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:-

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

Total surface area of a cuboid \(2(l \times b+b \times h+h \times l)=2(15 \times 10+10 \times 8+8 \times 15) \) \(=2(150+80+120) 2 \times 350=700 \mathrm{cm}^{2}\)

 (ii) l = 3.5m, b = 2.6m and h = 90cm,

Solution:-

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

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

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

=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

 We know

Length × Breadth × Height = Volume of a cuboid

24×18× Height =3456

Height \(=\frac{3456}{24 \times 18}\)

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

 We know 

Length x Breadth x Height = Volume of a cuboid 

3.2 × Breadth × 1.0=7.68

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

⇒ Breadth =2.4 m

(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

We know

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

Length × 1.20 × 0.8 = 1.92

Length × 0.96 = 1.92

\(=\frac{1.92}{0.96}\) \(=\frac{192}{96}\)

 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:-

Let length of the given cuboid =5x

Breadth of the given cuboid =3x

 Height of the given cuboid =2x

Volume of the given cuboid \(=\text { Length } \times \text { Breadth } \times {height}\) \(=5 x \times 3 x \times 2 x=30 x^{3}\)

But we are given volume \(=240 \mathrm{cm}^{3}\) \(30 x^{3}=240 \mathrm{cm}^{3} \) \(x^{3}=\frac{240}{30}\) \(x^{3}=8\) \(x=8^{\frac{1}{3}}\) \(x=(2 \times 2 \times 2)^{\frac{1}{3}}\)

⇒x=2 cm

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 \(=2 x=2 \times 2=4 \mathrm{cm} \)

 Total surface area of the given cuboid \(=2(1 \times b+b \times h+h \times 1) \) \(=2(10 \times 6+6 \times 4+4 \times 10)=2(60+24+40)=2 \times 124=248 \mathrm{cm}^{2}\)

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:-

Let length of the cuboid =6x

Breadth of the cuboid =5x

Height of the cuboid =3x

Total surface area of the given cuboid \(=2(1 \times b+b \times h+h \times l) \) \(=2(6 x \times 5 x+5 x \times 3 x+3 x \times 6 x)=2(30 \times 2+15 \times 2+18 \times 2) \) \(=2 \times 63 \times 2=126 x^{2}\)

But we are given total surface area of the given cuboid \(=504 \mathrm{cm}^{2} \) \(126 x^{2}=504 \mathrm{cm}^{2}\) \(\Rightarrow x^{2}=\frac{504}{126}\) \(\Rightarrow x^{2}=4\) \(\Rightarrow x=\sqrt{4}\)

⇒x=2 cm

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 \(=3 x=3 \times 2=6 \mathrm{cm}\)

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

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

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

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

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

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

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

Edge \(=3 \times 2\)

Ans. Edge =6 cm. 

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

Solution:-

\( (Edge)^3\) = Volume of a cube 

\(∴(\mathrm{Edge})^{3}=1.728 \mathrm{m}^{3}\) \(\Rightarrow(\text { Edge })^{3}=\frac{1.728}{1000}=\frac{1728}{1000}\) \(Edge =\left(\frac{1728}{1000}\right)^{1 / 3}\) \(\mathrm{Edge}=\left(\frac{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3}{10 \times 10 \times 10}\right)^{1 / 3}\) \(Edge =\frac{2 \times 2 \times 3}{10}\) \(Edge =\frac{12}{10} \mathrm{m}\)

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

\(6(\text {Edge})^{2}=216 \mathrm{cm}^{2}\) \( (\text {Edge})^{2}=\frac{216}{6} \) \( (\text {Edge})^{2}=36\)

Edge \(=\sqrt{36}\)

Edge \(=\sqrt{36}\)

 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:-

Length of the cuboid =24 cm

Breadth of the cuboid =18 cm

Height of the cuboid =4 cm

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

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:

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

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

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

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:-

Edge of the big solid cube = 14 cm

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

Edge of the small cube =2 cm

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

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:-

Length of closed box (1) =40cm

Breadth (b) =30cm

And height (h) =50cm

Class 8 Maths Selina Solutions for Chapter 21 Surface Area-1

Total surface area \(=2(l \times b+b \times h+h \times l) \) \(=2(40 \times 30+30 \times 50+50 \times 40) \mathrm{cm}^{2}\) \(=2(1200+1500+2000) \mathrm{cm}^{2}\) \(=2 \times 4700=9400 \mathrm{cm}^{2}\)

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

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

Total cost \(=18.8 \times 45=Rs.846\)

Question 12.

Four cubes, each of edge 9 cm, are joined as shown below:

Class 8 Maths Selina Solutions for Chapter 21 Surface Area-2

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) 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)

\(=2(36 \times 9+9 \times 9+9 \times 36) \mathrm{cm}^{2}\) \(=2(324+81+324) \mathrm{cm}^{2}\) \(=2 \times 729 \mathrm{cm}^{2}\) \(=1458 \mathrm{cm}^{2}\)

(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 \(=\frac{\text { Vol. of the hall }}{\text { Vol. of air required for each person }}\)

Length of the hall =40m

Breadth =25m

Height =15m

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 \(=5 \mathrm{m}^{3}\)

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:-

Length of the room =15m

Breadth of the room =12m

Height of the room =7.5m

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 \(=3.6 \mathrm{m}^{3}\)

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:-

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 \(=L \times B=6 \times 5.4=32.4 \mathrm{m}^{2}\)

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