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 triangle properties

In rectangle YVWX, triangle XVW is a right triangle with \(\angle W = 90^\circ\), \(\angle VXW = 30^\circ\), and \(YV = XW = 24\) (opposite sides of rectangle). Wait, no, \(YV\) is 24, and in triangle \(YXV\) or \(XVW\)? Wait, \(YV\) is length 24, and in triangle \(XVY\)? Wait, the rectangle has \(YV\) as top side, length 24. Then triangle \(XWV\) is right-angled at \(W\), with \(\angle VXW = 30^\circ\), so in a 30-60-90 triangle, the sides are in ratio \(1 : \sqrt{3} : 2\), where the side opposite 30° is the shortest. Wait, \(YV = XW = 24\)? No, wait, \(YV\) is horizontal, length 24, and \(YX\) is vertical, which is equal to \(VW\) (opposite sides of rectangle). In triangle \(XWV\), \(\angle XWV = 90^\circ\), \(\angle VXW = 30^\circ\), so \(VW\) is opposite 30°, \(XW\) is adjacent (length 24), and \(XV\) is hypotenuse. Wait, no, maybe I mixed up. Wait, \(YV = 24\), which is equal to \(XW\) (since YVWX is rectangle, \(YV \parallel XW\) and \(YV = XW\)). Then in triangle \(XWV\), right-angled at \(W\), \(\angle VXW = 30^\circ\), so \(\tan(30^\circ) = \frac{VW}{XW}\). Wait, \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), and \(XW = 24\), so \(VW = \frac{24}{\sqrt{3}} = 8\sqrt{3}\)? No, wait, no. Wait, maybe \(YV\) is the hypotenuse? Wait, no, the diagram: \(Y\) to \(V\) is 24, \(X\) to \(W\) is horizontal, \(Y\) to \(X\) is vertical, \(V\) to \(W\) is vertical. So triangle \(YXV\): no, triangle \(XWV\) is right-angled at \(W\), with \(XW\) horizontal, \(VW\) vertical, \(XV\) diagonal. Wait, \(\angle VXW = 30^\circ\), so in triangle \(XWV\), sides: \(VW\) (opposite 30°) is \(x\), \(XW\) (adjacent) is \(x\sqrt{3}\), hypotenuse \(XV\) is \(2x\). But \(YV = XW = 24\) (since YVWX is rectangle, \(YV = XW\)). So \(XW = 24 = x\sqrt{3}\), so \(x = \frac{24}{\sqrt{3}} = 8\sqrt{3}\)? No, wait, no. Wait, maybe \(YV\) is the hypotenuse? Wait, no, the problem says line segment \(YV\) is 24 units. Wait, maybe I made a mistake. Let's re-examine. The rectangle is YVWX, so vertices are Y, V, W, X in order. So Y to V is top side, V to W is right side (vertical), W to X is bottom side, X to Y is left side (vertical). So triangle \(XVW\) is right-angled at W. So \(XW\) is bottom side, \(VW\) is right side, \(XV\) is diagonal. \(\angle VXW = 30^\circ\), so in triangle \(XVW\), \(\cos(30^\circ) = \frac{XW}{XV}\), \(\sin(30^\circ) = \frac{VW}{XV}\). But \(YV = XW = 24\) (since YVWX is rectangle, \(YV = XW\)). Wait, no, \(YV\) is top side, so \(YV = XW\), yes. So \(XW = 24\). Then in triangle \(XVW\), \(\angle VXW = 30^\circ\), so \(\sin(30^\circ) = \frac{VW}{XV}\), and \(\cos(30^\circ) = \frac{XW}{XV}\). But we need \(YX\), which is equal to \(VW\) (since YVWX is rectangle, \(YX = VW\)). So let's find \(VW\). In triangle \(XVW\), \(\tan(30^\circ) = \frac{VW}{XW}\). So \(\tan(30^\circ) = \frac{1}{\sqrt{3}} = \frac{VW}{24}\), so \(VW = \frac{24}{\sqrt{3}} = 8\sqrt{3}\)? No, that can't be. Wait, maybe \(YV\) is the hypotenuse of triangle \(YXV\)? Wait, \(YV\) is 24, and triangle \(YXV\) is a triangle with \(YX\) vertical, \(YV\) horizontal, and \(XV\) diagonal. Wait, no, the angle at X is 30°, so maybe triangle \(YXV\) is a 30-60-90 triangle? Wait, no, the rectangle has right angles, so \(YX\) is perpendicular to \(YV\), so triangle \(YXV\) is right-angled at Y. Wait, that makes more sense! So \(Y\) is right angle, \(YV = 24\), \(\angle YXV = 30^\circ\), so in right triangle \(YXV\), right-angled at Y, \(\tan(30^\circ) = \frac{YV}{YX}\)? Wait, no, \(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\). So \(\angle YXV = 30^\circ\), so opposite side is \(YV = 24\), adjacent side is \(YX\). So \(\tan(30^\circ) = \frac{YV}{YX}\), so \(YX = \frac{YV}{\tan(30^\circ)} = \frac{24}{\frac{1}{\sqrt{3}}} = 24\sqrt{3}\)? No, that's not one of the options. Wait, maybe I got the angle wrong. Wait, the diagram shows \(\angle VXW = 30^\circ\), and \(W\) is right angle. So triangle \(XWV\) is right-angled at W, with \(\angle VXW = 30^\circ\), so \(VW\) is opposite 30°, \(XW\) is adjacent, \(XV\) is hypotenuse. Then \(YX = VW\) (since YVWX is rectangle, \(YX \parallel VW\) and \(YX = VW\)). \(YV = XW = 24\) (since \(YV \parallel XW\) and \(YV = XW\)). So in triangle \(XWV\), \(\sin(30^\circ) = \frac{VW}{XV}\), \(\cos(30^\circ) = \frac{XW}{XV}\). But we know \(XW = 24\), so \(\cos(30^\circ) = \frac{24}{XV}\), so \(XV = \frac{24}{\cos(30^\circ)} = \frac{24}{\frac{\sqrt{3}}{2}} = \frac{48}{\sqrt{3}} = 16\sqrt{3}\). Then \(\sin(30^\circ) = \frac{VW}{XV}\), so \(VW = XV \times \sin(30^\circ) = 16\sqrt{3} \times \frac{1}{2} = 8\sqrt{3}\). But that's not matching. Wait, maybe the triangle is 30-60-90 with hypotenuse 24? Wait, no, the options include \(8\sqrt{3}\), \(12\sqrt{3}\), etc. Wait, maybe I mixed up the sides. Let's recall: in a 30-60-90 triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\), where the side opposite 30° is the shortest (let's call it \(x\)), the side opposite 60° is \(x\sqrt{3}\), and the hypotenuse is \(2x\). Now, looking at the diagram, \(YV = 24\) is the side opposite 60°? Wait, \(\angle YXV = 30^\circ\), so in triangle \(YXV\) (right-angled at Y), \(\angle YXV = 30^\circ\), so the side opposite 30° is \(YV\), and the side adjacent is \(YX\). Wait, no: in right triangle \(YXV\), right angle at Y, so:
- \(\angle at Y: 90^\circ\)
- \(\angle at X: 30^\circ\)
- \(\angle at V: 60^\circ\)

