Question

A quadrilateral has vertices A(4, 9), B(2, 5), C(8, 2), and D(10, 6). Which statement about the quadrilateral is true?

A. ABCD is a parallelogram with non-perpendicular 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.

Answer 1

Ans: ABCD is a rectangle with non-congruent adjacent sides

Answer 2

Answer:

Option C is correct.

Step-by-step explanation:

Given a quadrilateral has vertices A(4, 9), B(2, 5), C(8, 2), and D(10, 6). we have to check the statements which are true.

Using distance formula,

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AB=\sqrt{(2-4)^2+(5-9)^2}=\sqrt{4+16}=\sqrt{20}units$

$BC=\sqrt{(8-2)^2+(2-5)^2}=\sqrt{36+9}=\sqrt{45}units$

$CD=\sqrt{(10-8)^2+(6-2)^2}=\sqrt{4+16}=\sqrt{20}units$

$AD=\sqrt{(10-4)^2+(6-9)^2}=\sqrt{36+9}=\sqrt{45}units$

Opposite sides of quadrilateral are equal.

Now, we have to find the slope of sides to find the angle between the sides of quadrilateral.

$\text{slope of AB=}\frac{y_2-y_1}{x_2-x_1}=\frac{5-9}{2-4}=2$

$\text{slope of BC=}\frac{y_2-y_1}{x_2-x_1}=\frac{2-5}{8-2}=\frac{-1}{2}$

$\text{slope of CD=}\frac{y_2-y_1}{x_2-x_1}=\frac{6-2}{10-8}=2$

$\text{slope of AD=}\frac{y_2-y_1}{x_2-x_1}=\frac{6-9}{10-4}=\frac{-1}{2}$

Adjacent sides has the slopes negative reciprocals.

⇒ Adjacent sides of the quadrilateral are perpendicular.

hence,  ABCD is a rectangle with non-congruent adjacent sides.

Option C is correct.