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
[tex]=\frac{6-2}{2-0} \times \frac{2-0}{0-4}\\\\=2 \times \frac{-1}{2}\\\\= -1[/tex]
Area of Rectangle=Length × Breadth
[tex]=\sqrt{(2-0)^2+(6-2)^2}\times \sqrt{(0-4)^2 \times (2-0)^2}}\\\\=\sqrt{20}\times \sqrt{20}\\\\=20[/tex] Square units
⇒Area of Parallelogram 2,
[tex]\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[/tex]
Height =2.6 units
[tex]=\sqrt{40} \times 2.6\\\\=6.23 \times 2.6\\\\=16 \text{Square units}[/tex]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.
