RS Aggarwal solution class 8 chapter 20 Volume and Surface Area of Solid Exercise 20A

Exercise 20A

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Q1 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 1:

Find the volume, lateral surface area and the total surface area of the cuboid whose dimensions are:
(i) length = 22 cm, breadth = 12 cm and height = 7.5 cm
(ii) length = 15 m, breadth = 6 m and height = 9 dm
(iii) length = 24 m, breadth = 25 cm and height = 6 m
(iv) length = 48 cm, breadth = 6 dm and height = 1 m

Answer 1:

Volume of a cuboid $=(Length\times Breadth\times Height)$ cubic units
Total surface area $=2(lb+bh+lh)$ sq units
Lateral surface area $=\left[2\left(l+b\right)\times h\right]$ sq units

(i) Length = 22 cm, 

breadth = 12 cm, 

height = 7.5 cm

Volume $=(Length\times Breadth\times Height)$
= $(22\times 12\times 7.5)=1980 c{m}^3$

Total surface area $=2(lb+bh+lh)$

$= 2\left[\left(22\times 12\right)+\left(22\times 7.5\right)+\left(12\times 7.5\right)\right]$

$=2\left[264+165+90\right]=1038 c{m}^2$

Lateral surface area $=\left[2\left(l+b\right)\times h\right]$

$=2\left(22+12\right)\times 7.5=510 c{m}^2$

(ii) Length = 15 m, 

breadth = 6 m, 

height = 9 dm = 0.9 m

Volume $=(Length\times Breadth\times Height)$ 

= $(15\times 6\times 0.9)=81 m^3$

Total surface area$=2(lb+bh+lh)$ 

$= 2\left[\left(15\times 6\right)+\left(15\times 0.9\right)+\left(6\times 0.9\right)\right]$

$=2\left[90+13.5+5.4\right]=217.8 m^2$

Lateral surface area $=\left[2\left(l+b\right)\times h\right]$

$=2\left(15+6\right)\times 0.9=37.8 m^2$

(iii) Length = 24 m, 

breadth = 25 cm = 0.25 m, 

height = 6 m

Volume $=(Length\times Breadth\times Height)$ = $(24\times 0.25\times 6)=36 m^3$

Total surface area$=2(lb+bh+lh)$ $= 2\left[\left(24\times 0.25\right)+\left(24\times 6\right)+\left(0.25\times 6\right)\right]$

$=2\left[6+144+1.5\right]=303 m^2$

Lateral surface area $=\left[2\left(l+b\right)\times h\right]$

$=2\left(24+0.25\right)\times 6$

$=291 m^2$

(iv) Length = 48 cm = 0.48 m, 

breadth = 6 dm = 0.6 m, 

height = 1 m

Volume $=(Length\times Breadth\times Height)$ 

= $(0.48\times 0.6\times 1)=0.288 m^3$

Total surface area $=2(lb+bh+lh)$

$= 2\left[\left(0.48\times 0.6\right)+\left(0.48\times 1\right)+\left(0.6\times 1\right)\right]$

$=2\left[0.288+0.48+0.6\right]$

$=2.736 m^2$

Lateral surface area $=\left[2\left(l+b\right)\times h\right]$

$=2\left(0.48+0.6\right)\times 1$

$=2.16 m^2$

Q2 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 2:

The dimensions of a rectangular water tank are 2 m 75 cm by 1 m 80 cm by 1 m 40 cm. How many litres of water does it hold when filled to the brim?

Answer 2:

 $1 m = 100 cm$
Therefore, dimensions of the tank are:
$2 m 75 cm\times 1 m 80 cm\times 1 m 40 cm$

$=275 cm \times 180 cm \times 140 cm$

∴ Volume =  $\mathrm{Length} \times \mathrm{Breadth}\times \mathrm{Height}$

$= 275\times 180\times 140=6930000 {\mathrm{cm}}^3$

Also, $1000c{m}^3=1L$

∴ Volume $=\frac{6930000}{1000}=6930 L$

Q3 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 3:

A solid rectangular piece of iron measures 1.05 m × 70 cm × 1.5 cm. Find the weight of this piece in kilograms if 1 cm³ of iron weighs 8 grams.

Answer 3:

$1m=100cm$
∴ Dimensions of the iron piece = $105 \mathrm{cm}\times 70 \mathrm{cm}\times 1.5 \mathrm{cm}$

Total volume of the piece of iron $=(105\times 70\times 1.5)$

$=11025 c{m}^3$

1 cm³ measures 8 gms.

∴Weight of the piece $=11025 \times 8 = 88200 g$

$=\frac{88200}{1000} = 88.2 kg \left(because 1 kg = 1000 g\right)$

Q4 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 4:

The area of a courtyard is 3750 m². Find the cost of covering it with gravel to a height of 1 cm if the gravel costs Rs 6.40 per cubic metre.

Answer 4:

$1 cm = 0.01 m$
Volume of the gravel used $= \mathrm{Area} \times H\mathrm{eight}$

$= (3750 \times 0.01)=37.5 {\mathrm{m}}^3$

Cost of the gravel is Rs 6.40 per cubic meter.

∴ Total cost $=(37.5\times 6.4)=$Rs 240

Q5 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 5:

How many persons can be accommodated in a hall of length 16 m. breadth 12.5 m and height 4.5 m, assuming that 3.6 m³ of air is required for each person?

Answer 5:

Total volume of the hall$=(16\times 12.5\times 4.5)=900 m^3$

It is given that $3.6 m^3$ of air is required for each person.

The total number of persons that can be accommodated in that hall 

$$\begin{gathered} =\frac{\mathrm{Total} \mathrm{volume}}{\mathrm{Volume} \mathrm{required} \mathrm{by} \mathrm{each} \mathrm{person}}= \frac{900}{3.6} \\ =250 \mathrm{people} \end{gathered}$$ 

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Question 6:

A cardboard box is 1.2 m long, 72 cm wide and 54 cm high. How many bars of soap can be put into it if each bar measures 6 cm × 4.5 cm × 4 cm?

Answer 6:

Volume of the cardboard box $=(120\times 72\times 54)$

$=466560 c{m}^3$

Volume of each bar of soap$=(6\times 4.5\times 4)$

$=108 c{m}^3$

Total number of bars of soap that can be accommodated in that box

$=\frac{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{box}}{\mathrm{Volume} \mathrm{of} \mathrm{each} \mathrm{soap}}=\frac{466560}{108}=4320$ bars

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Question 7:

The size of a matchbox is 4 cm × 2.5 cm × 1.5 cm. What is the volume of a packet containing 144 matchboxes? How many such packets can be placed in a carton of size 1.5 m × 84 cm × 60 cm?

Answer 7:

Volume occupied by a single matchbox$=(4\times 2.5\times 1.5)=15 c{m}^3$

Volume of a packet containing 144 matchboxes 

$=(15\times 144)=2160 c{m}^3$

Volume of the carton$=(150\times 84\times 60)$

$=756000 c{m}^3$

Total number of packets is a carton$=\frac{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{carton}}{\mathrm{Volume} \mathrm{of} \mathrm{a} \mathrm{packet} }$

$= \frac{75600}{2160}=350$ packets

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Question 8:

How many planks of size 2 m × 25 cm × 8 cm can be prepared from a wooden block 5 m long, 70 cm broad and 32 cm thick, assuming that there is no wastage?

Answer 8:

Total volume of the block $=(500\times 70\times 32)$

$=1120000 c{m}^3$
 

Total volume of each plank $=200\times 25\times 8=40000 c{m}^3$

$=200\times 25\times 8=40000 c{m}^3$

∴ Total number of planks that can be made

$=\frac{\mathrm{Total} \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{block}}{\mathrm{Volume} \mathrm{of} \mathrm{each} \mathrm{plank}}$

$= \frac{1120000}{40000}=28$ planks

Q9 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 9:

How many bricks, each of size 25 cm × 13.5 cm × 6 cm, will be required to build a wall 8 m long, 5.4 m high and 33 cm thick?

Answer 9:

Volume of the brick $=25\times 13.5\times 6$

$=2025 c{m}^3$

Volume of the wall $=800\times 540\times 33$

$=14256000 c{m}^3$

Total number of bricks $=\frac{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{wall}}{\mathrm{Volume} \mathrm{of} \mathrm{each} \mathrm{brick}}$

$=\frac{14256000}{2025}=7040$ bricks  

Q10 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 10:

A wall 15 m long, 30 cm wide and 4 m high is made of bricks, each measuring 22 cm × 12.5 cm × 7.5 cm. If $\frac{1}{12}$ of the total volume of the wall consists of mortar, how many bricks are there in the wall?

Answer 10:

Volume of the wall$=1500\times 30\times 400$

$=18000000 c{m}^3$

Total quantity of mortar$=\frac{1}{12}\times 18000000$

$=1500000 c{m}^3$

∴ Volume of the bricks$=18000000-1500000$

$=16500000 c{m}^3$

Volume of a single brick$=22\times 12.5\times 7.5$

$=2062.5 c{m}^3$

∴ Total number of bricks$=\frac{\mathrm{Total} \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{bricks}}{\mathrm{Volume} \mathrm{of} \mathrm{a} \mathrm{single} \mathrm{brick}}$

$=\frac{16500000}{2062.5}=8000$ bricks

Q11 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 11

Find the capacity of a rectangular cistern in litres whose dimensions are 11.2 m × 6 m × 5.8 m. Find the area of the iron sheet required to make the cistern.

Answer 11:

Volume of the cistern$=11.2\times 6\times 5.8$

$=389.76 m^3=389.76\times 1000$

$=389760$ litres

Area of the iron sheet required to make this cistern 

= Total surface area of the cistern
$=2(11.2\times 6+11.2\times 5.8+6\times 5.8)$

$=2(67.2+64.96+34.8)=333.92 m^2$

Q12 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 12:

The volume of a block of gold is 0.5 m³. If it is hammered into a sheet to cover an area of 1 hectare, find the thickness of the sheet.

Answer 12:

Volume of the block$=0.5 m^3$

We know:
 $1 hectare = 10000 m^2$

Thickness$=\frac{\mathrm{Volume}}{\mathrm{Area}}=\frac{0.5}{10000}$

$= 0.00005 \mathrm{m}= 0.005 \mathrm{cm} = 0.05 \mathrm{mm}$

Q13 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 13:

The rainfall recorded on a certain day was 5 cm. Find the volume of water that fell on a 2-hectare field.

Answer 13:

Rainfall recorded = 5 cm = 0.05 m

Area of the field = 2 hectare 

=  $2\times 10000 {\mathrm{m}}^{2 }$

$= 20000 {\mathrm{m}}^2$

Total rain over the field = $\mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{field} \times \mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{field}$

$= 0.05\times 20000 = 1000 {\mathrm{m}}^3$

Q14 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 14:

A river 2 m deep and 45 m wide is flowing at the rate of 3 km/h. Find the quantity of water that runs into the sea per minute.

Answer 14:

Area of the cross-section of river $=45\times 2=90 m^2$

Rate of flow$=3\raisebox{1ex}{$ km$}\!\left/ \!\raisebox{-1ex}{$hr$}\right.$

$=\frac{3\times 1000}{60}=50 \frac{m}{min}$

Volume of water flowing through the cross-section in one minute 

$=90\times 50=4500 m^3$ per minute

Q15 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 15:

A pit 5 m long and 3.5 m wide is dug to a certain depth. If the volume of earth taken out of it is 14 m³, what is the depth of the pit?

Answer 15:

Let the depth of the pit be d m.
$\mathrm{<br>}$

$\mathrm{Volume} = \mathrm{Length} \times \text{w}\mathrm{idth} \times \text{d}\mathrm{epth}$

$= 5 \mathrm{m} \times 3.5 \mathrm{m}\times d m$

But,

Given volume = 14 m³

∴ Depth $= d =\frac{\text{volume}}{\mathrm{length} \times \mathrm{width}}$

$=\frac{14}{5\times 3.5}=0.8 \mathrm{m}$ = 80 cm

Q16 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 16:

A rectangular water tank is 90 cm wide and 40 cm deep. If it can contain 576 litres of water, what is its length?

Answer 16:

Capacity of the water tank $=576\text{ litres}=0.576 {\text{m}}^3$
Width = 90 cm = 0.9 m
Depth = 40 cm = 0.4 m

Length = $=\frac{\text{capacity}}{\text{width×depth}}$

$=\frac{0.576}{0.9\times 0.4}=1.600 \text{m}$

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Q17 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 17:

A beam of wood is 5 m long and 36 cm thick. It is made of 1.35 m³ of wood. What is the width of the beam?

Answer 17:

Volume of the beam$=1.35 {\text{m}}^3$

Length = 5 m

Thickness = 36 cm = 0.36 m

Width = $=\frac{\text{volume}}{\text{thickness×length}}$

$=\frac{1.35}{5\times 0.36}$

$=0.75 \text{m}=75 \text{cm}$

Q18 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 18:

The volume of a room is 378 m³ and the area of its floor is 84 m². Find the height of the room.

