7. what is the length of $ \overline{xz} $?
32
44
22
8

Question

Accepted Answer

# Explanation:
## Step1: Use the mid - segment theorem
Since $AB$ is a mid - segment of $\triangle XYZ$ (because $A$ is the mid - point of $XY$ and $B$ is the mid - point of $XZ$), $AB\parallel YZ$ and $AB=\frac{1}{2}YZ$. Also, in the figure, $AC = 16$.
## Step2: Analyze the relationship between the segments
We know that $AC$ is related to $XZ$. Since $A$ is the mid - point of $XY$ and $C$ is the mid - point of $YZ$ (from the equal segment marks), we can use the properties of mid - segments and congruent triangles or parallel lines. But another approach:
We note that the length from $Y$ to $Z$ is $44$. But if we consider the triangle $AYC$ and the fact that $A$ is the mid - point of $XY$ and $C$ is the mid - point of $YZ$. However, looking at the segment $AC = 16$ is a distractor.
Since $A$ is the mid - point of $XY$ and $B$ is the mid - point of $XZ$, by the mid - segment theorem $AB=\frac{1}{2}YZ$. But we can also use the fact that in $\triangle AYC$ and $\triangle XYZ$ (by AA similarity, since $AB\parallel YZ$, $\angle XAB=\angle XYZ$ and $\angle XBA=\angle XZY$). But more simply, if we consider the length of $XZ$.
Since $A$ is the mid - point of $XY$ and $B$ is the mid - point of $XZ$, and we know that $AC = 16$ (where $AC$ is equal to $AB$ as $ABCA$ is a parallelogram (because $AB\parallel YC$ and $AY\parallel BC$)).
We can also use the Pythagorean theorem if we assume right - angled triangles. But since $A$ is the mid - point of $XY$ and $B$ is the mid - point of $XZ$, and $YC = CZ$ (from the marks), and $AY = AX$ (from the marks).
If we consider the fact that $AC = 16$ and using the mid - point properties. Wait, no, actually, since $A$ is the mid - point of $XY$ and $B$ is the mid - point of $XZ$, then $AB\parallel YZ$ and $AB=\frac{1}{2}YZ$. But we made a mistake above.
Let's use the property of congruent triangles. Since $A$ is the mid - point of $XY$ ($XA = AY$) and $C$ is the mid - point of $YZ$ ($YC = CZ$) and $AB\parallel YZ$, then $\triangle XAB\cong\triangle AYC$ (by ASA, $\angle XAB=\angle AYC$, $XA = AY$, $\angle XBA=\angle YCA$). So $AB = YC$. But $YC=\frac{1}{2}YZ$ (since $YC = CZ$ and $YZ=44$, no, wait no. Wait, the length $AC = 16$ is the length of $AB$ (parallelogram $ABCA$ - no, $AB\parallel YC$ and $AY\parallel BC$ so $ABCA$ is a parallelogram, so $AB = YC$). But we need $XZ$.
Since $B$ is the mid - point of $XZ$, and if we consider $\triangle XYZ$, $A$ is the mid - point of $XY$, $B$ is the mid - point of $XZ$, $C$ is the mid - point of $YZ$.
We know that $AC = 16$, and $AC = AB$ (parallelogram $ABCA$). Then, using the Pythagorean theorem in $\triangle XAB$ (assuming right - angled, but actually, since $A$ is the mid - point of $XY$ and $B$ is the mid - point of $XZ$, and $AB = 16$, then $XZ=32$ (because if we consider the fact that $AB$ is half of $YZ$ in terms of mid - segment, but no, wait, $AB$ is a mid - segment of $\triangle XYZ$ parallel to $YZ$ and $AB=\frac{1}{2}YZ$ is wrong. Wait, no, $AB$ is a mid - segment between $XY$ and $XZ$, so $AB\parallel YZ$ and $AB=\frac{1}{2}YZ$ is incorrect. The mid - segment between two sides of a triangle is parallel to the third side and half its length. But here $AB$ is a mid - segment between $XY$ and $XZ$, so $AB\parallel YZ$ and $AB = \frac{1}{2}YZ$. But $YZ = 44$ is wrong. Wait, no, the length $AC = 16$. Since $A$ is the mid - point of $XY$ and $C$ is the mid - point of $YZ$, then $AC\parallel XZ$ and $AC=\frac{1}{2}XZ$ (by the mid - segment theorem).

## Step3: Solve for $XZ$
Since $AC = 16$ and $AC=\frac{1}{2}XZ$ (mid - segment theorem: the segment connecting the mid - points of two sides of a triangle is parallel to the third side and half its length), then $XZ = 2\times AC$.

Substitute $AC = 16$ into the formula $XZ=2\times AC$. So $XZ = 2\times16=32$.

# Answer:
$32$

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QUESTION IMAGE

Question

Explanation:

Step1: Use the mid - segment theorem

Step2: Analyze the relationship between the segments

Answer: