Question

the triangle below is equilateral. find the length of side ( x ) to the nearest tenth.

Explanation:

Step1: Recall properties of equilateral triangle

In an equilateral triangle, the altitude (the perpendicular from a vertex to the opposite side) bisects the side. Also, in a 30 - 60 - 90 triangle (formed by the altitude, half of a side, and a side of the equilateral triangle), the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite \(30^{\circ}\) is the shortest one (let's say length \(a\)), the side opposite \(60^{\circ}\) is \(a\sqrt{3}\), and the hypotenuse (side of the equilateral triangle) is \(2a\).

Looking at the right - triangle in the figure, the side with length \(\sqrt{3}\) is opposite the \(60^{\circ}\) angle (since the altitude in an equilateral triangle creates a \(30 - 60 - 90\) triangle), and the side \(x\) is half of the side of the equilateral triangle (because the altitude bisects the base of the equilateral triangle).

Let's denote the side of the equilateral triangle as \(s\). In the 30 - 60 - 90 triangle, if the side opposite \(60^{\circ}\) is \(\sqrt{3}\), and we know that in a 30 - 60 - 90 triangle, the side opposite \(60^{\circ}\) is \(a\sqrt{3}\) (where \(a\) is the side opposite \(30^{\circ}\), which is equal to \(x\) in our case). So we have the equation:

\(a\sqrt{3}=\sqrt{3}\)

Step2: Solve for \(a\) (which is \(x\))

From the equation \(a\sqrt{3}=\sqrt{3}\), we can divide both sides by \(\sqrt{3}\):

\(a = \frac{\sqrt{3}}{\sqrt{3}}=1\)? Wait, no, wait. Wait, actually, the hypotenuse of the right - triangle is the side of the equilateral triangle. Wait, maybe I mixed up. Let's re - consider.

In an equilateral triangle with side length \(s\), the altitude \(h\) is given by \(h=\frac{\sqrt{3}}{2}s\). In our right - triangle, the vertical side (the altitude) is not given, but the side with length \(\sqrt{3}\) is the altitude? Wait, no, the side labeled \(\sqrt{3}\) is the hypotenuse? Wait, no, the right - angle is between \(x\) and the altitude. So the right - triangle has legs \(x\) (half of the side of the equilateral triangle) and the altitude \(h\), and hypotenuse \(s\) (side of the equilateral triangle). Also, we know that in an equilateral triangle, \(h = \frac{\sqrt{3}}{2}s\), and \(x=\frac{s}{2}\) (since the altitude bisects the base).

We are given that the length of the side (let's say the altitude - related side) is \(\sqrt{3}\). Wait, maybe a better approach: In the right - triangle, we can use trigonometric ratios. Let's consider the angle at the vertex of the equilateral triangle. In an equilateral triangle, each angle is \(60^{\circ}\), and when we draw the altitude, it splits the \(60^{\circ}\) angle into two \(30^{\circ}\) angles? No, wait, no. The altitude in an equilateral triangle makes a \(90^{\circ}\) angle with the base, so the two non - right angles of the right - triangle are \(30^{\circ}\) and \(60^{\circ}\).

Let's use the sine function. \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). The opposite side to the \(60^{\circ}\) angle is \(\sqrt{3}\), and the hypotenuse of the right - triangle is the side of the equilateral triangle, let's call it \(s\). But also, the adjacent side to the \(60^{\circ}\) angle is \(x\) (since \(x\) is half of the base of the equilateral triangle, and the base is equal to the side of the equilateral triangle).

Wait, another way: In a 30 - 60 - 90 triangle, the ratio of the sides is \(1:\sqrt{3}:2\). If the side with length \(\sqrt{3}\) is the longer leg (opposite \(60^{\circ}\)), then the shorter leg (opposite \(30^{\circ}\)) is \(x\), and we know that \(\frac{\text{longer leg}}{\text{shorter leg}}=\sqrt{3}\). So \(\frac{\sqrt{3}}{x}=\sqrt{3}\).

Step3: Solve for \(x\)

From \(\frac{\sqrt{3}}{x}=\sqrt{3}\), we can multiply both sides by \(x\) to get \(\sqrt{3}=x\sqrt{3}\). Then divide both sides by \(\sqrt{3}\) (assuming \(\sqrt{3} eq0\), which it isn't), we get \(x = 1\)? Wait, that can't be right. Wait, maybe I got the sides mixed up.

Wait, let's start over. Let the side of the equilateral triangle be \(s\). The altitude \(h\) of an equilateral triangle is given by \(h=\frac{\sqrt{3}}{2}s\). In the right - triangle, the hypotenuse is \(s\), one leg is \(h\) and the other leg is \(\frac{s}{2}\) (since the altitude bisects the base). We are given that one of the legs (the one with length \(\sqrt{3}\)) is the altitude? Wait, if \(h = \sqrt{3}\), and \(h=\frac{\sqrt{3}}{2}s\), then we can solve for \(s\):

\(\frac{\sqrt{3}}{2}s=\sqrt{3}\)

Multiply both sides by \(\frac{2}{\sqrt{3}}\):

\(s=\frac{2\sqrt{3}}{\sqrt{3}} = 2\)

Then, since \(x\) is half of the side of the equilateral triangle (because the altitude bisects the base), \(x=\frac{s}{2}=\frac{2}{2}=1\)? But that seems too simple. Wait, maybe the side with length \(\sqrt{3}\) is the side of the equilateral triangle? No, the right - triangle has hypotenuse equal to the side of the equilateral triangle. Wait, no, the right - triangle is formed by the altitude, half of the base, and a side of the equilateral triangle. So the hypotenuse of the right - triangle is the side of the equilateral triangle, one leg is the altitude (\(h=\frac{\sqrt{3}}{2}s\)), and the other leg is \(\frac{s}{2}\) (which is \(x\)).