So sides:
- Opposite 30° (angle at X): \(YV = 24\)
- Opposite 60° (angle at V): \(YX\)
- Hypotenuse: \(XV\)

In 30-60-90 ratio, opposite 30° is \(x\), opposite 60° is \(x\sqrt{3}\), hypotenuse \(2x\). So if \(YV = x = 24\)? No, that would make hypotenuse 48, and \(YX = 24\sqrt{3}\), which is not an option. Wait, maybe \(YV\) is the side opposite 60°, so \(YV = x\sqrt{3} = 24\), so \(x = \frac{24}{\sqrt{3}} = 8\sqrt{3}\)? No, that's not. Wait, the options are 8, \(8\sqrt{3}\), 12, \(12\sqrt{3}\). Wait, maybe the hypotenuse is 24? Wait, if hypotenuse \(XV = 24\), then opposite 30° is \(12\) (since \(2x = 24\) → \(x = 12\)), and opposite 60° is \(12\sqrt{3}\). Wait, that matches one of the options: \(12\sqrt{3}\)? No, wait, \(YX\) would be opposite 60°, so if hypotenuse is 24, then \(YX = 24 \times \sin(60^\circ) = 24 \times \frac{\sqrt{3}}{2} = 12\sqrt{3}\). Wait, but where is the hypotenuse? Wait, maybe \(XV\) is the hypotenuse with length 24? But the problem says \(YV = 24\). Wait, maybe the diagram has \(YV = 24\) as the side opposite 60°, and \(YX\) is opposite 30°? No, that would be \(YX = 24 / \sqrt{3} = 8\sqrt{3}\). Wait, let's check the options. The options are 8, \(8\sqrt{3}\), 12, \(12\sqrt{3}\). Let's re-express:

