Challenging Area Problem

well, you need to remember two things here..

1. Maximum area of a triangle with a given perimeter is when its an equilateral triangle.
2. Maximum area of a rectangle with a given perimeter is when its a square..

so let the length of the rope be 12a (choosing it for simple calculation)


Area of an equilateral triangle = root(3)*x^2/4
here x = 12a/3 = 4a

So area = 4root(3)a^2 ---------- (1)

Area of a square = x^2
here x = 12a/4 = 3a

So area = 9a^2 ----------- (2)

ratio of 1 and 2 = 4root(3)/9 = 0.76

The closest to this is C i.e. 3/4 or 0.75

So answer is C

I too got 4root(3)/9 and somehow put the answer as 2:3.

One should not go for shortcuts, never!!

well, you need to remember two things here..

1. Maximum area of a triangle with a given perimeter is when its an equilateral triangle.
2. Maximum area of a rectangle with a given perimeter is when its a square..

We can consolidate these 2 rules into 1:

in general, when we want to maximize the area of a 2-d figure, we need to equalize all of its sides.