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 opposite side and the angle. So, the right triangle formed has a 30 - 60 - 90 angle relationship. In a 30 - 60 - 90 triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite 30° is the shortest side (let's call it \(a\)), the side opposite 60° is \(a\sqrt{3}\), and the hypotenuse is \(2a\). Here, the side of length 5 is opposite the 30° angle? Wait, no. Wait, in the equilateral triangle, when we draw the altitude, it splits the equilateral triangle into two 30 - 60 - 90 triangles. The angle at the vertex of the equilateral triangle is 60°, so when we split it, the angle in the right triangle is 30°? Wait, no. Let's correct: In an equilateral triangle, each angle is 60°. When we draw the altitude from one vertex to the opposite side, it bisects the 60° angle into two 30° angles? No, wait, the altitude bisects the side and the angle. Wait, no, the angle at the vertex is 60°, so the altitude splits it into two 30° angles? Wait, no, 60° divided by 2 is 30°? Wait, no, 60° is the angle of the equilateral triangle. So the right triangle formed has angles 30°, 60°, 90°. The side adjacent to the 30° angle: Wait, let's denote the equilateral triangle as \(ABC\), with \(AB = BC=CA\). Let's draw the altitude from \(A\) to \(BC\), meeting at \(D\). Then \(AD\) is the altitude, \(BD = DC=\frac{BC}{2}\), and \(\angle BAD = 30°\), \(\angle ADB = 90°\), \(\angle ABD = 60°\). So in triangle \(ABD\), which is a 30 - 60 - 90 triangle, the side opposite 30° is \(BD\), the side opposite 60° is \(AD\), and the hypotenuse is \(AB\). Wait, but in our problem, the right triangle has one leg as 5, and the other leg as \(x\), and the hypotenuse? Wait, no, the diagram shows a right triangle with one leg 5, another leg \(x\), and the hypotenuse is a side of the equilateral triangle. Wait, maybe the 5 is the length of the shorter leg (opposite 30°), and \(x\) is the longer leg (opposite 60°)? Wait, no, in a 30 - 60 - 90 triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite 30° is the shortest (\(a\)), the side opposite 60° is \(a\sqrt{3}\), and the hypotenuse is \(2a\). Wait, but in our case, the right triangle: let's see, the angle with the right angle, and the other angle is 30°? Wait, maybe the 5 is the length of the side adjacent to the 30° angle? No, let's think again. The equilateral triangle has all sides equal, and when we draw the altitude, it creates two congruent 30 - 60 - 90 triangles. The side of length 5: is it the shorter leg (opposite 30°) or the longer leg (opposite 60°)? Wait, in the 30 - 60 - 90 triangle, the side opposite 30° is half the hypotenuse. Wait, the hypotenuse is the side of the equilateral triangle (let's call it \(s\)). Then the side opposite 30° is \(\frac{s}{2}\), and the side opposite 60° is \(\frac{s\sqrt{3}}{2}\). Wait, no, that's not right. Wait, the altitude (the height) of the equilateral triangle is \(\frac{s\sqrt{3}}{2}\), and the base is \(s\), so when we split it, each right triangle has base \(\frac{s}{2}\), height \(\frac{s\sqrt{3}}{2}\), and hypotenuse \(s\). Wait, maybe in our problem, the 5 is the length of the side opposite 30°, so \(\frac{s}{2}=5\)? No, that would make \(s = 10\), but then the other leg (the height) would be \(5\sqrt{3}\). Wait, but in the diagram, the right triangle has one leg 5, and the other leg \(x\), and the hypotenuse is a side of the equilateral triangle. Wait, maybe the angle in the right triangle is 60°, so the side adjacent to 60° is 5, and the side opposite is \(x\)? Wait, no, let's use trigonometry. In the right triangle, we have a right angle, and one angle is 60° (since the original triangle is equilateral, so the angle at the vertex is 60°, and the altitude splits it? Wait, no, maybe the right triangle is such that one angle is 60°, the right angle, and the other is 30°. Let's denote the angle opposite to side \(x\) as 60°, and the side adjacent to 60° is 5. Then \(\tan(60°)=\frac{x}{5}\). Since \(\tan(60°)=\sqrt{3}\), then \(x = 5\tan(60°)=5\sqrt{3}\)? Wait, no, that can't be. Wait, maybe the angle is 30°, so \(\tan(30°)=\frac{x}{5}\), but \(\tan(30°)=\frac{1}{\sqrt{3}}\), so \(x=\frac{5}{\sqrt{3}}\), but we need to rationalize the denominator. Wait, no, that would be if 5 is the adjacent side. Wait, maybe we should use the 30 - 60 - 90 triangle ratios. In a 30 - 60 - 90 triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite 30° is the shortest (\(a\)), the side opposite 60° is \(a\sqrt{3}\), and the hypotenuse is \(2a\). If the side of length 5 is opposite the 30° angle, then the side opposite 60° (which is \(x\)) is \(5\sqrt{3}\). Wait, no, if the side opposite 30° is \(a\), then the side opposite 60° is \(a\sqrt{3}\). So if \(a = 5\), then \(x = 5\sqrt{3}\)? Wait, no, that would be if 5 is the shorter leg. Wait, maybe the 5 is the longer leg. Wait, no, let's look at the diagram again. The right triangle has a right angle, one leg is 5, the other leg is \(x\), and the hypotenuse is a side of the equilateral triangle. Since the triangle is equilateral, all sides are equal, and the altitude (the perpendicular) splits the side into two equal parts, but also, the angle at the vertex is 60°, so the right triangle has angles 30°, 60°, 90°. So the side adjacent to 30° is \(x\), the side opposite 30° is 5, and the hypotenuse is the side of the equilateral triangle. Wait, no, in a 30 - 60 - 90 triangle, the side opposite 30° is the shortest, so if 5 is opposite 30°, then the hypotenuse is \(2\times5 = 10\), and the side opposite 60° is \(5\sqrt{3}\). But in our case, the side \(x\) is adjacent to 30°? Wait, maybe I got the angles wrong. Let's consider that in the equilateral triangle, each angle is 60°, so when we draw the altitude, it creates a right triangle where one angle is 60°, the right angle, and the other is 30°. So the side opposite 60° is the longer leg, and the side opposite 30° is the shorter leg. Let's say the shorter leg (opposite 30°) is \(x\), and the longer leg (opposite 60°) is 5. Then, in the 30 - 60 - 90 ratio, shorter leg : longer leg : hypotenuse = \(1:\sqrt{3}:2\). So if longer leg is \(5\), then shorter leg \(x=\frac{5}{\sqrt{3}}\), but we need to rationalize the denominator. Wait, no, that would be if the longer leg is \(a\sqrt{3}\), so \(a\sqrt{3}=5\), then \(a=\frac{5}{\sqrt{3}}=\frac{5\sqrt{3}}{3}\). But that seems complicated. Wait, maybe the 5 is the length of the altitude? No, the altitude of an equilateral triangle with side length \(s\) is \(\frac{s\sqrt{3}}{2}\). Wait, maybe the diagram is such that the right triangle has one leg 5, and the angle between the hypotenuse and the leg 5 is 60°, so we can use cosine or sine. Let's use cosine: \(\cos(60°)=\frac{5}{x}\)? No, \(\cos(60°)=\frac{\text{adjacent}}{\text{hypotenuse}}\). If the angle is 60°, adjacent side is 5, hypotenuse is \(x\), then \(\cos(60°)=\frac{5}{x}\), so \(x=\frac{5}{\cos(60°)}=\frac{5}{0.5}=10\). But that's not right. Wait, maybe the angle is 30°, so \(\cos(30°)=\frac{5}{x}\), so \(x=\frac{5}{\cos(30°)}=\frac{5}{\frac{\sqrt{3}}{2}}=\frac{10}{\sqrt{3}}=\frac{10\sqrt{3}}{3}\). No, this is getting confusing. Wait, let's start over. The triangle is equilateral, so all sides are equal, and all angles are 60°. The right triangle is formed by drawing a perpendicular from one vertex to the opposite side, so it splits the equilateral triangle into two congruent 30 - 60 - 90 triangles. In a 30 - 60 - 90 triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite 30° is the shortest (let's call it \(a\)), the side opposite 60° is \(a\sqrt{3}\), and the hypotenuse is \(2a\). Now, in the diagram, the right triangle has