4 the vertices of $\triangle map$ are $m(-3,1)$, $a(1,3)$, and $p(3,-2)$. find the perimeter of the shape in the coordinate plane. round to the nearest hundredth. $\overline{ma} = \square$ $\overline{mp} = \square$ $\overline{ap} = \square$ perimeter: $\square$

Question

Accepted Answer

# Explanation:
To find the lengths of the sides of the triangle $ \triangle MAP $ with vertices $ M(-3, 1) $, $ A(1, 3) $, and $ P(3, -2) $, we use the distance formula: $ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $.

## Step 1: Find $ \overline{MA} $
For points $ M(-3, 1) $ and $ A(1, 3) $:
$$
\begin{align*}
MA &= \sqrt{(1 - (-3))^2 + (3 - 1)^2} \
&= \sqrt{(4)^2 + (2)^2} \
&= \sqrt{16 + 4} \
&= \sqrt{20} \
&\approx 4.47
\end{align*}
$$

## Step 2: Find $ \overline{MP} $
For points $ M(-3, 1) $ and $ P(3, -2) $:
$$
\begin{align*}
MP &= \sqrt{(3 - (-3))^2 + (-2 - 1)^2} \
&= \sqrt{(6)^2 + (-3)^2} \
&= \sqrt{36 + 9} \
&= \sqrt{45} \
&\approx 6.71
\end{align*}
$$

## Step 3: Find $ \overline{AP} $
For points $ A(1, 3) $ and $ P(3, -2) $:
$$
\begin{align*}
AP &= \sqrt{(3 - 1)^2 + (-2 - 3)^2} \
&= \sqrt{(2)^2 + (-5)^2} \
&= \sqrt{4 + 25} \
&= \sqrt{29} \
&\approx 5.39
\end{align*}
$$

## Step 4: Find the Perimeter
The perimeter of a triangle is the sum of the lengths of its sides:
$$
\begin{align*}
\text{Perimeter} &= MA + MP + AP \
&\approx 4.47 + 6.71 + 5.39 \
&\approx 16.57
\end{align*}
$$

# Answer:
$ \overline{MA} \approx \boxed{4.47} $
$ \overline{MP} \approx \boxed{6.71} $
$ \overline{AP} \approx \boxed{5.39} $
Perimeter: $ \boxed{16.57} $

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QUESTION IMAGE

Question

Explanation:

Step 1: Find \( \overline{MA} \)

Step 2: Find \( \overline{MP} \)

Step 3: Find \( \overline{AP} \)

Step 4: Find the Perimeter

Answer: