Surface Areas and Volumes: 9 Math
NCERT Exercise 13.1 part 1: 9th math
Cube
Cube is a three dimensional object having six square faces. Each of the vertex of a cube has three meeting side. For example a dice. A dice is a typical example of a cube.
Area or Surface area or Lateral Surface Area of a Cube
Since, a cube has six square faces, thus area of one square multiplied by 6 gives the area or surface area of a cube.
Now, Area of a square having edge equal to `a` = a2
Thus, Area of a cube = 6 × a2
Thus, Area or Total Surface Area or Lateral Surface Area of a Cube = 6 a2
Cuboid
A Cuboid is a three dimensional shaped object having six rectangular faces. For example a brick or a match box. Generally brick and a match box are typical examples of cuboid.
Cuboid means object having cube like shape.
Surface Area or Lateral Surface Area or Area of Cuboid
Surface Area or Lateral Surface Area or Area of a Cuboid = `2(lb + bh + hl)`
Where '`l`' = length, 'b' = breadth, and 'h' = height. This means '`l`', 'b' and 'h' are respectively the three edges of the cuboid.
Surface Areas And Volumes Class nine Math NCERT Exercise 13.1 Question (1) A plastic box 1.5 long, 1.25 m wide and 65cm deep is to be made. It is opened 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 1m2costs Rs 20.
Solution
Given
Length (`l`) of the cuboid = 1.5m
Breadth (b) of the cuboid = 1.25 m
And, Height (h) of the cuboid = 65 cm
⇒ h = 65/100 =0.65m
(i) The area of the sheet required for making the box.
Calculation of Area of sheet required for making the box
Since, given box is open from its top,
Therefore, Area of sheet required for making the box = Area of 4 sides + Area of base
`= 2h(l+b) + lxxb`
=2×0.65 m (1.5 m + 1.25 m)+ 1.5 m× 1.25 m
=1.3 m× 2.75 m+1.875 m2
= 3.575 m2+ 1.875 m2
= 5.45 m2
Thus, Area of sheet required for making the box = 5.45 m2Answer
(ii) The cost of the sheet for it, if a sheet measuring 1 m2 costs Rs 20.
Calculation of cost of sheet
Since, as given cost of sheet for 1m2 = Rs 20
Therefore, cost of 5.45 m2 of sheet
= Rs 20 × 5.45
Thus, cost of sheet required for making of box = Rs109.00Answer
Surface Areas And Volumes Class nine Math NCERT Exercise 13.1 Question (2) The length, breadth and height of a room are 5m,4 m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs7.50 per m2.
Solution
Given, Length (`l`) of the room = 5m
Breadth (b) of the room = 4m
And, Height (h)of the room =3m
And, Cost of white washing = Rs 7.50 per m2
The cost of white washing of the walls and the ceiling of the room = ?
Calculation of Area of the walls and ceiling to be white washed
Now area of four walls + ceiling = `2h (l+b)+lxxb`
= 2 × 3m (5 m + 4 m)+5 m × 4 m
= 6 m × 9 m +20 m2
=54 m2+20 m2
= 74 m2
Thus, Area of walls and ceiling to be white washed = = 74 m2
Calculation of cost of white washing
As given, the cost of white washing for 1 m2 = Rs 7.50
Therefore, cost of 74 m2
= 74 × 7.50
=Rs 555.00
Thus, cost of white washing = Rs 555.00 Answer
Surface Areas And Volumes Class nine Math NCERT Exercise 13.1 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 m2 is Rs 15000, find the height of the hall.
[Hint: Area of the four walls = Lateral surface area.]
Solution
Given, Perimeter of the floor of the rectangular hall = 250m
Total cost of painting the walls = Rs 15000
And rate of painting of the walls = Rs10 per m2
Therefore, Height of the hall = ?
Now, Cost of painting of walls = Rate per meter × Area
⇒ Area = Cost/Rate
= 15000/10
Therefore, Area of four walls = 1500m2
And as given Perimeter of the floor of the hall = 250 m
`=>2(l+b)=250`
[∵ Perimeter = 2(Length + Breadth)]
Now, as calculated, Area of 4 walls = 1500m2
⇒ Lateral surface area = 1500 m2
[As given in hint Area of the four walls = Lateral surface area]
⇒ 2(l+b) h = 1500 m2
[∵ Lateral surface area `= 2(l+b)h`]
⇒ 250 m × h = 1500 m2
⇒ h = 1500/250
⇒ h = 6 m
Therefore, Height of the given rectangular hall (h) = 6 m Answer
Surface Areas And Volumes Class nine Math NCERT Exercise 13.1 Question(4) The paint in a certain container is sufficient to paint an area equal to 9375 m2. How many bricks of dimension 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?
Solution
[Strategy to solve this question: The number of bricks can be painted will be calculated by dividing the total area which can be painted by given pain by the total surface area of one brick. ]
Given,
Total area for which given volume of paint is sufficient to paint = 9.375 m2
Length of brick = 22.5 cm
= 22.5/100 = 0.225 m
Breadth of brick = 10 cm
= 10/100 = 0.1 cm
Height of brick = 7.6 cm
=7.6/100 = 0.075 cm
Now, Total surface area of brick `= 2(l b + b h + h l)`
= 2 [(0.225 m × 0.1 m) + (0.1 m × 0.075 m) + (0.075 m × 0.225 m)]
= 2 (0.0225 m2 + 0.0075 m2 + 0.01678 m2)
= 2 × 0.046875 m2
= 0.09375 m2
Now, as given total area can be painted by the given volume of paint = 9.375 m2
And Area of one brick = 0.09375 m2
Therefore, number of bricks = 9.373/0.09375
`= (9375 xx 100)/9375` = 100 bricks
Thus, total number of bricks which can be painted = 100 Answer
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