Surface Area and Volumes – Exercise 13.1 – Class IX

  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 1m2 costs Rs 20.

Solution:

Given: l = 1.5 m ; b = 1.25m ; h = 65cm = 0.65 m.

It is given that the box is open at the top, so it has only 5 faces.

Surface  area of the box = lb + 2(bh + hl)

= 1.5*1.25 + 2(1.25*0.65 + 0.65*1.5)

= 1.875 + 2(1.7875)

= 5.45 m2

Hence 5.45 m2 is the required sheet.

Cost of 1m2 costs Rs. 20, then cost of 5.45m2 sheet = Rs. 20 * 5.45m2 = Rs. 109.


  1. 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 m2.

Solution:

Given: l = 5m , b = 4m and h = 3m

Surface area of the wall = lb + 2(bh + hl)

= 5*4 + 2(4*3 + 3*5)

= 20 + 2(12 + 15)

= 20 + 2(27)

= 20 + 54

= 74 m2

The cost for 1m2 is Rs. 7.50 , then , the cost of 74m2 = 74m2*Rs.7.50 = Rs. 555


  1. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs 10 per m2 is Rs 15000, find the height of the hall.

[Hint : Area of the four walls = Lateral surface area.]

Solution:

Perimater of the rectangular hall 250m

i.e., 2(l +b) = 250 m

Area of the four walls of the hall = 2h(l + b)…………….(1)

It is given that, the cost of painting the four walls at the rate of Rs 10 per m2 is Rs 15000.

The area of the four walls of the hall = 15000/10 m2 = 1500m2………………..(2)

From (1) and (2), we have,

2h(l + b) = 1500

h*250 = 1500

h = 1500/250

h = 6

Hence the height o the wall is 6 m


  1. The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?

Solution:

Since, l = 22.5 cm ; b =  10 cm ; h = 7.5 cm

Total surface area of 1 brick = 2(lb + bh + lh)

= 2(22.5*10 + 10*7.5 + 7.5*22.5)

=2(225 + 75 + 168.75)

= 937.5 cm2

= 0.09375 m2

Therefore, no. of bricks can be painted = 9.375/0.09375 = 100


  1. 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:

(i)

Latteral surface area of the cube = 4a2 = 4(10)2 = 4*100 = 400 cm2

Lateral surface area of the cuboidal box = 2(lh + bh)

= 2(12.5*8 + 8*10)

= 2*8(22.5)

= 360 cm2

The difference in the lateral surface area of cube and cuboidal box =(400 – 360)cm2 = 40 cm2

Therefore, Cube has the greater lateral surface area.

(ii)

Total surface area of the cube = 6a2 = 6(10)2  = 6(100) = 600cm2

Total surface area of the cuboidal box = 2(lb + bh + lh)

= 2(12.5*10 + 10*8 + 8*12.5)

= 2(125 + 80 + 100)

= 2(305)

= 610 cm2

Difference in the total surface are of the cube and cuboidal box = (610 – 600)cm2 = 10cm2

Therefore, cube has the smaller total surface area.


  1. 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:

Here, l = 30 cm , b = 25 cm and h = 25 cm.

Total surface area of the glass = 2(lb + bh + lh)

= 2(30*25 + 25*25 + 25*30)

= 2*25(30 + 25 + 30)

= 50(85)

= 4250 cm2

(ii)

A cuboid has 12 edges i.e., 4lengths, 4 breadths and 4 heights.

Length of the tape required = 4(l+b+h)

= 4(30+25+25)

=4(80)

= 320 cm


  1. 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 cm2 , find the cost of cardboard required for supplying 250 boxes of each kind.

Solution:

Total surface area of the bigger box = 2(lb+hb+ lh)

= 2(25*20 + 20*5 + 5*25)

= 2*(500 + 100+ 125)

= 2(725)

= 1450 cm2

Area required for the overlaps is 5% of 1450cm2 = 1450*5/100 cm2 = 72.5 cm2

Total area of cardboard required for 1 bigger box = (1450+72.5)cm2 = 1522.5 cm2

Total area of cardboard needed for 250 bigger boxes = 1522.5 * 250 cm2 = 380625 cm2

 

Total surface area of the smaller box = 2(lb+hb+ lh)

= 2(15*12 + 12*5 + 5*15)

= 2(180 + 60 + 75)

= 630 cm2

Area required for the overlaps is 5% of 630cm2 = 630*5/100 cm2 = 31.5 cm2

Total area of cardboard required for 1 smaller boxes = (630+31.5)cm2 = 661.5 cm2

Total area of cardboard needed for 250 bigger boxes = 661.5 * 250 cm2 = 165375 cm2

Now, the total area of cardboard required for 250bigger boxes and 250 smaller boxes  = (380625+165375) cm2 = 546000 cm2

The cost of 1000 cm2 of cardboard = Rs. 4

Cost of 546000 cm2 of cardboard = Rs. 4/1000 *546000 = Rs. 2184


  1. 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:

h = 2.5 m , l = 4m and b = 3m

The tarpaulin required only 5 faces as excluding the floor area.

Surface area of the shelter for the car = lb + 2(bh + lh)

= 4*3 + 2(3*2.5 + 4*2.5)

= 12 + 2(7.5 + 10)

= 12 + 2(17.5)

= 12 + 35

= 47 m2


For next exercise – Surface Area and Volumes – Exercise 13.2 – Class IX


 

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