Answer:
Part A) The area of the figure is [tex]24\ units^{2}[/tex]
Part B) The perimeter of the figure is [tex]20\ units[/tex]
Step-by-step explanation:
step 1
Find the area of the figure
we know that
The area of the figure is equal to the area of triangle ABD plus the area of triangle BCD
The area of triangle is equal to
[tex]A=\frac{1}{2}bh[/tex]
Area of triangle ABD
Observing the graph
[tex]b=BD=(-2+8)=6\ units[/tex]
[tex]h=(9-5)=4\ units[/tex]
substitute
[tex]A=\frac{1}{2}(6)(4)=12\ units^{2}[/tex]
Area of triangle BCD
Observing the graph
[tex]b=BD=(-2+8)=6\ units[/tex]
[tex]h=(5-1)=4\ units[/tex]
substitute
[tex]A=\frac{1}{2}(6)(4)=12\ units^{2}[/tex]
The area of the figure is
[tex]12\ units^{2}+12\ units^{2}=24\ units^{2}[/tex]
step 2
Find the perimeter of the figure
we know that
The perimeter of the figure is equal to
[tex]P=AB+BC+CD+AD[/tex]
we have
[tex]A(-5,1),B(-8,5),C(-5,9),D(-2,5)[/tex]
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
Find the distance AB
[tex]d=\sqrt{(5-1)^{2}+(-8+5)^{2}}=5\ units[/tex]
Find the distance BC
[tex]d=\sqrt{(9-5)^{2}+(-5+8)^{2}}=5\ units[/tex]
Find the distance CD
[tex]d=\sqrt{(5-9)^{2}+(-2+5)^{2}}=5\ units[/tex]
Find the distance AD
[tex]d=\sqrt{(5-1)^{2}+(-2+5)^{2}}=5\ units[/tex]
substitute the values
[tex]P=5+5+5+5=20\ units[/tex]
