Question

How do the areas of the parallelogram compare?

Answer 1

Solution:

Area of Parallelogram = Base × Height

Opposite Sides of Parallelogram are equal.

⇒  Area of Parallelogram 1 ,

Since it is a Rectangle , because Adjacent sides have angle between their legs has measure equal to 90°.

Product of slopes

    $=\frac{6-2}{2-0} \times \frac{2-0}{0-4}\\\\=2 \times \frac{-1}{2}\\\\= -1$

Area of Rectangle=Length × Breadth

        $=\sqrt{(2-0)^2+(6-2)^2}\times \sqrt{(0-4)^2 \times (2-0)^2}}\\\\=\sqrt{20}\times \sqrt{20}\\\\=20$ Square units

⇒Area of Parallelogram 2,

$\text{Base}=\sqrt{(2-0)^2+(-8+2)^2}\\\\=\sqrt{40}\\\\=2\sqrt{5}\\\\Height=\sqrt{[2-(-0.4)]^2+[0-(-1)]^2}=\sqrt{5.76+1}\\\\=\sqrt{6.76}\\\\=2.6$

Height =2.6 units

   $=\sqrt{40} \times 2.6\\\\=6.23 \times 2.6\\\\=16 \text{Square units}$Approx

⇒≡ Difference in Area

=Area of Parallelogram 1 - Area of Parallelogram 2

    =20 -16

  =4 Square units  

Option A:

  Area of Parallelogram 1 is 4 unit greater than area of Parallelogram 2.

Answer 2

The first choice is right, area of 1 is 20, area of 2 is 16