NCERT Class 9 Maths Solutions Chapter - 13 Surface Areas and Volumes, Ex - 13.1

Ex - 13.1

Question 1.  A plastic box 1.5 m long, 1.25 m wide and 65 cm deep, is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine:
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1  costs Rs 20.

Solution

Length of box = 1.5 m
Breadth of box = 1.25 m
Depth of box = 65 cm = 0.65 m

(i) The box is open at the top.
    Area of sheet required = 2bh + 2lh + lb
    = [2  1.25 0.65 + 2 1.5 0.65 + 1.5 1.25] m²
    = (1.625 + 1.95 + 1.875)  = 5.45   
(ii) Cost of sheet of area 1 = Rs 20

     Cost of sheet of area 5.45  = Rs (5.45 20) = Rs 109

Question 2.  The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 per .

Solution

Length of room = 5 m
Breadth of room = 4 m
Height of room = 3 m
Area to be white washed = Area of walls + Area of ceiling of room
= 2lh + 2bh + lb 
= [2  5  3 + 2  4  3 + 5  4] 
= (30 + 24 + 20) 
= 74 

Cost of white washing 1  area = Rs 7.50
Cost of white washing 74  area = Rs (74 7.50) = Rs 555

Question 3.  The floor of a rectangular hall has a perimeter 250 m. If the cost of panting the four walls at the rate of Rs 10 per  is Rs 15000, find the height of the hall.

Solution

Let length, breadth and height of rectangular hall be l, b and h respectively.
Area of four walls = 2lh + 2bh = 2(l + b) h
Perimeter of floor of hall = 2(l + b) = 250 m
 Area of four walls = 2(l + b) h = 250h 
Cost of painting 1 area = Rs 10
Cost of painting 250h  area = Rs (250h  10) = Rs 2500h
It is given that the cost of paining the walls is Rs 15000.
 15000 = 2500h
h = 6
Thus, the height of hall is 6 m.

Question 4.  The paint in a certain container is sufficient to paint an area equal to 9.375 . How many bricks of dimensions 22.5 cm  10 cm 7.5 cm can be painted out of this container?

Solution

Total surface area of one brick = 2(lb + bh + lh)
 = [2(22.5 10 + 10  7.5 + 22.5  7.5)]
 = 2(225 + 75 + 168.75) 
          = (2  468.75) 
 = 937.5     
Let n number of bricks be painted by the container.
Area of n bricks = 937.5n     
Area that can be painted by the container = 9.375 m² = 93750 
 93750 = 937.5n
n = 100
Thus, 100 bricks can be painted out by the container.

Question 5.  A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.

(i)    Which box has the greater lateral surface area and by how much?
(ii)    Which box has the smaller total surface area and by how much?

Solution

Edge of the cubical box = 10 cm
    
Length of the cuboidal box = 12.5 cm
Breadth of the cuboidal box = 10 cm
Height of the cuboidal box = 8 cm

Lateral surface area of cubical box =  = =   
Lateral surface area of cuboidal box = 2[lh + bh] 
                            = [2(12.5  8 + 10  8)] 
                                                     = 360 
The lateral surface area of cubical box is greater than lateral surface area of cuboidal box.

Lateral surface area of cubical box - lateral surface area of cuboidal box = 400  - 360  = 40 

Thus, the lateral surface area of cubical box is greater than that of cuboidal box by 40 cm².

(ii)    Total surface area of cubical box =  =  = 600 cm²
Total surface area of cuboidal box = 2[lh + bh + lb]
= [2(12.5  8 + 10 8 + 12.5 10] = 

The total surface area of cubical box is smaller than that of cuboidal box

Total surface area of cuboidal box - total surface area of cubical box =  -  =

Thus, the total surface area of cubical box is smaller than that of cuboidal box by 

Question 6.  A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.

(i)    What is the area of the glass?
(ii)    How much of tape is needed for all the 12 edges?

Solution

(i)    Length of green house = 30 cm
Breadth of green house = 25 cm
Height of green house = 25 cm

Total surface area of green house = 2[lb + lh + bh]
= [2(30  25 + 30  25 + 25  25)] 
= [2(750 + 750 + 625)] 
= (2  2125) 
= 4250 

Thus, the area of the glass is 4250 .

(ii)    Total length of tape = 4(l + b + h)
        = [4(30 + 25 + 25)] cm
        = 320 cm
        
        Therefore, 320 cm tape is needed for all the 12 edges.

Question 7.  Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm  20 cm  5 cm and the smaller of dimensions 15 cm  12 cm  5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs 4 for 1000 , find the cost of cardboard required for supplying 250 boxes of each kind.

Solution

Length of bigger box = 25 cm
Breadth of bigger box = 20 cm
Height of bigger box = 5 cm

Total surface area of bigger box = 2(lb + lh + bh)
    = [2(25  20 + 25  5 + 20  5)] 
    = [2(500 + 125 + 100)] cm2 = 1450    

Extra area required for overlapping =  = 72.5             

Considering all overlaps, total surface area of 1 bigger box 
= (1450 + 72.5)  =1522.5 

Area of cardboard sheet required for 250 such bigger box 
= (1522.5 250)  = 380625   

Total surface area of smaller box = [2(15 12 + 15  5 + 12  5] = [2(180 + 75 + 60)]  = (2  315)  = 630 

Extra area required for overlapping =   = 31.5 

Considering all overlaps, total surface area of 1 smaller box 
= (630 + 31.5)  = 661.5 

Area of cardboard sheet required for 250 smaller box 
= (250 661.5)  = 165375  

Total cardboard sheet required = (380625 + 165375)  = 546000 

Cost of 1000  cardboard sheet = Rs 4
Cost of 546000  cardboard sheet = Rs  = Rs 2184 

So, cost of cardboard sheet required for 250 boxes of each kind will be Rs 2184.

Question 8.  Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m  3 m?

Solution

Length of shelter = 4 m
Breadth of shelter = 3 m
Height of shelter = 2.5 m

The tarpaulin will be required for top and four sides of the shelter.
Area of Tarpaulin required = 2(lh + bh) + lb
= [2(4  2.5 + 3  2.5) + 4  3] 
= [2(10 + 7.5) + 12] 
= 47 

Ex-12.2

Ex-13.2