# HBSE 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1

Haryana State Board HBSE 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes 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 opened 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 of Rs. 20.
Solution:
We have,
Length of the plastic box (l) = 1.5 m
Breadth of the plastic box (b)= 1.25 m
Depth of the plastic box (h)= 65 cm = $$\frac{65}{100}$$
m = 0.65 m
(i) The area of the sheet required for making the box = lb + 2(l + b) × h
= 1.5 × 1.25 + 2(1.5 + 1.25) × 0.65
= 1.875 + 2(2:75) × 0.65
= 1.875 +3.575
= 5.45m2

(ii) Cost of 1 m2 sheet = Rs. 20
Cost of 5.45 m2 sheet = Rs. (20 × 5.45)
= Rs. 109
Hence, (i) The area of the sheet required = 5.45 m2
(ii) Cost of sheet = 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 m2.
Solution:
We have,
Length of a room (l) = 5 m
Breadth of a room (b) = 4 m
Height of a room (h) = 3 m
Area for white washing = Area of four walls of a room + Area of ceiling of a room
= 2(l + b) × h + l × b
= 2(5 + 4) × 3 + 5 × 4
= 2 × 9 × 3 + 20
= 54 + 20 = 74 m2
Rate of white washing = Rs. 7.50 m2
Cost of white washing = Area × rate of white washing
= 74 × 7.50 = Rs. 555
Hence,cost of white washing = Rs. 555.

Question 3.
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:
We have,
Perimeter of a rectanglar hall = 250 m
⇒ 2(l + b) = 250 m …(1)
Let height of the hall be h m.
Area of four walls of a hall = 2(l + b) × h
Area of four walls of a hall = 250 × h, [using(1)] …(2)
Rate of painting = Rs. 10 per m2 (given)
Total cost of painting = Rs. 15000
∴ Area of four walls of a hall = $$\frac{\text { Total cost }}{1 m^2 \cos t}$$
⇒ Area of four walls of a hall = $$\frac{15000}{10}$$
⇒ Area of four walls of a hall = Rs. 1500 m2 …..(3)
From (2) and (3), we have
250 × h = 1500
⇒ h = $$\frac{1500}{250}$$
⇒ h = 6 m
Hence, height of the hall = 5 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?
Solution:
We have,
Length of brick (l) = 22.5 cm
Breadth of brick (b) = 10 cm
Height of brick (h) = 7.5 cm
Surface area of one brick = 2(lb + bh + hl)
= 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5)
= 2(225 + 75 + 168.75)
= 2(468.75) = 937.5 cm2
= $$\frac{937.5}{10000}$$ m2
= 0.09375 m2
Area for which the paint is just sufficient = 9.375 m2 (given)
∴ Number of bricks that can be painted with the available paint
= $$\frac{9.375}{0.09375}=\frac{937500}{9375}$$
Hence, number of bricks that can be painted = 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 the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?
Solution:
We have,
Each edge of cubical box = 10 cm
Cuboidal box length (l) = 12.5 cm
Cuboidal box breadth (b) = 10 cm
Cuboidal box height (h) = 8 cm
(i) Lateral surface area of the cubical box =
4a2 = 4 × (10)2
= 4 × 100 = 400 cm2
Lateral surface area of the cuboidal box = 2(l + b) × h
= 2(12.5 + 10) × 8
= 2 × 22.5 × 8 = 360 cm2
Thus, lateral surface area of the cubical box is greater and is more (400 – 360) cm2
i.e., 40 cm2.
(ii) Total surface area of cubical box = 6a2
= 6 × (10)2
= 6 × 100 = 600 cm2
Total surface area of the cuboidal box = 2(lb + bh + hl)
= 2(12.5 × 10 + 10 × 8 + 8 × 12.5)
= 2(125 + 80 + 100)
= 2(305) = 610 cm2
Thus, total surface area of cubical box is smaller by (610-600) i.e., 10 cm2.
Hence, (i) Lateral surface area of the cubical box is greater by 40 cm2.
(ii) Total surface area of the cubical box is smaller 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 ?
Solution:
The shape of the greenhouse is cuboid.
Length of the greenhouse (l) = 30 cm
Breadth of the greenhouse (b) = 25 cm
Height of the greenhouse (h) = 25 cm
(i) Area of the glass of the greenhouse=2(lb + bh + hl)
= 2(30 × 25 + 25 × 25 + 25 × 30)
= 2(750 + 625 + 750)
= 2(2125) = 4250 cm2
(ii) Length of required tape = Sum of length of 12 edges :
= 4l + 4b + 4h
= 4 × 30 + 4 × 25+ 4 × 25
= 120 + 100 + 100
= 320 cm
Hence, (i) Area of the glass = 4250 cm2
(i) Length of required tape = 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. 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:
Dimensions of the bigger box are
L = 25 cm, B = 20 cm and H = 5 cm
Surface area of cardboard for bigger box
= 2(LB + BH + HL)
= 2(25 × 20 + 20 × 5 + 5 × 25)
= 2(500 + 100 + 125)
= 2(725) = 1450 cm2
Surface area of the cardboard for 250 bigger box
= 1450 × 250
= 362500 cm2
Dimensions of the smaller box are l = 15 cm, b = 12 cm and h = 5 cm
Surface area of the cardboard for smaller box
= 2(lb + bh + hl)
= 2(15 × 12 + 12 × 5 + 5 × 15)
= 2(180 + 50 + 75)
= 2 × 315
= 630 cm2
Surface area of the cardboard for 250 smaller boxes
= 630 × 250 = 157500 cm2
Required total area of cardboard sheet
= 362500 + 157500
= 520000 cm2
Area of overlaps = 5% of total area of cardboard sheet
= 5% of 520000
= $$\frac{5}{100}$$ × 520000
= 26000 cm2
Total area of the sheet including overlapping area of sheet
= 520000 + 26000
= 546000 cm2
∵ 1000 cm2 cardboard sheet cost = Rs. 4
∴ 1 cm2 cardboard sheet cost = Rs. $$\frac{4}{1000}$$
∴ 546000 cm2 cardboard sheet cost
= Rs. $$\frac{4}{1000}$$ × 546000
= Rs. 2184.
Hence, cost of required cardboard sheet = 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: Dimensions of the base of car’s shelter are
l = 4 m and b = 3 m
Height of car’s shelter (h) = 2.5 m
Required area of the trapaulin for the shelter of the car
= 2(l + b) × h + l × b
= 2(4 + 3) 2.5 + 4 × 3
= 2 × 7 × 2.5 + 12
= 35 + 12 = 47 m2
Hence, required area of the trapaulin for the shelter of the car = 47 m2.