one leg 5, and the other leg \(x\), and the hypotenuse is a side of the equilateral triangle. Let's assume that the leg of length 5 is the side opposite 30°, so \(a = 5\). Then the side opposite 60° (which is \(x\)) is \(a\sqrt{3}=5\sqrt{3}\). Wait, but that would mean the hypotenuse is \(2a = 10\). But maybe the leg of length 5 is the side opposite 60°, so \(a\sqrt{3}=5\), then \(a=\frac{5}{\sqrt{3}}=\frac{5\sqrt{3}}{3}\), and the hypotenuse is \(2a=\frac{10\sqrt{3}}{3}\). But that doesn't seem right. Wait, maybe the diagram shows that the right triangle has a leg of 5, and the angle at the vertex (of the equilateral triangle) is 60°, so the right triangle has angles 60°, 30°, 90°, and the leg 5 is adjacent to 60°, so \(\tan(60°)=\frac{x}{5}\), so \(x = 5\tan(60°)=5\sqrt{3}\). Yes, that makes sense. Because \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\), so if \(\theta = 60°\), adjacent is 5, opposite is \(x\), then \(x = 5\tan(60°)=5\sqrt{3}\). Wait, but let's check with the 30 - 60 - 90 ratios. If the angle is 60°, then the opposite side is \(x\), adjacent is 5, hypotenuse is \(h\). Then \(\sin(60°)=\frac{x}{h}\), \(\cos(60°)=\frac{5}{h}\). Since \(\cos(60°)=0.5\), then \(h = \frac{5}{0.5}=10\), so \(x = h\sin(60°)=10\times\frac{\sqrt{3}}{2}=5\sqrt{3}\). Yes, that works. So the length of \(x\) is \(5\sqrt{3}\)? Wait, no, wait, if the hypotenuse is 10, then the side opposite 60° is \(10\times\frac{\sqrt{3}}{2}=5\sqrt{3}\), and the side opposite 30° is \(10\times\frac{1}{2}=5\). Ah! So the leg of length 5 is the side opposite 30°, and the leg \(x\) is the side opposite 60°, so \(x = 5\sqrt{3}\). Wait, but the problem says "Find the length of side \(x\)". So in the right triangle, the side \(x\) is opposite the 60° angle, and the side of length 5 is opposite the 30° angle. Therefore, using the 30 - 60 - 90 triangle ratios, the side opposite 60° is \(\sqrt{3}\) times the side opposite 30°. So \(x = 5\sqrt{3}\). Wait, but let's confirm with trigonometry. \(\tan(60°)=\frac{x}{5}\), so \(x = 5\tan(60°)=5\sqrt{3}\). Yes, that's correct.

Step2: Rationalize the denominator (if needed)

Wait, \(5\sqrt{3}\) already has a rational denominator? Wait, no, if we had a radical in the denominator, we would rationalize, but here the denominator is 1. Wait, maybe I made a mistake. Wait, maybe the side of length 5 is the adjacent side to the 30° angle, so the opposite side (x) is \(5\tan(30°)\). Wait, \(\tan(30°)=\frac{1}{\sqrt{3}}\), so \(x = 5\times\frac{1}{\sqrt{3}}=\frac{5}{\sqrt{3}}=\frac{5\sqrt{3}}{3}\). But that contradicts the earlier thought. Wait, now I'm confused. Let's look at the diagram again. The right triangle has a right angle, and one angle is 60° (from the equilateral triangle), so the other angle is 30°. So the sides: hypotenuse is the side of the equilateral triangle (let's call it \(s\)), one leg is 5 (let's say adjacent to 60°), and the other leg is \(x\) (opposite to 60°). Then, \(\cos(60°)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{5}{s}\), so \(s=\frac{5}{\cos(60°)}=\frac{5}{0.5}=10\). Then, \(\sin(60°)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{x}{s}\), so \(x = s\sin(60°)=10\times\frac{\sqrt{3}}{2}=5\sqrt{3}\). Alternatively, using the 30 - 60 - 90 ratio, the side opposite 30° is half the hypotenuse, so if the hypotenuse is \(s = 10\), then the side opposite 30° is 5 (which matches the diagram), and the side opposite 60° is \(5\sqrt{3}\) (which is \(x\)). So that's correct. So \(x = 5\sqrt{3}\). Wait, but the problem says "simplest radical form with a rational denominator". \(5\sqrt{3}\) is already in simplest radical form with a rational denominator because the denominator is 1. If we had a radical in the denominator, we would rationalize, but here we don't. So the length of \(x\) is \(5\sqrt{3}\).

Answer:

\(5\sqrt{3}\)