Question

$\overline{yz} \parallel \overline{vx}$. find $vz$.

Explanation:

Step1: Identify Similar Triangles

Since \(\overline{YZ} \parallel \overline{VX}\), by the Basic Proportionality Theorem (Thales' theorem), \(\triangle WVX \sim \triangle WZY\) (similar triangles) because corresponding angles are equal (AA similarity: \(\angle WVX = \angle WZY\) and \(\angle WXV = \angle WYZ\) as they are corresponding angles due to parallel lines).

Step2: Set Up Proportion

For similar triangles, the ratios of corresponding sides are equal. Let \(VZ = x\). Then \(WZ = WV + VZ = 36 + x\). The ratio of the sides of \(\triangle WVX\) to \(\triangle WZY\) is based on the lengths of \(WX\) and \(WY\). We know \(WX = 48 - 16 = 32\)? Wait, no, looking at the diagram, \(WY\) is the total length, and \(WX\) is the upper part. Wait, actually, \(WY = 48\) (the vertical segment), and \(WX = 48 - 16 = 32\)? Wait, no, the diagram shows \(WX\) (the segment from \(W\) to \(X\)) and \(XY = 16\), so \(WY = WX + XY = WX + 16\), but the vertical segment on the right is 48, so \(WY = 48\), so \(WX = 48 - 16 = 32\). Wait, no, maybe I misread. Let's re - examine: The vertical segment on the right is 48, and the segment from \(X\) to \(Y\) is 16, so \(WX\) (from \(W\) to \(X\)) is \(48 - 16 = 32\), and \(WY\) (from \(W\) to \(Y\)) is 48. The side \(WV = 36\), and \(WZ=WV + VZ=36 + VZ\).

Since \(\triangle WVX\sim\triangle WZY\), the ratio of corresponding sides is \(\frac{WV}{WZ}=\frac{WX}{WY}\)

Substitute the known values: \(\frac{36}{36 + VZ}=\frac{32}{48}\)

Step3: Solve the Proportion

Simplify \(\frac{32}{48}=\frac{2}{3}\). So we have the equation \(\frac{36}{36 + VZ}=\frac{2}{3}\)

Cross - multiply: \(3\times36 = 2\times(36 + VZ)\)

\(108=72 + 2VZ\)

Subtract 72 from both sides: \(108 - 72=2VZ\)

\(36 = 2VZ\)

Divide both sides by 2: \(VZ = 18\)? Wait, no, wait, maybe I mixed up the sides. Wait, maybe the ratio is \(\frac{WX}{WY}=\frac{WV}{WZ}\). Wait, \(WX\) is the length from \(W\) to \(X\), and \(WY\) is from \(W\) to \(Y\). If \(XY = 16\) and \(WY = 48\), then \(WX=48 - 16 = 32\). So \(\frac{WX}{WY}=\frac{32}{48}=\frac{2}{3}\), and \(\frac{WV}{WZ}=\frac{36}{36 + VZ}\). So \(\frac{36}{36 + VZ}=\frac{2}{3}\)

Cross - multiply: \(3\times36=2\times(36 + VZ)\)

\(108 = 72+2VZ\)

Subtract 72 from both sides: \(108 - 72=2VZ\)

\(36 = 2VZ\)

Divide both sides by 2: \(VZ = 18\)? Wait, that can't be right. Wait, maybe the ratio is reversed. Let's think again. Since \(\overline{VX}\parallel\overline{YZ}\), the ratio of \(WV\) to \(WZ\) should be equal to the ratio of \(WX\) to \(WY\). Wait, \(WX\) is the upper part, \(WY\) is the lower part? No, maybe the vertical sides: \(WX\) is the length from \(W\) to \(X\), and \(WY\) is from \(W\) to \(Y\), with \(XY = 16\), so \(WY=WX + XY\), and the right - hand vertical segment is 48, so \(WY = 48\), \(XY = 16\), so \(WX=48 - 16 = 32\). The side \(WV = 36\), \(WZ=WV + VZ\).

