you gotta draw it up Mister.
Scan it after drawing it, or else do it with Abobe Photoshop, or some software of Microsoft.

khizarj wrote:To satisfy largest area:

Is it going to be an equilateral triangle? (The question was about one string being made into either a square or a triangle and which gives bigger area, given length of string)

Or

However, if a triangle in inscribed in a circle then it's a right triangle? (the question was about a triangle with side 1, and 1, with vertex on center, and other two on the circle)

Not sure whats the question!! But from whatever i can interpret:

For a given length of string(L) the area of sqaure(L^2/16) will be more than the area of
Eq. Triangle (L^2/12 root 3).

khizarj wrote:To satisfy largest area:

Is it going to be an equilateral triangle? (The question was about one string being made into either a square or a triangle and which gives bigger area, given length of string)

Given a length of 12:

Square:
Side = 12/4 = 3.
Area = 3^2 = 9.

Equilateral triangle:
Base = 12/3 = 4.
Height = 2√3.
Area = (1/2)(4)(2√3) ≈ 7.

45-45-90 triangle:
Sides are x, x, x√2.
Thus, x+x+x√2 = 12.
x(2+√2) = 12.
x = 12/(2+√2) ≈ 3.5
Area = (1/2)(3.5)(3.5) ≈ 6.2.

The square has the largest area.

There is a Simple rule of thumb to answer this question....More uniform a figure is, the more area it will cover given the same perimeter of each figure.

So Ar(Circle) > Ar(Square) > Ar(Triangle) for a given perimeter...or in this case, fixed length string.

Considering Length of string to be 'L', we have already seen above, L^2/16 [Area of SQUARE] > (L^2*ROOT(3))/36 [Ar of Triangle]

Area of circle will be as - 2*Pi*R = L...Hence R = L/(2*Pi)...

Hence area of circle = Pi * [L/(2*Pi)]^2....or [(L^2)/(4*Pi)]......4* Pi will be definitely less than 16....

[(L^2)/(4*Pi)] [Area of Circle] > L^2/16 [Area of Square] > (L^2*ROOT(3))/36 [Area of Triangle]

The rule can be simply applied....even if the question contains options like Pentagon, Hexagon, Decagon........and a Beautiful Circle with the contant length string....One may not be able to apply area formulas for figures like Decagons etc....so, formula may even not be necessary...

Anurag