NCERT solution class 9 chapter Surface area and volume

Exercise 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 m2 cost 20 Rs

Sol :

It is given that, length (l) of box = 1.5 m

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 × 1.5 × 0.65 + 2 × 1.25 × 0.65 + 1.5 × 1.25] m2

= (1.95 + 1.625 + 1.875) m2 = 5.45 m2

(ii) Cost of sheet per m2 area = Rs 20

Cost of sheet of 5.45 m2 area = 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 m

Sol :

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

= [2 × 5 × 3 + 2 × 4 × 3 + 5 × 4] m2

= (30 + 24 + 20) m2

= 74 m2

Cost of white-washing per m2 area = Rs 7.50

Cost of white-washing 74 m2 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 m2 is Rs.15000, find the height of the hall.

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

Sol :

Let length, breadth, and height of the rectangular hall be l m, b m, and h m respectively.

Area of four walls = 2lh + 2bh

= 2(l + b) h

Perimeter of the floor of hall = 2(l + b)

= 250 m

∴ Area of four walls = 2(l + b) h = 250h m2

Cost of painting per m2 area = Rs 10

Cost of painting 250h m2 area = Rs (250h × 10) = Rs 2500h

However, it is given that the cost of paining the walls is Rs 15000.

∴ 15000 = 2500h

h = 6

Therefore, the height of the hall is 6 m.

 

Question 4

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?

Sol :

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

= [2(22.5 ×10 + 10 × 7.5 + 22.5 × 7.5)] cm2

= 2(225 + 75 + 168.75) cm2

= (2 × 468.75) cm2

= 937.5 cm2

Let n bricks can be painted out by the paint of the container.

Area of n bricks = (n ×937.5) cm2 = 937.5n cm2

Area that can be painted by the paint of the container = 9.375 m2 = 93750 cm2

∴ 93750 = 937.5n

n = 100

Therefore, 100 bricks can be painted out by the paint of 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?

Sol :

(i) Edge of cube = 10 cm

Length (l) of box = 12.5 cm

Breadth (b) of box = 10 cm

Height (h) of box = 8 cm

Lateral surface area of cubical box = 4(edge)2

= 4(10 cm)2

= 400 cm2

Lateral surface area of cuboidal box = 2[lh + bh]

= [2(12.5 × 8 + 10 × 8)] cm2

= (2 × 180) cm2

= 360 cm2

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 cm2 − 360 cm2 = 40 cm2

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

(ii) Total surface area of cubical box = 6(edge)2 = 6(10 cm)2 = 600 cm2

Total surface area of cuboidal box

= 2[lh + bh + lb]

= [2(12.5 × 8 + 10 × 8 + 12.5 × 10] cm2

= 610 cm2

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 cm2 − 600 cm2 = 10 cm2

Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm2.

 

 

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?

Sol :

(i) Length (l) of green house = 30 cm

Breadth (b) of green house = 25 cm

Height (h) of green house = 25 cm

Total surface area of green house

= 2[lb + lh + bh]

= [2(30 × 25 + 30 × 25 + 25 × 25)] cm2

= [2(750 + 750 + 625)] cm2

= (2 × 2125) cm2

= 4250 cm2

Therefore, the area of glass is 4250 cm2.

(ii)

It can be observed that tape is required along side AB, BC, CD, DA, EF, FG, GH, HE, AH, BE, DG, and CF.

Total length of tape = 4(l + b + h) which is equal to perimeter of cuboid

= [4(30 + 25 + 25)] cm

= 320 cm

Therefore, 320 cm tape is required 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 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.

Sol :

Length (l1) of bigger box = 25 cm

Breadth (b1) of bigger box = 20 cm

Height (h1) of bigger box = 5 cm

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

= [2(25 × 20 + 25 × 5 + 20 × 5)] cm2

= [2(500 + 125 + 100)] cm2

= 1450 cm2

Extra area required for overlapping =\dfrac{\left(1450\times5\right)}{100}~cm^2

= 72.5 cm2

While considering all overlaps, total surface area of 1 bigger box

= (1450 + 72.5) cm2 =1522.5 cm2

Area of cardboard sheet required for 250 such bigger boxes

= (1522.5 × 250) cm2 = 380625 cm2

Similarly, total surface area of smaller box = [2(15 ×12 + 15 × 5 + 12 × 5] cm2

= [2(180 + 75 + 60)] cm2

= (2 × 315) cm2

= 630 cm2

Therefore, extra area required for overlapping =\left(\dfrac{630\times5}{100}\right)cm2

Total surface area of 1 smaller box while considering all overlaps

= (630 + 31.5) cm2 = 661.5 cm2

Area of cardboard sheet required for 250 smaller boxes = (250 × 661.5) cm2

= 165375 cm2

Total cardboard sheet required = (380625 + 165375) cm2

= 546000 cm2

Cost of 1000 cm2 cardboard sheet = Rs 4

Cost of 546000 cm2 cardboard sheet =\left(\dfrac{546000\times4}{1000}\right)

= 2184

Therefore, the cost of cardboard sheet required for 250 such 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?

Sol :

Length (l) of shelter = 4 m

Breadth (b) of shelter = 3 m

Height (h) of shelter = 2.5 m

Tarpaulin will be required for the top and four wall sides of the shelter.

Area of Tarpaulin required = 2(lh + bh) + l b

= [2(4 × 2.5 + 3 × 2.5) + 4 × 3] m2

= [2(10 + 7.5) + 12] m2

= 47 m2

Therefore, 47 m2 tarpaulin will be required.

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