Question 320·Hard·Math

0 / 2000

In the figure shown, $\overline{WZ}$ and $\overline{XY}$ intersect at point $Q$ . The length of $\overline{WZ}$ is $120$ , the length of $\overline{XY}$ is $118$ , $XQ = 64$ , and $WX = 88$ . Point $Q$ is closer to $Z$ than to $W$ , and $\angle QWX = \angle QYZ$ .

Which choice is the length of $\overline{YZ}$ ?

Intersecting segments and a pair of equal angles often signal AA similarity: the opposite angles at the intersection supply the second pair. Match the triangles' vertices, then use corresponding sides to write a proportion. A Desmos regression can solve it, but a location clue may rule out one solution. Don't choose a value because it seems to fit the drawing.

Hints

Hint 1
By segment addition, $XY=XQ+QY$ because $Q$ lies between $X$ and $Y$ . Subtract the known piece from the whole segment. What length does that give you for $QY$ ?
Hint 2
The angles opposite each other at $Q$ are vertical angles, so they're equal. Together with the marked equal angles, they make the triangles similar. Which vertex matches $W$ , and which matches $X$ ?
Hint 3
The phrase closer to $Z$ than to $W$ tells you $ZQ<WQ$ . Since both pieces make up $WZ$ , which piece must be longer than half of $WZ$ ?

Step-by-step

Match the triangles and solve a proportion
Step 1
Find the missing piece of XY
$Q$ lies between $X$ and $Y$ , so the two pieces add to the whole segment. Subtract the given $XQ=64$ from $XY=118$ : $YQ=118-64=54$ .
Step 2
Show which vertices match
At $Q$ , $\angle WQX$ and $\angle YQZ$ are vertical angles, the opposite angles made by crossing lines, so they're equal. The given angles at $W$ and $Y$ are equal too. By AA similarity, two equal angle pairs make the triangles the same shape: $\triangle WQX\sim\triangle YQZ$ . The matches are $W\leftrightarrow Y$ , $Q\leftrightarrow Q$ , and $X\leftrightarrow Z$ . Match sides by these vertices, not by how the drawing looks.
Step 3
Write a proportion using matching sides
Let $w=WQ$ . The whole $WZ$ is $120$ , so $ZQ=120-w$ . The vertex matches pair $WQ$ with $YQ$ and $XQ$ with $ZQ$ . Similar triangles give equal ratios of matching sides: $\frac{w}{54}=\frac{64}{120-w}$ .
Step 4
Use the location clue
$Q$ is closer to $Z$ than to $W$ , so $ZQ<WQ$ . That makes $WQ$ longer than half of the $120$ -unit segment: $w>\frac{120}{2}=60$ . A solution below $60$ would put $Q$ closer to $W$ instead.
Step 5
Find WQ in Desmos
Type $\frac{x_1}{54}\sim\frac{64}{120-x_1}\left\{x_1>60\right\}$ . Replace $w$ with $x_1$ and $=$ with $\sim$ so Desmos fits the unknown; the restriction enforces the location clue. Under PARAMETERS, Desmos reports $x_1=72$ , so $WQ=72$ .
Step 6
Set up the length of YZ
The vertex matches pair $WX$ with $YZ$ and $WQ$ with $YQ$ . Keep second triangle over first triangle for both side pairs: $\frac{YZ}{88}=\frac{54}{72}$ .
Step 7
Calculate the requested side
Type $88\cdot54/x_1$ on the next Desmos line. It multiplies $WX$ by the matching-side scale factor, and Desmos prints $66$ . So the length of $\overline{YZ}$ is $66$ . Choice B.

Question 320·Hard·Math

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this