An isosceles triangle has area of 1m^2, what are the dimensions of the one with the shortest leg?

Question

An isosceles triangle has an area of 1m^2, what are the dimensions of the one with the shortest leg?

Answer 1

So we have an isosceles triangle who has two sides equal. Let's define the two sides which are equal to L. The short side will be called S. The area of a triangle is 1/2 * b * h. The base will be the short side. But what is the height?

/\

/ \

L / \ L

/___\

S

Let h be the bisector which makes a perpendicular to side S.

h² + (1/2*S)² = L²

h = sqrt(L² - S²/4)

Area = 1/2 * S * sqrt(L² - S²/4) = 1 (Here L is a constant)

d(Area)/dS = 0 = 1/2 * sqrt(L² - S²/4) + 1/2 * S * (-2S)/sqrt(L² - S²/4)

= 1/2 * sqrt(L² - S²/4) * [ 1 - 2S²/(L² - S²/4) ]

Is L = S/2 then the outer edges would not be shorter than the short side (squat like triangle) which does not match our problem, therefore L > S (otherwise it would be equilateral). So this leaves the quantity in square brackets needing to be zero to make the derivative of area equal zero.

1 - 2S²/(L² - S²/4) = 0 --> 1 = 2S²/(L² - S²/4) --> (L² - S²/4) = 2S²

L² = 2S² + S²/4 = 5S²/4

L = √5/2 * S. (since negative values don't make sense)

S = 2/√5 * L

Now we can calculate the area substituting the relationship of the longside to the shortside...

1/2 * S * sqrt(5S²/4 - S²/4) = 1 --> 1/2*S*S² --> 1/2*S³ = 1 --> S³ = 2 --> S = cube_root(2)