**NCERT Solutions Class 9 Maths Chapter 13 Surface area and volumes** – Here are all the NCERT solutions for Class 9 Maths Chapter 13. This solution contains questions, answers, images, explanations of the complete Chapter 13 titled Surface area and volumes of Maths taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Maths, then you must come across Chapter 13 Surface area and volumes. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 9 Maths Chapter 13 Surface area and volumes in one place.

## NCERT Solutions Class 9 Maths Chapter 13 Surface area and volumes

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Class | 9 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 13 |

Chapter Name |
Surface area and volumes |

### NCERT Solutions Class 9 Maths chapter 13 Surface area and volumes

Class 9, Maths chapter 13, Surface area and volumes solutions are given below in PDF format. You can view them online or download PDF file for future use.

### Surface area and volumes

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### Question & Answer

Q.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\( m^{2}\) costs ₹20.

Ans :It is given that, length (l) of box =1.5m Breadth (b) of box 1.25 m Depth (h) of box = 0.65 m (i) Box is to be open at top. Area of sheet required =2lh+2bh+lb =[2 x 1.5 x 0.65 +2 x 1.25 x 0.65 + 1.5 x 1.25] \( m^{2}\) =(1.95 + 1.625 + 1.875) m2 =5.45 \( m^{2}\) (ii) Cost of sheet per \( m^{2}\) area =Rs 20 Cost of sheet of 5.45 \( m^{2}\) area =Rs (5.45x20) =Rs 109

Q.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 ₹7.50 per \( m^{2}\) .

Ans :It is given that Length (l) of room = 5 m Breadth (b) of room = 4 m Height (h) of room = 3 m It can be observed that four walls and the ceiling of the room are to be white- washed. The floor of the room is not to be white-washed. Area to be white-washed = Area of walls + Area of ceiling of room = 2lh + 2bh + lb =[2x5x3+2x4x3+5x4] \( m^{2}\) (30 + 24 + 20) \( m^{2}\) = 74 \( m^{2}\) Cost of white-washing per \( m^{2}\) area = Rs 7.50 Cost of white-washlng 74 \( m^{2}\) area =Rs (74x7.50) = RS 555

Q.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 \( m^{2}\) is ₹ 15000, find the height of the hall. [Hint : Area of the four walls = Lateral surface area.]

Ans :Let length, breadth, and height of the rectangular hall be I m, b m, and h mrespectively. Area of four walls = 21h + 2bh =2( l + b )h Perimeter of the floor of hall =2(l + b) =250 m Area of four walls =2( l+b )h =250 \( m^{2}\) Cost of painting per \( m^{2}\) area = Rs 10 Cost of painting 250h \( m^{2}\) area = Rs (250h x 10) = Rs 2500h However, it is given that the cost of painting the walls is Rs 15000. 15000 = 2500h h=6 Therefore, the height of the hall is 6 m.

Q.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?

Ans :Total surface area of one brick = 2(lb + bh + /h) = [2(22.5 X10 + 10 x 7.5 + 22.5 x 7.5)] \( \mathrm{cm}^{2}\) = 2(225 + 75 + 168.75) \( \mathrm{cm}^{2}\) = (2 x 468.75) \( \mathrm{cm}^{2}\) = 937.5 \( \mathrm{cm}^{2}\) Let n bricks can be painted out by the paint of the container. Area of n bricks = (n x937.5) \( \mathrm{cm}^{2}\) = 937.5n \( \mathrm{cm}^{2}\) Area that can be painted by the paint of the container = 9.375 \( m^{2}\) = 93750 \(\mathrm{cm}^{2}\) = 93750 - 937.5n n = 100 Therefore, 100 bricks can be painted out by the paint of the container.

Q.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?

Ans :(i) Edge of cube = 10 cm Length (j) of box = 12.5 cm Breadth (b) of box = 10 cm Height (h) of box = 8 cm Lateral surface area of cubical box = \( 4(\text { edge })^{2}\) =400 \(\mathrm{cm}^{2}\) Lateral surface area of cuboidal box =2[lh+bh] = [2(12.5 x 10 x 8)] \(\mathrm{cm}^{2}\) = (2 x 180) \(\mathrm{cm}^{2}\) = 360 \(\mathrm{cm}^{2}\) Clearly, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box. Lateral surface area of cubical box - Lateral surface area of cuboidal box = 400 \(\mathrm{cm}^{2}\) - 360\(\mathrm{cm}^{2}\) = 40 \(\mathrm{cm}^{2}\) Therefore, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box by 40\(\mathrm{cm}^{2}\) (ii) Total surface area of cubical box =\( 6(\text { edge })^{2}\) =\( 6(10 \mathrm{cm})^{2}\)= 600 \(\mathrm{cm}^{2}\) Total surface area of cuboidal box = 2(lh + bh + lb] =[2(12.5 x 8 + 10 x 8+ 12.5 x 100]\(\mathrm{cm}^{2}\) 610 \(\mathrm{cm}^{2}\) Clearly, the total surface area of the cubical box is smaller than that of the cuboidal box. Total surface area of cuboidal box - Total surface area of cubical box = 610 \(\mathrm{cm}^{2}\) 600\(\mathrm{cm}^{2}\) = 10 \(\mathrm{cm}^{2}\) Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 \(\mathrm{cm}^{2}\).

## NCERT / CBSE Book for Class 9 Maths

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