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

Q1 (i) A plastic box $1.5m$ long, $1.25m$ wide and $65cm$ deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine: The area of the sheet required for making the box.

Answer:

Given, dimensions of the plastic box

Length, $l=1.5m$

Width, $b=1.25m$

Depth, $h=65cm=0.65m$

(i) The area of the sheet required for making the box (open at the top)= Lateral surface area of the box. + Area of the base.

= $2(bh+hl)+lb$

$$\begin{gathered} =2(1.25\times 0.65+0.65\times 1.5)+1.5\times 1.25 \\ =2(0.8125+0.975)+1.875=5.45 \end{gathered}$$

The required area of the sheet required for making the box is $5.45m^2$

Q1 (ii) A plastic box $1.5m$ long, $1.25m$ wide and $65cm$ deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine: The cost of sheet for it, if a sheet measuring $1m^2$ costs Rs 20.

Answer:

Given, dimensions of the plastic box

Length, $l=1.5m$

Width, $b=1.25m$

Depth, $h=65cm=0.65m$

We know, area of the sheet required for making the box is $5.45m^2$

(ii) Cost for $1m^2$ of sheet = Rs 20

$\therefore$ Cost for $5.45m^2$ of sheet = $5.45\times 20=109$

Required cost of the sheet is $Rs.109$

Q2 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 $Rs7.50$ per $m^2$ .

Answer:

Given,

Dimensions of the room = $5m\times 4m\times 3m$

Required area to be whitewashed = Area of the walls + Area of the ceiling

= $2(lh+bh)+lb$

$$\begin{gathered} =2(5\times 4+4\times 3)+5\times 4 \\ =2(32)+20=74m^2 \end{gathered}$$
Cost of white-washing per $m^2$ area = $Rs7.50$
Cost of white-washing $74m^2$ area = $Rs(74\times 7.50)$
$=Rs.555$

Therefore, the required cost of whitewashing the walls of the room and the ceiling is $Rs.555$

Q3 The floor of a rectangular hall has a perimeter $250m$ . 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.]

Answer:

Given,

The perimeter of rectangular hall = $250m$

Cost of painting the four walls at the rate of Rs 10 per $m^2$ = Rs 15000

Let the height of the wall be $hm$

$\therefore$ Area to be painted = $Perimeter\times height$

$=250h{m}^2$

$\therefore$ Required cost = $250h\times 10m^2=15000m^2$

$$\begin{gathered} ⟹2500h=15000 \\ ⟹h=\frac{150}{25}=6m \end{gathered}$$

Therefore, the height of the hall is $6m$

Q4 The paint in a certain container is sufficient to paint an area equal to $9.375m^2$ . How many bricks of dimensions $22.5cm\times 10cm\times 7.5cm$ can be painted out of this container?

Answer:

Given, dimensions of the brick = $22.5cm\times 10cm\times 7.5cm$

We know, Surface area of a cuboid = $2(lb+bh+hl)$

$\therefore$ The surface area of a single brick = $2(22.5\times 10+10\times 7.5+7.5\times 22.5)$

$=2(225+75+166.75)=937.5c{m}^2=0.09375m^2$

$\therefore$ Number of bricks that can be painted = $\frac{Totalareathecontainercanpaint}{Surfaceareaofasinglebrick}$

$=\frac{9.375}{0.09375}=100$

Therefore, the required number of bricks that can be painted = 100

Q5 (i) A cubical box has each edge $10cm$ and another cuboidal box is $12.5cm$ long, $10cm$ wide and $8cm$ high.

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

Answer:

Given,

Edge of the cubical box =

Dimensions of the cuboid =

(ii) The total surface area of the cubical box =

The total surface area of the cuboidal box =

Clearly, the total surface area of a cuboidal box is greater than the cubical box.

Difference between them =

Q5 (ii) A cubical box has each edge $10cm$ and another cuboidal box is $12.5cm$ long, $10cm$ wide and $8cm$ high.

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

Answer:

Given,

Edge of the cubical box =

Dimensions of the cuboid =

(ii) The total surface area of the cubical box =

The total surface area of the cuboidal box =

Clearly, the total surface area of a cuboidal box is greater than the cubical box.

Difference between them =

Q6 (i) A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is $30cm$ long, $25cm$ wide and $25cm$ high.What is the area of the glass?

Answer:

Given, dimensions of the greenhouse = $30cm\times 25cm\times 25cm$

Area of the glass = $2[lb+lh+bh]$
$$\begin{gathered} =[2(30\times 25+30\times 25+25\times 25)] \\ =[2(750+750+625)] \\ =(2\times 2125) \\ =4250c{m}^2 \end{gathered}$$
Therefore, the area of glass is $4250c{m}^2$

Q6 (ii) 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. How much of tape is needed for all the 12 edges?

Answer:

Given, dimensions of the greenhouse = $30cm\times 25cm\times 25cm$

(ii) Tape needed for all the 12 edges = Perimeter = $4(l+b+h)$

$4(30+25+25)=320cm$

Therefore, $320cm$ of tape is needed for the edges.

Q7 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 $25cm\times 20cm\times 5cm$ , and the smaller of dimensions $15cm\times 12cm\times 5cm$ . For all the overlaps, $5\mathrm{\%}$ 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.

Answer:

Given,

Dimensions of the bigger box = $25cm\times 20cm\times 5cm$ ,

Dimensions of smaller box = $15cm\times 12cm\times 5cm$

We know,

Total surface area of a cuboid = $2(lb+bh+hl)$

$\therefore$ Total surface area of the bigger box = $2(25\times 20+20\times 5+5\times 25)$

$=2(500+100+125)=1450c{m}^2$

$\therefore$ Area of the overlap for the bigger box = $5\mathrm{\%}of1450c{m}^2=\frac{5}{100}\times 1450=72.5c{m}^2$

Similarly,

Total surface area of the smaller box = $2(15\times 12+12\times 5+5\times 15)$

$=2(180+60+75)=630c{m}^2$

$\therefore$ Area of the overlap for the smaller box = $5\mathrm{\%}of630c{m}^2=\frac{5}{100}\times 630=31.5c{m}^2$

Since, 250 of each box is required,

$\therefore$ Total area of carboard required = $250[(1450+72.5)+(630+31.5)]=546000c{m}^2$

Cost of $1000c{m}^2$ of the cardboard = Rs 4

$\therefore$ Cost of $546000c{m}^2$ of the cardboard = $Rs.(\frac{4}{1000}\times 546000)=Rs.2184$

Therefore, the cost of the cardboard sheet required for 250 such boxes of each kind is $Rs.2184$

Q8 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 $2.5m$ , with base dimensions $4m\times 3m$ ?

Answer:

Given, Dimensions of the tarpaulin = $4m\times 3m\times 2.5m$

The required amount of tarpaulin = Lateral surface area of the shelter + Area of top

= $2(lh+bh)+lb$ Required

$$\begin{gathered} =2(4\times 2.5+3\times 2.5)+4\times 3 \\ =2(10+7.5)+12=47m^2 \end{gathered}$$

Therefore, $47m^2$ tarpaulin is required.