Question

what type of quadrilateral has the vertices a(3, 6), b(3, 3), c(6, 3), and d(6, 6)?
a. square
b. non - square parallelogram
c. non - square rhombus
d. non - special parallelogram

Explanation:

Step1: Calculate side - lengths

Use the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
For side $AB$ with $A(3,6)$ and $B(3,3)$: $AB=\sqrt{(3 - 3)^2+(3 - 6)^2}=\sqrt{0 + 9}=3$.
For side $BC$ with $B(3,3)$ and $C(6,3)$: $BC=\sqrt{(6 - 3)^2+(3 - 3)^2}=\sqrt{9+0}=3$.
For side $CD$ with $C(6,3)$ and $D(6,6)$: $CD=\sqrt{(6 - 6)^2+(6 - 3)^2}=\sqrt{0 + 9}=3$.
For side $DA$ with $D(6,6)$ and $A(3,6)$: $DA=\sqrt{(3 - 6)^2+(6 - 6)^2}=\sqrt{9+0}=3$.

Step2: Check angles

The slope of $AB$ is undefined (vertical line, $x = 3$), the slope of $BC$ is $0$ (horizontal line, $y = 3$). So, $\angle ABC=90^{\circ}$. Since adjacent sides are perpendicular and all sides are equal, it is a square.

Answer:

A. square