#### Class nine NCERT solutions Mathematics Chap 13

## Chapter 13: Surface Areas and Volumes

**TEXTBOOK’S
EXERCISE – 13.1 **

**1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep are 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 the sheet for it, if a sheet measuring 1 m ^{2} costs Rs 20. **

**Sol.
**Here, l = 1.5 m, b = 1.25 m,

h = 65 cm = 0.65 m.

Since the box is open at the top, it has only five faces.

(i) So, surface area of the box

= lb + 2(bh + hl)

=
1.5 × 1.25 m^{2} + 2 (1.25 × 0.65 + 0.65 × 1.5) m^{2}

=
1.875 m^{2} + 2 (1.7875) m^{2}

=
(1.875 + 3.575) m^{2} = 5.45 m^{2 }

Hence,
5.45 m^{2} of sheet is required.

(ii)
Cost of 1 m^{2} of the sheet = Rs20

∴ Cost of 5.45 m^{2} of the sheet

= Rs 20
× 5.45 m^{2} = Rs 109

**2. Length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of Rs 7.50 per m**^{2}**. **

**Sol.
**Here, l = 5 m, b = 4 m, h = 3 m

The surface area of the walls of the room and the ceiling

= 2h (l + b) + lb

= [2 × 3 (5 + 4) + 5
× 4] m^{2}

= (6 × 9 + 20) m^{2}
= 74 m^{2}

Cost of whitewashing
= Rs 7.50 per m^{2 }

∴ The total cost of whitewashing the walls and the ceiling of the room

= Rs 74 × 7.50 = Rs 555.

**TEXTBOOK’S EXERCISE – 13.1 **

**3. The floor of a rectangular hall has a perimeter 250 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. **

**Sol. **Let the length, breadth and height of the hall be l, b and h respectively.

Perimeter of the floor of the hall

= 2 (l + b) = 250 m.

Area of the four walls of the hall

= 2h (l + b) … (i)

Also, area of the four walls of the hall = 15000
10 m^{2} = 1500 m^{2} … (ii)

From (i) and (ii), we have

2h (l + b) = 1500

⇒ h × 250 = 1500 [since 2(l + b) = 250]

⇒ h = 1500/ 250 = 6

Hence, height of the hall is 6 m.

#### Class nine NCERT solutions Mathematics Chap 13

**4. 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? **

**Sol.
**Here, l = 22.5 cm,
b = 10 cm, h = 7.5 cm.

Total surface area of 1 brick = 2 (lb + bh + hl)

= 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm^{2}

= 2(225 + 75 + 168.75) cm^{2 }= 937.5 cm^{2}

∴ Required number of bricks

= 9 375/0.09375 = 100