Since \(\triangle WVX\sim\triangle WZY\), \(\frac{WV}{WZ}=\frac{WX}{WY}\)

\(\frac{36}{36 + VZ}=\frac{32}{48}\)

Simplify \(\frac{32}{48}=\frac{2}{3}\), so \(\frac{36}{36 + VZ}=\frac{2}{3}\)

Cross - multiply: \(36\times3=2\times(36 + VZ)\)

\(108 = 72+2VZ\)

\(2VZ=108 - 72\)

\(2VZ = 36\)

\(VZ = 18\)? Wait, that seems low. Wait, maybe I got the similar triangles wrong. Maybe \(\triangle WVX\) and \(\triangle WZY\) have the ratio of \(\frac{WX}{WY}=\frac{WV}{WZ}\), but if \(WX = 16\) and \(WY = 48\), then \(\frac{16}{48}=\frac{1}{3}\), and \(\frac{WV}{WZ}=\frac{36}{36 + VZ}\). Then \(\frac{36}{36 + VZ}=\frac{1}{3}\)

Cross - multiply: \(36\times3=36 + VZ\)

\(108=36 + VZ\)

\(VZ = 72\)? No, that doesn't make sense. Wait, maybe the parallel lines are \(VX\parallel YZ\), so \(\triangle WVX\sim\triangle WZY\) with \(\frac{WV}{WZ}=\frac{WX}{WY}\), where \(WX\) is the length from \(W\) to \(X\), \(WY\) is from \(W\) to \(Y\), and \(XY = 16\), \(WY = 48\), so \(WX=WY - XY=48 - 16 = 32\). \(WV = 36\), \(WZ=WV + VZ\)

\(\frac{WV}{WZ}=\frac{WX}{WY}\)

\(\frac{36}{36 + VZ}=\frac{32}{48}\)

\(\frac{32}{48}=\frac{2}{3}\), so \(\frac{36}{36 + VZ}=\frac{2}{3}\)

\(3\times36=2\times(36 + VZ)\)

\(108 = 72+2VZ\)

\(2VZ=108 - 72 = 36\)

\(VZ = 18\)? Wait, but let's check with another approach. Let's use the concept of similar triangles and the ratio of segments.

Alternatively, the ratio of the lengths of the sides of similar triangles is equal to the ratio of their corresponding segments. Since \(VX\parallel YZ\), the ratio of \(WV\) to \(WZ\) is equal to the ratio of \(WX\) to \(WY\).

We know that \(WX\) (the segment from \(W\) to \(X\)) and \(WY\) (from \(W\) to \(Y\)): the length of \(WY\) is 48, and \(XY = 16\), so \(WX=48 - 16 = 32\). \(WV = 36\), let \(VZ=x\), then \(WZ = 36 + x\)

\(\frac{WV}{WZ}=\frac{WX}{WY}\)

\(\frac{36}{36 + x}=\frac{32}{48}\)

\(\frac{32}{48}=\frac{2}{3}\), so \(\frac{36}{36 + x}=\frac{2}{3}\)

\(3\times36=2\times(36 + x)\)

\(108 = 72+2x\)

\(2x=36\)

\(x = 18\)

Wait, but maybe I made a mistake in identifying the segments. Let's look at the diagram again. The triangle \(WZY\) with base \(ZY\) and triangle \(WVX\) with base \(VX\), and \(VX\parallel ZY\). So the ratio of \(WV\) to \(WZ\) should be equal to the ratio of \(WX\) to \(WY\). If \(WX\) is the height of the smaller triangle and \(WY\) is the height of the larger triangle, then \(WX = 48 - 16=32\) and \(WY = 48\). Then the ratio of heights is \(\frac{32}{48}=\frac{2}{3}\), so the ratio of the sides is also \(\frac{2}{3}\). So \(\frac{WV}{WZ}=\frac{2}{3}\), \(WV = 36\), so \(\frac{36}{WZ}=\frac{2}{3}\), then \(WZ=\frac{36\times3}{2}=54\). Then \(VZ=WZ - WV=54 - 36 = 18\). Yes, that makes sense.

Answer:

\(18\)