Given that point A is(0,0) (from figure) and B (-3, -4) and C (-7, -7).
Now we know that area of a triangle \(|\frac{1}{2}(x1(y2-y3)+x2(y3-y1)+x3(y1-y2))|\) where we have (x1,y1), (x2, y2) and (x3, y3) as three vertices of the triangle.
Area of the triangle=\(|\frac{1}{2}(x1(y2-y3)+x2(y3-y1)+x3(y1-y2))|\)
=\(|\frac{1}{2}(0(-4-(-7))-3(-7-0)-7(0-(-4)))|\)
=\(|\frac{1}{2}(-3(-7)-7(4))|\)=3.5
So from geometry we can say that area of parallelogram ABCD = 2* area of the triangle ABC = 2*3.5 =7.
Hence
option C is the right answer.
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