Question

the height of trapezoid vwxz is ( 8sqrt{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 know the length of ( overline{yx} ), find the length of the lower base, ( overline{zx} ).
○ 14 units
○ ( 10 + 4sqrt{3} ) units
○ 18 units
○ ( 10 + 8sqrt{3} ) units

Explanation:

First, find the length of \(\overline{YX}\)

Step1: Recall 30 - 60 - 90 triangle ratios

In a \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\) (opposite \(30^{\circ}\), \(60^{\circ}\), \(90^{\circ}\) respectively). Let the side opposite \(30^{\circ}\) (shorter leg) be \(a\), opposite \(60^{\circ}\) (longer leg) be \(a\sqrt{3}\), and hypotenuse be \(2a\). Here, the height \(WY = 8\sqrt{3}\) is the longer leg (opposite \(60^{\circ}\)). So if \(a\sqrt{3}=8\sqrt{3}\), then \(a = 8\)? Wait, no, wait. Wait, in triangle \(WYX\), angle at \(X\) is \(60^{\circ}\), angle at \(W\) is \(30^{\circ}\), so the side \(WY\) (height) is opposite \(60^{\circ}\), so \(WY=a\sqrt{3}\), and \(YX\) is opposite \(30^{\circ}\), so \(YX = a\). So if \(WY = 8\sqrt{3}=a\sqrt{3}\), then \(a = 8\)? Wait, no, wait, maybe I mixed up. Wait, angle at \(W\) is \(30^{\circ}\), so the side opposite \(30^{\circ}\) is \(YX\), and the side opposite \(60^{\circ}\) is \(WY\). So in \(30 - 60 - 90\) triangle, \(\tan(60^{\circ})=\frac{WY}{YX}\), so \(\sqrt{3}=\frac{8\sqrt{3}}{YX}\), solving for \(YX\): \(YX=\frac{8\sqrt{3}}{\sqrt{3}} = 8\)? Wait, no, that can't be. Wait, maybe the angle at \(X\) is \(60^{\circ}\), so the triangle \(WYX\) has angle at \(X\) as \(60^{\circ}\), right angle at \(Y\), so angle at \(W\) is \(30^{\circ}\). So the sides: \(YX\) is adjacent to \(60^{\circ}\), \(WY\) is opposite to \(60^{\circ}\), and \(WX\) is hypotenuse. So \(\tan(60^{\circ})=\frac{WY}{YX}\), so \(\sqrt{3}=\frac{8\sqrt{3}}{YX}\), so \(YX = 8\)? Wait, but let's check the 30 - 60 - 90 ratios. If angle at \(W\) is \(30^{\circ}\), then the side opposite \(30^{\circ}\) is \(YX\), and the side opposite \(60^{\circ}\) is \(WY\). So in 30 - 60 - 90, the ratio of opposite \(30^{\circ}\) to opposite \(60^{\circ}\) is \(1:\sqrt{3}\). So \(YX:WY = 1:\sqrt{3}\), so \(YX=\frac{WY}{\sqrt{3}}=\frac{8\sqrt{3}}{\sqrt{3}} = 8\). Wait, but that seems off. Wait, maybe I made a mistake. Wait, the height is \(8\sqrt{3}\), which is the longer leg (opposite \(60^{\circ}\)), so the shorter leg (opposite \(30^{\circ}\)) is \(YX\), and longer leg is \(WY = 8\sqrt{3}\). So in 30 - 60 - 90, longer leg = shorter leg \(\times\sqrt{3}\), so \(8\sqrt{3}=YX\times\sqrt{3}\), so \(YX = 8\). Wait, that makes sense. So \(YX = 8\)? Wait, no, wait, the answer choices for the second part: ZX is upper base + YX? Wait, the trapezoid VWXZ: VW is upper base (10), ZY is equal to VW (since VZ and WY are both heights, so VZYW is a rectangle, so ZY = VW = 10). Then ZX = ZY + YX. So if YX is 8, then ZX = 10 + 8 = 18? Wait, but let's re - check.

Wait, the first part: find YX using 30 - 60 - 90. The triangle WYX is a 30 - 60 - 90 triangle with height WY = \(8\sqrt{3}\) (opposite 60°), so the side opposite 30° (YX) is \(\frac{WY}{\sqrt{3}}=\frac{8\sqrt{3}}{\sqrt{3}} = 8\). Then, for the lower base ZX: since VZYW is a rectangle, ZY = VW = 10. So ZX = ZY + YX = 10 + 8 = 18. So the length of ZX is 18 units.

Now, for the second part (finding ZX)

Step1: Identify the segments of the lower base

The lower base \(ZX\) is composed of \(ZY\) and \(YX\). Since \(VZYW\) is a rectangle (because \(VZ\perp ZX\), \(WY\perp ZX\), and \(VW\parallel ZX\)), \(ZY = VW = 10\) units. We found \(YX = 8\) units (from the 30 - 60 - 90 triangle calculation).

Step2: Calculate the length of \(ZX\)

To find \(ZX\), we add the lengths of \(ZY\) and \(YX\). So \(ZX=ZY + YX\). Substituting the known values, \(ZY = 10\) and \(YX = 8\), we get \(ZX = 10+8=18\) units.

Answer:

(for the length of \(\overline{ZX}\)):
18 units (corresponding to the option "18 units")