NCERT Solutions Class 9 Maths Chapter 13 Surface area and Volume Exercise 13.1

Introduction:

In this exercise/article we will learn about Surface Area And Volume. Surface Area of a cuboid and cube, Surface area of a cube . Lateral surface area and total surface area of the cuboid . this exercise based only on cube and cuboid formula . you know that cube and cuboid shows 3D picture of square and rectangle. In such a case , the area of these four faces is called the lateral surface area.

Class 9 Maths Chapter 13 Surface Area And Volume :

Class 9 Maths Exercise 13.1 (Page-213)

Q1. 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 1 m² costs Rs 20.

Solution :

Let us draw to the figure, according to the question,

According to the given figure plot in "cuboid shape"

(i) Given, Length = 1.5 m

Breadth = 1.25 m

Height = 0.65 m [ converted cm to m ]

So, area of the sheet = ( Lateral surface area ) + ( Base area )

= [ 2 ( l + b ) h + ( lb ) ]

= 2 ( 1.5 + 1.25 ) × 0.65 + ( 1.5 × 1.25 )

= 2 ( 2.75) × 0.65 + ( 1.875 )

= 5.50× 0.65 + ( 1.875 )

= 3.575 m² + 1.875 m²

= 5.45 m²

(ii) The cost of sheet 1 m² = Rs 20

So, Cost of 5.45 m² sheet = Rs 20 × 5.45

= Rs 109

∴ Total cost of sheet = 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 the ceiling at the rate of Rs 7.50 per m².

Solution :

According to the question,

Given, Length = 5 m

Breadth = 4 m

Height = 3 m

So, Area of four walls ceiling

Total surface area of room - area of floor

2 ( lb + bh + hl ) - lb

= [ 2 ( 5 × 4 + 4 × 3 + 3 × 5 ) - 5 × 4 ]

= 2 ( 20 + 12 + 15 ) - 20

= 2 ( 47 ) - 20

= 94 - 20

= 74 m²

∴ Area of four walls ceiling = 74 m²

The cost of white washing the wall of the room 1 m² = Rs 7.50

Cost of white washing 74 m² = Rs 7.50 × 74

= Rs 555

∴ Total cost of white washing of wall = Rs 555.

Q3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four wall at the rate of Rs 10 per m² is Rs 15000, find the height of the hall.

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

Solution :

According to the question,

Perimeter of floor = 2 ( l + b )

Area of the four walls = Lateral surface area = 2 ( l + b ) h = 250 h

The cost of paints the four wall at the rate of 10 per m² = 250 h × Rs 10

= Rs 2500 h

Now, Rs 2500 h = Rs 15000

⇒ h = \(\displaystyle \frac{{Rs\,15000}}{{Rs\,2500}}\)

⇒ h =

⇒ h = 6

∴ The height of the hall = 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?

Solution :

According to the question,

Total area can be painted on container = 9.375 m²

Given, Length = 22.5 cm

Breadth = 10 cm

Height = 7.5 cm

So, Total surface area of brick = 2 ( lb + bh + hl )

2 ( 22.5 × 10 + 10 × 7.5 + 7.5 × 22.5 )

= 2 ( 225 + 75 + 168.75 )

= 450 + 150 + 337.5

= 937.5 cm²

= 0.09375 m² [ converted cm to m ]

Number of painted bricks =

= 100 m²

∴ The required number of bricks = 100 .

Q5. 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 smallest total surface area and the by how much ?

Solution :

According to the question,

(i) The edge of cubical box a = 10 cm

So, Lateral surface area of cube = 4a²

= 4 × ( 10 )²

= 4 × 100

= 400 cm²

Now, Total surface area of cube = 6a²

= 6 × ( 10 )²

= 6 × 100

= 600 cm²

(ii) The cuboidal box of dimension

Here, Length = 12.5 cm

Breadth = 10 cm

Height = 8 cm

So, Lateral surface area of cuboidal box = 2 ( l + b )h

= 2 ( 12.5 + 10 ) × 8

= 2 ( 22.5 ) × 8

= 45 × 8

= 610

Now, Total surface area of cuboidal box = 2 ( lb + bh + hl )

= 2 ( 12.5 × 10 + 10 × 8 + 8 × 12.5 )

= 2 ( 125 + 80 + 100 )

= 2 ( 305 )

= 360 cm²

(i) The lateral surface area of cubical box is more than cuboidal box 400 - 360 = 40 cm²

∴ The lateral surface area of grater than is cubical box = 40 cm²

(ii) Total surface area of cuboidal box is more than cubical box = 610 - 600 = 10 cm²

∴ Total surface area of smallest than is cube box = 10 cm².

