## MP Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1

Question 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:

- The area of the sheet required for making the box.
- The cost of sheet for it, if a sheet measuring 1m
^{2}costs ₹ 20.

Solution:

Given

l = 1.5 m = 150 cm

b = 1.25 m = 125 cm

h = 65 cm

1. Total area of plastic sheet required = LSA + Area of base

= 2h(l + b) + l x b

= 2 x 65 (150 + 125) + 150 x 125

= 130 (275) + 18750

= 35750 + 18750

= 54500 cm^{2}

= 5.45 m^{2}

2. Cost of sheet = area of plastic sheet x rate = 5.45 x 20

= 109.00

= ₹ 109.

Question 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 ₹ 7.50 per m^{2}.

Solution:

Given

l = 5m

b = 4m

h = 3 m

Area of the room to be white washed = LSA + Area of ceiling

= 2h(l + b) + l x b

= 2 x 3 (5 + 4) + 5 x 4

= 6(9) + 20

= 54 + 20 = 74 m^{2}

Cost of painting = 7.50 x 74 m^{2}

= ₹ 555

Question 3.

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

Solution:

Given

P = 250 m

Rate of painting = ₹ 10/ m^{2}

Cost of painting = ₹ 15000

Area of walls to be painted = \(\frac{15000}{10}\) = 1500 m^{2}

Perimeter = 2 (l + b) = 250

∴ l + b = 250/2 = 125m

Area of walls = LSA = 2h (l + b) = 1500

= h (l + b) = \(\frac{1500}{10}\) = 750 m^{2} …..(i)

Putting the value of (l + b) in (i), we get

h(125) = 750

h = \(\frac{750}{125}\) = 6 m.

Question 4.

The paint in a certain container is sufficient to paint an area equal to 9315 m^{2}. How many bricks of dimensions 22.5 cm x 10 cm x 7.5 cm can be painted out of this container?

Solution:

Given

l = 22.5 cm

h = 7.5 cm = 0.075 m

b = 10 cm = 0.1 m

Area to be painted = 9.375 m^{2} = 93750 cm^{2}

Area of one brick = 2(lb + bh + hl)

= 2(22.5 x 10 + 10 x 7.5 + 7.5 x 22.5)

= 2(225 + 75 + 168.75)

= 2 x 468.75

= 937.5 cm^{2}

No. of bricks which can be painted = \(\frac{93750}{937.5}\) = 100

Question 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.

- Which box has the greater lateral surface area and by how much?
- Which box has the smaller total surface area and by how much? Sol.

Solution:

1. LSA of cubical box = 4a^{2}

= 4 x 10 x 10 = 400 cm^{2}

LSA of cuboidal box = 2h (l + b)

= 2 x 8 (12.5 x 10)

= 16 x 22.5

= 360 cm^{2}

LSA of cubical box is more than cuboidal box by 40 cm^{2}.

2. TSA of cubical box = 6a^{2} = 6 x 10 x 10

= 600 cm^{2}

TSA of cuboidal box = 2 (lb + bh + hl)

= 2(12.5 x 10 + 10 x 8 + 12.5 x 8)

= 2(125 + 80 + 100)

= 2 x 305 = 610 cm^{2}.

∴ TSA of cuboidal box is more than cubical box by 10 cm^{2}.

Question 6.

A small indoor green house (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.

- What is the area of the glass?
- How much of tape is needed for all the 12 edges?

Solution:

Given

l = 30 cm

b = 25 cm and

h = 25 cm

Area of glass required = 2(30 x 25 + 25 x 25 x 30)

= 2(750 + 625 + 750)

= 2 x 2125 = 4250 cm^{2}

Length of tape required = 4(l + b + h)

= 4(30 + 25 + 25)

= 4 x 80 = 320 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 25 cm x 20 cm x 5 cm and the smaller of dimensions 15 cm x 12 cm x 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is ₹ 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.

Solution:

Big Box :

l = 25 cm

b = 20 cm

h = 5 cm

Small Box:

l = 15 cm

b = 12 cm

h = 5 cm

Number of boxes required = 250

Rate = ₹ 4 per 1000 cm^{2}

TSA of bigger box = 2 (25 x 20 + 20 x 5 + 5 x 25)

= 2(500 + 100 + 125)

= 2 x 725 = 1450 cm^{2}

TSA of smaller box = 2 (15 x 12 +12 x 5 + 5 x 15)

= 2 (180 + 60 + 75)

= 2 x 315 = 630 cm^{2}

Area of cardboard required one bigger = TSA + 5% of TSA

= 1450 + \(\frac{5}{100}\) x 1450

= 1522.5 cm^{2}

Area of cardboard required for 250 boxes of bigger size = 250 x 1522.5

= 380625 cm^{2}

Area of cardboard required for one small box = 630 + \(\frac{5}{100}\) x 630

= 661.5 cm^{2}

Total area of cardboard required for 250 boxes of smaller size = 250 x 661.5

= 165375 cm^{2}

Total area = (165375 + 380625) cm^{2}

= 546000 cm^{2}

Total cost of each kind of cardboard = \(\frac{4}{1000}\) x 546000

= ₹ 2184/-

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 rollpd up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m x 3 m?

Solution:

Given

l = 4m

b = 3 m and

h = 2.5 m

Area of the tarpaulin required = 2h(l + b) + l x b

= 2 x 2.5(4 + 3) + 4 x 3

= 5 x 7 + 12

= 35 + 12= 47m