Surface Areas and Volumes

Class 9 NCERT Maths

NCERT

1   A plastic box $1.5 m$ long, $1.25 m$ wide and $65 cm$ 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.$\\$ (ii) The cost of sheet for it, if a sheet measuring $1 m^2$ costs Rs. $20.$

Solution :

It is given that, length (l) of box $= 1.5 m$$\\$ Breadth (b) of box $ = 1.25 m$$\\$ Depth (h) of box $= 0.65 m$$\\$ (i) Box is to be open at top.$\\$ Area of sheet required$\\$ $= 2lh + 2bh + lb$$\\$ $= [2 * 1.5 * 0.65 + 2 * 1.25 * 0.65 + 1.5 * 1.25] m^2$$\\$ $= (1.95 + 1.625 + 1.875) m^2 = 5.45 m^2$$\\$ (ii) Cost of sheet per $m^2$ area = Rs. $20$$\\$ Cost of sheet of $5.45 m^2$ area = Rs. $(5.45 * 20)$ = Rs.$ 109$

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 $m^2 .$

Solution :

It is given that$\\$ Length (l) of room $= 5 m$$\\$ Breadth (b) of room $= 4 m$$\\$ Height (h) of room $= 3 m$$\\$ It can be observed that four walls and the ceiling of the room are to be whitewashed. The floor of the room is not to be white-washed.$\\$ Area to be white-washed = Area of walls + Area of ceiling of room$\\$ $= 2lh + 2bh + lb$$\\$ $= [2 * 5 * 3 + 2 * 4 * 3 + 5 * 4] m^2$$\\$ $= (30 + 24 + 20) m^2$$\\$ $= 74 m^2$$\\$ Cost of white-washing per $m^2$ area = Rs. $7.50$$\\$ Cost of white-washing $74 m^2$ area = Rs. $(74 × 7.50)$ = Rs.$ 555$

3   The floor of a rectangular hall has a perimeter $250 m$. If the cost of panting 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.]

Solution :

Let length, breadth, and height of the rectangular hall be $l$ m, $b$ m, and $h$ m respectively.$\\$ Area of four walls $= 2lh + 2bh = 2(l + b) h$$\\$ Perimeter of the floor of hall $= 2(l + b) = 250 m$$\\$ $\therefore $ Area of four walls $= 2(l + b) h = 250h m^2$$\\$ Cost of painting per $m^2$ area = Rs. $10$$\\$ Cost of painting $250h m^2$ area = Rs. $(250h × 10)$ = Rs. $2500h$$\\$ However, it is given that the cost of paining the walls is Rs. $15000.$$\\$ $\therefore 15000 = 2500h$$\\$ $h = 6$$\\$ Therefore, the height of the hall is $6 m.$$\\$

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

Solution :

Total surface area of one brick $= 2(lb + bh + lh)$$\\$ $= [2(22.5 *10 + 10 * 7.5 + 22.5 * 7.5)] cm^2$$\\$ $= 2(225 + 75 + 168.75) cm^2$$\\$ $= (2 * 468.75) cm^2$$\\$ $= 937.5 cm^2$$\\$ Let $n$ bricks can be painted out by the paint of the container.$\\$ Area of $n$ bricks = $(n *937.5) cm^2 = 937.5n cm^2$$\\$ Area that can be painted by the paint of the container$ = 9.375 m^2 = 93750 cm^2$$\\$ $\therefore 93750 = 937.5n$$\\$ $n = 100$$\\$ Therefore, $100$ bricks can be painted out by the paint of the container.

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) Edge of cube $= 10 cm$$\\$ Length (l) of box $= 12.5 cm$$\\$ Breadth (b) of box $= 10 cm$$\\$ Height (h) of box $= 8 cm$$\\$ Lateral surface area of cubical box $= 4(\text{edge}) 2$$\\$ $= 4(10 cm)^2$$\\$ $= 400 cm^2$$\\$ Lateral surface area of cuboidal box $= 2[lh + bh]$$\\$ $= [2(12.5 * 8 + 10 * 8)] cm^2$$\\$ $= (2 * 180) cm^2$$\\$ $= 360 cm^2$$\\$ Clearly, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box.$\\$ Lateral surface area of cubical box - Lateral surface area of cuboidal box $\\$$= 400 cm^2 - 360 cm^2\\ = 40 cm^2 .$$\\$ Therefore, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box by $40 cm^2 .$$\\$ (ii) Total surface area of cubical box = $6(\text{edge}) 2 = 6(10 cm)^ 2 = 600 cm^2$$\\$ Total surface area of cuboidal box$\\$ $= 2[lh + bh + lb]$$\\$ $= [2(12.5 * 8 + 10 * 8 + 12.5 * 100] cm^2$$\\$ $= 610 cm^2$$\\$ Clearly, the total surface area of the cubical box is smaller than that of the cuboidal box.$\\$ Total surface area of cuboidal box - Total surface area of cubical box $\\$ $= 610 cm^2 -600 cm^2 = 10 cm^2$$\\$ Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by $10 cm^2 .$