Easy Rectangle

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?

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 ***

Originally posted by joe shmo
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)

What you did works well.

Starting with what I have quoted above there is a neat trick which avoids needing to complete the square.

If you square both sides of eq(1) you get something that involves a multiple of x² + y² (which is d² ) and a multiple of xy (which is the area we want).

Then rearrange to get your result.

Originally posted by Diapason
What you did works well.

Starting with what I have quoted above there is a neat trick which avoids needing to complete the square.

If you square both sides of eq(1) you get something that involves a multiple of x² + y² (which is d² ) and a multiple of xy (which is the area we want).

Then rearrange to get your result.

Wish I would have saw that,...by far less work!

Nicely done. And the squaring trick is what I was thinking too. 🙂

2x + 2y = p => x + y = p/2 => x^2 + 2xy + y^2 = p^2/4

x^2 + y^2 = d^2

combine for p^2/4 + 2xy = d^2 => xy = d^2/2 - p^2/8 = A

Originally posted by talzamir
Nicely done. And the squaring trick is what I was thinking too. 🙂

2x + 2y = p => x + y = p/2 => x^2 + 2xy + y^2 = p^2/4

x^2 + y^2 = d^2

combine for p^2/4 + 2xy = d^2 => xy = d^2/2 - p^2/8 = A

It's funny, the first numbers I chose showed me something about rectangles:

I chose d=10 and p=20 where A = zero🙂

which makes sense🙂

It also means you can get negative numbers: D=10 and P=22, A = minus 60.5)

How can you fix it so that doesn't happen?

Exploding ink?🙂

Originally posted by sonhouse
It's funny, the first numbers I chose showed me something about rectangles:

I chose d=10 and p=20 where A = zero🙂

which makes sense🙂

It also means you can get negative numbers: D=10 and P=22, A = minus 60.5)

How can you fix it so that doesn't happen?

Exploding ink?🙂

You happened to pick your variables outside of the domain of the equation.

In order for the equation to give valid results ( that is results that are physical ) the area of the rectangle must be strictly greater than zero. For if it were equal to zero, you don't have a rectangle.

So;

A > 0

P²/8 - d²/2 > 0

P²/8 > d²/2

P² > 4*d²

P > 2*d (this is the domain of the equation), which say that all perimeters must be greater than twice the diagonal.

You picked a perimeter that was exactly equal to twice the diagonal, which is out of the equations domain. See?

Originally posted by joe shmo
You happened to pick your variables outside of the domain of the equation.

In order for the equation to give valid results ( that is results that are physical ) the area of the rectangle must be strictly greater than zero. For if it were equal to zero, you don't have a rectangle.

So;

A > 0

P²/8 - d²/2 > 0

P²/8 > d²/2

P² > 4*d²

P > 2*d ...[text shortened]... imeter that was exactly equal to twice the diagonal, which is out of the equations domain. See?

I knew that, just pulling your chain🙂