Answer 18:

Volume $\text{= height × area}$
Given:
 Volume  = 378 m³
Area = 84 m²

∴ Height $=\frac{\text{volume}}{\text{area}}=\frac{378}{84}=4.5 \text{m}$

Q19 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 19:

A swimming pool is 260 m long and 140 m wide. If 54600 cubic metres of water is pumped into it, find the height of the water level in it.

Answer 19:

Length of the pool = 260 m
Width of the pool = 140 m

Volume of water in the pool = 54600 cubic metres

∴ Height of water $=\frac{\text{volume}}{\text{length×width}}$

$=\frac{54600}{260\times 140}=1.5$ metres

Q20 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 20:

Find the volume of wood used to make a closed box of outer dimensions 60 cm × 45 cm × 32 cm, the thickness of wood being 2.5 cm all around.

Answer 20:

External length = 60 cm
External width = 45 cm
External height = 32 cm

External volume of the box$=60\times 45\times 32$

$=86400 {\text{cm}}^3$

Thickness of wood = 2.5 cm

∴  Internal length $=60-(2.5\times 2)=55$ cm
Internal width $=45-(2.5\times 2)=40$ cm
Internal height $=32-(2.5\times 2)=27$ cm

Internal volume of the box$= 55 \times 40 \times 27$= 59400 cm3

Volume of wood = External volume - Internal volume

$= 86400 - 59400 = 27000{\text{ cm}}^3$

Q21 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 21:

Find the volume of iron required to make an open box whose external dimensions are 36 cm × 25 cm × 16.5 cm, the box being 1.5 cm thick throughout. If 1 cm³ of iron weighs 8.5 grams, find the weight of the empty box in kilograms.

Answer 21:

External length = 36 cm
External width = 25 cm
External height = 16.5 cm

External volume of the box$= 36 \times 25 \times 16.5 = 14850 {\text{cm}}^3$

Thickness of iron = 1.5 cm

∴ Internal length $=36-(1.5\times 2)=33$ cm
Internal width $=25-(1.5\times 2)=22$ cm
Internal height $=16.5- 1.5=15$ cm  (as the box is open)

Internal volume of the box$= 33 \times 22 \times 15 = 10890 {\mathrm{cm}}^3$

Volume of iron = External volume − Internal volume

$= 14850 - 10890 = 3960{\text{ cm}}^3$

Given: 
${\text{1 cm}}^3 \text{of iron} = 8.5 \text{grams}$

Total weight of the box $= 3960 \times 8.5 = 33660 \mathrm{grams} = 33.66 \mathrm{kilograms}$

Q22 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 22:

A box with a lid is made of wood which is 3 cm thick. Its external length, breadth and height are 56 cm, 39 cm and 30 cm respectively. Find the capacity of the box. Also find the volume of wood used to make the box.

Answer 22:

External length = 56 cm
External width = 39 cm
External height = 30 cm

External volume of the box$=56 \times 39 \times 30 = 65520 {\text{cm}}^3$

Thickness of wood = 3 cm

∴ Internal length $=56-(3\times 2)=50$ cm
Internal width $=39-(3\times 2)=33$ cm
Internal height $=30-(3\times 2)=24$ cm

Capacity of the box = Internal volume of the box

$= 50 \times 33 \times 24 = 39600{\text{ cm}}^3$

Volume of wood = External volume − Internal volume

$= 65520 - 39600 = 25920 {\text{cm}}^3$

Q23 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 23:

The external dimensions of a closed wooden box are 62 cm, 30 cm and 18 cm. If the box is made of 2-cm-thick wood, find the capacity of the box.

Answer 23:

External length = 62 cm
External width = 30 cm
External height = 18 cm

∴ External volume of the box$=62\times 30\times 18=33480 c{m}^3$

Thickness of the wood = 2 cm

Now, internal length $=62-(2\times 2)=58$ cm
Internal width $=30-(2\times 2)=26$ cm
Internal height $=18-(2\times 2)=14$ cm

∴ Capacity of the box = internal volume of the box

$=(58\times 26\times 14) c{m}^3$

$=21112 c{m}^3$

Q24 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 24:

A closed wooden box 80 cm long, 65 cm wide and 45 cm high, is made of 2.5-cm-thick wood. Find the capacity of the box and its weight if 100 cm³ of wood weighs 8 g.

