# NCERT Solutions for Exercise 13.1 Class 9 Maths Chapter 13 - Surface Area and Volumes

NCERT Solutions for Class 9 Maths exercise 13.1 deal with the concept of surface areas cuboid and cube. The cuboid is a three-dimensional object defined by six rectangular planes with varied magnitudes of length, breadth, and height. It has eight vertices and twelve edges, and opposite sides are always equal . In exercise 13.1 Class 9 Maths, The area of a cuboid is referred to as the surface area since the cuboid is a three dimensional solid. There are two types of surface area of the cuboid . They are

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• Total Surface Area

• Lateral Surface Area

The total surface area of the cuboid can be calculated by adding the areas of the 6 rectangular faces where the area is found by multiplying the length and breadth of each surface. TSA = 2(lb + bh + hl) . The lateral surface area of a cuboid can be calculated by adding the 4 planes of a rectangle, leaving the upper and the lower surface. LSA = 2h(l + b) . NCERT solutions for Class 9 Maths chapter 13 exercise 13.1 consists of 8 questions in which 6 of them are short and the remaining 2 might require some extra time to solve. The concepts related to the surface area are well explained in this Class 9 Maths chapter 13 exercise 13.1 . Along with Class 9 Maths chapter 13 exercise 13.1 the following exercises are also present in the NCERT book.

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• Surface Area and Volumes Exercise 13.2
• Surface Area and Volumes Exercise 13.3
• Surface Area and Volumes Exercise 13.4
• Surface Area and Volumes Exercise 13.5
• Surface Area and Volumes Exercise 13.6
• Surface Area and Volumes Exercise 13.7
• Surface Area and Volumes Exercise 13.8
• Surface Area and Volumes Exercise 13.9

## Surface Area and Volumes Class 9 Maths Chapter 13 Exercise: 13.1

Q1 (i) A plastic box long, wide and deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine: The area of the sheet required for making the box.

Given, dimensions of the plastic box

Length,

Width,

Depth,

(i) The area of the sheet required for making the box (open at the top)= Lateral surface area of the box. + Area of the base.

=

The required area of the sheet required for making the box is

Q1 (ii) A plastic box long, wide and deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine: The cost of sheet for it, if a sheet measuring costs Rs 20.

Given, dimensions of the plastic box

Length,

Width,

Depth,

We know, area of the sheet required for making the box is

(ii) Cost for of sheet = Rs 20

Cost for of sheet =

Required cost of the sheet is

Q2 The length, breadth and height of a room are , and respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of per .

Given,

Dimensions of the room =

Required area to be whitewashed = Area of the walls + Area of the ceiling

=

Cost of white-washing per area =
Cost of white-washing area =

Therefore, the required cost of whitewashing the walls of the room and the ceiling is

Q3 The floor of a rectangular hall has a perimeter . If the cost of painting the four walls at the rate of Rs 10 per is Rs 15000, find the height of the hall. [ Hint : Area of the four walls Lateral surface area.]

Given,

The perimeter of rectangular hall =

Cost of painting the four walls at the rate of Rs 10 per = Rs 15000

Let the height of the wall be

Area to be painted =

Required cost =

Therefore, the height of the hall is

Q4 The paint in a certain container is sufficient to paint an area equal to . How many bricks of dimensions can be painted out of this container?

Given, dimensions of the brick =

We know, Surface area of a cuboid =

The surface area of a single brick =

Number of bricks that can be painted =

Therefore, the required number of bricks that can be painted = 100

Q5 (i) A cubical box has each edge and another cuboidal box is long, wide and high.

Which box has the greater lateral surface area and by how much?

Given,

Edge of the cubical box =

Dimensions of the cuboid =

(ii) The total surface area of the cubical box =

The total surface area of the cuboidal box =

Clearly, the total surface area of a cuboidal box is greater than the cubical box.

Difference between them =

Q5 (ii) A cubical box has each edge and another cuboidal box is long, wide and high.

Which box has the smaller total surface area and by how much?

Given,

Edge of the cubical box =

Dimensions of the cuboid =

(ii) The total surface area of the cubical box =

The total surface area of the cuboidal box =

Clearly, the total surface area of a cuboidal box is greater than the cubical box.

Difference between them =

Q6 (i) A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is long, wide and high.What is the area of the glass?

Given, dimensions of the greenhouse =

Area of the glass =

Therefore, the area of glass is

Q6 (ii) 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. How much of tape is needed for all the 12 edges?

Given, dimensions of the greenhouse =

(ii) Tape needed for all the 12 edges = Perimeter =

Therefore, of tape is needed for the edges.

Q7 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 , and the smaller of dimensions . For all the overlaps, of the total surface area is required extra. If the cost of the cardboard is Rs 4 for , find the cost of cardboard required for supplying 250 boxes of each kind.

Given,

Dimensions of the bigger box = ,

Dimensions of smaller box =

We know,

Total surface area of a cuboid =

Total surface area of the bigger box =

Area of the overlap for the bigger box =

Similarly,

Total surface area of the smaller box =

Area of the overlap for the smaller box =

Since, 250 of each box is required,

Total area of carboard required =

Cost of of the cardboard = Rs 4

Cost of of the cardboard =

Therefore, the cost of the cardboard sheet required for 250 such boxes of each kind is

Q8 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 , with base dimensions ?

Given, Dimensions of the tarpaulin =

The required amount of tarpaulin = Lateral surface area of the shelter + Area of top

= Required

Therefore, tarpaulin is required.

## More About NCERT Solutions for Class 9 Maths Exercise 13.1 : Surface Area And Volumes

The NCERT solutions for Class 9 Maths exercise 13.1 also focused on the surface area of the cube . A three dimensional shape which has six faces, eight vertices and twelve edges is known as a cube in which the length, breadth, and height are equal . The cube's surface area is 6a2 , a is the length of one of its sides. Surface areas are measured in square units, but the volume of a cube is measured in cubic units. In NCERT solutions for Class 9 Maths Exercise 13.1 ,the formulas for computing surface areas for the cuboid and cube are thoroughly explored.

Also Read| Surface Areas And Volumes Class 9 Notes

## Benefits of NCERT Solutions for Class 9 Maths Exercise 13.1 :

• NCERT solutions for Class 9 Maths Exercise 13.1 enabled us to develop our basic concepts regarding the cuboid and its surface areas.

• By solving the NCERT solution for Class 9 Maths chapter 13 exercise 13.1 exercises, it helps us to score good marks in the first and second term examinations.

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• Exercise 13.1 Class 9 Maths, helps us to simplify complex forms by reducing them down into smaller, easier-to-understand individual objects.

Also See:

• NCERT Solutions for Class 9 Maths Chapter 13 – Surface Area and Volumes

• NCERT Exemplar Solutions Class 9 Maths Chapter 13 – Surface Area and Volumes

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