Answer:
Below.
Explanation:
I can do this using calculus but the algebra will be a bit messy.
Let the 2 equal sides of isosceles triangle be x and the base be y units long.
Then perimeter p = 2x + y.
Area A = 0.5yh where h is the altitude of the triangle.
h = sqrt [x^2 - (0.5y)^2]
= [x^2 - (0.5y)^2]^0.5
So A = 0.5y * [x^2 - (0.5y)^2]^0.5
From the perimeter formula y = p - 2x so:
A = 0.5(p-2x) * [x^2 - (0.5(p-2x)^2]^0.5
A = 0.5(p-2x) * [x^2 - 0.25(p^2 - 4xp + 4x^2)]^0.5
A = 0.5(p-2x) * [x^2 - 0.25p^2 + xp - x^2]^0.5
A = (0.5p-x) * [xp - 0.25p^2]^0.5
We now find the derivative and equate it to zero to find the value of x which corresponds to a maximum area.
We use the product rule:
dA/dx = (0.5p-x)* 0.5(xp - 0.25p^2)^(-0.5) * p - 1(xp - 0.25p^2)^0.5 = 0
p(0.5p - x) / (2((xp - 0.25p^2)^0.5 - (xp - 0.25p^2)^0.5 = 0
p(0.5p - x) / (2((xp - 0.25p^2)^0.5 = (xp - 0.25p^2)^0.5
Now multiplying though by (2((xp - 0.25p^2)^0.5 :-
p(0.5p - x) = 2 (xp - 0.25p^2)
0.5p^2 - px = 2px - 0.5p^2
p^2 = 3px
x = p^2/3p = p/3
So for a maximum value of the area x = p/3.
But y = p - 2x
= p - 2(p/3)
= p - 2/3p
= p/3
- but we've just shown that x = p/3
That means that x = y and the triangle is Equilateral.