Question

RectangleABCD has vertices at A(– 3, 1),B(– 2, – 1),C(2, 1), andD(1, 3). What is the area, in square units, of this rectangle? A.10 B.5 C.25 D.100

Answer 1

Answer:

Option A. $10\ units^{2}$

Step-by-step explanation:

we know that

The area of the rectangle is equal to

A=LW

where

L is the length of rectangle

W is the width of rectangle

we have

$A(-3,1),B(-2,-1),C(2,1),D(1,3)$

Plot the vertices

see the attached figure

L=AD=BC

W=AB=DC

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

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

Find the distance AD

$A(-3,1),D(1,3)$

substitute in the formula

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

$AD=\sqrt{(2)^{2}+(4)^{2}}$

$AD=\sqrt{20}$

$AD=2\sqrt{5}\ units$

Find the distance AB

$A(-3,1),B(-2,-1)$

substitute in the formula

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

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

$AB=\sqrt{5}$

$AB=\sqrt{5}\ units$

Find the area

$A=(2\sqrt{5})*(\sqrt{5})=10\ units^{2}$