Step 1
Find the perimeter of rectangle ABDE
we know that
the perimeter of rectangle is equal to
[tex]P=2b+2h[/tex]
In this problem
[tex]b=ED=2\ units[/tex]
[tex]h=AE=6\ units[/tex]
substitute
[tex]P=2*2+2*6=16\ units[/tex]
Step 2
Find the perimeter of triangle BCD
we know that
the perimeter of triangle is equal to
[tex]P=BD+DC+BC[/tex]
In this problem we have
[tex]BD=AE=6\ units[/tex]
[tex]DC=BC[/tex]
Applying the Pythagoras theorem
[tex]DC^{2}=4^{2}+3^{2}[/tex]
[tex]DC^{2}=25[/tex]
[tex]DC=5\ units[/tex]
substitute
[tex]P=6+5+5=16\ units[/tex]
Find the ratio of the perimeter of rectangle ABDE to the perimeter of triangle BCD
we have
the perimeter of rectangle is equal to
[tex]P=16\ units[/tex]
the perimeter of the triangle is
[tex]P=16\ units[/tex]
so
the ratio is equal to
[tex]\frac{16}{16} =1[/tex]
therefore
the answer Part 1) is the option B
[tex]1[/tex]
Step 3
Find the area of polygon ABCDE
we know that
The area of polygon is equal to the sum of the area of rectangle plus the area of triangle
Area of rectangle is equal to
[tex]A=AE*BD=6*2=12\ units^{2}[/tex]
Area of the triangle is equal to
[tex]A=\frac{1}{2}AEh[/tex]
the height h of the triangle is equal to [tex]4\ units[/tex]
substitute
[tex]A=\frac{1}{2}(6)(4)=12\ units^{2}[/tex]
The area of polygon is
[tex]12\ units^{2}+12\ units^{2}=24\ units^{2}[/tex]
therefore
the answer part 2) is the option C
[tex]24\ units^{2}[/tex]
