- Plane Figure
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- Solid Figure
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- Perimeter
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- Surface Area
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- Volume
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- Surface Area and Volume of Cube and Cuboid
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- Surface Area and Volume of Right circular cylinder
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- Surface Area and Volume of Right circular cone
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- Surface Area and Volume of sphere and hemisphere
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- How the Surface area and Volume are determined

- surfacearea_volume Formative Assignment
- |
- surfacearea_volume Assignments
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- surfacearea_volume Worksheet

In this page we have *NCERT Solutions for Class 9 Maths: Chapter 13 surface area and volume* for
EXERCISE 1 . Hope you like them and do not forget to like , social share
and comment at the end of the page.

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 1m

Length of open plastic box =

Breadth of open plastic box =b=1.25 m

Height of open plastic box =h=65 cm

Converting into meter

=65/100= 0.65 m

The area of the sheet required for making the open plastic box =lb+2bh+2hl

(Because it is open from the top)

=(1.5)(1.25)+2(1.25)(0.65)+2(0.65)(1.5)=1.875+1.625+1.95=5.45 m

Cost of one sheet of 1 m

Cost of 5.45 m

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 m

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 of walls + Area of ceiling of room

= 2lh + 2bh + lb

= [2 × 5 × 3 + 2 × 4 × 3 + 5 × 4] m

= (30 + 24 + 20) m

= 74 m

Now Cost of white-washing per m

So Cost of white-washing 74 m

= Rs 555

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 m2 is Rs 15000, find the height of the hall.

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

Let length, breadth and height of rectangular hall be L ,B and H respectively.

Area of four walls = 2LH + 2BH = 2(L + B) H

Perimeter of floor of hall = 2(L+ B ) = 250 m

So Area of four walls = 2(L + B) h = 250H m

Now Cost of painting 1 m

Cost of painting 250H m

Now

It is given that the cost of paining the walls is Rs 15000.

So

15000 = 2500H

H = 6

Thus, the height of hall is 6 m.

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?

We know from question Paint in a container can paint area =9.375 m

Converting into cm

=(9.375×100×100) cm

Now Dimensions of a brick is given as

22.5 cm × 10 cm × 7.5 cm

Surface Area of a brick can be calculated as( Brick is a cuboid)

=2(lb+bh+hl)=2((22.5)(10)+(10)(7.5)+(7.5)(22.5))

=2(225+75+168.75)=2(468.75)=937.5 cm

Now Number of bricks which can be painted=(Total area which can be painted)/(Surface area of a brick)

= 93750/937.5 =100

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?

(i)

Edge of cube (a) = 10 cm

Length (l) of box = 12.5 cm

Breadth (b) of box = 10 cm

Height (h) of box = 8 cm

Lateral surface area of cubical box is given by 4(a)

= 4(10 cm)

= 400 cm

Lateral surface area of cuboidal box is given by 2[lh + bh]

= [2(12.5 × 8 + 10 × 8)] cm

= (2 × 180) cm

= 360 cm

It is apparent from the data, 400 > 360

The difference =Lateral surface area of cubical box − Lateral surface area of cuboidal box = 400 cm

So Lateral surface area of cubical box is greater than Lateral surface area of cuboidal box by 40 cm

(ii) Total surface area of cubical box = 6(a)

Total surface area of cuboidal box

= 2[lh + bh + lb]

= [2(12.5 × 8 + 10 × 8 + 12.5 × 10] cm

= 610 cm

It is apparent from the data, 610 > 600

Difference=Total surface area of cuboidal box − Total surface area of cubical box = 610 cm

So Lateral surface area of cuboidal box is greater than Lateral surface area of cubical box by 10 cm

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 tape is needed for all the 12 edges?

We have

L=30 cm

B=25 cm

H=25 cm

Area of glass = 2(LB+BH+LH)= 4250 cm

Length of tape=4(L+B+H)=320 cm

Shanti Sweets Stall was placing an order for making cardboard boxes for packingtheir sweets. Two sizes of boxes were required. The bigger of dimensions

25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the verlaps, 5% of the total surface area is required extra. If the cost of the cardboard is s 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind

Surface area of bigger cardboard box =2(

=2(500+100+125)=2(725)=1450 cm

l=15 cm

b=12 cm

h=5 cm

Surface area of smaller cardboard box

=2(lb+bh+hl)=2((15)(12)+(12)(5)+(5)(15))

=2(180+60+75)=2(315)=630 cm

=surface area of 250 smaller and 250 bigger boxes

=362500+157500=520000 cm

Also It is given that 5 % extra of total surface area is needed for the overlaps.

Therefore,

=520000+5% of 520000

=520000+26000=546000 cm

Now lets us calculate the cost of the cardboard

Given 1000 cm

So 1 cm

Therefore

546000 cm

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

Length (l) of shelter = 4 m

Breadth (b) of shelter = 3 m

Height (h) of shelter = 2.5 m

Tarpaulin will be required for the top and four wall sides of the shelter.

Area of Tarpaulin required = 2(lh + bh) + l b

= [2(4 × 2.5 + 3 × 2.5) + 4 × 3] m

= [2(10 + 7.5) + 12] m

= 47 m

Download this assignment as pdf

Given below are the links of some of the reference books for class 9 Math.

- Mathematics for Class 9 by R D Sharma One of the best book for studying class 9 level mathematics. It has lot of problems to be solved.
- Secondary School Mathematics for Class 9 by R S Aggarwal This is also as good as R.D. Sharma. Either this or the book by R.D. Sharma will do. I find book R.S. Aggarwal little bit more challenging than the one by R.D. Sharma.
- Pearson IIT Foundation Series - Maths - Class 9 Buy this book if you want to challenge yourself further and want to prepare for JEE foundation.
- Pearson IIT Foundation Physics, Chemistry & Maths combo for Class 9 Only buy if you are prepared to study extra topics and want to take your studies a step further. You might need help to understand topics in these books.

You can use above books for extra knowledge and practicing different questions.

Class 9 Maths Class 9 Science

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