Question

How do the areas of the parallelograms compare?

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

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

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

What are parallelograms?

A parallelogram is a quadrilateral whose opposite sides are of equal length and parallel to each other.

What is the area of triangle and rectangle?

The area of a triangle can be found using the formula,

$\text{Area of triangle} = \rm \dfrac{1}{2} \times Height\times base\\\\\text{Area of triangle} = \rm 0.5 \times Height\times base$

The area of a rectangle is given by the formula,

$\text{Area of Rectangle} = \rm Length \times breadth$

To solve the problem we will calculate the area of each parallelogram first,

Area of Parallelogram 1,

Area of parallelogram 1= Area of the red-figure - Area ΔA - Area ΔB - Area ΔC - Area ΔD

Area of parallelogram 1= (6 x 6) - (0.5 x 4 x 2) - (0.5 x 2 x 4)- (0.5 x 4 x 2) - (0.5 x 2 x 4)

Area of parallelogram 1 = 36 - 4 - 4 - 4 - 4

Area of parallelogram 1 = 20 units²

Area of Parallelogram 2,

Area of parallelogram 2 = Area of the Blue-figure - Area ΔP - Area ΔQ - Area ΔR - Area ΔS

Area of parallelogram 2 = (8 x 4) - (0.5 x 2 x 2) - (0.5 x 6 x 2) - (0.5 x 2 x 2) - (0.5 x 6 x 2)

Area of parallelogram 1 = 32 - 2 - 6 - 2 - 6

Area of parallelogram 1 = 16 units²

The difference in the area of the parallelogram

= Area of parallelogram 1 - Area of parallelogram 2

= 20 - 16

= 4 units²

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

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