Rectangle ABCD has vertices at A(−3,1),B(−2,−1),C(2,1), and D(1,3). What is the area, in square units, of this rectangle?

A

5

B

10

C

25

D

100

B.  10

Step-by-step explanation:

To find the area of this rectangle, we will simply find the distance AB and then find the distance BC, then multiply both together.

First, we will find the distance AB

A(-3,1),      B(-2, -1)

Using the distance formula;

|AB| = √(x₂ - x₁)² + (y₂ - y₁)²

x₁= -3               y₁ = 1                     x₂ = -2                   y₂ = -1

|AB| = √(-2 + 3)² + (-1 - 1)²

        =√(1)² + (-2)²

       = √1 + 4

        = √5

Next, we will find the distance BC

B(-2, -1)          C(2, 1)

Using the distance formula;

|BC| = √(x₂ - x₁)² + (y₂ - y₁)²

x₁= -2               y₁ =- 1                     x₂ = 2                   y₂ = 1

|AB| = √(2 + 2)² + (-1 - 1)²

        =√(4)² + (-2)²

       = √16 + 4

        = √20

Area of the rectangle ABCD = |AB| . |BC|

                                                =  √5 × √20

                                                 = √5×20

                                                   =√100

                                                    =10

Area of the rectangle ABCD  =  10 square units

x = 25

step-by-step explanation:

a^2+b^2=c^2

7^2+24^2=c^2

49+576=625

sqrt625=25

x=25

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