Chapter 11: Surface Areas and Volumes

Q1Written answer

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 the sheet for it, if a sheet measuring 1m² costs Rs. 20.

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Worked solution

Ncert solutions class 9 chapter 13-1

Given: length (l) of box = 1.5m

Breadth (b) of box = 1.25 m

Depth (h) of box = 0.65m

(i) Box is to be open at the 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.

Q2Written answer

The length, breadth and height of a room are 5 m, 4 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 m².

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Worked solution

Length (l) of room = 5m

Breadth (b) of room = 4m

Height (h) of room = 3m

It can be observed that four walls and the ceiling of the room are to be whitewashed.

Total area to be whitewashed = Area of walls + Area of the ceiling of the room

= 2lh+2bh+lb

= [2×5×3+2×4×3+5×4]

= (30+24+20)

= 74

Area = 74 m²

Also,

Cost of whitewash per m²area = Rs.7.50 (Given)

Cost of whitewashing 74 m²area = Rs. (74×7.50)

= Rs. 555

Q3Written answer

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.15,000, find the height of the hall.

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

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Worked solution

Let the length, breadth, and height of the rectangular hall be l, b, and h, 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 square metre area = Rs.10

Cost of painting 250h square metre area = Rs (250h×10) = Rs.2500h

However, it is given that the cost of painting the walls is Rs. 15,000.

15000 = 2500h

Or h = 6

Therefore, the height of the hall is 6 m.

Q4Written answer

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?

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Worked solution

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

= [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

As per the given instructions, the area that can be painted by the paint of the container = 9.375 m²= 93750 cm²

So, we have 93750 = 937.5n

n = 100

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

Q5Written answer

A cubical box has each edge 10 cm, and another cuboidal box is 12.5cm 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?

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Worked solution

From the question statement, we have

Edge of a cube = 10cm

Length, l = 12.5 cm

Breadth, b = 10cm

Height, h = 8 cm

(i) Find the lateral surface area for both figures.

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

= 4(10)²

= 400 cm² …(1)

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

= [2(12.5×8+10×8)]

= (2×180) = 360

Therefore, the lateral surface area of the cuboidal box is 360 cm². …(2)

From (1) and (2), the lateral surface area of the cubical box is more than the lateral surface area of the cuboidal box. The difference between both lateral surfaces is 40 cm².

(Lateral surface area of the cubical box – Lateral surface area of cuboidal box=400cm²–360cm² = 40 cm²)

(ii) Find the total surface area for both figures.

The total surface area of the cubical box = 6(edge)² = 6(10 cm)² = 600 cm²…(3)

The total surface area of the cuboidal box

= 2[lh+bh+lb]

= [2(12.5×8+10×8+12.5×100)]

= 610

This implies that the total surface area of the cuboidal box is 610 cm²..(4)

From (3) and (4), the total surface area of the cubical box is smaller than that of the cuboidal box. And their difference is 10cm².

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

Q6Written answer

A small indoor greenhouse (herbarium) is made entirely of glass panes (including the base) held together with tape. It is 30cm long, 25 cm wide, and 25 cm high.

(i) What is the area of the glass?

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

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Worked solution

Length of the greenhouse, say l = 30cm

Breadth of the greenhouse, say b = 25 cm

Height of greenhouse, say h = 25 cm

(i) Total surface area of greenhouse = Area of the glass = 2[lb+lh+bh]

= [2(30×25+30×25+25×25)]

= [2(750+750+625)]

= (2×2125) = 4250

The total surface area of the glass is 4250 cm²

(ii)

Ncert solutions class 9 chapter 13-2

From the figure, the tape is required along sides AB, BC, CD, DA, EF, FG, GH, HE AH, BE, DG, and CF.

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

= [4(30+25+25)] (after substituting the values)

= 320

Therefore, 320 cm tape is required for all 12 edges.

Q7Written answer

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×20cm×5cm, and the smaller of dimension 15cm×12cm×5cm. 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.

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Worked solution

Let l, b and h be the length, breadth and height of the box.

Bigger Box

l = 25cm

b = 20 cm

h = 5 cm

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

= [2(25×20+25×5+20×5)]

= [2(500+125+100)]

= 1450 cm²

Extra area required for overlapping 1450×5/100 cm²

= 72.5 cm²

While considering all overlaps, the total surface area of the 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

Smaller Box

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

= [2(180+75+60)] cm²

= (2×315) cm²

= 630 cm²

Therefore, the extra area required for overlapping 630×5/100 cm² = 31.5 cm²

The 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

In Short

Box

Dimensions (in cm)

Total surface area (in cm² )

Extra area required for overlapping (in cm²)

Total surface area for all overlaps (in cm²)

Area for 250 such boxes (in cm²)

Bigger Box

l = 25

b = 20

c = 5

1450

1450×5/100

= 72.5

(1450+72.5) = 1522.5

(1522.5×250) = 380625

Smaller Box

l = 15

b = 12

h =5

630

630×5/100 = 31.5

(630+31.5) = 661.5

( 250×661.5) = 165375

Now, total cardboard sheet required = (380625+165375) cm²

= 546000 cm²

Given, cost of 1000 cm² cardboard sheet = Rs. 4

Therefore, the cost of 546000 cm²cardboard sheet =Rs. (546000×4)/1000 = Rs. 2184

Therefore, the cost of cardboard required for supplying 250 boxes of each kind will be Rs. 2,184.

Q8Written answer

Praveen wanted 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 4m×3m?

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Worked solution

Let l, b and h be the length, breadth and height of the shelter.

Given:

l = 4m

b = 3m

h = 2.5m

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

Using formula, area of tarpaulin required = 2(lh+bh)+lb

On putting the values of l, b and h, we get

= [2(4×2.5+3×2.5)+4×3] m²

= [2(10+7.5)+12]m²

= 47m²

Therefore, 47 m² of tarpaulin will be required.

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Exercise 13.1