NCERT solutions for class 9 Maths chapter 13 Surface Areas and Volumes (Exercise - 13.1)

NCERT solutions for class 9 Maths

Surface Areas and Volumes

Chapter 13 Exercise 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 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 1m2 costs ₹20.

Solution:

Length of plastic box l = 1.5 m

breadth b = 1.25 m

Height h = 65 cm = 0.65 m

Area of sheet required for making a plastic box = total surface area of box – area of top of box

=2(lb + bh + hl) – lb

= 2(1.50 * 1.25 + 1.25 * 0.65 +0.65 * 1.5) – (1.50 – 1.25) m²
= 2 (1.875 + 0.8125 + 0.975) – (1.875) m²
= 2 (3.6625) – 1.875 m² = 5.45 m²
∴ Area of the sheet required for making the box = 5.45 m²

EX 13.1 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 ₹17.50 per m².
Solution:
Length of the room (l) = 5 m
Breadth of the room (b) = 4 m
Height of the room (h) = 3 m
∴ Area of four walls and celling =[Total surface area of room] + [Area of the floor]
=2(lb + bh + hl) – lb
= 2(5 * 4 + 4 * 3 + 3 * 5) – (5 x 4)m² = 2(20 + 12 + 15 ) – 20 m² = 2(47) – 20m²
= 90 – 20 = 74m²
∴ The area of four walls and celling = 74 m²
Thus, the required cost of white washing the walls of the room and the cellings = ₹; 7.50 * 74 =₹555

EX 13.1 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 ₹10 per m² is ₹15000, find the height of the hall.
[Hint: Area of the four walls = Lateral surface area]
Solution:
Let, the length of the hall = l m, breadth= b m, and height = h m,respectively.
∴ Perimeter of the floor = 2(l + b)
⇒ 2(l + b) = 250 m
∵ Area of four walls = 2(l+b) h = 250h m²
Cost of painting the four walls = ₹10 x 250 h= ₹ 2500h
⇒ ₹2500 h = ₹15000 ⇒ h = 15000/ 2500= 6
Thus, the height of the hall = 6 m

EX 13.1 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 x 10 cm x 7.5 cm can be painted out of this container.
Solution:
Total area that can be painted = 9.375 m²
Here, Length of a brick (l) = 22.5 cm
Breadth of a brick (b) = 10 cm
Height of a brick (h) = 7.5 cm
Total surface area of a brick = 2[lb + bh + hl]
= 2[(225 * 1(0) + (10 * 7.5) + (7.5 * 22.5)] cm²
= 2[(225) + (75) + (168.75)] cm²
= 2[468.75] cm² = 937.5 cm² =0.09375m²
Number of bricks to be painted = total paint available / paint for one brick
= 9.375 m^{2 / 0.09375m2 = 100}
Thus, the paint of the container can paint = 100 bricks

EX 13.1 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:

Length (l) = 12.5 cm,
Breadth (b) = 10 cm,
Height (h) = 8 cm

(i) Side of cubical box (e) = 10 cm
Lateral surface area of a cubical box = 4e² = 4(10)²cm² = 4 (100)cm² = 400cm²
∴ Lateral surface area of cuboidal box = 2[l + b] x h = 2[12.5 + 10] x 8 cm² = 360 cm²
Lateral surface area of cubical box is greater then the lateral surface area of cuboidal box by (400 – 360) cm² = 40 cm².
(ii) Length (l) = 12.5 cm,
Breadth (b) = 10 cm,
Height (h) = 8 cm Total surface area = 6a² = 6 x 10² cm²= 600 cm²
For the cuboidal box with dimensions,

Total surface area = 2[lb + bh + hl]
= 2[(12.5 x 10) + (10 x 8) + (8 x 12.5)] cm²
= 2[125 + 80 + 100] cm²
= 2[305] cm²
= 610 cm²
(ii) Total surface area of a cubical box is smaller than the total surface area of cuboidal box by (610 – 600) cm² = 10 cm².

EX 13.1 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:
Length (l) = 30 cm,
Breadth (b) = 25 cm,
Height (h) = 25 cm
(i) Total Surface area of the herbarium (glass)
= 2[lb + bh + hl]
= 2[(30 x 25) + (25 x 25) + (25 x 30)] cm² – 2[750 + 625 + 750] cm²
= 2[2125] cm²
= 4250 cm²
Thus, the area of the glass for greenhouse = 4250 cm²

(ii) Total length of 12 edges = 4(l + b + h)
= 4(30 + 25 + 25) cm
= 4 x 80 cm = 320 cm
Thus, for all 12 edges, the required length of tape = 320 cm

EX 13.1 QUESTION 7.

Shanti Sweets Stalll was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm x 20 cm x 5 cm and the smaller of dimensions 15 cm x 12 cm x 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is ₹4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.
Solution:
For bigger box:
Length (l) = 25 cm,
Breadth (b) = 20 cm,
Height (h) = 5 cm
Total surface area of a box = 2(lb + bh + hl)
= 2[(25 x 20) + (20 x 5) + (5 x 25)] cm²
= 2 [500 + 100 + 125] cm²
= 2[725] cm²
= 1450 cm²
Area of cardboard for overlap = 5% of 1450 cm^{2 = 1450 *} $\frac { 5 }{ 100 }$ =72.5 cm²

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

∴The area of cardboard for 250 bigger boxes = (250 x 1522.5 cm² = 380625 cm²

For smaller box:
Length (l) = 15 cm,
Breadth (b) = 12 cm,
Height (h) = 5 cm
Total surface area of a box = 2 [lb + bh + hl]
= 2[(15 x 12) + (12 x 5) + (5 x 15)] cm²
= 2[180 + 60 + 75] cm² = 2[315] cm² = 630 cm²
Area of cardboard for overlap = 5% of 630 cm^{2 = 630 *} $\frac { 5 }{ 100 }$ =31.5 cm²

Total area of cardboard for 1 smaller box = 630 + 31.5 = 661.5 cm²
∴The area of cardboard for 250 smaller boxes = (250 x 661.5 cm² = 165375 cm²

The area of cardboard for 500 boxes = 380625 + 165375= 546000 cm²
Total cost of cardboard at the rate of ₹4 per 1000cm²= $\frac { 4\times 546000 }{ 1000 }$ = ₹2184

EX 13.1 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 x 3 m?
Solution:
Length (l) = 4 m,
Breadth (b) = 3m
Height (h) = 2.5 m
∴ Area of four walls and the top of the shelter = Total area of the shelter – area of floor
= [2(l + b)h] + [lb]
= [2(4 + 3) x 2.5] m² + [4 x 3] m²
= 35 m² + 12 m² = 47 m²
Thus, 47 m² tarpaulin is required to make this shelter.

NCERT solutions for class 9 Maths chapter 13 Surface Areas and Volumes (Exercise – 13.1)