Question

line segment yv of rectangle yvwx measures 24 units.
what is the length of line segment yx?
8 units
8√3 units
12 units
12√3 units

Explanation:

Step1: Identify the triangle type

In rectangle YVWX, triangle YXV is a 30 - 60 - 90 triangle (since angle at X is \(30^{\circ}\), angle at V is \(60^{\circ}\) and angle at Y is \(90^{\circ}\) as it's a rectangle). The side YV (opposite to \(30^{\circ}\) angle? Wait, no. Wait, YV is 24 units. Wait, in a 30 - 60 - 90 triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite \(30^{\circ}\) is the shortest one, let's call it \(x\), opposite \(60^{\circ}\) is \(x\sqrt{3}\) and hypotenuse is \(2x\). Wait, in triangle YXV, angle at X is \(30^{\circ}\), angle at V is \(60^{\circ}\), angle at Y is \(90^{\circ}\). So side YV is adjacent to \(60^{\circ}\)? Wait, no. Wait, YV is 24 units. Wait, YV is a side of the rectangle, so YV is equal to XW? Wait, no, in the rectangle YVWX, YV is the top side, XW is the bottom side, YX and VW are the vertical sides. The diagonal is XV. Wait, maybe I made a mistake. Let's re - draw: Y---V (length 24), V---W (vertical), W---X (horizontal), X---Y (vertical). Diagonal is X to V. So triangle XWV is a right triangle? No, triangle YXV: Y to X is vertical, X to V is diagonal, Y to V is horizontal (length 24). Angle at X is \(30^{\circ}\), angle at V is \(60^{\circ}\), angle at Y is \(90^{\circ}\). So in triangle YXV, angle at X is \(30^{\circ}\), so the side opposite to \(30^{\circ}\) is YV? Wait, no. Wait, angle at X is \(30^{\circ}\), so the side opposite is YV. Wait, YV is 24 units? Wait, no, if angle at X is \(30^{\circ}\), then in right - triangle YXV, \(\sin(30^{\circ})=\frac{YV}{XV}\), \(\cos(30^{\circ})=\frac{YX}{XV}\), \(\tan(30^{\circ})=\frac{YV}{YX}\). Wait, maybe it's a 30 - 60 - 90 triangle where YV is the side opposite \(30^{\circ}\). Wait, no, let's check the ratio. In a 30 - 60 - 90 triangle, if the side opposite \(30^{\circ}\) is \(a\), then the side opposite \(60^{\circ}\) is \(a\sqrt{3}\), and hypotenuse is \(2a\). Wait, maybe YV is the side opposite \(30^{\circ}\)? No, angle at X is \(30^{\circ}\), so the side opposite angle X (\(30^{\circ}\)) is YV. So \(YV = a\), \(YX=a\sqrt{3}\), and \(XV = 2a\). But we know that YV is 24? Wait, no, that can't be. Wait, maybe I got the angle wrong. Wait, the diagonal is XV. In the rectangle, YV is 24 (horizontal side). So in triangle YXV, angle at X is \(30^{\circ}\), so \(\tan(30^{\circ})=\frac{YV}{YX}\). Wait, \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}=\frac{YV}{YX}\), but YV is 24? No, that would make YX = 24\(\sqrt{3}\), which is not an option. Wait, maybe angle at V is \(60^{\circ}\), so in triangle YXV, angle at V is \(60^{\circ}\), so \(\tan(60^{\circ})=\sqrt{3}=\frac{YX}{YV}\). Wait, YV is 24? No, that would make YX = 24\(\sqrt{3}\), not an option. Wait, maybe YV is the hypotenuse? Wait, no, YV is a side of the rectangle, so it's a leg. Wait, maybe the length of YV is not 24 as the hypotenuse? Wait, no, the problem says "Line segment YV of rectangle YVWX measures 24 units". So YV is 24 units (horizontal side). Then in the right - triangle YXV (right - angled at Y), angle at X is \(30^{\circ}\), so \(\cos(30^{\circ})=\frac{YX}{XV}\), \(\sin(30^{\circ})=\frac{YV}{XV}\). Since \(\sin(30^{\circ})=\frac{1}{2}=\frac{YV}{XV}\), then \(XV = 2\times YV=48\)? No, that's not right. Wait, maybe I mixed up the sides. Wait, let's look at the options. The options are 8, \(8\sqrt{3}\), 12, \(12\sqrt{3}\). So maybe YV is not 24 as the horizontal side, but maybe the hypotenuse? Wait, no, the problem says "Line segment YV of rectangle YVWX measures 24 units". So YV is a side of the rectangle, so it's a leg. Wait, maybe the triangle is such that YV is the side adjacent to \(60^{\circ}\), and YX is the side opposite \(60^{\circ}\). In a 30 - 60 - 90 triangle, if the side adjacent to \(60^{\circ}\) is \(x\), then the side opposite \(60^{\circ}\) is \(x\sqrt{3}\), and the hypotenuse is \(2x\). Wait, if angle at V is \(60^{\circ}\), then in triangle YXV, \(\cos(60^{\circ})=\frac{YV}{XV}\), \(\sin(60^{\circ})=\frac{YX}{XV}\). But we know YV = 24? No, that doesn't fit. Wait, maybe the length of YV is not 24, but the length of the diagonal? No, the problem says YV is 24. Wait, maybe I made a mistake in the triangle. Let's consider that triangle YXV is a 30 - 60 - 90 triangle, and YV is the side opposite \(60^{\circ}\). Wait, in a 30 - 60 - 90 triangle, the side opposite \(60^{\circ}\) is \(a\sqrt{3}\), side opposite \(30^{\circ}\) is \(a\), hypotenuse is \(2a\). If YV (opposite \(60^{\circ}\)) is 24, then \(a\sqrt{3}=24\), \(a = \frac{24}{\sqrt{3}} = 8\sqrt{3}\)? No, that's not matching. Wait, maybe YV is the hypotenuse? If YV is the hypotenuse (length 24), then the side opposite \(30^{\circ}\) is \(\frac{24}{2}=12\), and the side opposite \(60^{\circ}\) is \(12\sqrt{3}\). Wait, YX is the side opposite \(60^{\circ}\)? Wait, angle at X is \(30^{\circ}\), so the side opposite angle X is YV, and the side opposite angle V (\(60^{\circ}\)) is YX. So if hypotenuse XV, then \(YV=\frac{1}{2}XV\) (since \(\sin(30^{\circ})=\frac{YV}{XV}\)), and \(YX=\frac{\sqrt{3}}{2}XV\). But we know YV is 24? No, that can't be. Wait, I think I messed up the side labels. Let's start over. The rectangle is YVWX, so the vertices are in order: Y, V, W, X. So YV is a horizontal side (length 24), VW is a vertical side, WX is a horizontal side (equal to YV), and XY is a vertical side (equal to VW). The diagonal is XV, connecting X to V. So triangle XYV is a right - triangle with right angle at Y. So angle at X is \(30^{\circ}\), angle at V is \(60^{\circ}\), right angle at Y. So in right - triangle XYV:
- \(\sin(30^{\circ})=\frac{YV}{XV}\)
- \(\cos(30^{\circ})=\frac{XY}{XV}\)
- \(\tan(30^{\circ})=\frac{YV}{XY}\)

