Question

Polygon ABCD is a rectangle. What is its area? Round your answer to the

Answer 1

Answer:

Option A. 20.0 square units

Step-by-step explanation:

Let

$A(2, 5),B(-4, 3),C(-3, 0),D(3, 2)$

Plot the points

see the attached figure

The area of the rectangle is equal to

$A=AB*AD$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance AB

$A(2, 5),B(-4, 3)$

substitute in the formula

$AB=\sqrt{(3-5)^{2}+(-4-2)^{2}}$

$AB=\sqrt{(-2)^{2}+(-6)^{2}}$

$AB=\sqrt{40}\ units$

step 2

Find the distance AD

$A(2, 5),D(3, 2)$

substitute in the formula

$AD=\sqrt{(2-5)^{2}+(3-2)^{2}}$

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

$AD=\sqrt{10}\ units$

step 3

Find the area of rectangle

$A=AB*AD$

we have

$AB=\sqrt{40}\ units$

$AD=\sqrt{10}\ units$

substitute

$A=(\sqrt{40})*(\sqrt{10})$

$A=20\ units^{2}$