a quadrilateral has vertices A(4,5), B(2,4), C(4,3), and D(6,4) which statement about the quadrilateral is true A ABCD is a parallelogram with noncongruent adj sides B ABCD is a trapezoid with only one pair of parallel sides C ABCD is a rectangle with noncongruent adj sides D ABCD is a square E ABCD is a rhombus with non perpendicular adj sides

The answer is E. If you draw the quadrilateral out, you can clearly see what it looks like. And you can know from the coordinates that the diagonal of this quadrilateral is not equal. So it can not be rectangle. And also AB//CD and AD//BC.

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frika frika · Answer 2 · 2018-07-31T14:06:41-04:00

Let's find the lengths of all sides:

[tex] AB=\sqrt{(4-2)^2+(5-4)^2}=\sqrt{4+1} =\sqrt{5} ,\\ BC=\sqrt{(2-4)^2+(4-3)^2}=\sqrt{4+1} =\sqrt{5},\\ CD=\sqrt{(4-6)^2+(3-4)^2} =\sqrt{4+1} =\sqrt{5},\\ AD=\sqrt{(6-4)^2+(4-5)^2}=\sqrt{4+1} =\sqrt{5} [/tex].

All sides are congruent, then this quadrilateral may occur to be square or rhombus. Check whether sides AB and AD are perpendicular. With this aim you should find vectors [tex] \vec{AB} [/tex] and [tex] \vec{AD} [/tex]:

[tex] \vec{AB}=(2-4,4-5) =(-2,-1)[/tex] ,

[tex] \vec{AD}=(6-4,4-5) =(2,-1)[/tex] .

The dot product: [tex] \vec{AB}\cdot \vec{AD}=-2\cdot 2+(-1)\cdot (-1)=-4+1=-3\neq 0 [/tex]. This means that sides AB and AD are not perpendicular and quarilateral ABCD is not square.

Answer: correct choice is E.