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

Explanation:

Step1: Recall equilateral triangle properties

In an equilateral triangle, all sides are equal, and the altitude (the perpendicular segment from a vertex to the opposite side) bisects the base. So the hypotenuse of the right triangle formed (the side ( x )) is equal to the side of the equilateral triangle, and one leg is 6 (half of the base? Wait, no—wait, the right triangle here: the leg with length 6 is adjacent to a 30° angle? Wait, no, in an equilateral triangle, each angle is 60°. When we draw the altitude, it splits the equilateral triangle into two 30-60-90 right 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 one (let's say length ( a )), the side opposite 60° is ( a\sqrt{3} ), and the hypotenuse (opposite 90°) is ( 2a ). Wait, but in our case, the leg of the right triangle (the one with the right angle) that is given is 6. Wait, actually, in the equilateral triangle, the altitude divides the base into two equal parts. Wait, no—wait, the side of the equilateral triangle is ( x ), and the segment with length 6 is half of the side? Wait, no, looking at the diagram: the right triangle has one leg 6, hypotenuse ( x ), and the other leg is the altitude. Wait, no—wait, the original triangle is equilateral, so all sides are ( x ). The right triangle is formed by drawing a perpendicular from one vertex to the opposite side, so the hypotenuse of the right triangle is ( x ) (a side of the equilateral triangle), one leg is 6 (half of the base? Wait, no, the base of the equilateral triangle is ( x ), so when we draw the altitude, it bisects the base into two segments of length ( \frac{x}{2} )? Wait, no, that can't be. Wait, maybe I got it reversed. Wait, the right triangle: the leg with length 6 is adjacent to the 60° angle? Wait, no, let's think again. In an equilateral triangle, all angles are 60°. When we drop a perpendicular from a vertex to the opposite side, we get two right triangles, each with angles 30°, 60°, 90°. The side opposite 30° is half of the hypotenuse (which is the side of the equilateral triangle). Wait, no: the hypotenuse of the right triangle is the side of the equilateral triangle (( x )), the side opposite 30° is half of the base (so length ( \frac{x}{2} )), and the side opposite 60° is the altitude. But in our diagram, the leg of the right triangle (the one with the right angle) is 6. Wait, maybe the 6 is the side opposite 30°? Wait, no, that would mean ( \frac{x}{2} = 6 ), so ( x = 12 ), but that doesn't involve the right triangle's other leg. Wait, no, maybe the right triangle has one leg 6, and the angle adjacent to that leg is 60°, so we can use trigonometry. Let's use cosine: ( \cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} ). The adjacent side to the 60° angle is 6, and the hypotenuse is ( x ). Wait, ( \cos(60^\circ) = 0.5 ), so ( 0.5 = \frac{6}{x} ), so ( x = \frac{6}{0.5} = 12 )? But that seems too simple. Wait, no, maybe the 6 is the side opposite 30°, so ( \sin(30^\circ) = \frac{6}{x} ), and ( \sin(30^\circ) = 0.5 ), so ( 0.5 = \frac{6}{x} ), so ( x = 12 ). Wait, but that would mean the side of the equilateral triangle is 12, and the base is 12, so half of the base is 6, which matches the segment with length 6. Oh! So that's the case. So in the equilateral triangle, the altitude bisects the base into two segments of length ( \frac{x}{2} ). Wait, no—wait, no, the segment with length 6 is half of the side? Wait, no, the side of the equilateral triangle is ( x ), so when we draw the altitude, it bisects the base (which is also length ( x )) into two segments of length ( \frac{x}{2} ). But in the right triangle, the leg adjacent to the 60° angle is ( \frac{x}{2} ), and the hypotenuse is ( x ). Wait, no, that can't be. Wait, I think I made a mistake. Let's look at the diagram again: the right triangle has one leg 6, hypotenuse ( x ), and the other leg is the altitude. The angle at the vertex of the equilateral triangle (the one connected to the hypotenuse ( x )) is 60°, so in the right triangle, the angle between the hypotenuse ( x ) and the leg 6 is 60°, so we can use the cosine of 60°: ( \cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{6}{x} ). Since ( \cos(60^\circ) = 0.5 ), then ( 0.5 = \frac{6}{x} ), so ( x = \frac{6}{0.5} = 12 )? But that seems too straightforward. Wait, no, maybe the 6 is the side opposite 60°, so we use sine: ( \sin(60^\circ) = \frac{6}{x} ), so ( x = \frac{6}{\sin(60^\circ)} ). Let's calculate that. ( \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 ), so ( x = \frac{6}{0.866} \approx 6.9 )? Wait, that's different. Wait, now I'm confused. Let's clarify the diagram: the right triangle has a right angle, one leg is 6, hypotenuse is ( x ), and the angle opposite the leg 6 is 60° (since the original triangle is equilateral, each angle is 60°). So in the right triangle, the angle at the