Question

How do the areas of the parallelograms compare?

Answer 1

see the attached figure to better understand the problem

Step $1$

Find the area of the parallelogram $1$

Find the area of the complete square and subtract the area of the four triangles

so

Area of the complete square

$A=6^{2}=36\ unit^{2}$

Area of the four triangles

$A=4*[\frac{1}{2}*2*4]=16\ unit^{2}$

Area of the parallelogram $1$

$A1=36\ unit^{2}-16\ unit^{2}=20\ unit^{2}$

Step $2$

Find the area of the parallelogram $2$

Find the area of the complete rectangle and subtract the area of the four triangles

so

Area of the complete rectangle

$A=4*8=32\ unit^{2}$

Area of the four triangles

$A=2*[\frac{1}{2}*2*6]+2*[\frac{1}{2}*2*2]=16\ unit^{2}$

Area of the parallelogram $2$

$A2=32\ unit^{2}-16\ unit^{2}=16\ unit^{2}$

Step $3$

Compare the areas

$A1=20\ unit^{2}$

$A2=16\ unit^{2}$

$A1-A2=20-16=4\ unit^{2}$

$A1=A2+4\ unit^{2}$

therefore

the answer is the option

The area of parallelogram 1 is 4 square units greater than the area of parallelogram 2.

Answer 2

The answer is A;

The area of parallelogram 1 is 4 square units greater than the area of parallelogram 2.