Question

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

Explanation:

Step1: Identify triangle properties

The triangle is equilateral, so all sides are equal, and the height (the segment of length 5) splits the base into two equal parts and forms a 30-60-90 right triangle with half the base and the side \( x \). In a 30-60-90 triangle, the sides are in the ratio \( 1 : \sqrt{3} : 2 \), where the side opposite 30° is the shortest, opposite 60° is \( \sqrt{3} \) times that, and the hypotenuse (which is \( x \) here) is twice the shortest side (half the base). Let the half - base length be \( y \). Then we know that \( \sin(60^{\circ})=\frac{5}{x} \) (since in the right triangle, the height is opposite the 60° angle and \( x \) is the hypotenuse). We know that \( \sin(60^{\circ})=\frac{\sqrt{3}}{2} \).

Step2: Solve for \( x \)

From \( \sin(60^{\circ})=\frac{5}{x} \), substitute \( \sin(60^{\circ})=\frac{\sqrt{3}}{2} \):
\[
\frac{\sqrt{3}}{2}=\frac{5}{x}
\]
Cross - multiply to get \( x\sqrt{3}=10 \). Then solve for \( x \) by dividing both sides by \( \sqrt{3} \):
\[
x = \frac{10}{\sqrt{3}}
\]
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3} \):
\[
x=\frac{10\sqrt{3}}{3}
\]
Wait, alternatively, since in an equilateral triangle, the height \( h \) of an equilateral triangle with side length \( s \) is given by \( h=\frac{\sqrt{3}}{2}s \). Here, \( h = 5 \), so we can also solve for \( s=x \) from \( 5=\frac{\sqrt{3}}{2}x \).
Multiply both sides by 2: \( 10=\sqrt{3}x \)
Then divide both sides by \( \sqrt{3} \): \( x=\frac{10}{\sqrt{3}}=\frac{10\sqrt{3}}{3} \) (after rationalizing). But wait, maybe I made a mistake in the angle. Wait, the height in an equilateral triangle bisects the angle, so the right triangle has angles 30°, 60°, 90°, with the side opposite 30° being \( \frac{x}{2} \), the side opposite 60° being 5, and the hypotenuse \( x \). So using the sine of 60°: \( \sin(60^{\circ})=\frac{5}{x} \), or using the cosine of 30°: \( \cos(30^{\circ})=\frac{5}{x} \), and \( \cos(30^{\circ})=\frac{\sqrt{3}}{2} \), so same as before. Or using the ratio of 30 - 60 - 90 triangle: the side opposite 60° is \( \frac{\sqrt{3}}{2} \) times the hypotenuse. Wait, no: in a 30 - 60 - 90 triangle, if the shorter leg (opposite 30°) is \( a \), the longer leg (opposite 60°) is \( a\sqrt{3} \), and the hypotenuse is \( 2a \). Here, the longer leg is 5 (opposite 60°), so \( a\sqrt{3}=5 \), so \( a = \frac{5}{\sqrt{3}} \), and the hypotenuse \( x = 2a=\frac{10}{\sqrt{3}}=\frac{10\sqrt{3}}{3} \).

Wait, but maybe the height is 5, and we can use the Pythagorean theorem. Let the side length be \( x \), then half of the side is \( \frac{x}{2} \). By Pythagoras: \( (\frac{x}{2})^2+5^2=x^2 \)
Expand: \( \frac{x^{2}}{4}+25=x^{2} \)
Subtract \( \frac{x^{2}}{4} \) from both sides: \( 25=x^{2}-\frac{x^{2}}{4}=\frac{3x^{2}}{4} \)
Multiply both sides by \( \frac{4}{3} \): \( x^{2}=\frac{100}{3} \)
Take square root: \( x = \frac{10}{\sqrt{3}}=\frac{10\sqrt{3}}{3} \) (since length is positive)

Answer:

\( \frac{10\sqrt{3}}{3} \)