We know that YV = 24. Wait, no, if angle at X is \(30^{\circ}\), then \(\tan(30^{\circ})=\frac{YV}{XY}\), so \(XY=\frac{YV}{\tan(30^{\circ})}\). But \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), so \(XY = YV\times\sqrt{3}\). But this doesn't match the options. Wait, maybe YV is not 24 as the horizontal side, but the hypotenuse is 24? Wait, the problem says "Line segment YV of rectangle YVWX measures 24 units". So YV is a side of the rectangle, not the hypotenuse. Wait, maybe the triangle is XVW? No, VW is vertical. Wait, I think the key is that in a 30 - 60 - 90 triangle, the sides are in the ratio \(1:\sqrt{3}:2\). If we consider that YV is the side opposite \(30^{\circ}\), then the side opposite \(60^{\circ}\) (which is YX) would be \(YV\times\sqrt{3}\). But if YV is 8, then YX is \(8\sqrt{3}\), no. Wait, maybe YV is the side adjacent to \(60^{\circ}\). Wait, let's look at the options. The options are 8, \(8\sqrt{3}\), 12, \(12\sqrt{3}\). Let's assume that the hypotenuse is related to 24. Wait, maybe YV is the side opposite \(30^{\circ}\), so if YV = 12, then the side opposite \(60^{\circ}\) (YX) is \(12\sqrt{3}\)? No. Wait, maybe the length of YV is 24, and it's the side opposite \(30^{\circ}\) in a 30 - 60 - 90 triangle? No, \(\sin(30^{\circ})=\frac{1}{2}\), so hypotenuse would be 48, and the other side would be \(24\sqrt{3}\), not an option. Wait, I think I made a mistake in the angle. Maybe angle at V is \(30^{\circ}\) and angle at X is \(60^{\circ}\). Let's try that. If angle at V is \(30^{\circ}\), then \(\sin(30^{\circ})=\frac{YX}{XV}\), \(\cos(30^{\circ})=\frac{YV}{XV}\). If YV = 24, then \(XV=\frac{YV}{\cos(30^{\circ})}=\frac{24}{\frac{\sqrt{3}}{2}}=\frac{48}{\sqrt{3}} = 16\sqrt{3}\), and \(YX = XV\times\sin(30^{\circ})=8\sqrt{3}\), which is one of the options. Ah! So maybe I had the angles reversed. So in triangle YXV, right - angled at Y, angle at V is \(30^{\circ}\), angle at X is \(60^{\circ}\). Then:
- \(\cos(30^{\circ})=\frac{YV}{XV}\)
- \(\sin(30^{\circ})=\frac{YX}{XV}\)

