NCERT Solutions for Class 9 Maths Chapter 13 Surface area and volumes

Ans : It is given that, length (l) of box =1.5m
Breadth (b) of box 1.25 m 
Depth (h) of box = 0.65 m 
(i) Box is to be open at top. 
Area of sheet required 
=2lh+2bh+lb
=[2 x 1.5 x 0.65 +2 x 1.25 x 0.65 + 1.5 x 1.25] \( m^{2}\)
 =(1.95 + 1.625 + 1.875) m2 =5.45  \( m^{2}\)
(ii) Cost of sheet per  \( m^{2}\) area =Rs 20
Cost of sheet of 5.45 \( m^{2}\) area =Rs (5.45x20)
=Rs 109
 

Ans : It is given that 
Length (l) of room = 5 m 
Breadth (b) of room = 4 m 
Height (h) of room = 3 m 
It can be observed that four walls and the ceiling of the room are to be white- washed. The floor of the room is not to be white-washed. 
Area to be white-washed = Area of walls + Area of ceiling of room 
= 2lh + 2bh + lb 
=[2x5x3+2x4x3+5x4]  \( m^{2}\)
(30 + 24 + 20)   \( m^{2}\)
 = 74   \( m^{2}\)
Cost of white-washing per   \( m^{2}\) area = Rs 7.50 
Cost of white-washlng 74   \( m^{2}\) area =Rs (74x7.50)
= RS 555 

Ans : Let length, breadth, and height of the rectangular hall be I m, b m, and h mrespectively. 
Area of four walls = 21h + 2bh 
=2( l + b )h
Perimeter of the floor of hall =2(l + b) 
=250 m 
Area of four walls =2( l+b )h =250  \( m^{2}\)
Cost of painting per  \( m^{2}\) area = Rs 10 
Cost of painting 250h  \( m^{2}\) area = Rs (250h x 10) = Rs 2500h 
However, it is given that the cost of painting the walls is Rs 15000. 
15000 = 2500h 
h=6
Therefore, the height of the hall is 6 m. 

Ans : Total surface area of one brick = 2(lb + bh + /h) 
= [2(22.5 X10 + 10 x 7.5 + 22.5 x 7.5)] \( \mathrm{cm}^{2}\)
= 2(225 + 75 + 168.75)  \( \mathrm{cm}^{2}\)
= (2 x 468.75)  \( \mathrm{cm}^{2}\)
= 937.5 \( \mathrm{cm}^{2}\)
Let n bricks can be painted out by the paint of the container. 
Area of n bricks = (n x937.5) \( \mathrm{cm}^{2}\) = 937.5n \( \mathrm{cm}^{2}\)
Area that can be painted by the paint of the container = 9.375    \( m^{2}\)
= 93750  \(\mathrm{cm}^{2}\) 
= 93750 - 937.5n 
n = 100 
Therefore, 100 bricks can be painted out by the paint of the container. 

Ans : (i) Edge of cube = 10 cm 
Length (j) of box = 12.5 cm 
Breadth (b) of box = 10 cm 
Height (h) of box = 8 cm 
Lateral surface area of cubical box = \( 4(\text { edge })^{2}\)
=400  \(\mathrm{cm}^{2}\) 
Lateral surface area of cuboidal box =2[lh+bh]
= [2(12.5 x 10 x 8)] \(\mathrm{cm}^{2}\) 
= (2 x 180)  \(\mathrm{cm}^{2}\) 
= 360 \(\mathrm{cm}^{2}\) 
Clearly, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box. 
Lateral surface area of cubical box - Lateral surface area of cuboidal box = 400 \(\mathrm{cm}^{2}\) - 360\(\mathrm{cm}^{2}\) = 40 \(\mathrm{cm}^{2}\) 
Therefore, the lateral surface area of the cubical box is greater than the lateral 
surface area of the cuboidal box by 40\(\mathrm{cm}^{2}\) 
(ii) Total surface area of cubical box =\( 6(\text { edge })^{2}\) =\( 6(10 \mathrm{cm})^{2}\)= 600 \(\mathrm{cm}^{2}\) 
Total surface area of cuboidal box 
= 2(lh + bh + lb] 
=[2(12.5 x 8 + 10 x 8+ 12.5 x 100]\(\mathrm{cm}^{2}\)
610 \(\mathrm{cm}^{2}\) 
Clearly, the total surface area of the cubical box is smaller than that of the cuboidal box. 
Total surface area of cuboidal box - Total surface area of cubical box = 610 \(\mathrm{cm}^{2}\)
600\(\mathrm{cm}^{2}\) = 10 \(\mathrm{cm}^{2}\)
Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 \(\mathrm{cm}^{2}\).