Question

triangle mrn is created when an equilateral triangle is folded in half.
what is the value of y?
2√3 units
4 units
4√3 units
8 units

Explanation:

Step1: Determine the side length of the equilateral triangle

Since triangle MRN is formed by folding an equilateral triangle in half, the side length of the equilateral triangle is \( 6 + 2=8 \)? Wait, no, wait. Wait, the segment NS is 6, and SM is 2, so the side length of the equilateral triangle should be \( 6 + 2 = 8 \)? Wait, no, maybe the original equilateral triangle has side length equal to \( MN + NS \)? Wait, no, when you fold an equilateral triangle in half, you get a 30 - 60 - 90 triangle. Wait, actually, in the right triangle MRN, angle at R is right, and since it's a folded equilateral triangle, angle at M should be 60 degrees, and angle at N should be 30 degrees. Wait, first, let's find the length of MR. Since the original triangle is equilateral, when folded, the length from M to S to N? Wait, no, the side length of the equilateral triangle: the segment from M to N (the hypotenuse of triangle MRN) should be equal to the side length of the equilateral triangle. Wait, MN is \( x + 2 \)? No, wait, the length of NS is 6, and SM is 2, so the total length of MN (the side of the equilateral triangle) is \( 6+2 = 8 \)? Wait, no, that can't be. Wait, maybe the side length of the equilateral triangle is \( 2 + 6=8 \)? Wait, no, let's think again. In a 30 - 60 - 90 triangle, the sides are in the ratio \( 1:\sqrt{3}:2 \), where the side opposite 30 degrees is the shortest one. Wait, triangle MRN is a right triangle (angle at R is right). Since it's formed by folding an equilateral triangle, the original equilateral triangle has all sides equal, so the length of MN (the hypotenuse of triangle MRN) should be equal to the side length of the equilateral triangle. Wait, but in the right triangle, if we consider angle at M: since the original triangle is equilateral, angle at M is 60 degrees, so in right triangle MRN, angle at M is 60 degrees, angle at N is 30 degrees. Then, the side opposite 30 degrees is MR, and the hypotenuse MN is twice MR. Wait, but MN is \( 6 + 2=8 \)? Wait, no, maybe the length of MN is 8? Wait, no, let's check the sides. Wait, the length of MS is 2, and NS is 6, so MN (the side of the equilateral triangle) is \( 2 + 6 = 8 \). So the hypotenuse of triangle MRN (which is MN) is 8? Wait, no, triangle MRN has hypotenuse MN? Wait, no, triangle MRN has vertices M, R, N, with right angle at R, so hypotenuse is MN. So MN is the hypotenuse, length equal to the side of the equilateral triangle. Since it's folded, the length from M to N is the side of the equilateral triangle, which is \( 2 + 6 = 8 \)? Wait, no, maybe I made a mistake. Wait, actually, when you fold an equilateral triangle in half, you get a 30 - 60 - 90 triangle, where the hypotenuse is the side of the equilateral triangle, the shorter leg (opposite 30 degrees) is half of the hypotenuse, and the longer leg (opposite 60 degrees) is \( \sqrt{3} \) times the shorter leg. Wait, let's find the length of MR first. Since the original triangle is equilateral, the length of MR should be equal to the length of MN? No, wait, no. Wait, when you fold the equilateral triangle along the altitude (from N to MR, let's say), then the altitude splits the equilateral triangle into two 30 - 60 - 90 triangles. So in triangle MRN, right - angled at R, angle at M is 60 degrees, angle at N is 30 degrees. The side MR is adjacent to the 60 - degree angle, RN (which is y) is opposite to the 60 - degree angle, and MN is the hypotenuse. Wait, the length of MN: from the diagram, MS is 2, and NS is 6, so MN = MS+NS = 2 + 6 = 8? Wait, no, that would mean the hypotenuse is 8, then the shorter leg (opposite 30 degrees) would be half of 8, which is 4, and the longer leg (opposite 60 degrees) would be \( 4\sqrt{3} \). Wait, but let's check the other way. Wait, maybe the side length of the equilateral triangle is 8? Then, when folded, the length from M to R is half of the base? No, in an equilateral triangle, the altitude splits the base into two equal parts. Wait, maybe I got the angles wrong. Let's re - examine: triangle MRN is right - angled at R. The original triangle is equilateral, so angle at M is 60 degrees, so in triangle MRN, angle at M is 60 degrees, so angle at N is 30 degrees. The hypotenuse MN is the side of the equilateral triangle. From the diagram, the length of MN is 8 (since MS = 2 and NS = 6, so 2+6 = 8). Then, in a 30 - 60 - 90 triangle, the sides are in the ratio \( 1:\sqrt{3}:2 \), where the side opposite 30 degrees is the shortest. The side opposite 30 degrees is MR, the side opposite 60 degrees is RN (y), and the hypotenuse MN is opposite the right angle. Wait, no: in triangle MRN, right - angled at R, angle at M is 60 degrees, so:
- Opposite 30 degrees (angle N): MR
- Opposite 60 degrees (angle M): RN (y)
- Hypotenuse (opposite right angle): MN

