The sum of the perimeters of an equlateral triangle and a rectangle is $90 \mathrm{~cm}$ the area, $T$ , of the triangle and the area, $R$ , of the rectangle, both in sq cm, satisfy the relationship $R=T^{2}$. If the sides of the rectangle are in the ratio $1: 3$, then the length, in cm , of the longer side of the rectangle, is

We need to set up equations using the given constraints about perimeters and areas.

Let us define:

Side of equilateral triangle = $a$ cm

Rectangle sides are in ratio 1:3, so if shorter side = $x$ cm, then longer side = $3x$ cm

---

Perimeter of equilateral triangle = $3a$

Perimeter of rectangle = $2(x + 3x) = 2(4x) = 8x$

Given that sum of perimeters = 90 cm:

$3a + 8x = 90 \quad \text{...(1)}$

---

Area of rectangle: $R = x \times 3x = 3x^2$

Area of equilateral triangle: $T = \frac{\sqrt{3}}{4}a^2$

The area formula for an equilateral triangle with side $a$ is $\frac{\sqrt{3}}{4}a^2$. This comes from using the general triangle area formula $\frac{1}{2} \times \text{base} \times \text{height}$, where the height of an equilateral triangle with side $a$ is $\frac{\sqrt{3}}{2}a$.

Given that $R = T^2$:

$3x^2 = \left(\frac{\sqrt{3}}{4}a^2\right)^2$

---

$3x^2 = \frac{3}{16}a^4$

Dividing both sides by 3:

$x^2 = \frac{a^4}{16}$

Taking square root:

$x = \frac{a^2}{4} \quad \text{...(2)}$

---

Substituting equation (2) into equation (1):

$3a + 8 \cdot \frac{a^2}{4} = 90$

$3a + 2a^2 = 90$

$2a^2 + 3a - 90 = 0$

---

Using the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 2$, $b = 3$, $c = -90$:

$a = \frac{-3 \pm \sqrt{9 + 720}}{4}$

$= \frac{-3 \pm \sqrt{729}}{4}$

$= \frac{-3 \pm 27}{4}$

This gives us $a = 6$ or $a = -7.5$

Since length cannot be negative, $a = 6$ cm.

---

From equation (2): $x = \frac{a^2}{4} = \frac{36}{4} = 9$ cm

Therefore, longer side of rectangle = $3x = 3 \times 9 = 27$ cm

---

Answer: 27 cm