Answer:
Option B. 13 square units
Step-by-step explanation:
Area of a parallelogram is defined by the expression
A = [tex]\frac{1}{2}(\text{Sum of two parallel sides)}[/tex] × (Disatance between them)
Vertices of A, B, C and D are (3, 6), (6, 5), (5, 1) and (2, 2) respectively.
Length of AB = [tex]\sqrt{(x-x')^{2}+(y-y')^{2}}[/tex]
= [tex]\sqrt{(5-6)^{2}+(6-3)^{2}}[/tex]
= [tex]\sqrt{10}[/tex]
Since length of opposite sides of a parallelogram are equal therefore, length of CD will be same as [tex]\sqrt{10}[/tex]
Now we have to find the length of perpendicular drawn on side AB from point D or distance between parallel sides AB and CD.
Expression for the length of the perpendicular will be = [tex]\frac{|Ax_{1}+By_{1}+C|}{\sqrt{A^{2}+B^{2}}}[/tex]
Slope of line AB (m) = [tex]\frac{y-y'}{x-x'}[/tex]
= [tex]\frac{6-5}{3-6}=-(\frac{1}{3} )[/tex]
Now equation of AB will be,
y - y' = m(x - x')
y - 6 = [tex]-\frac{1}{3}(x-3)[/tex]
3y - 18 = -(x - 3)
3y + x - 18 - 3 = 0
x + 3y - 21 = 0
Length of a perpendicular from D to side AB will be
= [tex]\frac{|(2+6-21)|}{\sqrt{1^{2}+3^{2}}}[/tex]
= [tex]\frac{13}{\sqrt{10}}[/tex]
Area of parallelogram ABCD = [tex]\frac{1}{2}(AB+CD)\times (\text{Distance between AB and CD})[/tex]
= [tex]\frac{1}{2}(\sqrt{10}+\sqrt{10})\times (\frac{13}{\sqrt{10} } )[/tex]
= [tex]\sqrt{10}\times \frac{13}{\sqrt{10} }[/tex]
= 13 square units
Option B. 13 units will be the answer.
