Consecutive Sides & Height

Equating Heron's Formula with the standard $\frac12bh$ formula for the area of the triangle we get: $\frac14 \sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)} = \frac12 \times \text{Base} \times \text{Perpendicular height}$ $\frac14 \sqrt{(3n + 6)(n)(n + 4)(n + 2)} = \frac12 \times (n + 2) \times (n)$ $\sqrt{(3n + 6)(n)(n + 4)(n + 2)} = 2 \times (n + 2) \times (n)$ $(3n + 6)(n)(n + 4)(n + 2) = 4 \times (n + 2) \times (n) \times (n+2) \times (n)$ $3(n + 2)(n + 4) = 4 \times (n + 2) \times (n)$ $3(n + 4) = 4n$ $\boxed{n = 12}$ (Notice that we can definitely cancel terms as above, since all of the factors are positive so not zero)

Therefore, $\text{Perimeter} = 3n + 6 = \boxed{42}$

Using Pythagorean theorem, we have:

$\begin{aligned} \sqrt{(n+1)^2-n^2} + \sqrt{(n+3)^2-n^2} & = n+2 \\ \sqrt{2n+1} + \sqrt{6n+9} & = n+2 & \small \color{#3D99F6} \text{Squaring both sides} \\ 2n+1 + 2\sqrt{(2n+1)(6n+9)} + 6n+9 & = n^2+4n+4 & \small \color{#3D99F6} \text{Rearranging} \\ n^2-4n-6 & = 2\sqrt{(2n+1)(6n+9)} & \small \color{#3D99F6} \text{Squaring both sides} \\ n^4 - 8n^3 + 4n^2 + 48n + 36 & = 48n^2+96n+36 \\ n^4 - 8n^3 - 44n^2 - 48n & = 0 & \small \color{#3D99F6} \text{Since }n=0 \text{ is not a solution} \\ (n+2)^2(n-12) & = 0 & \small \color{#3D99F6} \text{Since }n=-2 \text{ is not a solution} \\ \implies n & = 12 \end{aligned}$

Therefore the perimeter is $n+1+n+2+n+3 = 3(12)+6 = \boxed{42}$ .

divide the base into two parts: the left side = x; the right side = n + 2 - x. Then we apply the Pythagorean Theorem to each side, resulting in (1) n^2 + x^2 = (n + 1)^2. (2) n^2 + (n + 2 - x)^2 = ( n + 3)^2 Expanding these equations and substituting x^2 = 2n + 1, we straightforwardly get (3) n - 2 = 2x. Expanding, n^2 - 4n + 4 = 4x^2 =8n + 4. Then n^2 - 12n = 0, or n = 12. The perimeter is then = 13 + 14 + 15 = 42. Ed Gray