Question

find the perimeter and area of the figure below.
p = ___ units
a = ___ units²

Explanation:

Step1: Identify triangle vertices

Vertices: $(5,2)$, $(-6,7)$, $(-6,-7)$

Step2: Calculate side lengths (distance formula)

Distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

1. Length 1 (vertical side):

$\sqrt{(-6-(-6))^2+(-7-7)^2}=\sqrt{0+(-14)^2}=14$

1. Length 2:

$\sqrt{(5-(-6))^2+(2-7)^2}=\sqrt{11^2+(-5)^2}=\sqrt{121+25}=\sqrt{146}\approx12.08$

1. Length 3:

$\sqrt{(5-(-6))^2+(2-(-7))^2}=\sqrt{11^2+9^2}=\sqrt{121+81}=\sqrt{202}\approx14.21$

Step3: Calculate perimeter

Sum all side lengths:
$14 + \sqrt{146} + \sqrt{202} \approx14+12.08+14.21=40.29$

Step4: Calculate area (base-height formula)

Base = 14, height = $5-(-6)=11$
$A=\frac{1}{2}\times base\times height=\frac{1}{2}\times14\times11=77$

Answer:

$P = 14+\sqrt{146}+\sqrt{202}\approx40.29$ units
$A = 77$ units$^2$