Surface Area question (1 Viewer)

kpad5991

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1) A cuboid of length 5 m, breadth 3 m and depth 2 m required 4 tins of paint to give it three coats. How many tins of paint would be needed to give three coats of paint to a cuboid whose dimensions are doubled?

2) A rectangular prism has a Volume V. A second rectangular prism with edges twice the length of the edges of the first, has a volume which differs from that of the first prism by:

A) V
B) 3V
C) 5V
D) 7V

3) A hollow cylinder, open at one end and closed at the other, has a length of b cm and a base radius of r cm. If it is to be painted both inside and outside, the painted area to be covered is:

1623383927843.png

4) The volume of the right prism below is

1623383965080.png

5) The cube has edges 10 cm long. Calculate the volume of the pyramid PWXZ

1623384067411.png
 

Eagle Mum

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1) A cuboid of length 5 m, breadth 3 m and depth 2 m required 4 tins of paint to give it three coats. How many tins of paint would be needed to give three coats of paint to a cuboid whose dimensions are doubled?
Since the surface area of each face (& therefore total surface area of all faces combined) is quadrupled, four times more paint is required for the same number of coats. Therefore, 16 tins are required.

2) A rectangular prism has a Volume V. A second rectangular prism with edges twice the length of the edges of the first, has a volume which differs from that of the first prism by:

A) V
B) 3V
C) 5V
D) 7V
Since doubling of each edge dimension contributes to doubling of volume and a prism has three dimensions, the second prism is 2^3 (ie. 8 times) the volume of the first. The difference between the prisms is 8V - V = 7V. Option ‘D’.

3) A hollow cylinder, open at one end and closed at the other, has a length of b cm and a base radius of r cm. If it is to be painted both inside and outside, the painted area to be covered is:

View attachment 30828
The closed end (base) has an area of Pi.r^2.
The curved section has a surface area of 2pi.rb.

Since they don’t specify the thickness of the walls of the cylinder, we assume the thickness is negligible and therefore the internal and external surfaces are, for all intents & purposes, the same. Therefore the painted area is 2(Pi.r^2 + 2pi.rb). Option A


4) The volume of the right prism below is

View attachment 30829
The volume of a prism is the area of its face multiplied by its length/height. The area of the triangular face, which is shown to be a right angled triangle, is (4x6)/2= 12. Therefore the volume = 12x10= 120m^2
5) The cube has edges 10 cm long. Calculate the volume of the pyramid PWXZ

View attachment 30830
The volume of a pyramid is a third of its base area multiplied by its height. The area of its triangular base is half of the area of the cube’s face: (10x10)/2 = 50. Therefore the volume is 50x10/3 = 166.7cm^2 (to 1 decimal place).
 
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