Surface area and Volume (Part - 5)

1. Ramesh 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 * 4 m ?

Sol. Length of shelter l = 4 m , breadth b = 4 m

and height h = 2.5 m

Area of four walls and top of shelter = Total area of shelter – area of floor

Area of four walls and top of shelter = 2 ( l b + b h + h l ) – l b

Area of four walls and top of shelter = 2 ( 4 * 4 + 4 * 2.5 + 2.5 * 4 ) – 4 * 4 m^2

Area of four walls and top of shelter = 2 ( 16 + 10 + 10 ) – 16 m^2

Area of four walls and top of shelter = 2 ( 36 ) – 16 m^2

Area of four walls and top of shelter = 72 - 16 m^2

Area of four walls and top of shelter = 56 m^2

Hence , 56 m^2 tarpaulin is required to make this shelter .

2. Ramesh 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 3 m * 3 m ?

Sol. Length of shelter l = 3 m , breadth b = 3 m

and height h = 2.5 m

Area of four walls and top of shelter = Total area of shelter – area of floor

Area of four walls and top of shelter = 2 ( l b + b h + h l ) – l b

Area of four walls and top of shelter = 2 ( 3 * 3 + 3 * 2.5 + 2.5 * 3 ) – 3 * 3 m^2

Area of four walls and top of shelter = 2 ( 9 + 7.5 + 7.5 ) – 9 m^2

Area of four walls and top of shelter = 2 ( 24 ) – 9 m^2

Area of four walls and top of shelter = 48 - 9 m^2

Area of four walls and top of shelter = 39 m^2

Hence , 39 m^2 tarpaulin is required to make this shelter .

maths ncert solutions

3. Ramesh 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 m , with base dimensions 4 m * 3 m ?

Sol. Length of shelter l = 4 m , breadth b = 3 m

and height h = 2 m

Area of four walls and top of shelter = Total area of shelter – area of floor

Area of four walls and top of shelter = 2 ( l b + b h + h l ) – l b

Area of four walls and top of shelter = 2 ( 4 * 3 + 3 * 2 + 2 * 4 ) – 4 * 3 m^2

Area of four walls and top of shelter = 2 ( 12 + 6 + 8 ) – 12 m^2

Area of four walls and top of shelter = 2 ( 26 ) – 12 m^2

Area of four walls and top of shelter = 52 - 12 m^2

Area of four walls and top of shelter = 40 m^2

Hence , 40 m^2 tarpaulin is required to make this shelter .

4. Ramesh 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 1.5 m , with base dimensions 4 m * 3 m ?

Sol. Length of shelter l = 4 m , breadth b = 3 m

and height h = 1.5 m

Area of four walls and top of shelter = Total area of shelter – area of floor

Area of four walls and top of shelter = 2 ( l b + b h + h l ) – l b

Area of four walls and top of shelter = 2 ( 4 * 3 + 3 * 1.5 + 1.5 * 4 ) – 4 * 3 m^2

Area of four walls and top of shelter = 2 ( 12 + 4.5 + 6 ) – 12 m^2

Area of four walls and top of shelter = 2 ( 22.5 ) – 12 m^2

Area of four walls and top of shelter = 45 - 12 m^2

Area of four walls and top of shelter = 33 m^2

Hence , 33 m^2 tarpaulin is required to make this shelter .

Milan Tomic

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