# Surface Areas and Volumes - NCERT Class 9 Maths

## Question 1

Plastic box \(1.5 \, \rm{m}\) long, \(1.25 \, \rm{m}\) wide and \(65 \, \rm{cm}\) 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 \(\begin{align}1\, \rm{m^2} \end{align}\) costs \(\rm{Rs}. 20.\)

### Solution

**Video Solution**

**What is known?**

The length, breadth and depth of the plastic bag to be made opened at the top. Cost of \(1m^{2}\)^{ }sheet is ₹\( 20.\)

**What is unknown?**

Area of the sheet required for making the box and its cost.

**Reasoning:**

Since box is opened at the top, it has only \(5\) surfaces, including the \(4\) walls and the base. Area of the sheet required for making the cuboidal box includes the \(4\) walls of the box and the base.

Hence, area of the sheet can be obtained by adding area of the base to the lateral surface area of the cuboidal box.

Lateral surface area of cuboid \(=2(l+b)h\)

**Steps:**

\(\begin{align} \text{Length (l)} &= 1.5\, \rm{m} \\\text{Breadth(b)} &= 1.25\, \rm{m}\\\text{Height(h)} &= 65\, \rm{cm}\\&=\frac{{65}}{{100}}\, \rm{m} \\&= 0.65 \,\rm{m} \qquad \text{ }\end{align}\)

The area of the sheet required to make the box

\[\begin{align}&= \,\,lb + 2(l + b)h\\&= (1.5m \times 1.25)m + \\&\;\;\;\;\;2\times(1.25 m+ 1.5m)\!\times\!0.65m \\&=1.875m^2\!+\!2\!\times\!2.75m\!\times\!0.65m\\&=1.875m^2 + 3.575m^2\\&= 5.45 m^2\,\rm\end{align}\]

Therefore,

The cost of the sheet \(=\) Rate of the sheet \(\times\) Area of the sheet

\(\begin{align}&= \,₹20/m^2\times5.45m^2\\&= ₹109 \,\rm\end{align}\)

**Answer:**

The area of the sheet required for making the open box is \(5.45 m^2\) and

Cost of the sheet is ₹ \(109\)

## Question 2

The length, breadth and height of a room are \(5 \,\rm{m}, 4 \,\rm{m},\) and \(3\,\rm{ m}\) respectively. Find the cost of white washing the walls of the room and ceiling at the rate of \(\begin{align} \text{Rs 7.50 per }\rm{m^2}? \end{align}\)

### Solution

**Video Solution**

**What is known:**

Medium

**What is unknown:**

The lenght, breadth and height of a room are \(5m\), \(4m\), and \(3m\) respectively.

**Reasoning:**

Since the four walls and ceiling are to be whitewashed. So, it has \(5\) faces only, excluding the base.

Hence, area of the room to be whitewashed can be obtained by adding area of the ceiling to the lateral surface area of the cuboidal room.

Lateral surface area of cuboid \(=2(l+b)h\)

The cost of white washing the walls of the room and ceiling will be equal to area of the room to be whitewashed multiplied by rate of the whitewashing.

**Steps:**

\[\begin{align} \\\text{length} &= 5\,\rm{m} \\\text{breadth} &= \rm{4\,m} \\\text{height} &= \rm{3 \,\,m} \end{align}\]

Surface are of \(5\) faces \(=\) Area of the \(4\) walls and ceiling \(=\) \(lb+2(l+b)h\)

\[\begin{align}&=(5m\!\times\!4m)\!+\!2\!\times\!(5m\!+\!4m)\!\times\!3m\\ &= 20m^2+2\times9m\times3m \\&=20m^2+54m^2 \\&=74m^2\end{align}\]

The cost of white washing the walls of the room and ceiling = \(Rate\times area\)

\[\begin{align}&=₹ 7.50/m^2\times 74m^2\\ &= ₹\,555\end{align}\]

**Answer:**

Cost of white washing the walls of the room and the ceiling \(= Rs. 555\)

## Question 3

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

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

### Solution

**Video Solution**

**What is the known?**

Perimeter of the hall which is

\(\begin{align}2(l + b) = 250\,\rm{m}\end{align}\) and the cost of painting at the rate of \(\begin{align} \text{Rs.10 per}\,\rm {m^2} \end{align}\) is \(\rm{Rs}. 15000.\)

**What is the unknown?**

Height of the hall.

**Reasoning:**

Lateral surface area of the cuboid is only the area of \(4\) walls of the cuboid. Lateral surface area of cuboid \(\begin{align} = 2(l + b)h \end{align}\). So, the ratio between the total cost for painting and cost per \(\rm{m^2}\) will give the total lateral surface area painted.

**Steps:**

The ratio between the total cost for painting and cost per \(\rm{m^2}\) will give the total lateral surface area painted.

Area of four walls

\[\begin{align}&=\frac{{15000}}{{10}}\\&= 1500 \end{align}\]

Perimeter \(= [2\,\,(l + b)] = 250\,\rm{m}\)

\[\begin{align}2(l + b)h &= 1500 \\(250 \times h) &= 1500\\h &= \frac{{1500}}{{250}}\\&= 6\,\rm{m} \end{align}\]

The height of the hall is \(6\,\rm {m.}\)

## Question 4

The paint in a certain container is sufficient to paint an area equal to \(\begin{align} 9.375\,\rm{m^2} \end{align}\). How many

bricks of dimensions

\(\begin{align} 22.5 \rm{cm} \times 10 \rm{cm} \times 7.5 \rm{cm} \end{align}\) can be painted out of this container?

### Solution

**Video Solution**

**Reasoning:**

Brick is nothing but a cuboid having six faces. Surface area of the brick is the sum of the \(6\) faces. So, the total number of bricks that can be painted will be given by the ratio of total area of container divided by the surface area of per brick.

**What is the known?**

Dimensions of brick. The area can be painted with the paint.

**What is the unknown?**

Number of bricks can be painted.

**Steps:**

For a brick:

\[\begin{align} &\text {l = 22.5 cm}\\&\text{b = 10 cm}\\&\text{h = 7.5 cm}\end{align}\]

The total surface area of the brick

\[\begin{align}&={2(l b+b h+h l)} \\ &={2(22.5\!\times\!10\!+\!10\!\times\! 7.5 \!\times\! 7.5 \!\times\! 22.5)} \\ &={2(225+75+168.75)} \\ &=2(468.75) \\&= 937.5\,\mathrm{cm}^{2}\end{align}\]

Since the area given in \(\begin{align} \rm{m^2. }\end{align}\) So, the area of the brick has to be changed to \(\begin{align} \rm{m^2. }\end{align}\)

Surface area of the brick

\[\begin{align}&= 937.5 \rm{cm^2}\\&=\frac{{937.5}}{{100 \times 100}}\rm{m^2}\\&= 0.09375\,\rm{m^2} \end{align}\]

Number of bricks can be painted

\[\begin{align}&= \frac{{9.375}}{{0.09375}}\\ &= 100 \end{align}\]

Number of bricks that can be painted from the paint in the container \(= 100.\)

## Question 5

A cubical box has each edge \(10\,\rm{ cm}\) and another cuboidal box is \(12.5\,\rm{ cm}\) long, \(10\,\rm{ cm}\) wide and \(8\,\rm{ cm}\) 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

**Video Solution**

**Reasoning:**

A cube is cuboid whose length, breadth and height and equal. A cuboid has six faces and the total surface area is the sum of the surface area of the \(6\) faces.

**What is the known?**

(i) The length of the cube.

(ii) The length, breadth height of the cuboid.

**What is the unknown?**

Greater lateral surface area and by how much?

**Steps:**

Lateral surface area of the cube and cuboid is the sum of the area of the four faces.

**For cube:**

Edge of the cube is \(10\,\rm{ cm}.\)

Lateral surface area of the cube

\[\begin{align}&=4\,\rm{(edge)^2}\\&= 4 \times {10^2} \\&= 400\,\rm{cm^2} \end{align}\]

**For cuboid:**

\[\begin{align}\text{length(l)} &= 12.5\,\rm{cm}\\\text{breadth(b)} &= 10\,\rm{cm} \\\text{height(h)} &= 8\,\rm{cm} \end{align}\]

Lateral surface area of the cuboid

\(\begin{align} = 2\,(l + b)\,\,h \end{align}\)

Lateral surface area of the cuboid

\[\begin{align}&= 2\,\,(12.5 + 10) \times 8\\& = 2(12.5 + 10) \times 8\\& = 2 \times 22.5 \times 8\\ &= 16 \times 22.5\\ &= 360\,\rm{cm^2} \end{align}\]

Cubical box has the greater lateral surface area than the cuboidal box by

\(\begin{align}\,40\,\rm{cm^2}. \;(400 - 360 = 40\,\rm{cm^2}). \end{align}\)

(ii) Smaller total surface area

**What is the known?**

(i) The length of the cube.

