 # NCERT Solutions for Class 9 Maths: Chapter 13 surface area and volume

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.
Question 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 1m2 costs Rs 20.
Solution:
Length of open plastic box =l=1.5 m
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  m2
Cost of one sheet of 1 m2 = Rs 20
Cost of 5.45 m2    =20 X 5.45=Rs 109
Question 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 Rs 7.50 per m2
Solution:
The room dimension are given below
Length (l) of room = 5 m
Breadth (b) of room = 4 m
Height (h) of room = 3 m
Imp point:
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
= [2 × 5 × 3 + 2 × 4 × 3 + 5 × 4] m2
= (30 + 24 + 20) m2
= 74 m2
Now Cost of white-washing per m2 area = Rs 7.50
So Cost of white-washing 74 m2 area = Rs (74 × 7.50)
= Rs 555
Question 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 m2 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 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 m2
Now Cost of painting 1 m2area = Rs 10
Cost of painting 250H m2 area = Rs (250H X 10) = Rs 2500H
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.

Question 4
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:
We know from question Paint in a container can paint area =9.375 m2
Converting into cm2 as all other dimension are given in that
=(9.375×100×100) cm2=93750 cm2
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 cm2
Now Number of bricks which can be painted=(Total area which can be painted)/(Surface area of a brick)
= 93750/937.5 =100
Question 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?
Solution:
(i)
Dimension of Cube
Edge of cube (a) = 10 cm
Dimension of Cuboidal
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)2
= 4(10 cm)2
= 400 cm2
Lateral surface area of cuboidal box  is given by  2[lh + bh]
= [2(12.5 × 8 + 10 × 8)] cm2
= (2 × 180) cm2
= 360 cm2
It is apparent from the data, 400 > 360
The difference =Lateral surface area of cubical box − Lateral surface area of cuboidal box = 400 cm2 − 360 cm2 = 40 cm2
So Lateral surface area of cubical box is greater than Lateral surface area of cuboidal box by 40 cm2
(ii) Total surface area of cubical box = 6(a)2 = 6(10 cm)2 = 600 cm2
Total surface area of cuboidal box
= 2[lh + bh + lb]
= [2(12.5 × 8 + 10 × 8 + 12.5 × 10] cm2
= 610 cm2
It is apparent from the data, 610 > 600
Difference=Total surface area of cuboidal box − Total surface area of cubical box = 610 cm2 − 600 cm2 = 10 cm2
So Lateral surface area of cuboidal box is greater than Lateral surface area of cubical box by 10 cm2
Question 6
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?
Solution:
We have
L=30 cm
B=25 cm
H=25 cm
Area of glass = 2(LB+BH+LH)= 4250 cm2
Length of tape=4(L+B+H)=320 cm
Question 7
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
Solution:
Bigger cardboard dimension
L=25 cm,B=20 cm,H=5 cm
Surface area of bigger cardboard box =2(LB+BH+LH)=2((25)(20)+(20)(5)+(5)(25))
=2(500+100+125)=2(725)=1450 cm2                      Surface Area of 250 bigger cardboard boxes =250×1450=362500 cm2           (1)
Smaller Cardboard dimensions
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 cm2                      Surface Area of 250 smaller cardboard boxes =250×630=157500 cm2         (2)
Total surface area
=surface area of 250 smaller and 250 bigger boxes
=362500+157500=520000 cm2
Also It is given that 5 % extra of total surface area is needed for the overlaps.
Therefore,
Total surface area needed for making boxes including overlaps
=520000+5% of  520000
=520000+26000=546000 cm2
Now lets us calculate the cost of  the  cardboard
Given 1000 cm2 of cardboard costs Rs 4
So 1 cm2 of cardboard would cost Rs   4/1000
Therefore
546000 cm2 of cardboard would cost Rs (4×546000)/1000= Rs 2184
Question 8
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?
Solution:
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] m2
= [2(10 + 7.5) + 12] m2
= 47 m2

Reference Books for class 9 Math

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

1. 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.
2. 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.
3. Pearson IIT Foundation Series - Maths - Class 9 Buy this book if you want to challenge yourself further and want to prepare for JEE foundation.
4. 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.

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