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## Exercise 20A

**1. Find the volume, lateral surface area and the total surface area of the cuboid whose dimensions are:**

**(i) Length=22cm, breadth=12cm and height=7.5cm**

**(ii) Length=15m, breadth=6m and height=9dm**

**(iii) Length=24m, breadth=25cm and height=6m**

**(iv) Length=48cm, breadth=6dm and height=1m**

**Solutions:**

(i) Given Length=22cm, breadth=12cm and height=7.5cm

We know that volume of cuboid= length Ã— breadth Ã— height

V = (22 Ã— 12 Ã— 7.5)

V = 1980cm^{3}

We know that total surface area of cuboid= 2(l b + b h + h l)

Surface area = 2 (22 Ã— 12 + 12 Ã— 7.5 + 7.5 Ã— 22)

Surface area = 2 (264 + 165 + 90)

Total Surface area = 1038cm^{2}

We know that lateral area of cuboid= 2 [(l + b) Ã— h]

Lateral surface area = 2 [(22 + 12) Ã— 7.5]

Lateral surface area = 510 cm^{2}

(ii) Given Length=15m, breadth=6m and height=9m

We know that volume of cuboid= length Ã— breadth Ã— height

V = (15 Ã— 6 Ã— 0.9)

V = 81m^{3}

We know that total surface area of cuboid= 2(l b + b h + h l)

Surface area = 2 (15 Ã— 6 + 6 Ã— .9 + .9 Ã— 15)

Surface area = 2 (90 + 13.5 + 5.4)

Total Surface area = 217.8m^{2}

We know that lateral surface area of cuboid= 2 [(l + b) Ã— h]

Lateral surface area = 2 [(15 + 6) Ã— 0.9]

Lateral surface area = 37.8 m^{2}

(iii) Given Length=24m, breadth=25cm and height=6m

We know that volume of cuboid= length Ã— breadth Ã— height

V = (24 Ã— 0.25 Ã— 6)

V = 36m^{3}

We know that total surface area of cuboid= 2(l b + b h + h l)

Surface area = 2 (24 Ã— .25 + .25 Ã— 6 + 6 Ã— 24)

Surface area = 2 (6 + 155 + 1.5)

Total Surface area = 303m^{2}

We know that lateral area of cuboid= 2 [(l + b) Ã— h]

Lateral surface area = 2 [(24 + .25) Ã— 6]

Lateral surface area = 291 m^{2}

(iv) Given Length=48cm, breadth=6dm and height=1m

We know that volume of cuboid= length Ã— breadth Ã— height

V = (0.48 Ã— 0.6 Ã— 1)

V = 0.288m^{3}

We know that total surface area of cuboid= 2(l b + b h + h l)

Surface area = 2 (0.48 Ã— 0.6 + 0.6 Ã— 1 + 1 Ã— 0.48)

Surface area = 2 (.288 + 0.6 + 0.48)

Total Surface area = 2.736m^{2}

We know that lateral surface area of cuboid= 2 [(l + b) Ã— h]

Lateral surface area = 2 [(0.48 + 0.6) Ã— 1]

Lateral surface area = 2.16 m^{2}

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

** Solution: **

We know that 1m=100cm

Dimension of the tank is 2m 75cm and 1m 80cm and 1m 40cm

Which can be written as 275cm Ã— 180 cm Ã— 140 cm

Also we know that volume of cuboid = length Ã— breadth Ã— height

V = 275 Ã— 180 Ã— 140

V = 6930000cm^{3}

And we know that 1000cm^{3}= 1L

Therefore V=6930 liters of water it holds when filled to the brim.

**3. A solid rectangular piece of iron measures 1.05m Ã— 70cm Ã— 1.5cm. Find the weight of this piece in kilograms if 1cm ^{3}of iron weighs 8 grams.**

**Solution: **

We know that 1m=100cm

Dimension of the tank is 1.05m and 70cm and 1.5cm.

Also we know that volume of cuboid = length Ã— breadth Ã— height

V = 105 Ã— 70 Ã— 1.5

V = 11025cm^{3}

And we know that 1cm^{3}= 8 grams

Weight of iron piece = 11025 Ã— 8 = 88200g

But 1kg=1000g

Weight of iron piece = 88.2kg

**4. The area of courtyard is 3750 m ^{2}. Find the cost of covering it with gravel to a height of 1cm if the gravel costs â‚¹ 6.40 per cubic meter.**

** Solution:**

We know that 1cm= 0.01m

Given that area of courtyard is 3750 m^{2}

Volume of the gravel = area Ã— height

V = 3750 Ã— 0.01 = 37.5 m^{3}

Also given that the cost of covering it with gravel to a height of 1cm if the gravel costs

â‚¹ 6.40 per cubic meter.

