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