Question

the triangle below is equilateral. find the length of side x to the nearest tenth.
(image of an equilateral triangle with a perpendicular segment from a vertex to a side, forming a right angle, with the side of the right triangle labeled x and the hypotenuse labeled 12)
answer attempt 1 out of 9
x = blank submit answer

Explanation:

Step1: Define right triangle components

In the right triangle formed by the altitude, one leg is half of $x$ (let this be $\frac{x}{2}$), the other leg is the altitude, and hypotenuse = 12. Also, in an equilateral triangle, $\sin(60^\circ)=\frac{\text{altitude}}{12}$, and $\cos(60^\circ)=\frac{\frac{x}{2}}{12}$.

Step2: Solve for $x$ using cosine

$\cos(60^\circ)=\frac{x/2}{12}$
Since $\cos(60^\circ)=\frac{1}{2}$, substitute:
$\frac{1}{2}=\frac{x}{24}$
Multiply both sides by 24:
$x=24\times\frac{1}{2}=12$
Wait, correct approach: Use the right triangle where the side of the equilateral triangle is 12, and $x$ is the side, so the right triangle has hypotenuse 12, adjacent side $\frac{x}{2}$, angle $60^\circ$.
$\cos(60^\circ)=\frac{x/2}{12}$
$\frac{1}{2}=\frac{x}{24}$
$x=12$
Wait, no, error: The altitude creates a 30-60-90 triangle. The side opposite 30° is $\frac{x}{2}$, hypotenuse is 12. So $\frac{x}{2}=12\times\sin(30^\circ)=12\times\frac{1}{2}=6$, so $x=12$? No, wait, no: The hypotenuse of the right triangle is the side of the equilateral triangle, so if the equilateral triangle has side length $s$, then the altitude is $\frac{\sqrt{3}}{2}s$. But here, the right triangle has one leg $x$ (wait, no, the labeled side is 12, which is the side of the equilateral triangle, and $x$ is the side? No, re-read: The triangle is equilateral, so all sides are equal. The altitude is drawn to side $x$, so the right triangle has hypotenuse 12 (side of equilateral triangle), one leg $\frac{x}{2}$, and the altitude. Since it's equilateral, $x=12$? No, that can't be. Wait, no, the right triangle: hypotenuse is the side of the equilateral triangle (12), one leg is $\frac{x}{2}$, angle at the vertex is 60°, so $\cos(60^\circ)=\frac{x/2}{12}$
$\frac{1}{2}=\frac{x}{24}$
$x=12$
Wait, but that would mean all sides are 12, which makes sense for equilateral. But wait, maybe I misread: The side labeled 12 is the side, and $x$ is the side, so they are equal. But no, the altitude is drawn to $x$, so $x$ is a side, so $x=12$? But that seems too easy. Wait, no, maybe the right triangle has leg $x$ (the altitude), hypotenuse 12 (side of equilateral triangle). Then $x$ is the altitude: $x=12\times\sin(60^\circ)=12\times\frac{\sqrt{3}}{2}=6\sqrt{3}\approx10.4$
Ah, that's the mistake! The labeled side 12 is the side of the equilateral triangle, $x$ is the altitude (the perpendicular segment). So correct step:

Step1: Identify right triangle parts

Equilateral triangle side = 12, $x$ is the altitude. For equilateral triangle, altitude $h = s\times\sin(60^\circ)$, where $s=12$.

Step2: Calculate $x$

$x=12\times\frac{\sqrt{3}}{2}=6\sqrt{3}$
$6\sqrt{3}\approx6\times1.732=10.392$

Step3: Round to nearest tenth

$x\approx10.4$

Answer:

10.4