(ii) The length, breadth height of the cuboid.

**What is the unknown?**

Smaller total surface area and by how much?

**Steps:**

Total surface area of the cube

\[\begin{align}&= 6\,\rm{(edge)^2}\\&= 6 \times {10^2}\\&= 600\,\rm{cm^2} \end{align}\]

Total surface area of the cuboid

\[\begin{align}= 2(lb + bh + hl) \end{align}\]

\(\begin{align}&\text{length(l) = 12.5}\,\rm{cm}\\&\text{breadth(b) = 10}\,\rm{cm} \\&\text{height(h) = 8}\,\rm{cm} \end{align}\)

Total surface area:

\[\begin{align}&=\!2[(12.5\!\times\!10\!+\!10\!\times\!8\!+\!8\!\times\!12.5)]\\&= 2\,\,[125 + 80 + 100]\\ &= 610\,\rm{cm^2} \end{align}\]

Cubical box has the smaller total surface area than the cuboidal box by

\(\begin{align}(610 - 600) = 10\,\rm{cm^2} \end{align}\)

## Question 6

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

(i) What is the area of the glass?

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

### Solution

**Video Solution**

**Reasoning:**

A cuboid is enclosed by six rectangle regions called faces and it has \(12\) edges. The total surface area is the sum of the areas of the six faces which is nothing but the area of the glass.

(i) Area of the glass?

**What is the known?**

Measurements of herbarium.

**What is the unknown?**

The area of the glass.

**Steps:**

The area of the herbarium is the total surface area of the cuboid.

\[\begin{align}&\text{length(l) = 30}\,\rm{cm}\\&\text{breadth = 25}\,\rm{cm}\\&\text{height = 25}\,\rm{cm}\end{align}\]

Total surface area of the cuboid

\[\begin{align}&= 2(lb + bh + hl)\\ &= 2[25\!\times\!30\!+\!25 \!\times\! 25\!+\!30\!\times\!25]\\ &= 2[750 + 625 + 750]\\ &= 4250\,\rm{cm^2} \end{align}\]

Area of the glass \(\begin{align} = 4250\,\rm{cm^2}. \end{align}\)

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

**What is the unknown?**

Tape needed for all the \(12\) edge.

**Steps:**

Total length of the tape needed for \(12\) edges

\[\begin{align} & = 4(l + b + h)\\ &= 4(30 + 25 + 25)\\ &= 320\,\rm{cm}\end{align}\]

Tape required to cover all the \(12\) edges is \(320\,\rm{ cm}.\)

## 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

\(\begin{align}25 \,\rm{cm} \times 20 \,\rm{cm} \times 5 \,\rm{cm}\end{align}\) and the smaller of dimensions \(\begin{align}15 \,\rm{cm} \times 12\,\rm{cm} \times 5 \,\rm{cm}.\end{align}\) For all the overlaps, **\(5 \%\)** of the total surface area is required extra. If the cost of the cardboard is \( \rm{Rs.}\, 4\) for \(\begin{align}1000\,\rm{cm^2} \end{align}\), find the cost of cardboard required for supplying \(250\) boxes of each kind.

### Solution

**Video Solution**

**Reasoning:**

A cuboid has six faces and the total surface area is sum of the area of the six faces. So, the cost for supplying \(250\) boxes of each kind will be the summation of surface area of boxes multiplied by cost per cm square.

**What is the known?**

(i) Dimensions of the smaller and bigger boxes.

(ii) Cost of the card board.

**What is the unknown?**

Cost of the card board required for \(250\) boxes of each kind.

