Answer:
The statement that compares the area of two parallelograms is:
- The area of parallelogram ABCD is equal to the area of parallelogram EFGH.
Step-by-step explanation:
We know that the area of parallelogram is given by:
[tex]Area=bh[/tex]
where b is the base of the parallelogram and h is the height of the parallelogram.
Parallelogram ABCD
Base(b)=AB
and Height(h)=DE
we have the coordinates of A,B,C,D and E as:
A(4,2) B(7,2) C(4,6) D(1,6) E(1,2)
Hence,
[tex]AB=\sqrt{(7-4)^2+(2-2)^2}\\\\\\AB=\sqrt{3^2}\\\\\\AB=3\ units[/tex]
[tex]DE=\sqrt{(1-1)^2+(2-6 )^2}\\\\\\DE=\sqrt{4^2}\\\\\\DE=4\ units[/tex]
Hence, Area of parallelogram ABCD= 3×4=12 square units
Similarly,
In Parallelogram EFGH
we have:
Base(b)=EF
Height(h)=GI
The coordinates are:
E(-2,2) F(-5,2) G(-6,6) H(-3,6) and I(-6,2)
Hence,
[tex]EF=\sqrt{(-5-(-2))^2+(2-2)^2}\\\\\\EF=\sqrt{(-5+2)^2}\\\\\\EF=\sqrt{(-3)^2}\\\\\\EF=\sqrt{3^2}\\\\\\EF=3\ units[/tex]
and
[tex]GI=\sqrt{(-6-(-6))^2+(2-6)^2}\\\\\\GI=\sqrt{(-6+6)^2+(-4)^2}\\\\\\GI=\sqrt{4^2}\\\\\\GI=4\ units[/tex]
Hence,
Area of parallelogram EFGH= 3×4=12 square units
