find the perimeter and area of this figure.
p = ? units
a = units²

Question

Accepted Answer

# Explanation:
## Step1: Find the coordinates of the vertices
Let the vertices be $A(2,2)$, $B(5,9)$, $C(8,2)$.
## Step2: Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$
- Length of $AB$:
$$
\begin{align*}
AB&=\sqrt{(5 - 2)^2+(9 - 2)^2}\
&=\sqrt{3^2+7^2}\
&=\sqrt{9 + 49}\
&=\sqrt{58}\approx7.62
\end{align*}
$$
- Length of $BC$:
$$
\begin{align*}
BC&=\sqrt{(8 - 5)^2+(2 - 9)^2}\
&=\sqrt{3^2+(-7)^2}\
&=\sqrt{9 + 49}\
&=\sqrt{58}\approx7.62
\end{align*}
$$
- Length of $AC$:
$$
\begin{align*}
AC&=\sqrt{(8 - 2)^2+(2 - 2)^2}\
&=\sqrt{6^2+0^2}\
&=6
\end{align*}
$$
## Step3: Calculate the perimeter $P$
$$
\begin{align*}
P&=AB + BC+AC\
&=\sqrt{58}+\sqrt{58}+6\
&=2\sqrt{58}+6\approx2\times7.62 + 6\
&=15.24+6\
&=21.24
\end{align*}
$$
## Step4: Calculate the area $A$ using the formula for the area of a triangle $A=\frac{1}{2}\times base\times height$
Base $AC = 6$, height $h$ (vertical distance from $B$ to $AC$) is $9 - 2=7$
$$
\begin{align*}
A&=\frac{1}{2}\times6\times7\
&=21
\end{align*}
$$

# Answer:
$P = 21.24$ units, $A = 21$ units$^{2}$

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QUESTION IMAGE

Question

Explanation:

Step1: Find the coordinates of the vertices

Step2: Calculate the lengths of the sides using the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)

Step3: Calculate the perimeter \(P\)

Step4: Calculate the area \(A\) using the formula for the area of a triangle \(A=\frac{1}{2}\times base\times height\)

Answer: