 # ICSE Class 8 Maths Selina Solutions for Chapter 21 Surface Area,Volume and Capacity

### CHAPTER 21-SURFACE AREA, VOLUME AND CAPACITY

Question 1.

Find the volume and the total surface area of a cuboid, whose:

(i) Length = 15cm, breadth = 10cm and height = 8cm.

Solution:-

We know that

Volume of a cuboid $=\text {Length} \times \text { Breadth } \times \text { Height }=15 \times 10 \times 8=1200 \mathrm{cm}^{3}$

Here

Total surface area of a cuboid $=2(l \times b+b \times h+h \times l)=2(15 \times 10+10 \times 8+8 \times 15)$

By further calculation

$=2(150+80+120) 2 \times 350=700 \mathrm{cm}^{2}$

(ii) l = 3.5m, b = 2.6m and h = 90cm,

Solution:-

Length = 3.5m breadth = 2.6m, height = 90cm $=\frac{90}{100} m=0.9 m$.

We know that

Volume of a cuboid $=l \times b \times h=3.5 \times 2.6 \times 0.9=8.19 m^{3}$

Here

Total surface area of a cuboid $=2(l \times b+b \times h+h \times l)=2(3.5 \times 2.6+2.6 \times 0.9 \times 3.5)$

By further calculation

=2(910+2.34+3.15)=2(14.59) $=29.18 m^{2}$

Question 2.

(i)The volume of a cuboid is 3456 $\mathrm{cm}^{3}$. If its length =24 cm and breadth =18 cm; find its height.

Solution:

Volume of the given cuboid $=3456 \mathrm{cm}^{3}$.

Length of the given cuboid =24 cm

Breadth of the given cuboid =18 cm

Here

Length × Breadth × Height = Volume of a cuboid

Substituting the values

24×18× Height =3456

By further calculation

Height $=\frac{3456}{24 \times 18}$

So we get

Height $=\frac{3456}{432}$

Height =8cm

(ii) The volume of a cuboid is 7.68 $m^{3}$. If its length = 3.2m and height =1.0m; find its breadth.

Solution:-

Volume of a cuboid =7.68 $\mathrm{m}^{3}$

Length of a cuboid =3.2 m

Height of a cuboid =1.0 m

Here

Length x Breadth x Height = Volume of a cuboid

Substituting the values

By further calculation

$\Rightarrow \text { Breadth }=\frac{7.68}{3.2 \times 1.0}$

So we get

$\Rightarrow \text { Breadth }=\frac{7.68}{3.2}$

(iii) The breadth and height of a rectangular solid are 1.20 m and 80 cm respectively. If the volume of the cuboid is 1.92 $m^{3}$; find its length.

Solution:-

Volume of a rectangular solid =1.92 $\mathrm{m}^{3}$

Breadth of a rectangular solid = 1.20 m

Height of a rectangular solid =80 cm=0.8 m

Here

Length × Breadth × Height = Volume of a rectangular solid (cubical)

Substituting the values

Length × 1.20 × 0.8 = 1.92

By further calculation

Length × 0.96 = 1.92

$=\frac{1.92}{0.96}$

So we get

$=\frac{192}{96}$

Length =2 m

Question 3.

The length, breadth and height of a cuboid are in the ratio 5:3:2. If its volume is $240 \mathrm{cm}^{3}$, find its dimensions. (Dimensions means: its length, breadth and height). Also find the total surface area of the cuboid.

Solution:-

Consider length of the given cuboid =5x

Breadth of the given cuboid =3x

Height of the given cuboid =2x

We know that

Volume of the given cuboid $=\text { Length } \times \text { Breadth } \times {height}$

Substituting the values

$=5 x \times 3 x \times 2 x=30 x^{3}$

It is given that

Volume $=240 \mathrm{cm}^{3}$

Substituting the values

$30 x^{3}=240 \mathrm{cm}^{3}$

By further calculation

$x^{3}=\frac{240}{30}$ $x^{3}=8$

So we get

$x=8^{\frac{1}{3}}$ $x=(2 \times 2 \times 2)^{\frac{1}{3}}$

x=2 cm

Here

Length of the given cube $=5 x=5 \times 2=10 \mathrm{cm}$

Breadth of the given cube $=3 x=3 \times 2=6 \mathrm{cm}$

Height of the given cube $=2 x=2 \times 2=4 \mathrm{cm}$

We know that

Total surface area of the given cuboid $=2(1 \times b+b \times h+h \times 1)$

Substituting the values

$=2(10 \times 6+6 \times 4+4 \times 10)=2(60+24+40)=2 \times 124=248 \mathrm{cm}^{2}$

Question 4.

The length, breadth and height of a cuboid are in the ratio 6:5:3. If its total surface area is 504 c $m^{2}$; find its dimensions. Also, find the volume of the cuboid.