vertex (not the right angle) is 60°, so the side opposite 60° is 6, and the hypotenuse is ( x ). So ( \sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{6}{x} ). Therefore, ( x = \frac{6}{\sin(60^\circ)} ). Let's compute that. ( \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660 ), so ( x = \frac{6}{0.8660} \approx 6.928 ), which rounds to 6.9? Wait, no, that can't be. Wait, maybe the 6 is adjacent to the 60° angle. Let's try cosine: ( \cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{6}{x} ), so ( x = \frac{6}{\cos(60^\circ)} = \frac{6}{0.5} = 12 ). But then why is there a right triangle? Wait, maybe the diagram is such that the segment with length 6 is half of the side, and the hypotenuse is the side. Wait, no—wait, in an equilateral triangle, the altitude ( h ) can be calculated by ( h = \frac{\sqrt{3}}{2} x ), where ( x ) is the side length. But in the right triangle, the legs are ( \frac{x}{2} ) (half the base) and ( h ) (the altitude), and the hypotenuse is ( x ) (the side). So if the segment with length 6 is ( \frac{x}{2} ), then ( \frac{x}{2} = 6 ), so ( x = 12 ). But if the segment with length 6 is the altitude, then ( h = 6 = \frac{\sqrt{3}}{2} x ), so ( x = \frac{12}{\sqrt{3}} = 4\sqrt{3} \approx 6.9 ). Ah! Now I see the confusion. The diagram: the right triangle has a right angle, and the leg with length 6—if that leg is the altitude, then ( h = 6 ), so we can solve for ( x ). Let's check the diagram again: the right angle is between the segment of length 6 and the other leg (the altitude). Wait, the original triangle is equilateral, so all sides are ( x ). The right triangle is formed by drawing a perpendicular from one vertex to the opposite side, so the two legs of the right triangle are: one leg is the altitude (let's say length ( h )), the other leg is half of the base (length ( \frac{x}{2} )), and the hypotenuse is ( x ) (the side of the equilateral triangle). Now, in the diagram, the leg with length 6—if that's the half of the base, then ( \frac{x}{2} = 6 ), so ( x = 12 ). But if the leg with length 6 is the altitude, then ( h = 6 = \frac{\sqrt{3}}{2} x ), so ( x = \frac{12}{\sqrt{3}} = 4\sqrt{3} \approx 6.9 ). So which is it? Looking at the diagram: the right angle is between the segment of length 6 and the side ( x )? Wait, the diagram shows a right angle between the segment of length 6 and the other segment (the one going to the bottom vertex). So the right triangle has hypotenuse ( x ), one leg 6, and the other leg is the altitude. So the angle at the top (between the hypotenuse ( x ) and the leg 6) is 60°, because the original triangle is equilateral. So in the right triangle, angle = 60°, adjacent side = 6, hypotenuse = ( x ). So ( \cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{6}{x} ), so ( x = \frac{6}{\cos(60^\circ)} = \frac{6}{0.5} = 12 ). But that would mean the side is 12, and half the base is 6, so the altitude is ( \frac{\sqrt{3}}{2} \times 12 = 6\sqrt{3} \approx 10.4 ), which is the other leg. But in the diagram, the other leg (the altitude) is not labeled. Wait, maybe the diagram is such that the segment of length 6 is the altitude, so ( h = 6 ), and we need to find the side ( x ). Then, using the formula for the altitude of an equilateral triangle: ( h = \frac{\sqrt{3}}{2} x ), so ( x = \frac{2h}{\sqrt{3}} = \frac{2 \times 6}{\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3} \approx 6.9 ). Ah! That must be it. Because the right angle is between the altitude (6) and the base (half of ( x )), so the leg with length 6 is the altitude, and the other leg is ( \frac{x}{2} ), and the hypotenuse is ( x ). So using Pythagoras: ( \left( \frac{x}{2} \right)^2 + 6^2 = x^2 ). Let's solve that. ( \frac{x^2}{4} + 36 = x^2 ). Subtract ( \frac{x^2}{4} ) from both sides: ( 36 = \frac{3x^2}{4} ). Multiply both sides by 4: ( 144 = 3x^2 ). Divide by 3: ( 48 = x^2 ). Take square root: ( x = \sqrt{48} = 4\sqrt{3} \approx 6.928 ), which rounds to 6.9. Yes, that makes sense. So the correct approach is to use the Pythagorean theorem in the right triangle formed by the altitude, half the base, and the side of the equilateral triangle.

Step2: Apply Pythagorean theorem

Let the side length of the equilateral triangle be ( x ). When we draw the altitude, it bisects the base into two segments of length ( \frac{x}{2} ). The right triangle has legs ( \frac{x}{2} ) and 6 (the altitude), and hypotenuse ( x ) (the side of the equilateral triangle). By the Pythagorean theorem:
[ \left( \frac{x}{2} \right)^2 + 6^2 = x^2 ]
Expand and simplify:
[ \frac{x^2}{4} + 36 = x^2 ]
Subtract ( \frac{x^2}{4} ) from both sides:
[ 36 = x^2 - \frac{x^2}{4} = \frac{3x^2}{4} ]
Multiply both sides by ( \frac{4}{3} ):
[ x^2 = 36 \times \frac{4}{3} = 48 ]
Take the square root:
[ x = \sqrt{48} = 4\sqrt{3} \approx 6.928 ]

Step3: Round to the nearest tenth

( 6.928 ) rounded to the nearest tenth is ( 6.9 ).

Answer:

( 6.9 )