Surface Area and Volume (Part - 2)
1. The floor of a rectangular hall has a perimeter 200 m .
If the cost of painting the four walls at the rate of 10 Rs per m^2 is 12000 ,
find the height of the hall . [ Hint :- Area of the four walls = Lateral
surface area ]
Sol. The length of hall = l m , breath = b m and height = h m
Perimeter of floor = 2 ( l + b )
According to question
2 ( l + b ) = 200 m
….(1)
Area of four walls of hall = 2 ( l + b ) h
So the cost of painting four walls at the rate of Rs 10 per
m^2 = 2 ( l + b ) h * 10
= 200 h * 10
= Rs 2000 h
According to question ,
2000 h = 12000
h = 12000 / 2000
h = 6 m
Hence , the height of the hall is 6 m.
2. The floor of a rectangular hall has a perimeter 100 m .
If the cost of painting the four walls at the rate of 10 Rs per m^2 is 6000 ,
find the height of the hall . [ Hint :- Area of the four walls = Lateral
surface area ]
Sol. The length of hall = l m , breath = b m and height = h m
Perimeter of floor = 2 ( l + b )
According to question
2 ( l + b ) = 100 m
….(1)
Area of four walls of hall = 2 ( l + b ) h
So the cost of painting four walls at the rate of Rs 10 per
m^2 = 2 ( l + b ) h * 10
= 100 h * 10
= Rs 1000 h
According to question ,
1000 h = 6000
h = 6000 / 1000
h = 6 m
Hence , the height of the hall is 6 m.
3. The floor of a rectangular hall has a perimeter 150 m .
If the cost of painting the four walls at the rate of 10 Rs per m^2 is 9000 ,
find the height of the hall . [ Hint :- Area of the four walls = Lateral
surface area ]
Sol. The length of hall = l m , breath = b m and height = h m
Perimeter of floor = 2 ( l + b )
According to question
2 ( l + b ) = 150 m
….(1)
Area of four walls of hall = 2 ( l + b ) h
So the cost of painting four walls at the rate of Rs 10 per m^2
= 2 ( l + b ) h * 10
= 150 h * 10
= Rs 1500 h
According to question ,
1500 h = 9000
h = 9000 / 1500
h = 6 m
Hence , the height of the hall is 6 m.
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| maths ncert solutions |
4. The paint in a certain container is sufficient to paint
an area equal to 8.88 m^2 . How many bricks of dimensions 22 cm * 10 cm * 7 cm
can be painted out of this container ?
Sol. Length of the brick l = 22 cm , breath b = 10 cm and
height h = 7 cm
Total surface area of brick = 2 ( l b + b h + h l )
Total surface area of brick = 2 ( 22 * 10 + 10 * 7 + 7 * 22
)
Total surface area of brick = 2 ( 220 + 70 + 154 )
Total surface area of brick = 2 * 444
Total surface area of brick = 888
cm^2 = 0.0888 m^2
Number of bricks to be painted = (
Total paint available ) / ( Paint for one brick )
Number of bricks to be painted = 8.88 m^2 / 0.0888 m^2 = 100
Hence , the paint of this container can paint 100 bricks.
5. The paint in a certain container is sufficient to paint
an area equal to 7 m^2 . How many bricks of dimensions 20 cm * 10 cm * 5 cm can
be painted out of this container ?
Sol. Length of the brick l = 20 cm , breath b = 10 cm and
height h = 5 cm
Total surface area of brick = 2 ( l b + b h + h l )
Total surface area of brick = 2 ( 20 * 10 + 10 * 5 + 5 * 20
)
Total surface area of brick = 2 ( 200 + 50 + 100 )
Total surface area of brick = 2 * 350
Total surface area of brick = 700
cm^2 = 0.0700 m^2
Number of bricks to be painted = (
Total paint available ) / ( Paint for one brick )
Number of bricks to be painted = 7 m^2 / 0.0700 m^2 = 100
Hence , the paint of this container can paint 100 bricks.
6. The paint in a certain container is sufficient to paint
an area equal to 5 m^2 . How many bricks of dimensions 15 cm * 10 cm * 4 cm can
be painted out of this container ?
Sol. Length of the brick l = 15 cm , breath b = 10 cm and
height h = 4 cm
Total surface area of brick = 2 ( l b + b h + h l )
Total surface area of brick = 2 ( 15 * 10 + 10 * 4 + 4 * 15
)
Total surface area of brick = 2 ( 150 + 40 + 60 )
Total surface area of brick = 2 * 250
Total surface area of brick = 500
cm^2 = 0.0500 m^2
Number of bricks to be painted = (
Total paint available ) / ( Paint for one brick )
Number of bricks to be painted = 5 m^2 / 0.0500 m^2 = 100
Hence , the paint of this container can paint 100 bricks.
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