Question

Consider square ABCD with vertices at A(−8,19), B(2,−5), C(−22,−15), and D(−32,9). What is the area, in square units, of square ABCD?

Answer 1

Area of square = 676 square units

Explanation:

We have distance formula,

`AB=√((2-(-8))^2+(-5-19)^2)=26\\\\BC=√((-22-(-2))^2+(-15-(-5))^2)=26\\\\CD=√((-32-(-10))^2+(9-(-15))^2)=26\\\\DA=√((-8-(-32))^2+(9-19)^2)=26`

So side of square, a = 26 units

Area of square = a² = 26² = 676 unit²

Answer 2

Answer:
`676units^(2)`

Explanation:

1. To solve this problem you must apply the formula for calculate the area of a square, which is:

`A=s^(2)`

Where s is the lenght of a side of the square.

2. By definition, the sides of the square have equal lenghts. You can calculate the lenght of a side by calculating the distance between two vertices of the aquare:

`s_(AB)=\sqrt{(2-(-8))^(2)+(-5-19)^(2)}=26units`

Where
`s_(AB)` is the lenght of the side AB.

3. The area is:

`A=(26units)^(2)=676units^(2)`