Question

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: $\square$

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: \( \boxed{16.57} \)