Carcass wrote:
The area of rectangle ABCD is 12. What is its perimeter if BD = AC = 5?
Here's a diagram representing the given information:
Let
x = the length of the base
Let
y = the height
Given: The area of rectangle ABCD is 12Since the area of a rectangle = (base)(height), we can write
xy = 12Also, since BD is the hypotenuse of
right triangle ABD, we can apply the Pythagorean theorem to write:
x² + y² = 5²Evaluate to get:
x² + y² = 25We now have the following system of equations:
xy = 12x² + y² = 25Important: Notice that some of the terms in the above system closely resemble the special product x² + 2xy + y², which can be conveniently factored to become (x + y)².
We're going to use this property to answer our question. Take the top equation (of our system), and multiply both sides by 2 to get:
2xy = 24x² + y² = 25Now add the two equations to get:
x² + 2xy + y² = 49Factor the left side to get:
(x + y)² = 49Take the square root of both sides to get:
x + y = 7Since the perimeter of the rectangle = 2x + 2y, we can take
x + y = 7 and multiply both sides of the equation by 2 to get:
2x + 2y = 14So, the perimeter of the rectangle = 14
Answer: 14