There are two ways to solve the problem, one is of-course by going through the basic concepts and principles while there’s a one more method known as trial and error or approximation. Later method is more about applying common sense to solve a problem. Tough exams like IIT JEE often needs such capability to derive a solution in a very limited time period. It sounds scary but in reality its a fun and it makes solution easier.

Let me now explain above with an example. I am trying to find the formula for surface area of a frustum.

Frustum

Conventional process is to make it as full cone and try to use Pythagoras theorem in various triangles to come up with a formula for surface area.  Its good but quite a long way to solve this problem. Now let me tell you a shorter solution.

Surface area =  perimeter x height  – for any regular shape

How to use above simple formula for a irregular shape like frustum? Lets put it like below as an approximation:

Perimeter = average perimeter of frustum = (2 pi R + 2 pi r)/2 = 2 pi (R+r)/2

Height = slant height = Slant height.png

So surface area of frustum = average perimeter X slant height = 2 pi (R+r)/2   X Slant height.png

=>   pi (R+r)   X Slant height.png

Isn’t it simple? Lets see how similar reasoning helps solving IIT questions. Below question is from IIT 2004 paper:

The sides a, b and c of a triangle are in the ratio 1 : √3 : 2. Then the angles A,B, and C of the triangle are in the ratio?

 

But how do we proceed further? One way is to draw a triangle and try solving it but its a long and cumbersome process.Another easier way is:

We all know that SinA: SinB: SinC :: a:b:c

or  SinA: SinB: SinC :: 1 : √3 : 2

=> SinA: SinB: SinC :: 1/2 : √3/2 : 2/2    – Dividing by 2

=> Above gives A = 30, B = 60 and C = 90

Is above correct? one way to verify is A+B+C = 30+60+90 = 180, yes it looks right!

So answer is angles are in ratio of 1:2:3.

So next time you have limited time and trying to answer a rigorous question then it makes sense to give a thought on approximation method which yields a faster result.

 

 

 

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