NCERT solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.1, includes the solved problems from the NCERT textbook. The solutions are available in PDF format and students can download effortlessly. The NCERT solutions are created by Maths subject experts, along with proper geometric figures and explanations in a step by step procedure for good understanding.

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### Download PDF of NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.1

### Access other exercise solutions of Class 9 Maths Chapter 13 – Surface Areas and Volumes

Exercise 13.2 solution (8 questions)

Exercise 13.3 solution (9 questions)

Exercise 13.4 solution (5 questions)

Exercise 13.5 solution (5 questions)

Exercise 13.6 solution (8 questions)

Exercise 13.7 solution (9 questions)

Exercise 13.8 solution (10 questions)

Exercise 13.9 solution (3 questions)

### Access Answers of Maths NCERT Class 9 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.1

**1. 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 sheet for it, if a sheet measuring 1m ^{2} costs Rs. 20. **

**Solution:**

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 top

Area of sheet required.

= 2lh+2bh+lb

= [2×1.5×0.65+2×1.25×0.65+1.5×1.25]m^{2}

= (1.95+1.625+1.875) m^{2} = 5.45 m^{2 }

(ii) Cost of sheet per m^{2} area = Rs.20.

Cost of sheet of 5.45 m^{2} area = Rs (5.45×20)

= Rs.109.

**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 ceiling at the rate of Rs 7.50 per m ^{2}. **

**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 white washed.

Total area to be white washed = Area of walls + Area of ceiling of room

= 2lh+2bh+lb

= [2×5×3+2×4×3+5×4]

= (30+24+20)

= 74

Area = 74 m^{2}

Also,

Cost of white wash per m^{2 }area = Rs.7.50 (Given)

Cost of white washing 74 m^{2 }area = Rs. (74×7.50)

= Rs. 555

**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 m ^{2} is Rs.15000, find the height of the hall. **

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

**Solution:**

Let 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 m^{2}

Cost of painting per square meter area = Rs.10

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

However, it is given that the cost of paining the walls is Rs. 15000.

15000 = 2500h

Or h = 6

Therefore, the height of the hall is 6 m.

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

**Solution:**

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

= [2(22.5×10+10×7.5+22.5×7.5)] cm^{2}

= 2(225+75+168.75) cm^{2 }

= (2×468.75) cm^{2 }

= 937.5 cm^{2}

Let n bricks can be painted out by the paint of the container

Area of n bricks = (n×937.5) cm^{2 }= 937.5n cm^{2 }

As per given instructions, area that can be painted by the paint of the container = 9.375 m^{2 }= 93750 cm^{2 }

So, we have, 93750 = 937.5n

n = 100

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

**5. 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? **

**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 the figures

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

= 4(10)^{2}

= 400 cm^{2} …(1)

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

= [2(12.5×8+10×8)]

= (2×180) = 360

Therefore, Lateral surface area of cuboidal box is 360 cm^{2}. …(2)

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

(Lateral surface area of cubical box – Lateral surface area of cuboidal box=400cm^{2}–360cm^{2} = 40 cm^{2})

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

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

Total surface area of cuboidal box

= 2[lh+bh+lb]

= [2(12.5×8+10×8+12.5×100)]

= 610

This implies, Total surface area of cuboidal box is 610 cm^{2}..(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 10cm^{2}.

Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm^{2}

**6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including 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 of tape is needed for all the 12 edges?**

**Solution:**

Length of greenhouse, say l = 30cm

Breadth of 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

Total surface area of the glass is 4250 cm^{2}

(ii)

From figure, 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 the 12 edges.

**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×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 cm ^{2}, find the cost of cardboard required for supplying 250 boxes of each kind.**

**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 cm^{2}

Extra area required for overlapping 1450×5/100 cm^{2}

= 72.5 cm^{2}

While considering all over laps, total surface area of bigger box

= (1450+72.5) cm^{2} = 1522.5 cm^{2}

Area of cardboard sheet required for 250 such bigger boxes

= (1522.5×250) cm^{2} = 380625 cm^{2}

**Smaller Box: **

Similarly, total surface area of smaller box = [2(15×12+15×5+12×5)] cm^{2}

= [2(180+75+60)] cm^{2}

= (2×315) cm^{2}

= 630 cm^{2}

Therefore, extra area required for overlapping 630×5/100 cm^{2} = 31.5 cm^{2}

Total surface area of 1 smaller box while considering all overlaps

= (630+31.5) cm^{2} = 661.5 cm^{2}

Area of cardboard sheet required for 250 smaller boxes = (250×661.5) cm^{2} = 165375 cm^{2}

**In Short:**

Box | Dimensions (in cm) | Total surface area (in cm^{2} ) |
Extra area required for overlapping (in cm^{2}) |
Total surface area for all overlaps (in cm^{ 2}) |
Area for 250 such boxes (in cm^{2}) |

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) cm^{2}

= 546000 cm^{2}

Given: Cost of 1000 cm^{2} cardboard sheet = Rs. 4

Therefore, Cost of 546000 cm^{2 }cardboard sheet =Rs. (546000×4)/1000 = Rs. 2184

Therefore, the cost of cardboard required for supplying 250 boxes of each kind will be Rs. 2184.

**8. Praveen 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 2.5m, with base dimensions 4m×3m? **

**Solution:**

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

Given:

l = 4m

b = 3m

h = 2.5m

Tarpaulin 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] m^{2}

= [2(10+7.5)+12]m^{2}

= 47m^{2}

Therefore, 47 m^{2} tarpaulin will be required.

Exercise 13.1 of class 9 maths consists of problems which cover the concepts like the surface area of cube and cuboid. It involves application level of real-time problems that pushes students to think and apply the relevant formula.

It also explains how six rectangular pieces are used to cover the complete outer surface of the cuboid and how the surface area of a cuboid and cube is found.

Learn the entire NCERT solutions for class 9 Maths Chapter 13 along with other learning materials, notes provided by BYJU’S. The problems are solved in a detailed way with relevant formulas and figures to score well in the CBSE exams.

### Key Features of NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volume Exercise 13.1

- These NCERT Solutions let you solve and revise all questions of Exercise 13.1.
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