Question

the height of trapezoid vwxz is $8\sqrt{3}$ units. the upper base, $\overline{vw}$, measures 10 units. use the 30°-60°-90° triangle theorem to find the length of $\overline{yx}$. once you you know the length of $\overline{yx}$, find the length of the lower base, $\overline{zx}$. \bigcirc 14 units \bigcirc 10 + 4\sqrt{3} units \bigcirc 18 units \bigcirc 10 + 8\sqrt{3} units

Explanation:

Part 1: Find the length of \(\overline{YX}\)

Step 1: Recall 30 - 60 - 90 triangle ratios

In a \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite \(30^{\circ}\) is the shortest leg (\(x\)), the side opposite \(60^{\circ}\) is \(x\sqrt{3}\) (longer leg), and the hypotenuse is \(2x\). Here, the height of the trapezoid (longer leg of the right triangle \(WYX\)) is \(8\sqrt{3}\), which is opposite \(60^{\circ}\). Let the shorter leg (opposite \(30^{\circ}\), which is \(YX\)) be \(x\). So, \(x\sqrt{3}=8\sqrt{3}\).

Step 2: Solve for \(x\) (length of \(YX\))

Divide both sides of \(x\sqrt{3}=8\sqrt{3}\) by \(\sqrt{3}\): \(x = \frac{8\sqrt{3}}{\sqrt{3}}=8\)? Wait, no, wait. Wait, the angle at \(X\) is \(60^{\circ}\), so the angle at \(W\) in triangle \(WYX\) is \(30^{\circ}\). Wait, the height is \(8\sqrt{3}\), which is the side opposite \(60^{\circ}\) (since angle at \(X\) is \(60^{\circ}\)). So the side opposite \(30^{\circ}\) (which is \(YX\)) is \(x\), and the side opposite \(60^{\circ}\) is \(x\sqrt{3}\). So \(x\sqrt{3}=8\sqrt{3}\), so \(x = 8\)? Wait, no, maybe I mixed up. Wait, the height is \(WY = 8\sqrt{3}\), angle at \(X\) is \(60^{\circ}\), so in right triangle \(WYX\), \(\tan(60^{\circ})=\frac{WY}{YX}\). \(\tan(60^{\circ})=\sqrt{3}\), so \(\sqrt{3}=\frac{8\sqrt{3}}{YX}\). Then \(YX=\frac{8\sqrt{3}}{\sqrt{3}} = 8\)? Wait, but maybe another approach. Wait, the 30 - 60 - 90 triangle: if the longer leg (opposite 60°) is \(a\sqrt{3}\), then the shorter leg (opposite 30°) is \(a\). Here, longer leg is \(8\sqrt{3}\), so \(a\sqrt{3}=8\sqrt{3}\), so \(a = 8\)? Wait, no, that would mean shorter leg is 8. Wait, but let's check again. Wait, the height is \(8\sqrt{3}\), which is the vertical side (longer leg), and \(YX\) is the horizontal side (shorter leg) in the 30 - 60 - 90 triangle (since angle at \(X\) is 60°, so angle at \(W\) is 30°). So in 30 - 60 - 90, shorter leg (YX) : longer leg (WY) : hypotenuse (WX) = \(x : x\sqrt{3}: 2x\). Here, longer leg \(WY = 8\sqrt{3}=x\sqrt{3}\), so \(x = 8\). Wait, but that seems off. Wait, maybe I made a mistake. Wait, no, let's do it with trigonometry. \(\tan(60^{\circ})=\frac{WY}{YX}\), so \(YX=\frac{WY}{\tan(60^{\circ})}=\frac{8\sqrt{3}}{\sqrt{3}} = 8\). Wait, but the options for the second part: ZX is upper base (VW = 10) plus YX. Wait, ZX = ZY + YX, and ZY = VW = 10 (since VZYW is a rectangle). So ZX = 10 + YX. If YX is 8, then ZX = 18. Wait, that matches one of the options (18 units). Wait, maybe I messed up the first part. Wait, let's re - evaluate. Wait, the height is \(8\sqrt{3}\), angle at \(X\) is \(60^{\circ}\), so in triangle \(WYX\), \(\tan(60^{\circ})=\frac{WY}{YX}\), so \(YX=\frac{WY}{\tan(60^{\circ})}=\frac{8\sqrt{3}}{\sqrt{3}} = 8\). Then ZX = VW + YX = 10 + 8 = 18. So first, length of YX is 8? Wait, but the first question is to find YX. Wait, maybe my initial ratio was wrong. Wait, 30 - 60 - 90 triangle: the sides are \(x\) (opposite 30°), \(x\sqrt{3}\) (opposite 60°), \(2x\) (hypotenuse). If the side opposite 60° is \(8\sqrt{3}\), then \(x\sqrt{3}=8\sqrt{3}\), so \(x = 8\). So YX is 8. Then ZX = 10 + 8 = 18.

Part 2: Find the length of \(\overline{ZX}\)

Step 1: Recall the structure of the trapezoid

Trapezoid VWXZ has \(VW\) (upper base) = 10, and \(VZ\) and \(WY\) are the heights (so \(VZ = WY=8\sqrt{3}\)). The lower base \(ZX\) is equal to \(VW + YX\) (since \(VZYW\) is a rectangle, \(VZ = WY\) and \(VW = ZY\)).

Step 2: Substitute the value of \(YX\)

We found \(YX = 8\) (from the 30 - 60 - 90 triangle calculation). So \(ZX=VW + YX=10 + 8 = 18\) units.

For the first part (length of \(YX\)):

The trapezoid \(VWXZ\) has upper base \(VW = 10\). The segment \(ZY\) is equal to \(VW\) (as \(VZYW\) is a rectangle). The segment \(YX\) is found to be 8 units (from the \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle). So the lower base \(ZX=ZY + YX=10 + 8 = 18\) units.

Answer:

8 units (but since the second part's options include 18, and the process for YX: from \(30 - 60 - 90\) triangle, longer leg \(=8\sqrt{3}\) (opposite 60°), so shorter leg (YX) \(=\frac{8\sqrt{3}}{\sqrt{3}} = 8\))

For the second part (length of \(ZX\)):