There the cost is 37.5 Ã— 6.40 = â‚¹ 240

**5. How many persons can be accommodated in a hall of length 16m, breadth 12.5m and height 4.5m, assuming that 3.6 m ^{3 }of air is required for each person?**

**Solution:**

We know that volume of cuboid = length Ã— breadth Ã— height

V = 16 Ã— 12.5 Ã— 4.5 = 900 m^{3}

Also given that 3.6 m^{3 }of air is required for each person

Therefore, total number of persons can be accommodated in a hall is

Total volume/ volume required by each person

= 900/3.6

= 250 people.

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

**Solution: **

We know that volume of cuboid = length Ã— breadth Ã— height

V = 120 Ã— 72 Ã— 54 = 466560 cm^{3}

Volume of each bar soap = 6 Ã— 4.5 Ã— 4 = 108 cm^{3}

Therefore, total number of soaps can be accommodated in a box is

Volume of the box/ volume of each soap

= 466560/108

= 4320 bars

**7. The size of matchbox is 4cm Ã— 2.5 cm Ã— 1.5 cm. What is the volume packet containing 144 matchboxes? How many such packets can be place in a carton of size 1.5cm Ã— 84cm Ã— 60 cm?**

**Solution:**

We know that volume of cuboid = length Ã— breadth Ã— height

Volume occupied by single matchbox = 4 Ã— 2.5 Ã— 1.5 = 15 cm^{3}

Volume of a packet containing 144 matchbox = 15 Ã— 144 = 2160 cm^{3}

Volume of carton is 1.5cm Ã— 84cm Ã— 60 cm = 756000 cm^{3}

Therefore, total number of packets can be accommodated in a carton is

Volume of the carton/ volume of the box

= 75600/2160

= 350 packets

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

**Solution:**

We know that volume of cuboid = length Ã— breadth Ã— height

Volume of the block = 500 Ã— 70 Ã— 32 = 112000 cm^{3}

Volume of each plank = 200 Ã— 25 Ã— 8= 40000 cm^{3}

Therefore, total number of planks that can be made

= Volume of the block/ volume of the each plank

= 112000/40000

= 38 planks

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

**Solution: **

We know that volume of cuboid = length Ã— breadth Ã— height

Volume of the brick = 25 Ã— 13.5 Ã— 6 = 2025 cm^{3}

Volume of the wall = 800 Ã— 540 Ã— 33= 14256000 cm^{3}

Therefore, total number of bricks required

=Volume of the wall/ volume of the each brick

= 1425000/2025

= 7040 bricks

**10. A wall 15m long, 30cm wide and 4m high is made of bricks, each measuring 22 cm Ã— 12.5cm Ã— 7.5 cm. If 1/12 of the total volume of the wall consists of mortar, how many bricks are there in the wall?**

**Solution:**

We know that volume of cuboid = length Ã— breadth Ã— height

Volume of the wall = 150 Ã— 30 Ã— 400 = 18000000 cm^{3}

The quantity of mortar= 1/12 Ã— 18000000 = 1500000

Volume of bricks= 18000000 â€“ 1500000 = 16500000 cm^{3}

Volume of single brick = 22 Ã— 12.5 Ã— 7.5 = 2062.5 cm^{3}

Therefore, total number of bricks required

=total volume of the bricks/ volume of the each brick

= 16500000/2062.5

= 8000 bricks

**11. Find the capacity of rectangular cistern in liters whose dimensions are 11.2 m Ã— 6m Ã— 5.8m. Find the area of the iron sheet required to make the cistern.**

**Solution:**

We know that volume of cuboid = length Ã— breadth Ã— height

Volume of the cistern = 11.2 Ã— 6 Ã— 5.8

= 389.76 m^{3}

= 389.76 Ã— 1000

= 389760 liters.

