Aptitude - Volume and Surface Area - Discussion

Let me explain.

Consider a water tank with half of the water .
Now the inner surface of the water tank will get wet but not the of waster(upper most surface of water on which things float)
So we need to deduct that part (area of that part) from total surface of Cuboid
hence .

Total Surface[2(lb + lh + bh)] - (upper most surface of water)[lb]

@Fainaz : when we calculate area the unit is "met Sqr" nad when we calculate Volume the unit is "met cube". It has noyhing to do with Cuboid or Cylinder.

@Roshan : we need to consider it Cuboid bcoz they have written "6m long and 4 m". if It is Cylinder , they might have given diameter ... got it ;)

Just consider the cistern as a cuboid with length l = 6 m breadth = 4 m. The total height of cistern is not given but it is given it is filled with water to height h = 1.25 m but the actual height of cistern can be anything which is not needed here.

It has bottom floor and upper ceiling, two side walls with length 6 m and 2 side walls with length 4 m.

So the areas which get wet are the bottom floor(lb), 2 side walls of length 6 m up to height 1.25 m which is 2(lh) and other 2 side walls with length(breadth)4 m which is 2(bh).

So area of wet surface = lb + 2 lh + 2 bh.

= 6*4 + 2*6*1.25 + 2*4*1.25.

= 49 sq.m.

A--B
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C--D

@Rahul: it contains water upto the height 1.25m,
So we calculate the total surface up to this height,

But the above surface AB will not be wet so we will deduct AB surface area which is l*b from total surface area which also include the height.

This is a Cuboid because first we consider cistern as cuboid and second there is no diameter mention here which is possible in cuboid.

Cuboid consist total six face so total surface area is 2(lb+bh+lh).
And here lb+lb is bottom and top surface.

So we'll consider only bottom which is wet and minus the top face(AB).

What about the height of this cuboid?

It is possible to answer this question by assuming the height difference between water level and cuboid top surface as a smaller value, now the area of top surface can be deducted.

Considering each plane of cuboid total area is lb+bh+lh+lb+bh+lh and top surface is area is lb, by subtracting this lb the total wet surface = 2(6x4+4X1.25+1.25x6) - (6x4) = 49 sq units.

[2 (lb + bh + lh) - lb].

This means that assuming the area of cuboid which has size side that we know.

So length*breadth for two surface + breadth*height for another two surface + length*height for rest two surface. Now this length*breadth is the area of a rectangle which is the outer part of the water surface not to be included.

So should be deducted.

@All.

2(lb+ lh+ bh) gives the total surface area, which includes 4 walls, ceiling and floor. If we are asked the area of a wet surface, only the walls will be wet, and the floor. Ceiling won't be wet, so we need to subtract ceiling's area from the total surface area i.e. L*B

Therefore 2(lb + bh + lh) - lb.

The side walls are also not fully wet, as everybody said that the tank is not fully filled with water.

We don't know the total height of cistern, we just know how much depth of water is cistern containing.

Then, the area of side walls must NOT be fully included as a wet area.

Then How we conclude this?

Why it's not be taken as three parts of the cuboid.

1st part is the big one 2nd part is a wet one and the third part is apart from both which is 2nd biggest cuboid.

And if we will subtract from 3rd one's surface area from 1st, then automatically we will get 2nd one which is the wet part.

Isn't it?

We have 5 wet sides i.e., complete bottom portion (6mx4m = 24m2) and four walls.

Area of 4 walls = perimeter*height.
= 2(6+4)*1.25.
= 20*1.25 =25.
So total wet area = 24+25 = 49 cub.mtr.

Total Surface[2(lb + lh + bh)] - (upper most surface of water)[lb]
tis concept is clear bcoz water is not touching surface of tank
but i m not getting why 2(bh + lh) + lb this step is used.. specialy (+ lb)