We know YV = 24. So \(XV=\frac{YV}{\cos(30^{\circ})}=\frac{24}{\frac{\sqrt{3}}{2}}=\frac{48}{\sqrt{3}} = 16\sqrt{3}\)? No, that's not right. Wait, \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(XV=\frac{24}{\frac{\sqrt{3}}{2}}=24\times\frac{2}{\sqrt{3}} = 16\sqrt{3}\), and \(YX = XV\times\sin(30^{\circ})=16\sqrt{3}\times\frac{1}{2}=8\sqrt{3}\). Yes! That gives YX = \(8\sqrt{3}\) units, which is one of the options.

Step2: Confirm the 30 - 60 - 90 triangle ratios

In a 30 - 60 - 90 right - triangle, the ratio of the sides is \( \text{opposite }30^{\circ}:\text{opposite }60^{\circ}:\text{hypotenuse}=1:\sqrt{3}:2\). Let's assume that angle at V is \(30^{\circ}\), angle at X is \(60^{\circ}\), right angle at Y. Then:
- Side opposite \(30^{\circ}\) (which is YX) is \(x\)
- Side opposite \(60^{\circ}\) (which is YV) is \(x\sqrt{3}\)
- Hypotenuse (XV) is \(2x\)

We know that YV = 24, so \(x\sqrt{3}=24\), then \(x=\frac{24}{\sqrt{3}}=\frac{24\sqrt{3}}{3}=8\sqrt{3}\). Wait, no, if YV is opposite \(60^{\circ}\), then \(x\sqrt{3}=24\), so \(x = \frac{24}{\sqrt{3}} = 8\sqrt{3}\)? No, that's the side opposite \(30^{\circ}\). Wait, I'm getting confused. Let's use trigonometric ratios correctly. In right - triangle \( \triangle YXV\) with \( \angle Y = 90^{\circ}\), \( \angle V = 30^{\circ}\), \( \angle X = 60^{\circ}\):
- \(\sin(\angle V)=\sin(30^{\circ})=\frac{YX}{XV}=\frac{1}{2}\)
- \(\cos(\angle V)=\cos(30^{\circ})=\frac{YV}{XV}=\frac{\sqrt{3}}{2}\)

Given \(YV = 24\), from \(\cos(30^{\circ})=\frac{YV}{XV}\), we have \(XV=\frac{YV}{\cos(30^{\circ})}=\frac{24}{\frac{\sqrt{3}}{2}}=\frac{48}{\sqrt{3}} = 16\sqrt{3}\). Then from \(\sin(30^{\circ})=\frac{YX}{XV}\), \(YX = XV\times\sin(30^{\circ})=16\sqrt{3}\times\frac{1}{2}=8\sqrt{3}\).

Answer:

\(8\sqrt{3}\) units (the option: \(8\sqrt{3}\) units)