Question

the triangle below is equilateral. find the length of side x in simplest radical form with a rational denominator.
answer attempt 1 out of 2
x =

Explanation:

Step1: Use properties of equilateral triangle

In an equilateral triangle, the altitude bisects the base. Let the side of the equilateral triangle be \(a = 11\). After drawing the altitude, we get a right - triangle with hypotenuse \(c = 11\) and one leg \(y=\frac{x}{2}\).

Step2: Apply Pythagorean theorem

By the Pythagorean theorem \(a^{2}=b^{2}+c^{2}\) (in the right - triangle, \(a = 11\), \(b\) is the altitude, \(c=\frac{x}{2}\)). Also, for an equilateral triangle with side \(s\), if we consider the right - triangle formed by the altitude \(h\), half of the side \(\frac{s}{2}\) and the side \(s\) of the equilateral triangle. Using \(\cos60^{\circ}=\frac{\frac{x}{2}}{11}\). Since \(\cos60^{\circ}=\frac{1}{2}\), we have \(\frac{\frac{x}{2}}{11}=\frac{1}{2}\) (another way: using the right - triangle and Pythagorean theorem \(11^{2}=h^{2}+(\frac{x}{2})^{2}\), and also for an equilateral triangle, the altitude \(h\) and the side \(s\) relationship: if we know that in a \(30 - 60-90\) triangle (formed by the altitude of the equilateral triangle), the sides are in the ratio \(1:\sqrt{3}:2\). Here, the side of the equilateral triangle is the hypotenuse of the \(30 - 60 - 90\) triangle. If the hypotenuse \(s = 11\), and the side opposite to \(30^{\circ}\) is \(\frac{x}{2}\). We know that \(\sin60^{\circ}=\frac{\text{altitude}}{11}\) and \(\cos60^{\circ}=\frac{\frac{x}{2}}{11}\). Since \(\cos60^{\circ}=\frac{1}{2}\), we can solve for \(x\) directly. \(\frac{x}{2}=11\times\cos60^{\circ}\), \(\frac{x}{2}=\frac{11}{2}\), \(x = 11\).

Answer:

\(x = 11\)