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

Question

nexus9112 [7]

4 years ago

15

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?

Mathematics

2 answers:

Dmitrij [34] · Answer 1 · 2022-03-07T22:02:20+03:00

Dmitrij [34]4 years ago

7 0

Answer:

Area of square = 676 square units

Step-by-step explanation:

We have distance formula,

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

So side of square, a = 26 units

Area of square = a² = 26² = 676 unit²

Evgesh-ka [11] · Answer 2 · 2022-03-07T20:39:50+03:00

Answer: $676units^{2}$

Step-by-step 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:

$A=(26units)^{2}=676units^{2}$