*Originally posted by talzamir*

**This is trivial enough, but I'll probably use it in a high school someday.
**

A rectangle has circumference length of p. The diagonal has length d. What is the area of the rectangle?

I don't know if I did this the hard way, because it didn't seem obvious to me...

Let the sides of the rectangle be x, and y respectively.

Perimeter = P = 2*x + 2*y ....eq (1)

from the Pythagorean theorem

d² = x² + y² ...eq(2)

from eq(1)

y = 1/2*P - x ...eq(1)'

from eq(2)

y = Sqrt ( d² - x² )

Set eq(1)' = eq(2)

1/2*P - x = Sqrt ( d² - x² )

solve for x in terms of d, and P by completing the square

x = Sqrt(d²/2 - 1/16*P² ) + P/4 ...eq(3)

Sub eq(3) into eq(1)'

y = 1/2*P - [Sqrt(d²/2 - 1/16*P² ) + P/4]

= 1/4*P - Sqrt(d²/2 - 1/16*P² )

then Area = A = x*y

= (P/4 + Sqrt(d²/2 - 1/16*P² ))(P/4 - Sqrt(d²/2 - 1/16*P² ))

which is of the form (a + b)(a - b) = a² - b² : giving after simplification

A = P²/8 - d²/2 ***