Solution:-

Consider length of the cuboid =6x

Height of the cuboid =3x

We know that

Total surface area of the given cuboid $=2(1 \times b+b \times h+h \times l)$

Substituting the values

$=2(6 x \times 5 x+5 x \times 3 x+3 x \times 6 x)=2(30 \times 2+15 \times 2+18 \times 2)$

We get

$=2 \times 63 \times 2=126 x^{2}$

It is given that

Total surface area of the given cuboid $=504 \mathrm{cm}^{2}$

Substituting the values

$126 x^{2}=504 \mathrm{cm}^{2}$

By further calculation

$\Rightarrow x^{2}=\frac{504}{126}$

So we get

$\Rightarrow x^{2}=4$ $\Rightarrow x=\sqrt{4}$

x=2 cm

Here

Length of the cuboid $=6 x=6 \times 2=12 \mathrm{cm}$

Breadth of the cuboid $=5 x=5 \times 2=10 \mathrm{cm}$

Height of the cuboid $=3 x=3 \times 2=6 \mathrm{cm}$

We get

Volume of the cuboid $=l \times b \times h=12 \times 10 \times 6=720 \mathrm{cm}^{3}$

Question 5.

Find the volume and total surface area of a cube whose edge is:

(i) 8 cm

Solution:-

Edge of the given cube =8cm

We know that

Volume of the given cube $=(\text { Edge })^{3}=(8)^{3}=8 \times 8 \times 8=512 \mathrm{cm}^{3}$

Total surface area of a cube $=6(\text { Edge })^{2}=6 \times(8)^{2}=384 \mathrm{cm}^{2}$

(ii) 2m 40 cm.

Solution:-

(ii)Edge of the given cube =2 m 40 cm=2.40 m

We know that

Volume of a cube $=(\text { Edge })^{3}$

Substituting the values

Volume of the given cube $=(2.40)^{3}=2.40 \times 2.40 \times 2.40=13.824 \mathrm{m}^{2}$

Total surface area of the given cube $=6 \times 2.4 \times 2.4=34.56 \mathrm{m}^{2}$

Question 6.

Find the length of each edge of a cube, if its volume is:

(i) $216 \mathrm{cm}^{3}$

Solution:-

$(\text { Edge })^{3}$=Volume of a cube

Substituting the values

$(\text { Edge })^{3}=216 \mathrm{cm}^{3}$

It can be written as

Edge $=(216)^{1 / 3}$

Edge $=(3 \times 3 \times 3 \times 2 \times 2 \times 2)^{1 / 3}$

We get

Edge $=3 \times 2$

Ans. Edge =6 cm.

(ii) $1.728 \mathrm{m}^{3}$

Solution:-

$(Edge)^3$ = Volume of a cube

Substituting the values

$(\mathrm{Edge})^{3}=1.728 \mathrm{m}^{3}$ $\Rightarrow(\text { Edge })^{3}=\frac{1.728}{1000}=\frac{1728}{1000}$

It can be written as

$Edge =\left(\frac{1728}{1000}\right)^{1 / 3}$

By further calculation

$\mathrm{Edge}=\left(\frac{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3}{10 \times 10 \times 10}\right)^{1 / 3}$ $Edge =\frac{2 \times 2 \times 3}{10}$

So we get

$Edge =\frac{12}{10} \mathrm{m}$

Edge =1.2 m.

Question 7.

The total surface area of a cube is 216 cm2. Find its volume.

Solution:-

$6(\text {Edge})^{2}$= Total surface area of a cube

Substituting the values

$6(\text {Edge})^{2}=216 \mathrm{cm}^{2}$ $(\text {Edge})^{2}=\frac{216}{6}$

By further calculation

$(\text {Edge})^{2}=36$

Edge $=\sqrt{36}$

Edge =6 cm

We know that

Volume of the given cube $=(E d g e)^{3}=(6)^{3}=6 \times 6 \times 6=216 \mathrm{cm}^{3}$

Question 8.

A solid cuboid of metal has dimensions 24 cm, 18 cm and 4 cm. Find its volume.

Solution:-

It is given that

Length of the cuboid =24 cm

Breadth of the cuboid =18 cm

Height of the cuboid =4 cm

We know that

$Volume of the cuboid =l \times \mathrm{b} \times \mathrm{h}=24 \times 18 \times 4=1728 \mathrm{cm}^{3}$

Question 9.

A wall 9 m long, 6 m high and 20 cm thick, is to be constructed using bricks of dimensions 30 cm, 15 cm and 10 cm. How many bricks will be required?

Solution:

It is given that

Length of the wall $=9 \mathrm{m}=9 \times 100 \mathrm{cm}=900 \mathrm{cm}$

Height of the wall $=6 \mathrm{m}=6 \times 100 \mathrm{cm}=600 \mathrm{cm}$

Breadth of the wall =20 cm

We know that

Volume of the wall $=900 \times 600 \times 20 \mathrm{cm}^{3}=10800000 \mathrm{cm}^{3}$

Volume of one Brick $=30 \times 15 \times 10 \mathrm{cm}^{3}=4500 \mathrm{cm}^{3}$

So we get

Number of bricks required to construct the wall $=\frac{\text { Volume of wall }}{\text { Volume of one brick}}$ $=\frac{10800000}{4500}$

=2400

Question 10.

