triangle pqr is formed by connecting the midpoints of the side of triangle mno. the lengths of the sides of triangle pqr are shown. find the perimeter of triangle mno. figures not necessarily drawn to scale.

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# Explanation:
## Step1: Recall Midsegment Theorem
The Midsegment Theorem (also known as the Midline 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 this case, triangle $PQR$ is formed by connecting the midpoints of triangle $MNO$, so each side of $PQR$ is half the length of the corresponding side of $MNO$.

## Step2: Find lengths of $MNO$ sides
- For the side corresponding to $PR = 2$ in $PQR$, the side in $MNO$ (let's say $NO$) will be $2\times2 = 4$? Wait, no, wait. Wait, actually, let's identify the midsegments. Wait, $P$, $Q$, $R$ are midpoints. So:
  - If $PR = 2$, then the side of $MNO$ parallel to $PR$ (which is $NO$)? Wait, no, maybe better to look at the given lengths in $PQR$. Wait, the triangle $PQR$ has sides: let's see, $PR = 2$, $RQ = 3$, $QP = 3$? Wait, no, the diagram: $PR$ is labeled 2, $RQ$ is 3, $QP$ is 3? Wait, no, the triangle $PQR$: $PR = 2$, $RQ = 3$, $QP = 3$? Wait, no, actually, since $P$, $Q$, $R$ are midpoints, then:
    - $PR$ is midsegment, so $PR\parallel NO$ and $PR=\frac{1}{2}NO$, so $NO = 2\times PR = 2\times2 = 4$? Wait, no, wait the other sides: $RQ = 3$, so the side parallel to $RQ$ (let's say $MN$) would be $2\times3 = 6$? Wait, no, maybe I got the labels wrong. Wait, the triangle $MNO$, with midpoints $P$ (midpoint of $MN$), $Q$ (midpoint of $NO$), $R$ (midpoint of $MO$)? Wait, no, the problem says "Triangle $PQR$ is formed by connecting the midpoints of the side of triangle $MNO$". So by Midsegment Theorem, each side of $PQR$ is half the length of the corresponding side of $MNO$. So:

- Let's denote:
    - $PR$ is midsegment, so $PR = \frac{1}{2}NO$, so $NO = 2\times PR = 2\times2 = 4$
    - $RQ$ is midsegment, so $RQ = \frac{1}{2}MN$, so $MN = 2\times RQ = 2\times3 = 6$
    - $QP$ is midsegment, so $QP = \frac{1}{2}MO$, so $MO = 2\times QP = 2\times3 = 6$

Wait, no, maybe I mixed up. Wait, actually, the perimeter of $PQR$ is $2 + 3 + 3 = 8$. Then, since $PQR$ is the midsegment triangle, its perimeter is half the perimeter of $MNO$. Wait, is that a property? Yes! The perimeter of the midsegment triangle is half the perimeter of the original triangle. Wait, let's check:

If $P$, $Q$, $R$ are midpoints, then $PQ=\frac{1}{2}MO$, $QR=\frac{1}{2}MN$, $RP=\frac{1}{2}NO$. Therefore, perimeter of $PQR$ is $\frac{1}{2}(MO + MN + NO)$, which is half the perimeter of $MNO$.

Wait, but first, let's find the perimeter of $PQR$. From the diagram, $PR = 2$, $RQ = 3$, $QP = 3$. So perimeter of $PQR$ is $2 + 3 + 3 = 8$. Then, since perimeter of $PQR$ is half of perimeter of $MNO$, then perimeter of $MNO$ is $2\times8 = 16$? Wait, no, that can't be. Wait, maybe I got the sides wrong. Wait, let's re-examine.

Wait, the triangle $PQR$: let's see, the sides are $PR = 2$, $RQ = 3$, and $QP = 3$. Wait, but if $P$, $Q$, $R$ are midpoints, then:

- $PR$ is midsegment, so $PR \parallel NO$ and $PR = \frac{1}{2}NO$ ⇒ $NO = 2\times2 = 4$
- $RQ$ is midsegment, so $RQ \parallel MN$ and $RQ = \frac{1}{2}MN$ ⇒ $MN = 2\times3 = 6$
- $QP$ is midsegment, so $QP \parallel MO$ and $QP = \frac{1}{2}MO$ ⇒ $MO = 2\times3 = 6$

Then, perimeter of $MNO$ is $MN + NO + MO = 6 + 4 + 6 = 16$? Wait, but let's check again. Wait, maybe the sides of $PQR$ are: $PR = 2$, $RQ = 3$, $QP = 3$. So perimeter of $PQR$ is $2 + 3 + 3 = 8$. Then, by Midsegment Theorem, the perimeter of $MNO$ should be twice that, so $16$. Wait, but let's confirm the Midsegment Theorem for perimeter. The Midsegment Theorem says each side is half, so sum of sides (perimeter) is half the sum of the original triangle's sides. So if $PQR$ has perimeter $P_{PQR} = a + b + c$, then $MNO$ has perimeter $P_{MNO} = 2a + 2b + 2c = 2(a + b + c) = 2P_{PQR}$. So if $P_{PQR} = 2 + 3 + 3 = 8$, then $P_{MNO} = 2\times8 = 16$. Wait, but let's check the sides again. Wait, maybe the labels are different. Wait, the triangle $MNO$: $M$, $N$, $O$. $P$ is midpoint of $MN$, $Q$ is midpoint of $NO$, $R$ is midpoint of $MO$. Then:

- $PR$ connects midpoints of $MN$ and $MO$, so $PR \parallel NO$ and $PR = \frac{1}{2}NO$ ⇒ $NO = 2\times PR = 2\times2 = 4$
- $RQ$ connects midpoints of $MO$ and $NO$, so $RQ \parallel MN$ and $RQ = \frac{1}{2}MN$ ⇒ $MN = 2\times RQ = 2\times3 = 6$
- $QP$ connects midpoints of $NO$ and $MN$, so $QP \parallel MO$ and $QP = \frac{1}{2}MO$ ⇒ $MO = 2\times QP = 2\times3 = 6$

Then, perimeter of $MNO$ is $MN + NO + MO = 6 + 4 + 6 = 16$. Yes, that matches. So the perimeter of triangle $MNO$ is 16.

Wait, but let's check again. The triangle $PQR$ has sides 2, 3, 3. Perimeter 8. Then, since each side of $PQR$ is half of $MNO$'s sides, then $MNO$'s sides are 4, 6, 6. Perimeter 4 + 6 + 6 = 16. Yes, that's correct.

## Step1: Identify Midsegments
Since $P$, $Q$, $R$ are midpoints of $MNO$'s sides, by the Midsegment Theorem, each side of $PQR$ is half the length of the corresponding side of $MNO$.

## Step2: Calculate Perimeter of $PQR$
From the diagram, $PR = 2$, $RQ = 3$, $QP = 3$.
Perimeter of $PQR$ = $2 + 3 + 3 = 8$.

## Step3: Calculate Perimeter of $MNO$
By the Midsegment Theorem, the perimeter of $MNO$ is twice the perimeter of $PQR$ (since each side of $MNO$ is twice the corresponding side of $PQR$).
Thus, Perimeter of $MNO$ = $2 \times 8 = 16$.

# Answer:
The perimeter of triangle $MNO$ is $\boxed{16}$.

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QUESTION IMAGE

Question

Explanation:

Step1: Recall Midsegment Theorem

Step2: Find lengths of \(MNO\) sides

Answer: