On a coordinate plane, parallelogram A B C D has points (3, 6), (6, 5), (5, 1), and (2, 2).

What is the area of parallelogram ABCD?

13 square units

14 square units

15 square units

16 square units

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

Step-by-step explanation:

Point of intersections of the graph with x-axis are not clear.

Let these points are x = -a, -b and c

Therefore, function representing the graph will be,

f(x) = p(x + a)(x + b)(x - c)

Since, this graph crosses y-axis at (0, -6),

f(0) = p(0 + a)(0 + b)(0 - c) = -6

-pabc = -6

p =

Therefore, equation of the given function will be,

f(x) =