In rectangle YVWX, \(YV = XW = 24\) (opposite sides). Triangle \(XWV\) is right-angled at W, with \(\angle VXW = 30^\circ\). So in triangle \(XWV\):

- \(\angle W = 90^\circ\)
- \(\angle VXW = 30^\circ\)
- \(XW = 24\) (adjacent to 30°)
- \(VW = YX\) (opposite sides of rectangle)

We can use \(\tan(30^\circ) = \frac{VW}{XW}\) → \(VW = XW \times \tan(30^\circ) = 24 \times \frac{1}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8\sqrt{3}\). Wait, but that's \(8\sqrt{3}\), which is an option. Wait, but earlier I thought maybe hypotenuse, but no, \(XW\) is adjacent, \(VW\) is opposite. So \(\tan(30^\circ) = \text{opposite}/\text{adjacent} = VW/XW\), so \(VW = 24 \times (1/\sqrt{3}) = 8\sqrt{3}\). So \(YX = VW = 8\sqrt{3}\) units. Wait, but let's check with sine: \(\sin(30^\circ) = VW / XV\), so \(XV = VW / \sin(30^\circ) = 8\sqrt{3} / 0.5 = 16\sqrt{3}\). Then \(\cos(30^\circ) = XW / XV = 24 / 16\sqrt{3} = 3/(2\sqrt{3}) = \sqrt{3}/2\), which is correct. So that works. So the length of \(YX\) is \(8\sqrt{3}\) units? Wait, no, wait, maybe I mixed up the angle. Wait, the angle at V is 60°, so in triangle \(XVW\), \(\angle V = 60^\circ\), so \(\tan(60^\circ) = XW / VW\) → \(\sqrt{3} = 24 / VW\) → \(VW = 24 / \sqrt{3} = 8\sqrt{3}\). Yes, that's the same. So \(YX = VW = 8\sqrt{3}\) units. Wait, but the option is \(8\sqrt{3}\) units. So that's the answer.

Wait, but let's confirm again. Rectangle YVWX: Y to V is 24 (top), V to W is vertical (right), W to X is 24 (bottom), X to Y is vertical (left). Triangle XWV: right-angled at W, with angle at X is 30°, angle at V is 60°. So sides: XW = 24 (adjacent to 30°), VW = YX (opposite to 30°), XV is hypotenuse. So \(\sin(30°) = VW / XV\), \(\cos(30°) = XW / XV\). But we can use \(\tan(30°) = VW / XW\) → \(VW = XW \times \tan(30°) = 24 \times (1/\sqrt{3}) = 8\sqrt{3}\). So \(YX = 8\sqrt{3}\) units.

Step2: Confirm with 30-60-90 ratios

In a 30-60-90 triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\) (short leg : long leg : hypotenuse). Here, the short leg is opposite 30°, long leg opposite 60°, hypotenuse opposite 90°. If \(XW = 24\) is the long leg (opposite 60°), then the short leg (opposite 30°) is \(24 / \sqrt{3} = 8\sqrt{3}\)? No, wait, no: long leg is \(x\sqrt{3}\), short leg is \(x\), hypotenuse \(2x\). So if long leg (opposite 60°) is \(24 = x\sqrt{3}\), then \(x = 24 / \sqrt{3} = 8\sqrt{3}\)? No, that's not. Wait, I think I messed up which side is which. Let's define:

- Let \(x\) be the length of the short leg (opposite 30°), so \(x = VW = YX\)
- Then the long leg (opposite 60°) is \(x\sqrt{3} = XW = 24\)
- So \(x\sqrt{3} = 24\) → \(x = 24 / \sqrt{3} = 8\sqrt{3}\)

Yes! That's correct. So the short leg (opposite 30°) is \(x = 8\sqrt{3}\), which is \(YX\) (since \(YX = VW\), the short leg). Wait, no, if \(XW\) is the long leg (opposite 60°), then \(XW = x\sqrt{3} = 24\), so \(x = 24 / \sqrt{3} = 8\sqrt{3}\), which is the short leg (opposite 30°), which is \(VW = YX\). So that's correct. So \(YX = 8\sqrt{3}\) units.

Answer:

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