Posers and Puzzles
10 Mar 14
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11 Mar 14
3 edits
Originally posted by talzamirI don't know if I did this the hard way, because it didn't seem obvious to me...
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?
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 shmoWhat you did works well.
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)
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.
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11 Mar 14
Originally posted by DiapasonWish I would have saw that,...by far less work!
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.
21 Mar 14
2 edits
Originally posted by talzamirIt's funny, the first numbers I chose showed me something about rectangles:
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
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?🙂
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21 Mar 14
1 edit
Originally posted by sonhouseYou happened to pick your variables outside of the domain of the equation.
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?🙂
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?
22 Mar 14
Originally posted by joe shmoI knew that, just pulling your chain🙂
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?