**Steps:**

For bigger box:

\[\begin{align}&\text{length (l)} = 25\,\rm{cm}\\&\text{breadth (b)} = 20\,\rm{cm}\\&\text{height (h) }= 5\,\rm{cm} \end{align}\]

Total surface area

\[\begin{align}&= 2\,\,(lb + bh + hl)\\&= 2\,\,[25 \times 20 + 25 \times 20 + 5 \times 25]\\ &= 2\,\,[500 + 100 + 125]\\ &= 1450\,\rm{cm^2} \end{align}\]

Card board required for all the overlaps is **\(5\%\)** of their total surface area.

\[\begin{align}∴ \frac{5}{{100}} \times 1450 = 72.5\,\rm{cm} \end{align}\]

Net cardboard required for bigger box

\[\begin{align}&= 1450 + 72.5\\&=1522.5\,\rm{cm^2} \end{align}\]

Card board required for \(250\) such boxes

\[\begin{align}&= 1522.5 \times 250 \\&= 380625\,\rm{cm^2} \end{align}\]

Cost for 1000 \(\rm{m^2} = \,\,Rs.\,4\)

∴ Cost for \(380625\) \(\rm{cm^2}\)

\[\begin{align} &= \frac{4}{{1000}} \times \,\,380625\\&= \rm{Rs.}\,1522.5 \end{align}\]

For smaller box:

\[\begin{align}&\text{length (l) }= 15\,\rm{cm} \\&\text{breadth (b)} = 12\,\rm{cm}\\&\text{height (h)} = 5\,\rm{cm} \end{align}\]

Total surface area

\[\begin{align}&= 2(lb + bh + hl)\\&= 2[50 \times 12 + 12 \times 5 + 5 \times 15]\\&= 2[180 + 60 + 75]\\&= 630\,\rm{cm^2} \end{align}\]

Card board required for all the overlaps is **\(5\%\)** of their total surface area.

\[\begin{align}∴ \frac{5}{{100}} \times 630 = 31.5\,\,c{m^2} \end{align}\]

Net surface area of the smaller box

\[\begin{align} &= 630 + 31.5\\ &= 661.5\,\rm{cm^2} \end{align}\]

Card board required for \(250\) such boxes

\[\begin{align} &= 661.5 \times 250 \\ &= 165375\,\rm{cm^2} \end{align}\]

Cost of the card board is **\(\rm{Rs}. 4\)** for \(\begin{align}1000\,\rm{cm^2} \end{align}\)

For \(165375\) \(\rm{cm^2}\) cost is

\[\begin{align}&= \frac{4}{{1000}} \times 16537\\&= \rm{Rs.}\,661.50 \end{align}\]

Cost of card board required for supplying **\(250\)** boxes of each kind

\[\begin{align}&= 1522.50 + 661.50\\&= Rs. 2184 \end{align}\]

## 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 \(\begin{align}2.5 \rm{m} \end{align}\), with base dimensions \(\begin{align}4\,\rm{m} \times 3\,\rm{m}?\end{align}\)

### Solution

**Video Solution**

**Reasoning:**

The surface area of the cuboid is the sum of the surface area of the faces. Since Parveen wanted to make a cuboid with **\(5\)** outer faces covered with tarpaulin that covers all the four side and the top.

**What is the known?**

Measurements of the shelter needed is given.

**What is the unknown?**

Area of the tarpaulin required to make the shelter.

**Steps:**

\[\begin{align}\text{length(l)} &= 4\rm{m} \\\text{breadth(b)} &= 3\rm{m}\\\text{height(h) }&= 2.5\rm{m} \end{align}\]

Total surface area with five faces

\[\begin{align} &= lb + 2(bh + hl) \\& = 4 \times 3 + 2(3 \times 2.5 + 2.5 \times 4)\\ &= 12 + 2[7.5{\rm{ + }}10]\\ &= 47\,\rm{cm^2} \end{align}\]

Hence\(\begin{align}\,47\,\rm{cm^2} \end{align}\) of tarpaulin will be required.

The chapter 13 begins with surface areas of the cube, cuboid, right circular cylinder, right circular cone and sphere followed by their curved surface area and total surface area. The next unit focuses on volumes of cuboid, cube, right circular cylinder, right circular cone and sphere. Many examples are shown for better understanding.

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