Chapter-13 - Surface Area and Volumes - Exercise 13.1 - Open box, whitewashing, painting, bricks

Sep 18, 2026

Uploaded by Bintang

Exercise 13.1 under Chapter 13 focuses on surface area and volumes of cuboids, cubes, and related composite shapes. Problems include finding the sheet area for an open-top plastic box, computing whitewashing cost for room walls and ceiling using surface area of five faces, determining hall height from perimeter and total painting cost via lateral surface area, and calculating the number of bricks that can be painted given paint coverage by surface area. It also compares lateral and total surface areas of a cube and a cuboid, requiring unit conversions and systematic application of formulas.

LanguageEnglish

CategoryExam

FormatPDF

Pages81

Reading Time3 h 24 min

AccessFree to download

Exercise 13.1 (Page 213)

Q1: Open-top plastic box sheet area and cost

Q2: Whitewashing walls and ceiling cost

Q3: Hall height from painting cost and perimeter

Q4: Number of bricks paintable from container coverage

Q5: Compare lateral and total surface areas of cube vs cuboid

Chapter-13: Surface Area and Volumes Exercise 13.1 (Page 213 of Grade 9 NCERT Textbook) Q1. 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 Rs. 20. Difficulty Level: Medium Reasoning: (i) The outer surface of a cuboid is made up of six rectangles called the faces whose areas can be found by multiplying length and breadth for each of them separately and then adding six areas together. Since the box is opened at the top, it has only 5 surfaces. (ii) The cost of the sheet to create the box will be equal to total surface area of the box multiplied by cost per meter squared of the sheet. Known: Area of the sheet required for making the box. Unknown: The length, breadth and depth of the plastic bag to be made. Solution: Length(l) = 1.5 m Breadth(b) = 1.25 m Height(h) = 65 cm = 0.65m [100 cm = 1 m] 􀀿 All the units should be the same The area of the open box = lb + 2(bh + hl) = (1.5􀁵1.25) + 2 [(1.25􀁵 .65) + (.65􀁵1.5)] =1.875 + 2(.8125 + .975) =1.875 + 2(1.7875) = 1.875 + 3.575 = 5.45 m2 Answer: The area of the sheet required for making the open box in = 5.45 m2 Unknown: Cost of the sheet if 1m 2 costs Rs. 20 | Solution:
1m2 costs = Rs. 20 | | | --- | --- | | 5.45m2 | costs = Rs. 20 􀁵 5.45
= Rs. 109 | | Answer:
Cost of the sheet is Rs. 109
Q2. 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 ceiling at the rate of Rs 7.50 per m 2 ?
Reasoning:
A cuboid is enclosed by six rectangular regions called the faces of the cuboid. Surface area of the cuboid of length l, breadth b and height h is 2(lb + bh + hl) The four walls and ceiling are to be white washed. So it is also having 5 faces only. For finding the cost of white washing, the surface area should be found and then should be multiplied by the cost for whitewashing per meter square.
Known:
The length, breadth and height of the cuboid.
Unknown:
Cost of white washing the ceiling and the walls Rs 7.50 per m2 .
Solution:
Surface area of 5 faces = lb + 2(bh + hl) length = 5m
breadth = 4m height = 3 m
Area of the 4 walls and ceiling = 5􀁵 4 + 2(5 + 4)3
= 20 + 54
= 74 m2
Cost of white washing the walls of the room and the room and the ceiling
= 74 􀁵 7.50
= Rs. 555
Answer:
Cost of white washing the walls of the room and the ceiling = Rs. 555 | | Q3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate ofRs. 10 per m2 is Rs. 15000, find the height of the hall.[Hint: Area of the four walls = Lateral surface area.] Difficulty Level: Medium Reasoning: Lateral surface area of the cuboid is only the area of 4 walls of the cuboid. Lateral surface area of cuboid = 2(l + b)h . So, the ratio between the total cost for painting and cost perm2 will give the total lateral surface area painted. Known: i. Perimeter of the hall which is 2(l + b) = 250 m ii. Cost of painting at the rate ofRs.10 per m2 is Rs. 15000. Unknown: Height of the hall. Solution: The ratio between the total cost for painting and cost per m2 will give the total lateral surface area painted. 15000 Area of four walls = = 1500 10 Perimeter = [2 (l + b) ] = 250 m 2(l + b)h =1500 (250 􀁵 h) = 1500 1500 h = 250 = 6 m Answer: The height of the hall is 6 m. Q4. 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? Reasoning: Brick is nothing but a cuboid having six faces. Surface area of the brick is the sum of the 6 faces. So, the total number of bricks that can be painted will be given by the ratio of total area of

FAQs

Why is the open-top box considered to have five surfaces?

Because opening at the top removes one face of the cuboid, leaving five rectangular faces whose areas are combined.

How is the cost of whitewashing computed in Q2?

Find the total surface area of the four walls and the ceiling (five faces), then multiply by the given cost per square meter.

How is the hall height determined in Q3?

Use the perimeter relation for the base (2(l+b)), convert the painting cost to lateral surface area via the rate, and solve 2(l+b)h for height.

Sep 18, 2026

Uploaded by Bintang