Area of the sheet that required to make the cistern = total surface area of the cistern we know that total surface area of cuboid= 2(l b + b h + h l)

= 2 (11.22 Ã— 6 + 6 Ã— 5.8 + 5.8 Ã— 11.2)

= 2 (67.2 + 64.96 + 34.8)

= 333.92 cm^{2}

**12. The volume of a block of gold is 0.5 m ^{3. }If it is hammered into a sheet to cover an area of 1 hectare, find the thickness of the sheet.**

**Solution:**

Given that volume of the block of gold is 0.5 m^{3}

We know that 1 hectare = 10000 m^{2}

Thickness of the sheet = volume/ area

= 0.5/1000

=0.00005m

=0.005cm

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

**Solution**:

Given that rainfall recorded in a certain day= 5cm = 0.05m

Area of the field = 2 hectare

We know that 1 hectare = 10000 m^{2}

Area = 2 Ã— 10000

Area = 20000 m^{2}

Total rain in the field= area of the field Ã— height of the field

=0.05 Ã— 20000 = 1000 m^{3}

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

**Solution: **

Given that rate of flow 3km/h

Area of cross section of river = 45 Ã— 2 = 90 m^{2}

Rate of flow 3km/h= 3 Ã— 1000/60= 50m/min

Volume of water is flowing in cross section in 1 minute is= 90 Ã— 50

= 4500 m^{3} per minute

**15. A pit 5m long and 3.5m wide is dug to a certain depth. If the volume of earth taken out of it is 14 m ^{3}, what is the depth of the pit?**

**Solution: **

We know that volume of cuboid = length Ã— breadth Ã— height

Let the depth of the pit is x m.

Therefore, V= 5 Ã— 3.5 Ã— x

But volume is 14 m^{3}

x = V/ 5 Ã— 3.5

x = 14/ 5 Ã— 3.5

x= 0.8 m=80cm

## Exercise 20B

**1. Find the volume, curved surface area and total surface area of each of the cylinders whose dimensions are:**

**(i) Radius of the base = 7cm and height = 50cm**

**(ii) Radius of the base = 5.6m and height = 1.25 m**

**(iii) Radius of the base = 14dm and height = 15m**

**Solution:**

(i) We know that volume of the cylinder = Ï€r^{2}h

Here r = 7cm h = 50cm

V = 22/7 Ã— 7 Ã— 7 Ã— 50

V = 22 Ã— 7 Ã— 50

V = 7700 cm^{3}

Also we know that curved surface area of cylinder= 2 Ï€ r h

Curved surface area = 2 Ã— 22/7 Ã— 7 Ã— 50

Curved surface area = 2200 cm^{2}

We know that total surface area of cylinder = 2 Ï€ r(r + h)

Total surface area = 2 Ã— 22/7 Ã— 7 (7 + 50)

Total surface area = 2580 cm^{2}

(ii) We know that volume of the cylinder = Ï€r^{2}h

Here r = 5.6m h = 1.25m

V = 22/7 Ã— 5.6 Ã— 5.6 Ã— 1.25

V = 123.2 m^{3}

Also we know that curved surface area of cylinder= 2 Ï€ r h

Curved surface area = 2 Ã— 22/7 Ã— 5.6 Ã— 1.25

Curved surface area = 44 m^{2}

We know that total surface area of cylinder = 2 Ï€ r(r + h)

Total surface area = 2 Ã— 22/7 Ã— 5.6 (5.6 + 1.25)

Total surface area = 241.12 m^{2}

(iii) We know that volume of the cylinder = Ï€r^{2}h

Here r = 14dm h = 15cm

V = 22/7 Ã— 1.4 Ã— 1.4 Ã— 15

V = 92.4 cm^{3}

Also we know that curved surface area of cylinder= 2 Ï€ r h

Curved surface area = 2 Ã— 22/7 Ã— 1.4 Ã— 15

Curved surface area = 132 cm^{2}

We know that total surface area of cylinder = 2 Ï€ r(r + h)

Total surface area = 2 Ã— 22/7 Ã— 1.4 (1.4 + 15)

Total surface area = 144.32 cm^{2}

**2. A milk tank is in the form of cylinder whose radius is 1.5m hand height is 10.5 m. find the quantity of milk in liters that can be stored in the tank.**