A solid cube of edge 14 cm is melted down and recasted into smaller and equal cubes each of edge 2 cm; find the number of smaller cubes obtained.

Solution:-

We know that

Edge of the big solid cube = 14 cm

Volume of the big solid cube $=14 \times 14 \times 14 \mathrm{cm}^{3}=2744 \mathrm{cm}^{3}$

Similarly

Edge of the small cube =2 cm

Volume of one small cube $=2 \times 2 \times 2 \mathrm{cm}^{3}=8 \mathrm{cm}^{3}$

So we get

Number of smaller cubes obtained $=\frac{\text { Volume of big cube }}{\text { Volume of one small cube }}=\frac{2774}{8}=343$

Question 11.

A closed box is cuboid in shape with length =40cm, breadth =30cm and height =50cm. It is made of thin metal sheet. Find the cost of metal sheet required to make 20 such boxes, if 1 $m^{2}$ of metal sheet costs Rs. 45.

Solution:-

It is given that

Length of closed box (1) =40cm

And height (h) =50cm We know that

Total surface area $=2(l \times b+b \times h+h \times l)$

Substituting the values

$=2(40 \times 30+30 \times 50+50 \times 40) \mathrm{cm}^{2}$

By further calculation

$=2(1200+1500+2000) \mathrm{cm}^{2}$

So we get

$=2 \times 4700=9400 \mathrm{cm}^{2}$

Here

Surface area of sheet used for 20 such boxes $=9400 \times 20=188000 \mathrm{cm}^{2}=18.8 m^{2}$

Cost of $1 \mathrm{m}^{2} sheet = Rs.45$

We get

Total cost $=18.8 \times 45=Rs.846$

Question 12.

Four cubes, each of edge 9 cm, are joined as shown below: Write the dimensions of the resulting cuboid obtained. Also, find the total surface area and the volume of the resulting cuboid.

Solution:-

Edge of each cube =9cm

(i) We know that

Length of the cuboid formed by 4 cubes (1) $=9 \times 4=36 \mathrm{cm}$

Breadth (b) =9cm and height (h) = 9cm

(ii) Total surface area of the cuboid = 2(lb + bh + hl)

Substituting the values

$=2(36 \times 9+9 \times 9+9 \times 36) \mathrm{cm}^{2}$

By further calculation

$=2(324+81+324) \mathrm{cm}^{2}$

So we get

$=2 \times 729 \mathrm{cm}^{2}$ $=1458 \mathrm{cm}^{2}$

(iii) $Volume =\quad l \times b \times h=36 \times 9 \times 9 cm^{2}=2916 cm^{3}$

Question 13.

How many persons can be accommodated in a big-hall of dimensions 40 m, 25m and 15m; assuming that each person requires $5 m^{3}$ of air?

Solution:-

No. of persons $=\frac{\text { Vol. of the hall }}{\text { Vol. of air required for each person }}$

It is given that

Length of the hall =40m

Height =15m

Here

Volume of the hall$=1 \times \mathrm{b} \times \mathrm{h}=40 \times 25 \times 15=15000 \mathrm{m}^{3}$

Volume of the air required for each person $=5 \mathrm{m}^{3}$

So we get

No. of persons who can be accommodated $=\frac{\text { Volume of the hall }}{\text { Volume of air required for each person }}=\frac{15000 \mathrm{m}^{3}}{5 \mathrm{m}^{3}}=3000$

Question 14.

The dimension of a class-room are; length = 15m, breadth =12m and height =7.5m. Find, how many children can be accommodated in this class-room; assuming 3.6 $m^{3}$ of air is needed for each child.

Solution:-

It is given that

Length of the room =15m

Height of the room =7.5m

We know that

Volume of the room$=L \times B \times H=15 \times 12 \times 7.5 \mathrm{m}^{3}=1350 \mathrm{m}^{3}$

Volume of air required for each child $=3.6 \mathrm{m}^{3}$

So we get

No. of children who can be accommodated in the class room. $=\frac{\text { Volume of class room }}{\text { Volume of air needed for each child }}=\frac{1350 \mathrm{m}^{3}}{3 \cdot 6 \mathrm{m}^{3}}$

=375.

Question 15.

The length, breadth and height of a room are 6m, 5.4m and 4 m respectively. Find the area of:

(i) Its four-walls

(ii) Its roof.

Solution:-

It is given that

Length of the room = 6m

Breadth of the room = 5.4m

Height of the room = 4m

(i) Area of four walls $=2(L+B) \times H=2(6+5.4) \times 4=2 \times 11.4 \times 4=91.2 \mathrm{m}^{2}$

(ii) Area of the roof $=L \times B=6 \times 5.4=32.4 \mathrm{m}^{2}$