Answer:
Area of parallelogram ABCD=13.05=13 Sq.Units
Step-by-step explanation:
Given;
A(3,6) B(6,5) C(5,1) and D(2,2)
are points for parallelogram
To Find:
Area of Parallelogram ABCD
Solution:
By using distance formula we can calculate for each length of parallelogram
But by property of parallelogram
Opposite side are parallel and equal in length
So AB|| DC i.e AB=DC
And AC|| BD i.e BD=AC
Hence I.e Dist(AB)=Dist(DC)
Now construct the ABCD parallelogram ,on graph so as to find angle made by parallelogram with plane .
(Refer the attachment)
Now
Distance of AD=Sqrt[(2-3)^2+(2-6)^2]
=Sqrt[1+16]
=4.123
Similarly for AB=Sqrt[(6-3)^2+(5-6)^2]
=Sqrt[9+1]
=3.16
ABOVE VALUES ARE SAME AS GRAPH (REFER THE ATTACHMENT)
Now ,angle made by parallelogram with plane and it is 90 degree i.e the
Now Area of parallelogram(ABCD)=a*b*sinФ
here a=4.123 units and b=3.16 units and Ф=90
Area of parallelogram=3.167*4.123*sin90
=3.167*4.123
=13.05 sq. units
(divided both sides by -3z)