Question

the figure below shows the midpoints of each side of \\( \triangle a b c \\) connected to form \\( \triangle x y z \\). \\( a b = 12 \\) in, \\( b c = 14 \\) in, and \\( a c = 8 \\) in. find the length of \\( \overline{z y} \\).

Explanation:

Step1: Recall Midline Theorem

The Midline Theorem (also known as the Midsegment Theorem) states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half as long.

Step2: Identify the sides and midpoints

In $\triangle ABC$, $X$, $Y$, $Z$ are midpoints of the sides. For $\overline{ZY}$, we need to see which side of $\triangle ABC$ it is related to. Looking at the triangle, $Z$ is the midpoint of $AC$ and $Y$ is the midpoint of $BC$? Wait, no, let's check the sides. Wait, $AB = 12$ in, $BC = 14$ in, $AC = 8$ in. Wait, the midline connecting midpoints of $AB$ and $AC$? Wait, no, let's see the labels. Wait, $X$ is on $AB$, $Y$ on $BC$, $Z$ on $AC$. Wait, the segment $ZY$: $Z$ is midpoint of $AC$, $Y$ is midpoint of $BC$? No, wait, the Midline Theorem: the segment connecting midpoints of two sides is half the third side. Wait, actually, in the triangle formed by midpoints (the medial triangle), each side is half the length of the corresponding side of the original triangle. Wait, let's see: $\triangle XYZ$ is the medial triangle of $\triangle ABC$, so each side of $\triangle XYZ$ is parallel to and half the length of a side of $\triangle ABC$. Now, which side is $\overline{ZY}$ parallel to? Let's see the points: $Z$ is on $AC$, $Y$ is on $BC$? Wait, no, maybe $Z$ is on $AC$, $Y$ is on $BC$, but actually, the midline between midpoints of $AB$ and $AC$? Wait, no, let's check the lengths. Wait, $AB = 12$ in. Wait, the midline corresponding to $AB$: the segment connecting midpoints of $AC$ and $BC$? No, wait, the Midline Theorem: if we have midpoints of $AB$ and $AC$, the midline is parallel to $BC$ and half its length. Wait, maybe I got the points wrong. Wait, the problem says "the midpoints of each side of $\triangle ABC$ connected to form $\triangle XYZ$". So $X$ is midpoint of $AB$, $Y$ midpoint of $BC$, $Z$ midpoint of $AC$. Then, the segment $ZY$: $Z$ is midpoint of $AC$, $Y$ is midpoint of $BC$? No, $ZY$ would be connecting midpoint of $AC$ and midpoint of $BC$? Wait, no, that would be parallel to $AB$? Wait, no, the midline between midpoints of $AC$ and $BC$: the third side is $AB$. So by Midline Theorem, the length of $ZY$ would be half of $AB$. Wait, $AB = 12$ in, so half of that is $6$ in? Wait, no, wait: let's clarify. Let's denote:

Let $X$ be midpoint of $AB$, $Y$ midpoint of $BC$, $Z$ midpoint of $AC$. Then:

• $XY$ is midline parallel to $AC$, so $XY = \frac{1}{2}AC = \frac{1}{2} \times 8 = 4$ in.

• $YZ$ is midline parallel to $AB$, so $YZ = \frac{1}{2}AB = \frac{1}{2} \times 12 = 6$ in.

• $ZX$ is midline parallel to $BC$, so $ZX = \frac{1}{2}BC = \frac{1}{2} \times 14 = 7$ in.

Ah, there we go. So $ZY$ (which is $YZ$) is parallel to $AB$ and half its length. So $AB = 12$ in, so $YZ = \frac{1}{2} \times 12 = 6$ in.

Step1: Apply Midline Theorem

The segment connecting the midpoints of two sides of a triangle is half the length of the third side. Here, $Z$ is the midpoint of $AC$ and $Y$ is the midpoint of $BC$? Wait, no, earlier correction: $X$ (midpoint of $AB$), $Y$ (midpoint of $BC$), $Z$ (midpoint of $AC$). Then $YZ$ connects midpoint of $AC$ (Z) and midpoint of $BC$ (Y)? No, that can't be. Wait, no, $Y$ is midpoint of $BC$, $Z$ is midpoint of $AC$, so $YZ$ is the segment between midpoints of $AC$ and $BC$, which would be parallel to $AB$ (the third side) and half its length. So $AB = 12$ in, so $YZ = \frac{1}{2} \times AB = \frac{1}{2} \times 12 = 6$ in.

Answer:

The length of $\overline{ZY}$ is $\boxed{6}$ inches.