These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1

Question 1.

A plastic box 13 m long, 1.25 wide and 65 cm deep are to be made. It is to be open at the top. Ignoring the thickness of a plastic sheet, determine.

(i) The area of the sheet required for making the box.

(ii) The cost of the sheet for it, if a sheet measuring 1m costs Rs. 20.

Solution:

(i) We have given,

length of box = 1.5 m

width of the box = 1.25 m

height of the box = 65 cm = 0.65 m

∴ Surface area of box = 2(lb + bh + lh)

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

= 2(1.875 + 0.8125 + 0.975)

= 2 × 3.6625

= 7.325 m^{2}

But we have given box is open at the top.

Area of top of the box = 1.875 m^{2}

∴ Surface area of required box = 7.325 – 1.875 = 5.45 m^{2}

So, the area of sheet required for making the box is 5.45 m^{2}

(ii) Rate of sheet = Rs. 20/cm^{2}

∴ Total cost of sheet for box = 20 × 5.45 = Rs. 109.

Question 2.

The length, breadth, and height of a room are 5 m, 3 m, and 3 m respectively. Find the cost of whitewashing the walls of the room and ceiling at the rate of Rs. 7.50 per m^{2}.

Solution:

We have given

length of room = 5 m

breadth of a room = 4 m

height of a room = 3 m

Surface area of 4 walls and ceiling of the room

= 2(l + b)h + lb

= 2(5 + 4)3 + 5 × 4

= 54 + 20

= 74 m^{2}

Cost of whitewashing the walls and ceiling of the room at Rs. 7.50 per m^{2} is = 74 × 7.50 = Rs. 555.

Question 3.

The floor of the rectangular hall has a perimeter of 250 m. If the cost of painting the four walls at the rate of Rs. 10 per m^{2} is Rs. 15000, find the height of the hall.

Solution:

We have given that

Perimeter of floor = 250 m

2(l + b) = 250

l + b = 125 m

Now,

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

= 2 × 125 × h

= 250h m^{2}.

Rate of painting the four walls is Rs. 10 per m^{2} is Rs. 250h × 10

But we have given that the total cost of painting of four walls = 15000

i.e. 2500h = 15000

or, h = \(\frac{15000}{2500}\) = 6 m

∴ Height of the hall = 6 m

Question 4.

The paint in a certain container is sufficient to paint an area equal to 9.375 m^{2}. How many bricks of dimension 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?

Solution:

We have given that the container contains sufficient paint for the area of 9.375 m^{2}.

Again

length of the bricks = 22.5 cm = 0.225 m

breadth of the bricks = 10 cm = 0.10 m

height of the bricks = 7.5 cm = 0.075 m

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

= 2(0.225 × 0.10 + 0.10 × 0.075 + 0.075 × 0.225)

= 2(0.0225 + 0.0075 + 0.016875)

= 0.09375 m^{2}.

Area of one brick = 0.09375 m^{2}.

Let x bricks can be painted out of this container

∴ 0.9375 × x = 9.375

or, x = \(\frac{9.375}{0.09375}\) = 100

Total number of bricks can be painted out of this container is 100.

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 a greater lateral surface area and by how much?

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

Solution:

(i) Length of edge of cubical box = 10 cm

The lateral surface area of cubical box = 4a^{2}

= 4 × (10)^{2}

= 400 cm^{2}

Again, length of cuboidal box = 12.5 cm

breadth of cuboidal box = 10 cm

and height of cuboidal box = 8 cm.

∴ Lateral surface area of cubodical box = 2(l + b)h

= 2(12.5 + 10)8

= 360 cm^{2}

Lateral surface area of cubical box is greater than cubodical box by = (400 – 360) = 40 cm^{2}

Now,

Total surface ara of cubical box = 6a^{2}

= 6 × (10)^{2}

= 600 cm^{2}

and total surface area of cubodical box = 2 (lb + bh + lh)

= 2(12.5 × 10 + 10 × 8 + 8 × 12.5)

= 2(125 + 80 + 100)

= 610 cm^{2}

Total surface area of cubodical box is greater than cubical box by = (610 – 600) = 10 cm^{2}.

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 glass?

(ii) How much tape is needed for all the 12 edges?

Solution:

(i) We have given that

length of small indoor greenhouse = 30 cm.

the breadth of small indoor greenhouse = 25 cm.

and height of small indoor greenhouse = 25 cm.

Total surface area of indoor greenhouse (herbarium)

= 2(lb + bh + lh)

= 2(30 × 25 + 25 × 25 + 25 × 30)

= 2(750 + 625 + 750)

= 4250 cm^{2}

∴ Area of glass to make herbarium is 4250 cm^{2}.

(ii) Total length of tape is needed = 4(l + b + h)

= 4(30 + 25 + 25)

= 320 cm

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. 5% of the total surface area is required extra, for all the overlaps. If the cost of cardboard is Rs. 4 for 1000 cm^{2}, find the cost of cardboard required for supplying 250 boxes of each kind.

Solution:

Dimensions of the bigger box are 25 cm × 20 cm × 5 cm

Total surface ara of one big box = 2(lb + bh + lh)

= 2(25 × 20 + 20 × 5 + 5 × 25)

= 2(500 + 100 + 125)

= 1450 cm^{2}.

∴ Total surface area of 250 such boxes are 250 × 1450 = 3,62,600 cm^{2}.

Again,

Dimension of smaller box is 15 cm × 12 cm × 5 cm.

∴ Total surface area of one small box = 2(lb + bh + lh)

= 2(15 × 12 + 12 × 5 + 5 × 15)

= 2(180 + 60 + 75)

= 630 cm^{2}.

∴ Total surface area of 250 such boxes = 250 × 630 = 1,57,500 cm^{2}.

∴ Total surface area of both type of boxes are = 3,62,500 + 1,57,500 = 5,20,000 cm^{2}

Now, 5% of the total surface area is required for the overlaps.

5% of 5,20,000 = \(\frac{5}{100}\) × 5,20,000 = 26,000

∴ Total area of card board required = 5,20,000 + 26000 = 5,46,000

∴ Total cost of cardboard at the rate of Rs. 4 per 1000 cm^{2} = \(\frac{546000}{1000} \times 4\) = Rs. 2184

Question 8.

Praveen wants to make a temporary shelter for her car, by making a box-like structure with a tarpaulin that covers all 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 many tarpaulins would be required to make the shelter of height 2.5m, with base dimensions 4 m × 3 m?

Solution:

We have given that,

length of temporary shelter for car = 4 m

the breadth of temporary shelter for car = 3 m

and height of temporary shelter for car = 2.5 m

∴ Total tarpaulin be required to make the shelter = Area of 4 wall + Area of roof

= 2(l + b)h + l × b

= 2(4 + 3) × 2.5 + 4 × 3

= 2 × 7 × 2.5 + 4 × 3

= 35 + 12

= 47 m^{2}

Therefore total tarpaulin be required to make temporary shelter for the car is 47 m^{2}.