Question

the figure below shows the midpoints of each side of \\( \triangle a b c \\) connected to form \\( \triangle x y z \\).
find the perimeter of \\( \triangle a b c \\).

Explanation:

Step1: Recall the 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.

In $\triangle ABC$, $X$, $Y$, $Z$ are midpoints of the sides. So, for example, $XZ$ is a midline, $XY$ is a midline, and $YZ$ is a midline.

Step2: Determine the lengths of the sides of $\triangle ABC$

- From the figure, we see that the lengths of the sides of $\triangle XYZ$ are $7$, $7$, and $11$.
- By the Midline Theorem, each side of $\triangle ABC$ is twice the length of the corresponding side of $\triangle XYZ$.
- Side corresponding to length $7$ in $\triangle XYZ$: The side of $\triangle ABC$ will be $2\times7 = 14$.
- Side corresponding to length $11$ in $\triangle XYZ$: The side of $\triangle ABC$ will be $2\times11 = 22$.

Step3: Calculate the perimeter of $\triangle ABC$

The perimeter of a triangle is the sum of its three sides. Since two sides are $14$ and one side is $22$ (wait, no, let's correct: Wait, $\triangle XYZ$ has sides $7$, $7$, $11$. So the sides of $\triangle ABC$:
- For the side parallel to the side of length $7$ in $\triangle XYZ$: length is $2\times7 = 14$
- For the other side parallel to the side of length $7$ in $\triangle XYZ$: length is $2\times7 = 14$
- For the side parallel to the side of length $11$ in $\triangle XYZ$: length is $2\times11 = 22$

Wait, actually, let's check the figure again. Wait, the midpoints: So $X$ is midpoint of $AB$, $Y$ midpoint of $BC$, $Z$ midpoint of $AC$. Then $XZ$ is midline, so $XZ \parallel BC$ and $XZ = \frac{1}{2}BC$. Similarly, $XY \parallel AC$ and $XY = \frac{1}{2}AC$, $YZ \parallel AB$ and $YZ = \frac{1}{2}AB$.

From the figure, $XZ = 7$, $XY = 7$, $YZ = 11$? Wait, no, the figure shows $XZ = 7$, $XY = 7$, $YZ = 11$? Wait, the labels: $X$ is on $AB$, $Y$ on $BC$, $Z$ on $AC$. Then $XZ$ connects midpoints of $AB$ and $AC$, so $XZ \parallel BC$ and $XZ = \frac{1}{2}BC$. $XY$ connects midpoints of $AB$ and $BC$, so $XY \parallel AC$ and $XY = \frac{1}{2}AC$. $YZ$ connects midpoints of $BC$ and $AC$, so $YZ \parallel AB$ and $YZ = \frac{1}{2}AB$.

So:
- $BC = 2\times XZ = 2\times7 = 14$
- $AC = 2\times XY = 2\times7 = 14$
- $AB = 2\times YZ = 2\times11 = 22$

Wait, no, that can't be. Wait, maybe I mixed up. Wait, the figure: Let's see the markings. The sides of $AB$ have one mark, $AC$ has two marks, $BC$ has three marks. The midpoints: $X$ is midpoint of $AB$, $Y$ midpoint of $BC$, $Z$ midpoint of $AC$. Then $XZ$ is between $X$ (mid $AB$) and $Z$ (mid $AC$), so $XZ$ is midline, so $XZ \parallel BC$ and $XZ = \frac{1}{2}BC$. $XY$ is between $X$ (mid $AB$) and $Y$ (mid $BC$), so $XY \parallel AC$ and $XY = \frac{1}{2}AC$. $YZ$ is between $Y$ (mid $BC$) and $Z$ (mid $AC$), so $YZ \parallel AB$ and $YZ = \frac{1}{2}AB$.

From the figure, $XZ = 7$, $XY = 7$, $YZ = 11$. So:
- $BC = 2\times XZ = 2\times7 = 14$
- $AC = 2\times XY = 2\times7 = 14$
- $AB = 2\times YZ = 2\times11 = 22$

Wait, but then the perimeter of $\triangle ABC$ is $AB + BC + AC = 22 + 14 + 14 = 50$? Wait, no, wait: Wait, maybe the sides of $\triangle XYZ$ are $7$, $11$, $7$? Wait, the figure shows $XZ = 7$, $XY = 7$, $YZ = 11$. So then:

$BC = 2\times XZ = 14$

$AC = 2\times XY = 14$

$AB = 2\times YZ = 22$

Perimeter = $22 + 14 + 14 = 50$? Wait, no, that seems off. Wait, maybe I made a mistake. Wait, let's re-express:

Wait, the Midline Theorem: The segment connecting the midpoints of two sides of a triangle is parallel to the third side and half as long. So if $X$, $Y$, $Z$ are midpoints, then:

- $XZ \parallel BC$ and $XZ = \frac{1}{2}BC$
- $XY \parallel AC$ and $XY = \frac{1}{2}AC$
- $YZ \parallel AB$ and $YZ = \frac{1}{2}AB$

So from the figure, $XZ = 7$, so $BC = 2\times7 = 14$

$XY = 7$, so $AC = 2\times7 = 14$

$YZ = 11$, so $AB = 2\times11 = 22$

Then perimeter of $\triangle ABC$ is $AB + BC + AC = 22 + 14 + 14 = 50$? Wait, no, that can't be. Wait, maybe the sides of $\triangle XYZ$ are $7$, $11$, $7$? Wait, the figure has $XZ = 7$, $XY = 7$, $YZ = 11$. So then the sides of $\triangle ABC$ are $22$, $14$, $14$. So perimeter is $22 + 14 + 14 = 50$? Wait, but let's check again. Wait, maybe the length of $YZ$ is $11$, so $AB = 2\times11 = 22$. $XZ = 7$, so $BC = 14$. $XY = 7$, so $AC = 14$. Then perimeter is $22 + 14 + 14 = 50$. Wait, but maybe I misread the figure. Wait, the figure shows $XZ = 7$, $XY = 7$, $YZ = 11$. So yes, that's correct. So the perimeter is $22 + 14 + 14 = 50$? Wait, no, wait: Wait, $AB$ is twice $YZ$, $BC$ is twice $XZ$, $AC$ is twice $XY$. So $AB = 2\times11 = 22$, $BC = 2\times7 = 14$, $AC = 2\times7 = 14$. Then perimeter is $22 + 14 + 14 = 50$. Wait, but let's check with another approach. Wait, $\triangle XYZ$ has perimeter $7 + 7 + 11 = 25$. Then the perimeter of $\triangle ABC$ should be twice that? Wait, no, because each side of $\triangle ABC$ is twice the corresponding side of $\triangle XYZ$, so perimeter of $\triangle ABC$ is $2\times(7 + 7 + 11) = 2\times25 = 50$. Yes, that's a better way. So perimeter of $\triangle XYZ$ is $7 + 7 + 11 = 25$, so perimeter of $\triangle ABC$ is $2\times25 = 50$.

Answer:

The perimeter of $\triangle ABC$ is $\boxed{50}$.