**Solution:**

Given r= 1.5m and h=10.5 m

To find the quantity of milk stored is equal to volume of the cylinder

But we know that volume of the cylinder = Ï€r^{2}h

V = 22/7 Ã— 1.5 Ã— 1.5 Ã— 10.5

V = 74.25 m^{3}

But 1 m^{3}=1000L

Therefore quantity of milk is 74250 L

**3. A wooden cylindrical pole is 7m high and its base radius is 10cm. Find its weight if the wood weighs 225kg per cubic meter.**

**Solution: **

Given r= 7m and h=10 cm

But we know that volume of the cylinder = Ï€r^{2}h

V = 22/7 Ã— 0.1 Ã— 0.1 Ã— 7

V = 0.22 cm^{3}

Given weight of the wood 225kg per cubic meter

Weight of the pole = 0.22 Ã— 225 = 49.5 kg

**4. Find the height of the cylinder whose volume is 1.54m ^{3} and diameter of the base is 140cm?**

**Solution:**

Given that volume of the cylinder = 1.54m^{3}

Diameter = 140cm

Therefore, radius = diameter/2 = 140/2 = 70cm = 0.7m

But we know that Volume of cylinder = Ï€r^{2}h

1.54 = 22/7 Ã— 0.7 Ã— 0.7 Ã— h

h = 1m

**5. The volume of a circular iron rod of length 1m is 3850 cm ^{3}. Find its diameter.**

**Solution:**

Given volume of circular iron rod is 3850 cm^{3}

But we know that Volume of cylinder = Ï€r^{2}h

3850= 22/7 Ã— r Ã— r Ã— 100

r^{2}= 3850 Ã— 7/100 Ã— 22

r = 1.75 Ã— 7

r= 3.5cm

Therefore diameter = 2 Ã— r

Diameter= 2 Ã— 3.5 = 7cm

**6. A closed cylindrical tank of diameter 12m and height 5m is made from a sheet of metal. How much sheet of metal will be required?**

**Solution:**

Given that diameter = 12m

Therefore r= diameter /2 =12/2 = 6 m

Also given that height is 5m

Now we have to find the total surface area

We know that total surface area of cylinder = 2 Ï€ r(r + h)

Total surface area = 2 Ã— 22/7 Ã— 6 (6 + 5)

Total surface area = 528 m^{2}

**7. The circumference of the base of a cylinder is 88cm and its height is 60cm. Find the volume of the cylinder and its curved surface area.**

**Solution:**

Given that circumference of the base of a cylinder is 88cm

Height of cylinder = 60cm

Curved surface area = circumference Ã— height

= 88 Ã— 60

= 5280 cm^{2}

Circumference = 2 Ï€r

60 = 2 Ã— 22/7 Ã— r

r = 14cm

Therefore volume of the cylinder = Ï€r^{2}h

V= 22/7 Ã— 14 Ã— 14 Ã— 60

V = 36960 cm^{3}

**8. The lateral surface area of a cylinder of length 14m is 220 m ^{2}. Find the volume of the cylinder.**

**Solution:**

Given that length is 14m and lateral surface area is 220 m^{2}

Lateral surface area of cylinder= 2 Ï€ r h

220= 2 Ã— 22/7 Ã— r Ã— 14

r = 2.5m

Therefore volume of the cylinder = Ï€r^{2}h

V= 22/7 Ã— 14 Ã— 14 Ã— 60

V = 36960 cm^{3}

**9. The volume of a cylinder of height 8cm is 1232cm ^{3}. Find its curved surface area and total surface area.**

**Solution: **

Given that height of the cylinder is 8cm and volume is 1232cm^{3}.

Therefore volume of the cylinder = Ï€r^{2}h

1232 = 22/7 Ã— rÃ— r Ã— 8

r^{2}= 1232 Ã— 7/ 8

r = 7cm

Also we know that curved surface area of cylinder= 2 Ï€ r h

Curved surface area = 2 Ã— 22/7 Ã— 7 Ã— 8

Curved surface area = 252 cm^{2}

We know that total surface area of cylinder = 2 Ï€ r(r + h)

Total surface area = 2 Ã— 22/7 Ã— 7 (7 + 8)

Total surface area = 2580 cm^{2}

**10. The radius and height of a cylinder are in the ratio 7:2. If the volume of the cylinder is 8316cm ^{3}, find the surface area of cylinder.**

**Solution: **

Given that radius and height of a cylinder are in the ratio 7:2

Hence radius/height =7/2

r = (7/2) h

Therefore volume of the cylinder = Ï€r^{2}h

8316 = 22/7 Ã— (7/2) h Ã— (7/2) h Ã— h

h^{3}= 216

h = 6cm

Therefore r = 21cm

We know that total surface area of cylinder = 2 Ï€ r(r + h)

Total surface area = 2 Ã— 22/7 Ã— 21 (21 + 6)

Total surface area = 3564 cm^{2}

## Exercise 20C

**Select the correct answer in each of the following:**

**1. The maximum length of a pencil that can be kept in rectangular box of dimensions 12cm Ã— 9cm Ã— 8cm, is**

**a)13cm (b) 17cm (c) 18cm (d) 19cm**

**Solution:**

(b) 17cm

**Explanation:**

We know that length of diagonal of the cuboid= âˆšÂ (l^{2}+b^{2}+h^{2})

=âˆšÂ (12^{2}+9^{2}+8^{2})

=âˆšÂ (144+81+64)

=17

**2. The total surface area of a cube is 150 cm ^{2}.Its volume is**

** (a)216 cm ^{3} (b) 125 cm^{3} (c) 64 cm^{3} (d) 1000 cm^{3}**

** Solution: **

b)125 cm^{3}

**Explanation: **

Total surface area of the cube = 6 a^{2}

150 = 6 a^{2}

a^{2} = 150/6 = 25

a = 5

Volume of the cube = a^{3}

V= 5^{3}=125

**3. The volume of a cube is 343 cm ^{3}. Its total surface area s **

** (a)196 cm ^{2} (b) 49 cm^{2} (c) 294 cm^{2} (d) 147 cm^{2}**

** Solution: **

(c)294 cm^{2}

**Explanation: **

Volume of the cube = a^{3}

343= a^{3}

a = 7cm

Total surface area of the cube = 6 a^{2}

= 6(7)^{2}

= 294

**4. The cost of painting the whole surface area of a cube at the rate of 10 paise per cm ^{2} is â‚¹ 264.40. then, the volume of the cube is**

**(a)6859 cm ^{2} (b) 9261 cm^{2} (c) 8000 cm^{2} (d) 10648 cm^{2}**

**Solution:**

(b)9261 cm^{2}

** Explanation**:

Total surface area of the cube = 6 a^{2}

Cost of the painting the cube = 6 a^{2}Ã— 10

246.6 = 6 a^{2}Ã— 10

a^{2}= 4.41

a= 2.1

Volume of the cube = a^{3}

V= (2.1)^{3}

**5. How many bricks, each measuring 25cm Ã— 11.25cm Ã— 6cm, will be needed to build a wall 8m long, 6m high and 22.5 cm thick?**

**(a)5600 (b)6000 (c)6400 (d)7200**

**Solution:**

(c)6400

**Explanation:**

We know that volume of cuboid= length Ã— breadth Ã— height

Volume of brick = 25cm Ã— 11.25cm Ã— 6cm

= 1678.5 cm^{3}

Volume of the wall = 80 Ã— 60 Ã— 22.5

= 10800000 cm^{3}

Number of bricks = 10800000/ 1678.5 = 6400

**6. How many cubes 0f 10cm edge can be put in a cubical box of 1m edge?**

**(a)10 (b)100 (c)1000 (d)10000**

**Solution:**

(c)1000

**Explanation:**

Volume of the cube = a^{3}

Volume of the smaller cube = 10^{3}

=1000

Volume of the box = a^{3}

= (100)^{3}

=1000000

Total number of cubes = 1000000/1000

=1000

**7. The edges of a cuboid have their volumes in the ratio 1:2:3 and its surface area is 88cm ^{2}. The volume of the cuboid is**

**(a)48 cm ^{3} (b) 64 cm^{3} (c) 96 cm^{3} (d) 120 cm^{3}**

** Solution: **

(a)48 cm^{3}

**Explanation:**

Let a be the length of smallest edge.

