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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?

Respuesta :

carlosego

Answer: [tex]676units^{2}[/tex]


Step-by-step explanation:

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

[tex]A=s^{2}[/tex]

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:

[tex]s_{AB}=\sqrt{(2-(-8))^{2}+(-5-19)^{2}}=26units[/tex]

Where [tex]s_{AB}[/tex] is the lenght of the side AB.

3. The area is:

[tex]A=(26units)^{2}=676units^{2}[/tex]


Blacklash

Answer:

Area of square = 676 square units

Step-by-step explanation:

We have distance formula,

[tex]AB=\sqrt{(2-(-8))^2+(-5-19)^2}=26\\\\BC=\sqrt{(-22-(-2))^2+(-15-(-5))^2}=26\\\\CD=\sqrt{(-32-(-10))^2+(9-(-15))^2}=26\\\\DA=\sqrt{(-8-(-32))^2+(9-19)^2}=26[/tex]

So side of square, a = 26 units

Area of square = a² = 26² = 676 unit²

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