Question

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

Explanation:

Step1: Use properties of equilateral triangle

In an equilateral triangle, the altitude bisects the base. Let the side of the equilateral triangle be \(a\). Here, the hypotenuse of the right - triangle (formed by the altitude, half of the base and the side of the equilateral triangle) is \(a = 11\), and the base of the right - triangle is \(\frac{x}{2}\).

Step2: Apply the Pythagorean theorem

For a right - triangle with hypotenuse \(c = 11\) and one side \(b=\frac{x}{2}\), and the other side (altitude) \(h\). Using the Pythagorean theorem \(c^{2}=h^{2}+(\frac{x}{2})^{2}\). In a \(30 - 60-90\) triangle (since an equilateral triangle can be divided into two \(30 - 60 - 90\) triangles), if the hypotenuse \(c = 11\) and the side opposite to \(30^{\circ}\) is \(\frac{x}{2}\). We know that \(\cos60^{\circ}=\frac{\frac{x}{2}}{11}\) (alternatively, using the Pythagorean theorem: let the side of the equilateral triangle be \(s = 11\), and if we divide the equilateral triangle into two right - triangles, by the Pythagorean theorem \(s^{2}=(\frac{x}{2})^{2}+h^{2}\), and also for an equilateral triangle \(h=\frac{\sqrt{3}}{2}s\). But using trigonometry: \(\cos60^{\circ}=\frac{\text{adjacent}}{\text{hypotenuse}}\), where the adjacent side to \(60^{\circ}\) in the right - triangle is \(\frac{x}{2}\) and hypotenuse is \(11\). Since \(\cos60^{\circ}=\frac{1}{2}\), we have \(\frac{\frac{x}{2}}{11}=\frac{1}{2}\). Cross - multiplying gives \(\frac{x}{2}=\frac{11}{2}\).

Another way: In a right - triangle (formed by bisecting the equilateral triangle), let the side of the equilateral triangle be \(a = 11\), and the base of the right - triangle is \(\frac{x}{2}\). Using the Pythagorean theorem \(a^{2}=(\frac{x}{2})^{2}+(\frac{\sqrt{3}x}{2})^{2}\) (but this is more complex).

Using the property of \(30 - 60 - 90\) triangle: In a \(30 - 60 - 90\) triangle, if the hypotenuse \(c = 11\) and the side opposite \(30^{\circ}\) is \(\frac{x}{2}\) (because the altitude of an equilateral triangle bisects the base). Since in a \(30 - 60 - 90\) triangle, if the hypotenuse \(c\), the side opposite \(30^{\circ}\) is \(\frac{c}{2}\) when we consider the relationship between the sides of a \(30 - 60 - 90\) triangle (\(1:\sqrt{3}:2\)). So \(\frac{x}{2}=\frac{11}{2}\), then \(x=\frac{11}{\cos60^{\circ}}\). Since \(\cos60^{\circ}=\frac{1}{2}\), \(x = \frac{11}{\frac{1}{2}}=11\).

Wait, no, let's start over.

We know that in an equilateral triangle, when we draw an altitude, we get two congruent right - triangles. Let the side of the equilateral triangle be \(s\) (here \(s = 11\)) and the base of the right - triangle be \(\frac{x}{2}\). Using the Pythagorean theorem \(s^{2}=(\frac{x}{2})^{2}+h^{2}\), and also for an equilateral triangle \(h=\frac{\sqrt{3}}{2}s\). But we can use the fact that \(\cos60^{\circ}=\frac{\frac{x}{2}}{s}\) (where \(s = 11\)).

Since \(\cos60^{\circ}=\frac{1}{2}\), we have \(\frac{\frac{x}{2}}{11}=\frac{1}{2}\). Cross - multiplying: \(\frac{x}{2}=\frac{11}{2}\), so \(x=\frac{11}{2}\times2 = 11\).

Wait, no, wrong approach. Let's use the right - triangle.

The right - triangle has hypotenuse \(11\) (side of the equilateral triangle) and one angle \(60^{\circ}\). The side adjacent to \(60^{\circ}\) is \(\frac{x}{2}\).

We know that \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\theta = 60^{\circ}\), \(\cos60^{\circ}=\frac{1}{2}\), so \(\frac{\frac{x}{2}}{11}=\frac{1}{2}\), then \(\frac{x}{2}=\frac{11}{2}\), \(x = 11\).

Wait, no! Wait, the side of the equilateral triangle is \(11\)? No, no, wait the problem is to find \(x\). Wait, no, the side of the equilateral triangle is \(2x\) (because the altitude bisects the base). Wait, no, the hypotenuse of the right - triangle (which is the side of the equilateral triangle) is \(11\). Let the base of the right - triangle be \(x\) (half of the base of the equilateral triangle). Using the Pythagorean theorem: \(11^{2}=x^{2}+(\frac{\sqrt{3}}{2}\times2x)^{2}\) (no, wrong).

Wait, correct approach:

An equilateral triangle with side \(a\) can be divided into two right - triangles with hypotenuse \(a\), base \(\frac{a}{2}\) and height \(\frac{\sqrt{3}}{2}a\).

In our case, the hypotenuse of the right - triangle (side of the equilateral triangle) is \(11\). Let the base of the right - triangle be \(x\) (half of the base of the equilateral triangle). Using the Pythagorean theorem:

\(11^{2}=x^{2}+(\sqrt{11^{2}-x^{2}})^{2}\) (not helpful).

Using trigonometry: \(\sin60^{\circ}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\sqrt{3}}{2}\), \(\cos60^{\circ}=\frac{\text{adjacent}}{\text{hypotenuse}}\)

If the hypotenuse of the right - triangle (side of equilateral triangle) is \(11\), and the adjacent side to \(60^{\circ}\) is \(x\) (half of the base of the equilateral triangle)

\(\cos60^{\circ}=\frac{x}{11}\), since \(\cos60^{\circ}=\frac{1}{2}\), then \(x=\frac{11}{2}\)

But the base of the equilateral triangle is \(2x\). Wait, no, wait the problem says "the triangle is equilateral", and we have a right - triangle with hypotenuse \(11\) (side of equilateral triangle) and base \(x\) (half of the base of the equilateral triangle).

Wait, no, the side of the equilateral triangle is \(11\), and we need to find \(x\) (half of the base). No, no, the problem is misread.

Wait, the side of the equilateral triangle is \(11\), and when we bisect it (draw an altitude), the base of the equilateral triangle is \(2x\). Using the Pythagorean theorem for the right - triangle:

\(11^{2}=x^{2}+h^{2}\), and for an equilateral triangle \(h = \sqrt{11^{2}-x^{2}}\) and also \(h=\frac{\sqrt{3}}{2}(2x)=\sqrt{3}x\)

So \(11^{2}=x^{2}+(\sqrt{3}x)^{2}\)

\(121=x^{2}+3x^{2}\)

\(121 = 4x^{2}\)

\(x^{2}=\frac{121}{4}\)

\(x=\frac{11}{2}\)

But the base of the equilateral triangle is \(2x\). Wait, no, the problem says "find the length of side \(x\)" (assuming \(x\) is half of the base). No, wait the problem is: the triangle is equilateral. When we draw an altitude, we get two right - triangles. Let the side of the equilateral triangle be \(s\), then \(s = 11\). The base of the equilateral triangle is \(b\), and when we bisect it (altitude), the base of the right - triangle is \(\frac{b}{2}\). Using the Pythagorean theorem \(s^{2}=(\frac{b}{2})^{2}+h^{2}\), and \(h=\frac{\sqrt{3}}{2}s\)

\(s^{2}=(\frac{b}{2})^{2}+(\frac{\sqrt{3}}{2}s)^{2}\)

\(s^{2}=\frac{b^{2}}{4}+\frac{3s^{2}}{4}\)

\(4s^{2}=b^{2}+3s^{2}\)

\(b^{2}=s^{2}\)

\(b = s\) (no, wrong).

Wait, correct formula:

In a right - triangle (formed by bisecting the equilateral triangle) with hypotenuse \(c = 11\) (side of equilateral triangle), and one side \(a=x\) (half of the base of equilateral triangle)

Using \(\cos60^{\circ}=\frac{a}{c}\) (since in the right - triangle, the angle adjacent to \(a\) is \(60^{\circ}\))

\(\cos60^{\circ}=\frac{1}{2}=\frac{x}{11}\)

\(x=\frac{11}{2}\)

The base of the equilateral triangle (which is \(2x\)) is \(11\). But the problem says "find the length of side \(x\)" (assuming \(x\) is half of the base). No, wait the problem is:

The triangle is equilateral. When we draw an altitude, we have a right - triangle with hypotenuse \(11\) (side of equilateral triangle) and base \(x\) (half of the base of the equilateral triangle)

Using the Pythagorean theorem:

\(11^{2}=x^{2}+(\sqrt{11^{2}-x^{2}})^{2}\)

Also, since it's a \(30 - 60 - 90\) triangle (angles \(30^{\circ},60^{\circ},90^{\circ}\)), the sides are in ratio \(1:\sqrt{3}:2\)

If the hypotenuse \(= 11\) (corresponds to \(2\) in the ratio \(1:\sqrt{3}:2\)), then the side opposite \(30^{\circ}\) (which is \(x\), half of the base of the equilateral triangle) is \(\frac{11}{2}\)

The base of the equilateral triangle (which is \(2x\)) is \(11\). But the problem asks for \(x\) (assuming \(x\) is half of the base).

Wait, no, re - reading the problem: "the triangle below is equilateral. Find the length of side \(x\)". When we bisect an equilateral triangle, the right - triangle has hypotenuse \(11\) (side of equilateral triangle), and \(x\) is half of the base.

Using \(\cos60^{\circ}=\frac{x}{11}\), \(x = 11\times\cos60^{\circ}=11\times\frac{1}{2}=\frac{11}{2}\)

But the base of the equilateral triangle is \(2x = 11\) (which is the side of the equilateral triangle, which is correct as it's equilateral)

Answer:

\(x=\frac{11}{2}\)