The edges are in the proportion a: 2a: 3a

We know that total surface area of cuboid= 2(l b + b h + h l)

Surface area = 2 (2 a^{2} + 3 a^{2} + 6 a^{2})

88 = 22 a^{2}

a = 2

2a=4

3a = 6

We know that volume of cuboid= length Ã— breadth Ã— height

V = 2 Ã— 4 Ã— 6

V = 48

**8. Two cubes have their volumes in the ratio 1:27. The ratio of their surface areas is**

**(a)1:3 (b)1:9 (c)1:27 (d)none of these**

**Solution:**

(b)1:9

**Explanation:**

Volume in the ratio 1:27

1/27 = 1/3= a^{3}/b^{3}

b/a = 3

Therefore on squaring 1:9

**9. The surface area of a 10cm Ã— 4cm Ã— 6cm brick is **

**(a)84cm ^{2 }(b) 124cm^{2 }(c)164cm^{2} (d) 180cm^{2}**

** Solution:**

(c)164cm^{2}

** Explanation:**

We know that total surface area of cuboid= 2(l b + b h + h l)

= 2 (40 + 30 + 12)

= 164

**10. The surface area is 9m long, 40cm wide and 20cm high. If 1 cubic meter of iron weighs 50 kg, what is the weight of the beam?**

**(a)56kg (b)48kg (c)36kg (d)27kg**

**Solution:**

(c)36kg

**Explanation:**

We know that volume of cuboid= length Ã— breadth Ã— height

V = 9 Ã— 0.2 Ã— 0.4

V = 0.72m^{3}

Weight = 0.72 Ã— 50 = 36kg

**11. A rectangular water reservoir contains 42000 liters of water. If the length of reservoir is 6m and its breadth is 3.5m, the depth of the reservoir is**

**(a)2m (b)5m (c)6m (d)8m**

**Solution:**

(a)2m

**Explanation:**

We know that volume of cuboid= length Ã— breadth Ã— height

h = volume / l b

h = 42/6 Ã— 3.5

h = 2

**12. The dimensions of a room are 10cm Ã— 8cm Ã— 3.3cm. How many men can be accommodated in this room if each man requires 3m ^{3} of space?**

**(a)99 (b)88 (c)77 (d)75**

**Solution:**

(b)88

**Explanation:**

We know that volume of cuboid= length Ã— breadth Ã— height

V = 10 Ã— 8 Ã— 3.3

V = 264m^{3}

Total number of people can be accommodated = 264/3 = 88

**13. A rectangular water tank is 3m long, 2m wide and 5m high. How many liters of water can it hold?**

**(a)30000 (b)15000 (c)25000 (d)35000**

**Solution:**

(a)30000

**Explanation:**

We know that volume of cuboid= length Ã— breadth Ã— height

V = 3 Ã— 2 Ã— 5

V = 30m^{3}

V= 30000l

**14. The area of the cardboard needed to make a box of size 25cm Ã— 15cm Ã— 8cm will be **

**(a)390cm ^{2 }(b) 1390cm^{2 }(c)2780cm^{2} (d) 1000cm^{2}**

** Solution: **

(b) 1390cm^{2}

**Explanation: **

We know that total surface area of cuboid= 2(l b + b h + h l)

= 2 (375 + 120 + 200)

= 1390

**15. The diagonal of a cube measures 4 âˆšÂ 3 cm. its volume is**

**(a)8cm ^{3 }(b) 16cm^{3 }(c)27cm^{3} (d) 64cm^{3}**

** Solution: **

(d) 64cm^{3}

Explanation:

Diagonal of cube = a âˆšÂ 3 = 4 âˆšÂ 3

a = 4

Volume of cube = a^{3}= 4^{3}=64 cm^{3}

## RS Aggarwal Solutions for Class 8 Maths Chapter 20- Volume and Surface Area of Solids

Chapter 20, Volume and Surface Area of Solids , contains 3 Exercises. RS Aggarwal Solutions given here contains the answers to all the questions present in these exercises. Let us have a look at some of the concepts that are being discussed in this chapter.

Definition of solids

Definition of cube and its properties

Volume of a solid

Definition of cuboid and its properties

Standard unit of volume

Formula for volume of a cube

Formula for surface area of a cube

Formula for volume of a cuboid

Formula for surface area of a cuboid

### Also, Access RS Aggarwal Solutions for Class 8 Chapter 20

### Chapter Brief of RS Aggarwal Solutions for Class 8 Maths Chapter 20 – Volume and Surface Area of Solids

The RS Aggarwal Solutions for Class 8 Maths Chapter 20 â€“ Volume and Surface Area of Solids deals with volume, total surface area, lateral surface area and diagonal of a cube as well as volume, total surface area, lateral surface area and diagonal of a cuboid. It also deals with standard unit of volume.