Q6. 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.

(i) What is the area of the glass ?

(ii) How much of tapes is needed for all the 12 edges ?

Solution :

According to the question,

(i) Given, Length = 30 cm

Breadth = 25 cm

Height = 25 cm

So, Total surface area of herbarium = 2 ( lb + bh + hl )

= 2 ( 30 × 25 + 25 × 25 + 25 × 30 )

= 2 ( 750 + 625 + 750 )

= 2 ( 2125 )

= 4250 cm²

The required area of the glass = 4250 cm²

(ii) Length of the Total edges is 12

So, Length of tape for all edges = 4 ( l + b + h )

= 4 ( 30 + 25 + 25 )

= 4 ( 80 )

= 320 cm

∴ The required tape = 320 cm .

Q7. Shanti sweets stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The biggest of dimension 25 cm × 20 cm × 5 cm and the smaller of dimension 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is ₹4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.

Solution :

According to the question,

The biggest box,

Given, Length = 25 cm

Breadth = 20 cm

Height = 5 cm

So, Total surface area of a biggest box = 2 ( lb + bh + hl )

= 2 ( 25 × 20 + 20 × 5 + 5 × 25 )

= 2 ( 500 + 100 + 125 )

= 2 ( 725 )

= 1450 cm²

Extra area is required for overlapping 5% = 1450 × \(\displaystyle \frac{5}{{100}}\)

= 1450 ×

= \(\displaystyle \frac{{1450}}{{20}}\)

= 72.5 cm²

Total area of cardboard of biggest box = 1450 + 72.5 cm²

= 1522.5 cm²

The area of cardboard of 250 boxes = 25 × 1522.5 cm²

= 380625 cm²

∴ The cost of biggest box = 380625 cm²

The smallest box,

Given, Length = 15 cm

Breadth = 12 cm

Height = 5 cm

So, Total surface area of smallest box = 2 ( lb + bh + hl )

= 2 ( 15 × 12 + 12 × 5 + 5 × 15 )

= 2 ( 180 + 60 + 75 )

= 2 ( 315 )

= 630 cm²

Extra area is required for overlapping 5% = 630 × \(\displaystyle \frac{5}{{100}}\)

= 630 ×

= \(\displaystyle \frac{{630}}{{20}}\)

= 31.5 cm²

Total area of cardboard of smallest box = 630 + 31.5 cm²

= 661.5 cm²

The area of cardboard of 250 boxes = 250 × 661.5 cm²

= 165375 cm²

∴ The cost of smallest box = 165375 cm²

Now, The cardboard sheet required = 380625 + 165375 cm²

= 546000 cm²

The cost of the cardboard is Rs 4 for 1000 cm² = 546000 × \(\displaystyle \frac{4}{{1000}}\) cm²

= Rs 546000 ×

= \(\displaystyle \frac{{546000}}{{250}}\)

= Rs 2184

∴ Total cost of cardboard = Rs 2184 .

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 flag 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 2.5 m, with base dimensions 4 m × 3 m ?

Solution :

According to the question,

Given, Length = 4 m

Breadth = 3 m

Height = 2.5 m

So, tarpaulin will be required to make for top and four wall sides of the shelter.

The surface area of cuboid + including Base area

∴ Lateral surface area + Area of floor = 2 ( l + b ) h + lb

= 2 ( 4 + 3 ) 2.5 + 4 × 3

= 2 × 7 × 2.5 + 12

= 35 + 12

= 47 m²

∴ Tarpaulin would be required to make the shelter = 47 m² .

Class 9 Maths Chapter 13 Surface Area And Volume :

NCERT Solutions Class 9 Maths Chapter 13 Surface Area And Volume Exercise 13.1

NCERT Solutions Class 9 Maths Chapter 13 Surface area and Volume Exercise 13.1

Introduction:

Class 9 Maths Exercise 13.1 (Page-213)

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