NCERT Solutions for Class 9 Maths Chapter 13: Surface Areas and Volumes - Exercise 13.1

Exercise 13.1

1. A plastic box $1.5\text{ m}$ long, \[\mathbf{1}.\mathbf{25}\text{ }\mathbf{m}\] wide and \[\mathbf{65}\text{ }\mathbf{cm}\] 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.

Ans: Length of the box = $1.5m$

Breadth of the box = $1.25m$

Depth of the box = \[0.65m\]

Since the box is to be open at top, therefore the area of sheet required

$=2\text{lh}+2\text{bh}+\text{lb}$

$=2\times 1.5\times 0.65+2\times 1.25\times 0.65+1.5\times 1.25$

$=1.95+1.625+1.875$

$=5.45~{{\text{m}}^{2}}$

Hence, $5.45~{{\text{m}}^{2}}$ of sheet is required.

ii) The cost of sheet for it, if a sheet measuring $1{{m}^{2}}$ costs \[\mathbf{Rs}.\text{ }\mathbf{20}\].

Ans: Cost of sheet per ${{m}^{2}}$ area = \[Rs.20\]

Cost of sheet of $5.45{{m}^{2}}$ area \[=Rs.20\times 5.45\]\[=Rs.109\]

Hence, the cost of the sheet is \[Rs.109\].

2. The length, breadth and height of a room are $5m$, $4m$ and $3m$ respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs. $7.50$ per ${{m}^{2}}$.

Ans: Length of the room = $5m$

Breadth of the room = $4m$

Height of the room = $3m$

Since four walls and the ceiling of the room are to be whitewashed, hence the floor of the room will not be considered.

Therefore, area to be white-washed = Area of walls $+$ Area of ceiling of room

$=2\text{lh}+2\text{bh}+\text{lb}$
$=2\times 5\times 3+2\times 4\times 3+5\times 4$

$=30+24+20$

$=74~{{\text{m}}^{2}}$
Cost of white-washing per ${{m}^{2}}$ area = $Rs.7.50$

Cost of white-washing $74{{m}^{2}}$ area $=Rs.7.50\times 74=Rs.555$
Hence, the cost of white washing the walls of the room and the ceiling is $Rs.555$.

3. The floor of a rectangular hall has a perimeter $250\text{ m}$. If the cost of painting the four walls at the rate of Rs.$10$ per ${{m}^{2}}$ is Rs. $15000$, find the height of the hall.

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

Ans: Perimeter of the floor of hall\[=2\left( l+b \right)=250\text{ }m\]

Area of four walls\[=2lh+2bh=2\left( l+b \right)h=250h\text{ }{{m}^{2}}\]
Cost of painting per ${{m}^{2}}$ area \[=Rs.\text{ }10\] 

Cost of painting \[250h\text{ }{{m}^{2}}\] area \[=Rs.\text{ }\left( 250h\times 10 \right)=Rs.\text{ }2500h\]

Since, it is given that the cost of paining the walls \[=Rs.\text{ }15000\] 

Therefore, 

\[15000=2500h\]

\[\Rightarrow h=6\]

Hence, the height of the hall is $6m$.

4. The paint in a certain container is sufficient to paint an area equal to $9.375\text{ }{{m}^{2}}$. How many bricks of dimensions \[\mathbf{22}.\mathbf{5}\text{ }\mathbf{cm}\times \mathbf{10}\text{ }\mathbf{cm}\times \mathbf{7}.\mathbf{5}\text{ }\mathbf{cm}\] can be painted out of this container?

Ans: Total surface area of one brick\[=2\left( lb+bh+lh \right)\]

\[=2\left( 22.5\times 10+10\times 7.5+22.5\times 7.5 \right)\]

\[=2\left( 225+75+168.75 \right)\]

\[=2\times 468.75\]

\[=937.5\text{ }c{{m}^{2}}\]
Let $n$ bricks be painted out by the paint of the container.

Area of n bricks \[=n\times 937.5\text{ }c{{m}^{2}}\] 

Area that can be painted by the paint of the container\[=93750\text{ }c{{m}^{2}}\] 

Therefore, \[93750=n\times 937.5\]

$\Rightarrow n=100$

Hence, $100$ bricks can be painted out by the container.

5. A cubical box has each edge \[\mathbf{10}\text{ }\mathbf{cm}\] and another cuboidal box is \[\mathbf{12}.\mathbf{5}\text{ }\mathbf{cm}\] long, \[\mathbf{10}\text{ }\mathbf{cm}\] wide and \[\text{8 }\mathbf{cm}\] high.

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

Ans: Edge of the cube \[=10\text{ }cm\]

Length of the box $=12.5\text{ }cm$

Breadth of the box \[=10\text{ }cm\]

Height of the box \[=8\text{ }cm\]

Lateral surface area of cubical box \[=4{{a}^{2}}\]

\[=4{{a}^{2}}\]

\[=4{{(10)}^{2}}\]

\[=400\text{ c}{{\text{m}}^{2}}\]

Lateral surface area of cuboidal box\[=2\left[ lh+bh \right]\]

\[=2\left( 12.5\times 8+10\times 8 \right)\]

$=2\times 180$

\[=360\text{ }c{{m}^{2}}\]

Since $400>360$, therefore, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box.
The difference between the lateral surface area of both the boxes$=400-360=40\text{ c}{{\text{m}}^{2}}$
Hence, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box by $40\text{ c}{{\text{m}}^{2}}$.

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

Ans: Total surface area of cubical box \[=6{{a}^{2}}=6{{\left( 10 \right)}^{2}}=600\text{ }c{{m}^{2}}\]

Total surface area of cuboidal box\[=2\left[ lh+bh+lb \right]\]

\[=2(12.5\times 8+10\times 8+12.5\times 10)\]

\[=2(100+80+125)\]

\[=2\times 305\]

\[=610\text{ }c{{m}^{2}}\]
Since $600<610$, the total surface area of the cubical box is smaller than that of the cuboidal box.
The difference between the lateral surface area of both the boxes$=610-600=10\text{ c}{{\text{m}}^{2}}$
Hence, the lateral surface area of the cubical box is smaller than the lateral surface area of the cuboidal box by $10\text{ c}{{\text{m}}^{2}}$.

6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is \[\mathbf{30}\text{ }\mathbf{cm}\] long, \[\mathbf{25}\text{ }\mathbf{cm}\] wide and \[\mathbf{25}\text{ }\mathbf{cm}\] high.

i) What is the area of the glass?

Ans: Length of the greenhouse\[=30\text{ }cm\]

Breadth of the greenhouse \[=25\text{ }cm\]

Height of the greenhouse \[=25\text{ }cm\]

Total surface area of the greenhouse\[=2\left[ lh+bh+lb \right]\]

\[=2\left( 30\times 25+30\times 25+25\times 25 \right)\]

\[=2\left( 750+750+625 \right)\]

\[=2\times 2125\]

\[=4250\text{ }c{{m}^{2}}\]

Therefore, the area of the glass is \[4250\text{ }c{{m}^{2}}\].

ii) How much of tape is needed for all the $12$ edges?

Ans: Since the greenhouse has $4$edges of the length, $4$edges of the breadth and $4$edges of the height, therefore, total length of the tape \[=4\left( l+b+h \right)\]

\[=4\left( 30+25+25 \right)\] 

\[=4(80)\]

\[=320\text{ cm}\]

Hence, \[320\text{ cm}\] of tape is required for all the $12$ edges.

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 $25~\text{cm}\times 20~\text{cm}\times 5~\text{cm}$ and the smaller of dimensions $15~\text{cm}\times 12~\text{cm}\times 5~\text{cm}$. For all the overlaps, $5%$ of the total surface area is required extra. If the cost of the cardboard is Rs. $4$ for $1000c{{m}^{2}}$, find the cost of cardboard required for supplying $250$ boxes of each kind.

