Triangle and Rectangle Area

Let the side of the equilateral triangle be $a$. Let the sides of the rectangle be $x$ and $3x$. Sum of perimeters: $3a + 2(x + 3x) = 90 \implies 3a + 8x = 90$. Area of triangle $T = \frac{\sqrt{3}}{4} a^2$. Area of rectangle $R = 3x^2$. Given $R = T^2 \implies 3x^2 = \frac{3}{16} a^4 \implies 16x^2 = a^4 \implies a^2 = 4x \implies a = 2\sqrt{x}$. Substitute $a = 2\sqrt{x}$ into $3a + 8x = 90$: $6\sqrt{x} + 8x = 90 \implies 4x + 3\sqrt{x} - 45 = 0$ Let $u = \sqrt{x}$: $4u^2 + 3u - 45 = 0 \implies (4u + 15)(u - 3) = 0$ Since $u > 0$, $u = 3 \implies x = 9$. The longer side of the rectangle is $3x = 3(9) = 27\text{ cm}$.