NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1.
- Surface Areas and Volumes Class 9 Ex 13.1
- Surface Areas and Volumes Class 9 Ex 13.2
- Surface Areas and Volumes Class 9 Ex 13.3
- Surface Areas and Volumes Class 9 Ex 13.4
- Surface Areas and Volumes Class 9 Ex 13.5
- Surface Areas and Volumes Class 9 Ex 13.6
- Surface Areas and Volumes Class 9 Ex 13.7
- Surface Areas and Volumes Class 9 Ex 13.8
- Surface Areas and Volumes Class 9 Ex 13.9
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 9 |
| Subject | Maths |
| Chapter | Chapter 13 |
| Chapter Name | Surface Areas and Volumes |
| Exercise | Ex 13.1 |
| Number of Questions Solved | 8 |
| Category | NCERT Solutions |
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1
Ex 13.1 Class 9 Maths 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 1 m2 costs ?
Solution.
We have, a plastic box whose length (l) = 1.5 m, width (b) = 1.25 m
and depth (b) = 65 cm = \(\cfrac { 65 }{ 100 } \) = 0.65 m
[ ∵ 1 cm = \(\cfrac { 65 }{ 100 } \)
Surface area of the box = 2(lb + bh + hi)
= 2 (1.5 x 1.25 +1.25 x 0.65 + 0.65 x 1.5)
= 2(1.875 + 0.8125 + 0.975)
= 2 (3.6625) = 7.325 m2
(i) Area of sheet required for making the box
= Surface area of the box – Area of opened the top
= 7.325 -l x b [since, box is opened at the top]
= 7.325 – (1.5 x 1.25)
= 7.325 – 1.875= 5.45m2
(ii) Given, cost of 1 m2 sheet = Rs. 20
∴ Cost of 5.45 m2 sheet = 20 x 5.45 = Rs. 109
Ex 13.1 Class 9 Maths 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.
Given, dimensions of room are l = 5 m, b = 4m and h = 3 m
Required area for white washing = Area of the four walls + Area of ceiling = 2 (l + b) x h + (l x b)
= 2 (5 + 4) x 3 + (5 x 4) = 2 x 9 x 3 + 20 = 54 + 20 = 74 m2
Given, cost of white washing 1 m2 wall = RS. 7.50
Cost of white washing of 74 m2 wall = 7.50 x 74 = RS. 555
Ex 13.1 Class 9 Maths 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 ? 10 per m2 is Rs. 15000, find the height of the hall. [Hint: Area of the four walls = Lateral surface area]
Solution.
Let l, b and h be the length, breadth and height of rectangular hall respectively.
Ex 13.1 Class 9 Maths 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 x 10 cm x 7.5 cm can be painted out of this container?
Solution.
Given, dimensions of a brick are l = 22.5 cm, b = 10 cm and h = 7.5 cm
and total surface area of container = 9.375 m2
Total surface area of brick = 2 (l x b + b x h + h x l)
= 2(22.5 x 10 +10 x 7.5 + 7.5 x 22.5)
= 2(225 + 75 + 168.75)
= 2 x 468.75 = 937.5 cm2
Ex 13.1 Class 9 Maths 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.
Given, length of the cubical box (l) = 10 cm
For cuboidal box, length (Z) = 12.5 cm, breadth (b) = 10 cm and height (h) = 8 cm
(i) Lateral surface area of cubical box
= 4l2 =4 (10)2 = 4 x 100 = 400 cm2
Lateral surface area of cuboidal box
= 2(l + b)x h = 2 (12.5 +10) x 8
= 2 (22.5) x 8 = 45 x 8 = 360 cm2
(Lateral surface area of cubical box) >(Lateral surface area of cuboidal box) [∵ 400 > 360]
∴ Required difference = 400 -360 = 40 cm2
Hence, cubical box has 40 cm2 more lateral surface area.
(ii) Total surface area of cubical box
= 6l2 = 6(10)2 = 6 x 100 = 600 cm2
Total surface area of cuboidal box
= 2(l x b + b x h + h x l)
= 2 (12.5 x 10+10 x 8+8 x 12.5)
= 2 (125 + 80 +100) = 2 x 305 = 610 cm2
(Total surface area of cuboidal box) > (Total surface area of cubical box) [∵ 610 >600]
∴ Required difference = 610 – 600 = 10 cm2
Hence, cubical box has 10 cm2 less total surface area.
Ex 13.1 Class 9 Maths 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.
Dimensions of herbarium are Z = 30 cm, b = 25 cm and h = 25 cm
(i) Area of the glass = Total surface area of herbarium
= 2 (Z x b + b x h + h x l)
= 2 (30 x 25 + 25 x 25 + 25 x 30)
= 2(750 + 625 + 750) = 2 (2125) = 4250 cm2
(ii) Herbarium is a shape of cuboid.
So, length of required tape will be equal to total length of edges of cuboid.
∴ Length of the tape = 4 (l + b + h) = 4(30 + 25 + 25)
= 4 x 80 = 320 cm
Ex 13.1 Class 9 Maths Question 7.
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 25 cm x 20 cm x 5 cm and the smaller of dimensions 15 cm x 12 cm x 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs. 4 for 1000 cm2, then find the cost of cardboard required for supplying 250 boxes of each kind.
Solution.
Dimensions of bigger box are l = 25 cm, b = 20 cm and h = 5 cm
∴ Total surface area of the bigger box = 2 (l x b + b x h + h x l)
= 2 (25 x 20 + 20 x 5 + 5 x 25)
= 2 (500 +100 +125) = 2(725) = 1450 cm2
Dimensions of smaller box are 1 = 15 cm, b = 12 cm and h = 5 cm
∴ Total surface area of the smaller box
= 2(15 x 12+12 x 5 + 5 x 15)
= 2(180 + 60 + 75) = 2 (315) = 630 cm2
Total surface area of both boxes = 1450 + 630 = 2080 cm2
Area for all the overlaps = 5 % of total surface area
= 5% x 2080 = \(\cfrac { 5 }{ 100 } \) x 2080 = 104 cm2
Total surface area of both boxes with area of overlaps = 2080 +104 = 2184 cm2
∴ Total surface area for such 250 boxes = 2184 x 250 cm2
Now, cost of the cardboard for 1000 cm2 = Rs. 4
Then, cost of the cardboard for 1 cm 2 = Rs. \(\cfrac { 4 }{ 1000 } \)
∴ Cost of the cardboard for (2184 x 250) cm2
= \(\cfrac { 4\times 2184\times 250 }{ 1000 } \) = Rs.2184
Hence, required cost of cardboard is Rs. 2184.
Ex 13.1 Class 9 Maths 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 he 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 x 3 m?
Solution.
Dimensions of shelter are l = 4 m, b = 3 m and h = 2.5 m Required area of tarpaulin to make the shelter
= Area of 4 sides + Area of the top of the car = 2(l + b) x h + (l x b)
= 2 (4 + 3) x 2.5 + (4 x 3)
= (2 x 7 x 2.5)+12 = 35+ 12 = 47m2
We hope the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1, drop a comment below and we will get back to you at the earliest.