} × \frac{4}{3}{\pi} × {d}^3}{\frac{1}{8} × \frac{4}{3}{\pi} × {d}^3} \)

= \(\frac{1}{64}\)

Therefore, the volume of moon = \(\frac{1}{64}\) is of the volume of earth.

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

Answer :

Given :

Radius (r) of hemisphere bowl = \(\frac{10.5}{2} cm = 5.25 cm\)

We know that,

Volume of hemisphere

= \(\frac{2}{3}{\pi}{r}^3\)

= \(\frac{2}{3} × \frac{22}{7} × {5.25}^3\) \({cm}^3\)

= \(303.1875\) \({cm}^3\)

= \(\frac{303.1875}{1000} l\)

= 0.303 litre (approximately)

Therefore ,the volume hemisphere bowl is 0.303 litres.

Q6 ) 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.

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

= \(\frac{2}{3}{\pi}({R}^3 - {r}^3)\)

= \(\frac{2}{3} × \frac{22}{7} × ({1.01}^3 - {1}^3)\) \({m}^3\)

= \(\frac{44}{21} × (1.030301 - 1)\) \({m}^3\)

= 0.06348 \({m}^3\) (approx.)

Therefore, the volume of the iron used to make the tank is 0.06348 \({m}^3\).

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

Answer :

Given :

Surface area of sphere = 154 \({cm}^2\).

Let radius of sphere be r.

But, we also know that,

Surface area of sphere = \(4{\pi}{r}^2\)

\(\Rightarrow \) 154 \({cm}^2\)= \(4 × \frac{22}{7} × {r}^2\)

\(\Rightarrow \) \({r}^2 = (\frac{154 × 7}{4 × 22})\) \({cm}^2\)

\(\Rightarrow \) r = \(\frac{7}{2} cm\)

\(\Rightarrow \) r = 3.5 cm

Now, Volume of sphere

= \(\frac{4}{3}{\pi}{r}^3\)

= \(\frac{4}{3} × \frac{22}{7} × {3.5}^3\) \({cm}^3\)

= \(179\frac{2}{3}\) \({cm}^3\)

Therefore , the volume of the sphere is

= \(179\frac{2}{3}\) \({cm}^3\).

Q8 )
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.

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 \({m}^2\) area = Rs. 2.00

Therefore, CSA of the inner side of dome

= Rs. \(\frac{498.96}{2} {m}^2 \)

= \( 249.48 {m}^2\)

ii) Let the inner radius of the hemispherical dome be r.

Now, CSA of inner side of dome = 249.48 \({m}^2\)

But, CSA = \(2{\pi}{r}^2\)

\(\Rightarrow \) 249.48 \({m}^2\) = \(2 × \frac{22}{7} × {r}^2\)

\(\Rightarrow \) \({r}^2 = (\frac{249.48 × 7}{2 × 22})\) \({m}^2\)

\(\Rightarrow \) r = 6.3 m.

Also, we have,

Volume of air inside the dome = Volume of hemispherical dome

= \(\frac{2}{3}{\pi}{r}^3\)

= \(\frac{2}{3} × \frac{22}{7} × {6.3}^3\) \({m}^3\)

= \(523.908\) \({m}^3\) (approx.)

Therefore, volume of the air inside the dome is \(523.908\) \({m}^3\).

Q9 )
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'.

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 = \(\frac{4}{3}{\pi}{r}^3\)

\(\therefore \) Volume of 27 solid iron spheres

= 27 × \(\frac{4}{3}{\pi}{r}^3\)

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 = \(\frac{4}{3}{\pi}{r'}^3\)

\(\Rightarrow \) \(\frac{4}{3}{\pi}{r'}^3\) = 27 × \(\frac{4}{3}{\pi}{r}^3\)

\(\Rightarrow \) \({r'}^3 = 27 × {r}^3\)

\(\Rightarrow \) r' = 3r.

ii) Now, we have,

surface area of 1 solid iron sphere of radius r = \(4{\pi}{r}^2\)

So, Surface area of iron sphere of radius r'

= \(4{\pi}{r'}^2\)

= \(4{\pi}{3r}^2\)

= \(36{\pi}{r}^2\)

Therefore, ratio of S and S'

= \(\frac{S}{S'} \)

= \(\frac{4{\pi}{r}^2}{36{\pi}{r}^2}\)

= \(\frac{1}{9}\)

= 1 : 9

Q10 ) A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in \({mm}^3\)) is needed to fill this capsule?

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

Answer :

Given :

Radius (r) of capsule = \(\frac{3.5}{2} mm = 1.75 mm\)

We know that,

Volume of spherical capsule

= \(\frac{4}{3}{\pi}{r}^3\)

= \(\frac{4}{3} × \frac{22}{7} × {1.75}^3\) \({mm}^3\)

= \(22.458\) \({mm}^3\)

= \(22.46\) \({mm}^3\)(appro×.)

Therefore ,the volume of the spherical capsule is \(22.46\) \({mm}^3\).

Q1 )
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 \(cm^2 \)and the rate of pointing is 10 paise per \(cm^2 \) find the total expenses required for palishing and painting the surface of the bookshelf.

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

Therefore External surface area of shelf is 19100 \(cm^2 \)

Area of front face

= [85×110-75×100+2(75×5)]

= 1850+750

= 2600 \( cm^2 \)

\(\therefore \) area is 2600 \( cm^2 \)

Area to be polished

= (19100+2600) \( cm^2 \)

= 21700 \( cm^2 \).

Cost of polishing 1 \( cm^2 \) area = Rs 0.20

Cost of polishing 21700 \( cm^2 \) 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 \( cm^2 \).

Area to be painted in 3 rows

= (3×6450)\( cm^2 \)

= 19350 \( cm^2 \).

Cost of painting 1 \( cm^2 \) area = Rs. 0.10

Cost of painting 19350 \( cm^2 \) area = Rs (19350 x 0.1) = Rs 1935

Total expense required for polishing and painting = Rs. (4340+1935) = Rs. 6275

Q2 )
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 \( cm^2 \) and black paint costs 5 paise per \( cm^2 \).

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

Answer :

Diameter of wooden sphere = 21 cm

Radius of wooden sphere, (r)

= \(\frac{diameter}{2} \)

= (\(\frac{21}{2} \)) cm

= 10.5 cm

Surface area of a wooden sphere

\( = 4\pi r^2 \)

\( = 4×(\frac{22}{7} )×(10.5)^2 = 1386 \)\( cm^2\)

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\(\pi \)rh

= 2×(\(\frac{22}{7} \))×1.5×7

= 66 \( cm^2 \)

Now,

Area of the circular end of cylindrical support

= \( \pi r^2 \)

= \(\frac{22}{7} ×(1.5)^2 = 7.07 \)

Again,

Area to be painted silver

= [8 ×(1386-7.07)]

= 8×1378.93

= 11031.44

Area to be painted is 11031.44 \( cm^2 \)

Cost for painting with silver colour

= Rs(11031.44×0.25)

=Rs 2757.86

Area to be painted black

= (8×66) \( cm^2 \)

= 528 \( cm^2 \)

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

Q3 ) The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?

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 – \(\frac{25}{100} \) x d

= d –\(\frac{1}{4} d = \frac{3}{4} d \)

Since, surface area of a sphere

= \( 4\pi r^2 \)

= \( \pi (2r)^2 \)

= \( \pi d^2 \)

Surface area of a sphere, when diameter of the sphere is

= \(\pi (\frac{3}{4} d)^2 \)

= \(\pi \frac{9}{16} d^2 \)

Now, decrease percentage in curved surface area

= \(\frac{\pi d^2 - \pi \frac{9}{16} d^2 }{\pi d^2} × 100 \)

= \(\frac{16-9}{16} × 100 \)

= \( \frac{7}{16} × 100 \)

= 43.75%

There are total 76 questions present in ncert solutions for class 9 maths chapter 13 surface areas and volumes

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There are total 9 exercise present in ncert solutions for class 9 maths chapter 13 surface areas and volumes