151k views
2 votes
What is the area of rectangle ABCD, if points A, B, C, and D have the following coordinates?


* A(2, 5), B(2 (2)/(3) , 5) C(2 (2)/(3) , 5 (3)/(5) ), D(2, 5 (3)/(5) )

1 Answer

2 votes

Answer:

The area of the rectangle ABCD is
(2)/(5) square units.

Explanation:

From statement we have that rectangle ABCD is formed by the following points:
A(x,y) = (2, 5), B(x,y) = \left((8)/(3), 5 \right), C(x,y) = \left((8)/(3), (28)/(5) \right), D(x,y) = \left(2, (28)/(5) \right). First, we calculate the length of each side by the Pythagorean Theorem:


AB = \sqrt{\left((8)/(3)-2 \right)^(2)+ (5-5)^(2)}


AB = (2)/(3)


BC = \sqrt{\left((8)/(3)-(8)/(3)\right)^(2)+\left((28)/(5)-5\right)^(2) }


BC = (3)/(5)


CD = \sqrt{\left(2-(8)/(3) \right)^(2)+\left((28)/(5)-(28)/(5) \right)^(2)}


CD = (2)/(3)


DA = \sqrt{\left(2-2\right)^(2)+\left(5-(28)/(5) \right)^(2)}


BC = (3)/(5)

Which satisfies all minimum characteristics for a rectangle. The area of the rectangle ABCD is the product of its base and its height, that is:


A = \left((2)/(3)\right)\cdot \left((3)/(5) \right)


A = (2)/(5)

The area of the rectangle ABCD is
(2)/(5) square units.

User Andlabs
by
3.9k points