Question

the triangle below is equilateral. find the length of side x in simplest radical form with a rational denominator.

Explanation:

Step1: Recall properties of equilateral triangles

In an equilateral triangle, all sides are equal, and the altitude (the perpendicular from a vertex to the opposite side) bisects the side and the angle. So, the right triangle formed has a hypotenuse equal to the side of the equilateral triangle (let's call it \( s \)), one leg \( x \) (half of the side, so \( x=\frac{s}{2} \)) and the other leg 6 (the altitude). Also, in a 30 - 60 - 90 triangle (which this right triangle is, since the angle opposite \( x \) is \( 30^\circ \) because the altitude bisects the \( 60^\circ \) angle of the equilateral triangle), the ratios of the sides are \( 1:\sqrt{3}:2 \), where the side opposite \( 30^\circ \) (which is \( x \)) is the shortest side, the side opposite \( 60^\circ \) (which is 6) is \( \sqrt{3} \) times the shortest side, and the hypotenuse is twice the shortest side.

We can also use trigonometry. In the right triangle, \( \cos(30^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}} \), but maybe easier to use the 30 - 60 - 90 ratios. Alternatively, use the Pythagorean theorem. Let the hypotenuse of the right triangle be \( s \) (which is equal to the side of the equilateral triangle, and also \( s = 2x \) because the altitude bisects the base). So by Pythagorean theorem, \( s^2=x^2 + 6^2 \), but since \( s = 2x \), substitute:

\( (2x)^2=x^2+36 \)

\( 4x^2=x^2 + 36 \)

\( 4x^2-x^2=36 \)

\( 3x^2=36 \)

\( x^2 = 12 \)

\( x=\sqrt{12}=2\sqrt{3} \)? Wait, no, wait. Wait, maybe I mixed up the sides. Wait, in the equilateral triangle, when we draw the altitude, it splits the triangle into two 30 - 60 - 90 triangles. The angle at the vertex is \( 60^\circ \), so the angle in the right triangle at the top is \( 30^\circ \)? Wait, no. Wait, the equilateral triangle has all angles \( 60^\circ \). When we draw the altitude from a vertex to the opposite side, it bisects the angle, so the angle in the right triangle at the vertex is \( 30^\circ \), the right angle, and the other angle is \( 60^\circ \). So the side opposite \( 60^\circ \) is 6, and the side opposite \( 30^\circ \) is \( x \), and the hypotenuse is \( 2x \).

In a 30 - 60 - 90 triangle, the side opposite \( 60^\circ \) is \( \sqrt{3} \) times the side opposite \( 30^\circ \). So if the side opposite \( 60^\circ \) is 6, and the side opposite \( 30^\circ \) is \( x \), then \( 6=\sqrt{3}x \)

Step2: Solve for \( x \)

From \( 6=\sqrt{3}x \), we solve for \( x \) by dividing both sides by \( \sqrt{3} \):

\( x = \frac{6}{\sqrt{3}} \)

To rationalize the denominator, multiply numerator and denominator by \( \sqrt{3} \):

\( x=\frac{6\sqrt{3}}{\sqrt{3}\times\sqrt{3}}=\frac{6\sqrt{3}}{3}=2\sqrt{3} \)? Wait, no, wait, maybe I had the sides reversed. Wait, let's re - examine. Let's say the right triangle has angle \( 30^\circ \), \( 60^\circ \), \( 90^\circ \). 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 \( 2a \).

In our case, the altitude (length 6) is opposite the \( 60^\circ \) angle? Wait, no. The side \( x \) is adjacent to the \( 60^\circ \) angle? Wait, maybe better to use the Pythagorean theorem correctly. Let the side of the equilateral triangle be \( s \). Then the altitude \( h \) of an equilateral triangle with side \( s \) is given by \( h=\frac{\sqrt{3}}{2}s \). But in our right triangle, the altitude is 6? Wait, no, in the diagram, the right triangle has one leg 6, one leg \( x \), and hypotenuse \( s \) (the side of the equilateral triangle). Also, since the triangle is equilateral, when we draw the altitude, it bisects the base, so \( x=\frac{s}{2} \) (because the base is \( s \), and it's bisected into two segments of length \( x \)? Wait, no, maybe the side of the equilateral triangle is \( 2x \), so the base of the right triangle is \( x \), the height is 6, and the hypotenuse is \( 2x \). Then by Pythagorean theorem:

\( (2x)^2=x^2 + 6^2 \)

\( 4x^2=x^2+36 \)

\( 3x^2 = 36 \)

\( x^2=12 \)

\( x = \sqrt{12}=2\sqrt{3} \)? Wait, but let's check with the 30 - 60 - 90 ratios. If the hypotenuse is \( 2x \), the side opposite \( 60^\circ \) is \( x\sqrt{3} \). In our right triangle, the side opposite \( 60^\circ \) is 6, so \( x\sqrt{3}=6 \), then \( x=\frac{6}{\sqrt{3}} = 2\sqrt{3} \) (after rationalizing). Yes, that's correct.

Answer:

\( 2\sqrt{3} \)