Ans: Length of the bigger box $=25\text{ cm}$

Breadth of the bigger box \[=20\text{ }cm\]

Height of the bigger box \[=5\text{ }cm\]

Total surface area of bigger box \[=2\left[ lh+bh+lb \right]\]

\[=2\left( 25\times 20+25\times 5+20\times 5 \right)\]

\[=2\left( 500+125+100 \right)\]

\[=2\left( 725 \right)\]

\[=1450\text{ }c{{m}^{2}}\]

Extra area required for overlapping $=5%$ of the total surface area

\[=\dfrac{5}{100}(1450)\]

\[=72.5\text{ c}{{\text{m}}^{2}}\]

Total surface area of the bigger box considering all overlaps

\[=1450+72.5\]

\[=1522.5\text{ }c{{m}^{2}}\]

Area of cardboard sheet required for $250$ such bigger boxes

\[=1522.5\times 250\]

\[=380625\text{ }c{{m}^{2}}\]
Now, 

Length of the smaller box $=15\text{ cm}$

Breadth of the smaller box \[=12\text{ }cm\]

Height of the smaller box \[=5\text{ }cm\]

Hence, total surface area of the smaller box\[=2\left[ lh+bh+lb \right]\] 

\[=2\left( 15\times 12+15\times 5+12\times 5 \right)\]

\[=2\left( 180+75+60 \right)\]

\[=2\left( 315 \right)\]

\[=630\text{ }c{{m}^{2}}\]

Therefore, extra area required for overlapping$=5%$ of the total surface area

\[=\dfrac{5}{100}(630)\]

\[=31.5\text{ c}{{\text{m}}^{2}}\]

Total surface area of the smaller box considering all overlaps

\[=630+31.5\]

\[=661.5\text{ }c{{m}^{2}}\]

Area of cardboard sheet required for $250$ such bigger boxes

\[=250\times 661.5\]

\[=165375\text{ }c{{m}^{2}}\]

Hence, total cardboard sheet required \[=380625+165375\]

\[=546000\text{ }c{{m}^{2}}\]

Cost of $1000\text{ c}{{\text{m}}^{2}}$ sheet of cardboard \[=Rs.\text{ }4\]

Cost of $546000\text{ c}{{\text{m}}^{2}}$ sheet of cardboard \[=Rs.\text{ 4}\times 546\]

\[=Rs.\text{ 2184}\]

Therefore, the cost of cardboard sheet required for $250$ such boxes of each kind is \[Rs.\text{ 2184}\].

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 \[\mathbf{2}.\mathbf{5}\text{ }\mathbf{m}\], with base dimensions $4~\text{m}\times 3~\text{m}$?

Ans: Length of the shelter = 4 m

Breadth of the shelter = 3 m

Height of the shelter = 2.5 m

Since the tarpaulin will be cover the top and four wall sides of the shelter, therefore, area of the tarpaulin \[=2\left( lh+bh \right)+lb\]

\[=2\left( 4\times 2.5+3\times 2.5 \right)+4\times 3\]

\[=2\left( 10+7.5 \right)+12\]

\[=2\left( 17.5 \right)+12\]

\[=35+12\]

\[=47\text{ }{{m}^{2}}\]

Hence, \[47\text{ }{{m}^{2}}\] tarpaulin will be required.

NCERT Solutions For Class 9 Maths Chapter 13 Surface Area And Volume

If you download NCERT Solutions and understand the concept, you can easily solve all the questions from any book. Students can download NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volume and understand the detailed concept about the Surface Area and Volume and how to apply the formulas. By solving the Ch 13 Maths Class 9 ex 13.1, students will:

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NCERT Solutions For Class 9  Maths Chapter 13 Exercise 13.1

Chapter 13 of Class 9th Mathematics is Surface Areas and Volume. In this chapter, students will understand how to calculate the surface area of objects and the volume occupied using the given data. In Ex 13.1 Class 9 Maths, students will also learn about the lateral surface area of a cube or curved surface area of a cylindrical pipe. Get NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.1 PDF download to understand the concept better.

Indirect questions about calculating the surface areas will be asked in the exams. So students should be well versed with the concepts. This will ensure that even if the question is twisted, they can solve them and focus on just the important information and not get confused with the twists and turns of the question. If students have NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volume, then they will not face any problems during the exam.

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Exercise 13.2 - 13.5:

In these exercises, you will learn how to evaluate the surface area of three-dimensional objects and the curved surface area of the cylinder objects.

Exercise 13.6 - 13.9:

In these exercises, students will learn about volumes of different geometrical objects and how to calculate the volume of a cone, cylinder, cuboid, sphere, etc. with the help of particular formulas.

CBSE Class 9 Maths Chapter 13 Other Exercises

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Chapter wise NCERT Solutions for Class 9 Maths

• Chapter 1 - Number Systems

• Chapter 2 - Polynomials

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introduction to Euclids Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelograms and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Heron's Formula

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability