Answer:
The statement that compares the area of two parallelograms is:
- The area of parallelogram ABCD is equal to the area of parallelogram EFGH.
Explanation:
We know that the area of parallelogram is given by:
![Area=bh](https://img.qammunity.org/2019/formulas/mathematics/high-school/kqsknsvlig5wmm05nwseroh2pru9y5y82b.png)
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,
![AB=√((7-4)^2+(2-2)^2)\\\\\\AB=√(3^2)\\\\\\AB=3\ units](https://img.qammunity.org/2019/formulas/mathematics/high-school/3nmpaftbf54y9ioj447tvs2h3opfqu3ehu.png)
![DE=√((1-1)^2+(2-6 )^2)\\\\\\DE=√(4^2)\\\\\\DE=4\ units](https://img.qammunity.org/2019/formulas/mathematics/high-school/9fznwk61mbqwu63rxhcqrv7y54xs8nt05r.png)
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,
![EF=√((-5-(-2))^2+(2-2)^2)\\\\\\EF=√((-5+2)^2)\\\\\\EF=√((-3)^2)\\\\\\EF=√(3^2)\\\\\\EF=3\ units](https://img.qammunity.org/2019/formulas/mathematics/high-school/ni6wqrj6hu8mp3x3k4cjhokwzbdqpq1bhr.png)
and
![GI=√((-6-(-6))^2+(2-6)^2)\\\\\\GI=√((-6+6)^2+(-4)^2)\\\\\\GI=√(4^2)\\\\\\GI=4\ units](https://img.qammunity.org/2019/formulas/mathematics/high-school/qbyvs59y4uqj7sopaims4xetz8izrc5rvd.png)
Hence,
Area of parallelogram EFGH= 3×4=12 square units