Surface Areas and Volumes

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Questions:

Generic 2

9 x 2 = 18

The volume of a cylinder is 448\(\pi \) cubic cm and the height is 7 cm. Find its total surface area.

A storage tank is in the form of a cube, when full has the volume of water as 15.625 m³. If the present depth of water is 1.3 m, find the volume of water already used from the tank.

The height, breadth and length of a cuboidal box are in the ratio 1 : 2 : 3. Find the volume of the box if its surface area is 1078 dm².

The total surface area of a solid hemisphere is 5940 cm², Find the diameter of the hemisphere.

How many metres of cloth \(1\frac { 4 }{ 7 } m\) wide will be 7 required to make a conical tent whose base diameter is 10 m and whose vertical height is 12 cm?

The circumference of the base of a 24 m high solid wooden cone is 44 m. Find its curved surface area.

The floor of a rectangular hall has a perimeter of 250 m and its length and breadth are in the ratio of 13: 12. If the cost of painting the four walls and ceiling at the rate of Rs. 5 per m² is Rs.27000, find the height of the hall.

Three cubes each of the side 3 cm are joined end to end. Find the surface area of the resulting cuboid.

The edge of a cube is 10.5 mm. Find its total surface area in cm².

Generic 3

4 x 3 = 12

10.

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 area of the balloon in the two cases.

11.

In figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

12.

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 neligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m \(\times\) 3 m?

13.

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.

Generic 4

6 x 4 = 24

14.

The teacher asked students to prepare project related to Diwali. Lipsa prepared 12 Cylindrical candles each having radius of base 2 cm and height 7 cm. Himanshu prepared 14 fire crackers each of spherical shape of radius 1.50 cm.
(i) find the volume of candles and fire crakers.
(ii) According to you which has better project work and why?
(iii) Which message has been conveyed in above information?

15.

Arihant builds a room measuring roof 22 m by 20 m. He also builds a cylindrical tank having diameter of base 2 m and height 3.5 m sdjoining the room to collect the rain water of roof for harvesting.
(i) If the tank is just filled with rain water, find the rainfall in cm.
(ii) What values are depicted in Arihant's plan?

16.

Student A stated the formula for the volume of cone as "V=\(\frac { 1 }{ 3 } (\pi { l }^{ 2 }h)-\frac { 1 }{ 3 } \pi { h }^{ 3'' }\)and the student B as "V = \(\frac { 1 }{ 3 } \pi { r }^{ 2 }h"\)
(a) Are the formulae stated by A and B true? If yes, why?
(b) By stating formula different than as usual which of the following values is deicted.
(i) Truth value
(ii) Social value
(iii) Respect for other views
(iv) Equality

17.

A right angled \(\Delta \)ABC with sides 3 cm, 4 cm and 5 cm is revolved about the fixed side of 4 cm. Find the volume of the solid generated. Also, find the total surface area of the solid.

18.

The diameter of roller 1.5 m long is 84 cm. If it takes 100 revolutions to level a playground, find the cost of levelling this ground at the rate of 50 paise per square metre.

19.

The dimensions of a rectangular box are in the ratio of 2 : 3 : 4 and the difference between the cost of covering it with sheet of paper at the rates of Rs 8 and Rs 9.50 per m² is Rs 1248. Find the dimensions of the box.

Multiple Choice Question

6 x 1 = 6

20.

The area of the four walls of a room is 80 cm² and its height is 4 m. Then, the perimeter of the floor of the room is

(a)

16 m

(b)

5 m

(c)

20 m

(d)

10 m

21.

The area of the four walls of a room is 300 m². Its length and height are 15 m and 6 m respectively. Find its breadth.

(a)

10 m

(b)

5 m

(c)

20 m

(d)

15 m

22.

A brick measures 25 cm \(\times\) 12 cm \(\times\) 10 cm. Its surface area is

(a)

670 cm²

(b)

1340 cm²

(c)

3000 cm²

(d)

1500 cm²

23.

The side of a cube is 1 cm. The total surface area of the figure formed by joining two such cubes is

(a)

2(2 + 1 + 2) cm²

(b)

2(2 + 2 + 2) cm²

(c)

2(1 + 1 + 1) cm²

(d)

2(1 + 1 + 2) cm²

24.

The total surface area of a cube of side a is

(a)

4a²

(b)

6a²

(c)

3a²

(d)

8a².

25.

Which of the following is a plane figure?

(a)

Cone

(b)

Square

(c)

Cylinder

(d)

Cube.

1 Marks

4 x 1 = 4

26.

The radii of two right circular cylinders are in the ratio 2:3 and their heights are in the ratio 5:4, then the ratio of their volumes will be _______________

27.

The diameter of a football is five times the diameter of a criket ball. Ratio of surface areas of football and criket ball is _____________

28.

If the number of square centimetres in the surface area of a shpere is equal to the number of cubic cm in its volume. find the diameter of the sphere?

29.

Find the capacity of a tank of demensions 8 am \(\times\)6 cm \(\times\)2.5 cm.

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Answers Key:

Generic 2

9 x 2 = 18

240 \(\pi \) cm²

7.5 m³

2058 dm³

50.18 cm

130 m

550 m²

21.6 m

126 cm²

6.615 cm²

Generic 3

4 x 3 = 12

10.

Case I. r = 7 cm
\(\therefore\) Surface area = \(4\pi { r }^{ 2 }\)
\(=4\times \frac { 22 }{ 7 } \times { \left( 7 \right) }^{ 2 }=616{ cm }^{ 2 }\)
Case II. r = 14 cm
\(\therefore\) Surface area = \(4\pi { r }^{ 2 }\)
\(=4\times \frac { 22 }{ 7 } \times { \left( 14 \right) }^{ 2 }\)
\(\\ =2464{ cm }^{ 2 }\)
\(\therefore\) Ratio of surface area of the balloon
\(=616:2464\)
\(\\ =\frac { 616 }{ 2464 } =\frac { 1 }{ 4 } =1:4\)

11.

2r = 20 cm
\(\Rightarrow\) r = 10 cm
h = 30 cm
\(\therefore \) Cloth required = \(2\pi r\left( h+2.5+2.5 \right) \)
\(=2\pi r\left( h+5 \right) =2\times \frac { 22 }{ 7 } \times 10\times \left( 30+5 \right) \)
\(\\ =2200{ cm }^{ 2 }.\)

12.

For shelter
l = 4 m, b = 3 m,
h = 2.5 m
\(\therefore\) Total surface area of the shelter
= lb + 2(bh + hl)
= (4)(3) + 2[(3)(2.5) + (2.5)(4)]
= 12 + 2[7.5 + 10]
= 47 m²
Hence, 47 m² of tarpaulin will be required.

13.

Let the length, breadth and height of the rectangular hall be l m, b m and h m respectively.
Perimeter = 250 m
\(\Rightarrow\) 2(l + b) = 250
\(\Rightarrow\)  l + b = 125 ...(1)
Area of the four walls
\(=\frac { 15000 }{ 10 } =1500\)m²
\(\Rightarrow\) 2(l + b)h = 1500
\(\Rightarrow\) (l + b)h = 750
\(\Rightarrow\) 125 h = 750    Using (1)
\(\Rightarrow\)   \(h=\frac { 750 }{ 125 } \)
\(\Rightarrow\)   h = 6 m
Hence, the height of the hall is 6 m.

Generic 4

6 x 4 = 24

14.

(i) Radius of a cylindrical candle(r) = 2 cm
Height of a cylindrical candle (h) = 7 cm
\(\therefore\) Volume of 12 Cylindrical candles = 12 \(\times\) 88
= 1056 cm³
Also, radius of a spherical fire cracker
r = 1.5 cm = \(\frac { 3 }{ 2 } \)cm
\(\therefore\) Volume of a spherical fire cracker
\(=\frac { 4 }{ 3 } \pi { r }^{ 3 }\)
\(=\frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times \frac { 3 }{ 2 } \times \frac { 3 }{ 2 } \times \frac { 3 }{ 2 } \)
\(=\frac { 99 }{ 7 } { cm }^{ 3 }\)
Volume of 14 spherical fire crackers
\(=\frac { 99 }{ 7 } \times 14=198\quad { cm }^{ 3 }\)
(ii) Lipsa has better project because candles do not pollute the environment.
(iii) Avoid pollution and save energy.

15.

(i) We have, radius of cylindrical tank
r = 1 m
and height of cylindrical tank h = 3.5 m
Volume of cylindrical tank = \(\pi\)r²h
\(=\frac { 22 }{ 7 } \times 1\times 1\times 3.5\)
= 11 m³
Let the rainfall be h m, then
Volume of water on the roof = Volume of cylindrical tank
\(\Rightarrow\) 22 \(\times\) 20 \(\times\) h = 11
\(\Rightarrow \ h=\frac { 11 }{ 22\times 20 } \)
\(=\frac { 1 }{ 40 } m\)
\(=\frac { 100 }{ 40 } \)
= 2.5 cm
(ii) Save water to save earth.

16.

(a) Volume of cone = \(\frac { 1 }{ 3 } \pi { r }^{ 2 }h\) ......(i)
Also, l²= r²+ h² ............(ii)
r²= l²- h²
Substituting (ii), in (i), we get
Volume of cone (v) = \(\frac { 1 }{ 3 } \pi ({ l }^{ 2 }-{ h }^{ 2 })h\)
\(=\frac { 1 }{ 3 } \pi { l }^{ 2 }h-\frac { 1 }{ 3 } \pi { h }^{ 3 }\)
So, answer is yes.
(b) Truth value

17.

r_cone = 3 cm
h_cone = 4 cm
l_cone= 5 cm

Above given cone is formed with radius 3 cm, height 4 cm and slant height 5 cm when revolved about the fixed side of 4 cm.
\(V=\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
\(=\frac { 1 }{ 3 } .\frac { 22 }{ 7 } .(3)(3)(4)\)
= 37.71 cm³
Total surface area=\(\pi\)rl+\(\pi\)r²
\(=\frac { 22 }{ 7 } \times 3(5+3)\)
= 75.43 cm²

18.

For roller
\(r=\frac { 1.5 }{ 2 } m=0.75m\)
\(\\ h=84cm=0.84m\)
\(\therefore \) Curved surface area = \(2\pi rh\)
\(=2\times \frac { 22 }{ 7 } \times 0.75\times 0.84\)
\(\\ =3.96{ m }^{ 2 }\)
\(\therefore \)  Area of the ground levelled in 1 revolution
= 3.96 m²
\(\therefore \)  Area of the ground levelled in 100 revolutions
= 3.96 100 m²= 396 m²
\(\therefore \)  Cost of levelling
= Rs \(396\times \frac { 50 }{ 100 } =\) Rs 198

19.

Let the dimensions of the box be 2k, 3k and 4k.
\(\therefore \) Total surface area = 2(lb + bh + hl)
= 2(2k.3k + 3k.4k + 4k.2k)
= 52k²m²
Cost of covering at the rate of Rs 8 per m²
= 52k² \(\times\) 8 = Rs 416k²
Cost of covering at the rate of Rs 9.50 per m²
= 52k² \(\times\) 9.50 = Rs 494k²
Difference between the costs
= Rs 494k² - Rs 416k² = Rs 78k²
According to the question,
78k² = 1248
\(\Rightarrow\) k²= 16 \(\Rightarrow\) k = 4
Hence, the dimensions of the box are 8 m, 12 m and 16 m.

Multiple Choice Question

6 x 1 = 6

20.

Required number \(=\frac { 60\times 30\times 30 }{ 15\times 6\times 4 } =150\)

21.

Number of cubes = \(\frac { { \left( 20 \right) }^{ 3 } }{ { \left( 5 \right) }^{ 3 } } =64\)

22.

\(\frac { 2 }{ 3 } \times \left( 6\times 5\times 4 \right) 80{ m }^{ 3 }\)

23.

v = 5 \(\times\) (6 \(\times\) 2 \(\times\) 1.5)

24.

(b)
6a²

25.

(b)
Square

1 Marks

4 x 1 = 4

26.

()
Let radii of cylinders be 2x and 3x and heights be 5y and 3y respectively.
\(\therefore\) Ratio of volumes = \(\frac { \pi { (2x) }^{ 2 }\times 5y }{ \pi { (3x) }^{ 2 }\times 3y } \)
\(=\frac { { 4x }^{ 2 }\times 5 }{ { 9x }^{ 2 }\times 3 } \)
= 20:27.

27.

()
Given, diameter of football = 5 \(\times\) diameter of cricket ball
If r denotes radius of a football and r' that of a criket ball, then we have
2r = 5\(\times\)(2r')
\(\frac { 2r }{ 2r' } =5\)
or \(\frac { r }{ r' } =5\)
Now, ratio of surface areas\(=\frac { 4\pi { r }^{ 2 } }{ 4\pi { (r') }^{ 2 } } ={ \left( \frac { r }{ r' } \right) }^{ 2 }=\frac { 25 }{ 1 } \)
= 25 : 1

28.

()
Given, Area of Sphere=Volume of sphere
\(4\pi { r }^{ 2 }=\frac { 4 }{ 3 } \pi { r }^{ 3 }\)
where r is the radius of sphere
\(\Rightarrow\) r = 3 cm [on solving]
\(\therefore\) Diameter = 2r = 6 cm.

29.