Triangles... In Progression!

The formula for the area of an equilateral triangle is

$\dfrac {\sqrt{3}a^{2}} {4}$

where $a$ is the side length. Since we are given that the area of the $k$ th triangle is $A_{k} = (k+1)^{2}\sqrt{3}$ , we can equate them together and obtain and expression of $a$ in terms of $k$ .

$\dfrac {\sqrt{3}a^{2}} {4} = (k+1)^{2}\sqrt{3}$

$a^{2} = 4(k+1)^{2}$

$a = 2k + 2$ ( $a$ must be positive)

Thus we can conclude that the side length of each triangle is $2 cm$ longer than the preceding triangle. Now we consider the perimeter of each equilateral triangle:

$P_{1} = 12$

$P_{2} = 18$

$P_{3} = 24$

$P_{k} = 6k + 6$

Since the $n$ th triangle has perimeter $90$ ,

$P_{n} = 6n + 6$

$6n + 6 = 90$

$n = 14$ .

An equilateral triangle with a perimeter of $90$ has a side of $s = \cfrac{90}{3} = 30$ and an area of $A = \cfrac{\sqrt{3}}{4}s^2 = \cfrac{\sqrt{3}}{4}(30)^2 = 15^2\sqrt{3}$ .

Since the $n^{\text{th}}$ triangle is $A_n = (n + 1)^2 \sqrt{3} = 15^2\sqrt{3}$ , that makes $n + 1 = 15$ , so $n = \boxed{14}$ .