In a 30 - 60 - 90 triangle, the ratio of sides is: opposite 30° : opposite 60° : hypotenuse = \( 1:\sqrt{3}:2 \)

So if hypotenuse MN = 8, then the side opposite 60° (y) is \( \frac{\sqrt{3}}{2}\times\) hypotenuse? Wait, no, wait: let's denote the side opposite 30° as \( a \), opposite 60° as \( a\sqrt{3} \), and hypotenuse as \( 2a \). So if hypotenuse \( 2a=8 \), then \( a = 4 \), and the side opposite 60° (y) is \( a\sqrt{3}=4\sqrt{3} \)? Wait, no, that's not right. Wait, no: in a right - triangle, \( \sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}} \). So for angle at M (60°), \( \sin(60^{\circ})=\frac{y}{MN} \). We know that \( \sin(60^{\circ})=\frac{\sqrt{3}}{2} \), and if MN = 8, then \( y = MN\times\sin(60^{\circ})=8\times\frac{\sqrt{3}}{2}=4\sqrt{3} \). Wait, but let's check the other way. Wait, maybe the length of MN is 8? Wait, from the diagram, the segment from M to N: MS is 2, NS is 6, so MN = 2 + 6 = 8. Then, using the 30 - 60 - 90 triangle ratios, since it's a folded equilateral triangle, the length of MR should be 4 (half of 8, because in an equilateral triangle, the altitude splits the side into two equal parts? Wait, no, when you fold an equilateral triangle in half, the altitude (which is RN, y) splits the base (MR) into two equal parts? No, wait, in the equilateral triangle, the altitude from N to MR would split MR into two equal parts, but in this case, triangle MRN is right - angled at R, so R is on the base, and N is the top vertex. Wait, maybe I made a mistake in the length of MN. Wait, the length of NS is 6, and SM is 2, so the length of MN (the side of the equilateral triangle) is 6 + 2 = 8? Wait, no, that can't be, because in the diagram, the length from N to M is labeled with 6 and 2, so total length is 8. Then, in the right triangle MRN, right - angled at R, with hypotenuse MN = 8, and angle at M is 60 degrees (since it's an equilateral triangle), then we can use trigonometry. \( \cos(60^{\circ})=\frac{MR}{MN} \), and \( \sin(60^{\circ})=\frac{y}{MN} \). Since \( \cos(60^{\circ})=\frac{1}{2} \), then \( MR = MN\times\cos(60^{\circ})=8\times\frac{1}{2}=4 \). Then, \( \sin(60^{\circ})=\frac{y}{8} \), so \( y = 8\times\frac{\sqrt{3}}{2}=4\sqrt{3} \). Wait, but let's check the other option. Wait, maybe the side length of the equilateral triangle is 8? Then, when folded, the length of MR is 4 (half of 8, because in an equilateral triangle, the altitude splits the base into two equal parts), and then using Pythagoras' theorem: \( y=\sqrt{MN^{2}-MR^{2}} \). If MN = 8 and MR = 4, then \( y=\sqrt{8^{2}-4^{2}}=\sqrt{64 - 16}=\sqrt{48}=4\sqrt{3} \). Yes, that makes sense. So \( y = 4\sqrt{3} \) units.

Step2: Verify using Pythagoras' theorem

We know that in right triangle MRN, \( MN^{2}=MR^{2}+RN^{2} \). We found that \( MN = 8 \) (since \( 2 + 6=8 \)), \( MR = 4 \) (because in the equilateral triangle, the altitude splits the base into two equal parts, so MR is half of the side length? Wait, no, the side length of the equilateral triangle is 8, so the base (MR) when folded? Wait, no, MR is one of the legs. Wait, we found MR = 4 using \( \cos(60^{\circ})=\frac{MR}{MN} \), so \( MR = 8\times\frac{1}{2}=4 \). Then, using Pythagoras: \( y=\sqrt{MN^{2}-MR^{2}}=\sqrt{8^{2}-4^{2}}=\sqrt{64 - 16}=\sqrt{48}=4\sqrt{3} \).

Answer:

\( 4\sqrt{3} \) units (the option "4√3 units")