Answered

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

Aquadrilateral

Bparallelogram

Crhombus

Drectangle

Verified Answer :

D a rectangle because it has four sides easy for abcd

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.

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