Answer 24:

External length = 80 cm
External width = 65 cm
External height = 45 cm

∴ External volume of the box$=80\times 65\times 45$

$=234000 c{m}^3$

Thickness of the wood = 2.5 cm

Then internal length$=80-(2.5\times 2)=75$ cm
Internal width $=65-(2.5\times 2)=60$ cm
Internal height $=45-(2.5\times 2)=40$ cm

Capacity of the box = internal volume of the box

$=(75\times 60\times 40) c{m}^3=180000 c{m}^3$

Volume of the wood = external volume − internal volume

$=(234000-180000) c{m}^3=54000 c{m}^3$

It is given that $100 {\mathrm{cm}}^3 \mathrm{of} \mathrm{wood} \mathrm{weighs} 8 \mathrm{g}.$

∴ Weight of the wood $=\frac{54000}{100}\times 8 \mathrm{g}$

$=4320 \mathrm{g}=4.32 \mathrm{kg}$

Q25 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 25:

Find the volume, lateral surface area and the total surface area of a cube each of whose edges measures:
(i) 7 m
(ii) 5.6 cm
(iii) 8 dm 5 cm

Answer 25:

(i) Length of the edge of the cube = a = 7 m
Now, we have the following:
​Volume$=a^3=7^3=343 m^3$
Lateral surface area $=4a^2=4\times 7\times 7=196 m^2$
Total Surface area$=6a^2=6\times 7\times 7=294 m^2$

(ii) Length of the edge of the cube = a = 5.6 cm
​Now, we have the following:
Volume$=a^3=5.6^3=175.616 c{m}^3$
Lateral surface area $=4a^2=4\times 5.6\times 5.6=125.44 c{m}^2$
Total Surface area$=6a^2=6\times 5.6\times 5.6=188.16 c{m}^2$

(iii) Length of the edge of the cube = a = 8 dm 5 cm = 85 cm
​Now, we have the following:
Volume$=a^3={85}^3=614125 c{m}^3$
Lateral surface area $=4a^2=4\times 85\times 85=28900 c{m}^2$
Total Surface area$=6a^2=6\times 85\times 85=43350 c{m}^2$

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Question 26:

The surface area of a cube is 1176 cm². Find its volume.

Answer 26:

Let a be the length of the edge of the cube.
Total surface area$=6a^2=1176 c{m}^2$
$\Rightarrow a=\sqrt{\frac{1176}{6}}=\sqrt{196}=14 cm$
 ∴ Volume$=a^3={14}^3=2744 c{m}^3$

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Question 27:

The volume of a cube is 729 cm³. Find its surface area.

Answer 27:

Let a be the length of the edge of the cube.

Then volume $=a^3=729 c{m}^3$

Also, $a=\sqrt[3]{729}=9 cm$

∴ Surface area$=6a^2=6\times 9\times 9=486 c{m}^2$

Q28 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 28:

The dimensions of a metal block are 2.25 m by 1.5 m by 27 cm. It is melted and recast into cubes, each of side 45 cm. How many cubes are formed?

Answer 28:

1 m = 100 cm
Volume of the original block $=225\times 150\times 27=911250 c{m}^3$

Length of the edge of one cube = 45 cm
Then volume of one cube$={45}^3=91125 c{m}^3$

∴ Total number of blocks that can be cast $=\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{block}}{\mathrm{volume} \mathrm{of} \mathrm{one} \mathrm{cube}}$

$=\frac{911250}{91125}=10$

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Question 29:

If the length of each edge of a cube is doubled, how many times does its volume become? How many times does its surface area become?

Answer 29:

Let a be the length of the edge of a cube.
Volume of the cube$=a^3$
Total surface area$=6a^2$

If the length is doubled, then the new length becomes 2a.

Now, new volume $=(2a{)}^3=8a^3$

Also, new surface area=$=6(2a{)}^2=6\times 4a^2=24a^2$ 

∴ The volume is increased by a factor of 8, while the surface area increases by a factor of 4.

Q30 | Ex-20A | Class 8 | RS AGGARWAL | chapter 20 | Volume and Surface Area of Solid | myhelper

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Question 30:

A solid cubical block of fine wood costs Rs 256 at Rs 500 per m³. Find its volume and the length of each side.

Answer 30:

Cost of wood = Rs $500/{\mathrm{m}}^3$

Cost of the given block = Rs 256

∴ Volume of the given block $={\mathrm{a}}^3=\frac{256}{500}$

$=0.512 {\mathrm{m}}^3 = 512000 {\mathrm{cm}}^3$

Also, length of its edge = a $=\sqrt[3]{0.512}=0.8 m$ = 80 cm