Question

What is the most precise name for quadrilateral ABCD with vertices A(-4, -4), B(-4, -2), C(-1, -2), and D(-1, -4)?

A) quadrilateral

B) parallelogram

C) rhombus

D) rectangle

Answer 1

D a rectangle because it has four sides easy for abcd

Answer 2

Answer:

Option D.

Step-by-step explanation:

The given vertices of quadrilateral ABCD are A(-4, -4), B(-4, -2), C(-1, -2), and D(-1, -4).

Distance formula:

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

Using distance formula, we get

$AB=\sqrt{\left(-4-\left(-4\right)\right)^2+\left(-2-\left(-4\right)\right)^2}=\sqrt{0^2+(2)^2}=\sqrt{4}=2$

Similarly,

$BC=\sqrt{\left(-1-\left(-4\right)\right)^2+\left(-2-\left(-2\right)\right)^2}=3$

$CD=\sqrt{\left(-1-\left(-1\right)\right)^2+\left(-4-\left(-2\right)\right)^2}=2$

$AD=\sqrt{\left(-1-\left(-4\right)\right)^2+\left(-4-\left(-4\right)\right)^2}=3$

$AB = CD$

$BC = AD$

Measure of diagonals:

$AC=\sqrt{\left(-1-\left(-4\right)\right)^2+\left(-2-\left(-4\right)\right)^2}=\sqrt{13}$

$BD=\sqrt{\left(-1-\left(-4\right)\right)^2+\left(-4-\left(-2\right)\right)^2}=\sqrt{13}$

$AC = BD$

Since measure of opposite sides are equal and measure of diagonals are equal, therefore the given quadrilateral ABCD is a rectangle.

Hence, the correct option is D.