Question

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 adjacent sides.

b. ( abcd ) is a trapezoid with only one pair of parallel sides.

c. ( abcd ) is a rectangle with noncongruent adjacent sides.

d. ( abcd ) is a square.

e. ( abcd ) is a rhombus with non - perpendicular adjacent sides.

Explanation:

Step1: Calculate the slopes of the sides

The slope formula is \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
For \(AB\) with \(A(4,5)\) and \(B(2,4)\):
\(m_{AB}=\frac{4 - 5}{2 - 4}=\frac{-1}{-2}=\frac{1}{2}\)
For \(BC\) with \(B(2,4)\) and \(C(4,3)\):
\(m_{BC}=\frac{3 - 4}{4 - 2}=\frac{-1}{2}=-\frac{1}{2}\)
For \(CD\) with \(C(4,3)\) and \(D(6,4)\):
\(m_{CD}=\frac{4 - 3}{6 - 4}=\frac{1}{2}\)
For \(DA\) with \(D(6,4)\) and \(A(4,5)\):
\(m_{DA}=\frac{5 - 4}{4 - 6}=\frac{1}{-2}=-\frac{1}{2}\)
Since \(m_{AB}=m_{CD}=\frac{1}{2}\) and \(m_{BC}=m_{DA}=-\frac{1}{2}\), \(AB\parallel CD\) and \(BC\parallel DA\), so \(ABCD\) is a parallelogram.

Step2: Calculate the lengths of the sides

The distance formula is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For \(AB\):
\(d_{AB}=\sqrt{(2 - 4)^2+(4 - 5)^2}=\sqrt{(-2)^2+(-1)^2}=\sqrt{4 + 1}=\sqrt{5}\)
For \(BC\):
\(d_{BC}=\sqrt{(4 - 2)^2+(3 - 4)^2}=\sqrt{2^2+(-1)^2}=\sqrt{4+1}=\sqrt{5}\)
For \(CD\):
\(d_{CD}=\sqrt{(6 - 4)^2+(4 - 3)^2}=\sqrt{2^2+1^2}=\sqrt{4 + 1}=\sqrt{5}\)
For \(DA\):
\(d_{DA}=\sqrt{(4 - 6)^2+(5 - 4)^2}=\sqrt{(-2)^2+1^2}=\sqrt{4+1}=\sqrt{5}\)
All sides are congruent (\(d_{AB}=d_{BC}=d_{CD}=d_{DA}=\sqrt{5}\)), so it's a rhombus.
For a rhombus to be a square, adjacent sides should be perpendicular. The product of slopes of \(AB\) and \(BC\) is \(\frac{1}{2}\times(-\frac{1}{2})=-\frac{1}{4}
eq - 1\), so adjacent sides are not perpendicular.

Answer:

E. \(ABCD\) is a rhombus with non - perpendicular adjacent sides.