Question

the triangle below is equilateral. find the length of side ( x ) in simplest radical form with a rational denominator.
image of an equilateral triangle with a perpendicular segment of length 6 to one side, forming a right triangle with side ( x )

Explanation:

Step1: Analyze the equilateral triangle

In an equilateral triangle, all sides are equal, and the altitude (the perpendicular from a vertex to the opposite side) also bisects the opposite side. So, the right - triangle formed has one angle of $30^{\circ}$ (since the altitude bisects the $60^{\circ}$ angle of the equilateral triangle), one angle of $90^{\circ}$, and the hypotenuse of the right - triangle is equal to the side of the equilateral triangle. Let the side of the equilateral triangle be $s$. The leg adjacent to the $30^{\circ}$ angle is $\frac{s}{2}$, and the leg opposite to the $30^{\circ}$ angle is the altitude. But in our case, we know that the length of the altitude (the side with length 6) is related to the side $x$ (which is half of the side of the equilateral triangle) by the trigonometric ratio. In a $30 - 60 - 90$ triangle, the ratio of the sides is $1:\sqrt{3}:2$, where the side opposite $30^{\circ}$ is the shortest side (let's call it $a$), the side opposite $60^{\circ}$ is $a\sqrt{3}$, and the hypotenuse is $2a$.

Step2: Use trigonometric ratio or Pythagorean theorem

We know that in the right - triangle, if we consider the angle of $60^{\circ}$, $\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}$. The opposite side to the $60^{\circ}$ angle is 6, and the hypotenuse is the side of the equilateral triangle, but we can also use the fact that if we let the side of length $x$ be the adjacent side to the $60^{\circ}$ angle, and the side of length 6 be the opposite side to the $60^{\circ}$ angle. We know that $\tan(60^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\sqrt{3}=\frac{6}{x}$.

Step3: Solve for $x$

From $\tan(60^{\circ})=\sqrt{3}=\frac{6}{x}$, we can cross - multiply to get $x\sqrt{3}=6$. Then, to solve for $x$, we divide both sides by $\sqrt{3}$: $x = \frac{6}{\sqrt{3}}$. To rationalize the denominator, we multiply the numerator and denominator by $\sqrt{3}$: $x=\frac{6\sqrt{3}}{3}=2\sqrt{3}$.

Or we can use the Pythagorean theorem. Let the side of the equilateral triangle be $2x$ (since the right - triangle's leg $x$ is half of the side of the equilateral triangle). Then, by the Pythagorean theorem, $(2x)^{2}=x^{2}+6^{2}$. Expanding, we get $4x^{2}=x^{2}+36$. Subtract $x^{2}$ from both sides: $3x^{2}=36$. Divide both sides by 3: $x^{2}=12$. Take the square root of both sides: $x = \sqrt{12}=2\sqrt{3}$ (we take the positive value since length can't be negative).

Answer:

$2\sqrt{3}$