EXERCISE 13.1
Introduction - Surface Areas & Volumes Class 9
Q 1, Ex 13.1, Page No 213, Surface Areas & Volumes, Class IXth Maths
Page No 213:
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:(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1 m2 costs Rs 20.
Answer:
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] m2
= (1.95 + 1.625 + 1.875) m2 = 5.45 m2
(ii) Cost of sheet per m2 area = Rs 20
Cost of sheet of 5.45 m2 area = Rs (5.45 × 20)
= Rs 109
Q 2, Ex 13.1, Page No 213, Surface Areas & Volumes, NCERT Maths Solutions Class 9th
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 Rs 7.50 per m2.Answer:
It is given thatLength (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 white-washed. 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] m2
= (30 + 24 + 20) m2
= 74 m2
Cost of white-washing per m2 area = Rs 7.50
Cost of white-washing 74 m2 area = Rs (74 × 7.50)
= Rs 555
Q 3, Ex 13.1, Page No 213, Surface Areas & Volumes, NCERT Mathematics Class 9th
Question 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 m2 is Rs.15000, find the height of the hall.[Hint: Area of the four walls = Lateral surface area.]
Answer:
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
∴ Area of four walls = 2(l + b) h = 250h m2
Cost of painting per m2 area = Rs 10
Cost of painting 250h m2 area = Rs (250h × 10) = Rs 2500h
However, it is given that the cost of paining the walls is Rs 15000.
∴ 15000 = 2500h
h = 6
Therefore, the height of the hall is 6 m.
Q 4, Ex 13.1, Page No 213, Surface Areas & Volumes, Maths CBSE Class 9th
Question 4:
The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?Answer:
Total surface area of one brick = 2(lb + bh + lh)= [2(22.5 ×10 + 10 × 7.5 + 22.5 × 7.5)] cm2
= 2(225 + 75 + 168.75) cm2
= (2 × 468.75) cm2
= 937.5 cm2
Let n bricks can be painted out by the paint of the container.
Area of n bricks = (n ×937.5) cm2 = 937.5n cm2
Area that can be painted by the paint of the container = 9.375 m2 = 93750 cm2
∴ 93750 = 937.5n
n = 100
Therefore, 100 bricks can be painted out by the paint of the container.
Q 5, Ex 13.1, Page No 213, Surface Areas & Volumes, NCERT Maths Solutions
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.(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?
Answer:
(i) Edge of cube = 10 cmLength (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(edge)2
= 4(10 cm)2
= 400 cm2
Lateral surface area of cuboidal box = 2[lh + bh]
= [2(12.5 × 8 + 10 × 8)] cm2
= (2 × 180) cm2
= 360 cm2
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 cm2 − 360 cm2 = 40 cm2
Therefore, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box by 40 cm2.
(ii) Total surface area of cubical box = 6(edge)2 = 6(10 cm)2 = 600 cm2
Total surface area of cuboidal box
= 2[lh + bh + lb]
= [2(12.5 × 8 + 10 × 8 + 12.5 × 10] cm2
= 610 cm2
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 cm2 − 600 cm2 = 10 cm2
Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm2.
Q 6, Ex 13.1, Page No 213, Surface Areas & Volumes, Class 9th Mathematics
Question 6:
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.(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?
Answer:
(i) Length (l) of green house = 30 cmBreadth (b) of green house = 25 cm
Height (h) of green house = 25 cm
Total surface area of green house
= 2[lb + lh + bh]
= [2(30 × 25 + 30 × 25 + 25 × 25)] cm2
= [2(750 + 750 + 625)] cm2
= (2 × 2125) cm2
= 4250 cm2
Therefore, the area of glass is 4250 cm2.
(ii)
It can be observed that tape is required along side AB, BC, CD, DA, EF, FG, GH, HE, AH, BE, DG, and CF.
Total length of tape = 4(l + b + h)
= [4(30 + 25 + 25)] cm
= 320 cm
Therefore, 320 cm tape is required for all the 12 edges.
Q 7, Ex. 13.1, Page No 213, Surface Areas & Volumes - NCERT Class 9th Maths
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 × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.Answer:
Length (l1) of bigger box = 25 cmBreadth (b1) of bigger box = 20 cm
Height (h1) of bigger box = 5 cm
Total surface area of bigger box = 2(lb + lh + bh)
= [2(25 × 20 + 25 × 5 + 20 × 5)] cm2
= [2(500 + 125 + 100)] cm2
= 1450 cm2
Extra area required for overlapping $=\left(\frac{1450 \times 5}{100}\right) \mathrm{cm}^{2}$
= 72.5 cm2
While considering all overlaps, total surface area of 1 bigger box
= (1450 + 72.5) cm2 =1522.5 cm2
Area of cardboard sheet required for 250 such bigger boxes
= (1522.5 × 250) cm2 = 380625 cm2
Similarly, total surface area of smaller box = [2(15 ×12 + 15 × 5 + 12 × 5] cm2
= [2(180 + 75 + 60)] cm2
= (2 × 315) cm2
= 630 cm2
Therefore, extra area required for overlapping$=\left(\frac{630 \times 5}{100}\right) \mathrm{cm}^{2}$
=31.5cm2
Total surface area of 1 smaller box while considering all overlaps
= (630 + 31.5) cm2 = 661.5 cm2
Area of cardboard sheet required for 250 smaller boxes = (250 × 661.5) cm2
= 165375 cm2
Total cardboard sheet required = (380625 + 165375) cm2
= 546000 cm2
Cost of 1000 cm2 cardboard sheet = Rs 4
Cost of 546000 cm2 cardboard sheet $=\operatorname{Rs}\left(\frac{546000 \times 4}{1000}\right)$
=Rs 2184
Therefore, the cost of cardboard sheet required for 250 such boxes of each kind will be Rs 2184.
Q 8, Ex 13.1, Page No 213, Surface Areas & Volumes, NCERT Solutions Class 9
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 2.5 m, with base dimensions 4 m × 3 m?Answer:
Length (l) of shelter = 4 mBreadth (b) of shelter = 3 m
Height (h) of shelter = 2.5 m
Tarpaulin will be required for the top and four wall sides of the shelter.
Area of Tarpaulin required = 2(lh + bh) + l b
= [2(4 × 2.5 + 3 × 2.5) + 4 × 3] m2
= [2(10 + 7.5) + 12] m2
= 47 m2
Therefore, 47 m2 tarpaulin will be required.
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