Question

How do the areas of the parallelograms compare?

Answer 1

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:

          $Area=bh$

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,

$AB=\sqrt{(7-4)^2+(2-2)^2}\\\\\\AB=\sqrt{3^2}\\\\\\AB=3\ units$

$DE=\sqrt{(1-1)^2+(2-6 )^2}\\\\\\DE=\sqrt{4^2}\\\\\\DE=4\ units$

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,

$EF=\sqrt{(-5-(-2))^2+(2-2)^2}\\\\\\EF=\sqrt{(-5+2)^2}\\\\\\EF=\sqrt{(-3)^2}\\\\\\EF=\sqrt{3^2}\\\\\\EF=3\ units$

and

$GI=\sqrt{(-6-(-6))^2+(2-6)^2}\\\\\\GI=\sqrt{(-6+6)^2+(-4)^2}\\\\\\GI=\sqrt{4^2}\\\\\\GI=4\ units$

Hence,

Area of parallelogram EFGH= 3×4=12 square units

Answer 2

Check the picture below.

get the area of each, compare them away.

notice though, regardless of the slantness, the height and base are the same length.