Wait, maybe the problem is that the side labeled \(\sqrt{3}\) is the altitude. So \(h = \sqrt{3}\), and we know that \(h=\frac{\sqrt{3}}{2}s\), so \(\sqrt{3}=\frac{\sqrt{3}}{2}s\), then \(s = 2\). Then \(x=\frac{s}{2}=1\)? But 1 to the nearest tenth is 1.0. Wait, maybe I made a mistake in the ratio.

Wait, let's use trigonometry. Let's consider the angle at the vertex of the equilateral triangle. The angle is \(60^{\circ}\), and the right - triangle has an angle of \(30^{\circ}\) at the base (since the altitude splits the equilateral triangle into two 30 - 60 - 90 triangles). So in the right - triangle, \(\sin(30^{\circ})=\frac{x}{s}\), and \(\cos(30^{\circ})=\frac{\sqrt{3}}{s}\). Since \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), we have \(\frac{\sqrt{3}}{s}=\frac{\sqrt{3}}{2}\), so \(s = 2\). Then \(\sin(30^{\circ})=\frac{1}{2}=\frac{x}{2}\), so \(x = 1\). Rounding to the nearest tenth, \(x = 1.0\).

Wait, but maybe the side with length \(\sqrt{3}\) is the hypotenuse? No, because the right - angle is between \(x\) and the altitude, so the hypotenuse should be the side of the equilateral triangle, which is longer than \(x\) and longer than the altitude. If \(\sqrt{3}\) is the hypotenuse, then \(s=\sqrt{3}\), and \(x=\frac{s}{2}=\frac{\sqrt{3}}{2}\approx0.9\)? Wait, now I'm confused. Let's re - examine the figure.

The figure shows an equilateral triangle with a perpendicular from a vertex to the opposite side, creating a right - triangle. The right - triangle has one leg as \(x\) (half of the base of the equilateral triangle), one leg as the altitude \(h\), and hypotenuse as the side of the equilateral triangle \(s\). The side labeled \(\sqrt{3}\) is either \(h\) or \(s\) or \(x\).

Case 1: If \(\sqrt{3}\) is the altitude (\(h\)):

We know that for an equilateral triangle, \(h=\frac{\sqrt{3}}{2}s\), so \(s = \frac{2h}{\sqrt{3}}\). Substituting \(h = \sqrt{3}\), we get \(s=\frac{2\sqrt{3}}{\sqrt{3}}=2\). Then \(x=\frac{s}{2}=1\).

Case 2: If \(\sqrt{3}\) is the side of the equilateral triangle (\(s\)):

Then \(x=\frac{s}{2}=\frac{\sqrt{3}}{2}\approx0.866\approx0.9\) (to the nearest tenth). But we need to check which case is correct.

In an equilateral triangle, the altitude \(h=\frac{\sqrt{3}}{2}s\), so \(h\) is less than \(s\) (since \(\frac{\sqrt{3}}{2}\approx0.866\lt1\) when \(s = 2\), \(h=\sqrt{3}\approx1.732\), and \(s = 2\), so \(h\lt s\) is true. Wait, when \(s=\sqrt{3}\), \(h=\frac{\sqrt{3}}{2}\times\sqrt{3}=\frac{3}{2}=1.5\), and \(x=\frac{\sqrt{3}}{2}\approx0.866\). But the length of \(\sqrt{3}\approx1.732\), and if \(\sqrt{3}\) is the hypotenuse (side of equilateral triangle), then the altitude \(h = 1.5\) which is less than \(\sqrt{3}\), and \(x=\frac{\sqrt{3}}{2}\approx0.866\). But the problem is that in the right - triangle, the side labeled \(\sqrt{3}\) is one of the sides. Let's use the Pythagorean theorem.

Let the side of the equilateral triangle be \(s\), so \(x=\frac{s}{2}\), and the altitude \(h=\sqrt{s^{2}-x^{2}}=\sqrt{s^{2}-\frac{s^{2}}{4}}=\sqrt{\frac{3s^{2}}{4}}=\frac{\sqrt{3}s}{2}\).

If we assume that the side with length \(\sqrt{3}\) is the altitude \(h\), then \(h=\frac{\sqrt{3}s}{2}=\sqrt{3}\), so \(s = 2\), and \(x=\frac{s}{2}=1\).

If we assume that the side with length \(\sqrt{3}\) is the hypotenuse \(s\), then \(x=\frac{s}{2}=\frac{\sqrt{3}}{2}\approx0.9\), and \(h=\frac{\sqrt{3}s}{2}=\frac{\sqrt{3}\times\sqrt{3}}{2}=\frac{3}{2}=1.5\). But in the figure, the side labeled \(\sqrt{3}\) is adjacent to the right - angle? No, the right - angle is between \(x\) and the altitude, so the hypotenuse is the side of the equilateral triangle, and the other two sides are \(x\) and the altitude.

Wait, the key is that in the 30 - 60 - 90 triangle, the ratio of the sides is \(1:\sqrt{3}:2\). The side opposite \(30^{\circ}\) is the shortest side (let's say length \(a\)), the side opposite \(60^{\circ}\) is \(a\sqrt{3}\), and the hypotenuse is \(2a\).

Looking at the right - triangle, if the side with length \(\sqrt{3}\) is the side opposite \(60^{\circ}\), then \(a\sqrt{3}=\sqrt{3}\), so \(a = 1\) (where \(a\) is the side opposite \(30^{\circ}\), which is \(x\)). So \(x = 1\), and the hypotenuse (side of the equilateral triangle) is \(2a=2\).

So \(x = 1\), and to the nearest tenth, \(x = 1.0\).

Answer:

\(1.0\)