How Do You Calculate the Area of a Rectangle with Given Perimeter and Diagonal?

[LeadyKnight]

Hi all, I am new to physicsforums, so i just want to get to know yall. I am a sophomore and takin alg II, i have an AMC test coming up on February 9. and my teacher gave me a review packet. there are only 25 questions but they are like the hardest questions I've ever seen. So if there's someone has already taken the test, and can help me out with those questions, i would rly appreciate it. i don't need the answer, i just want to know how to solve the problems.

Homework Statement

The perimeter of a rectangle is 100 and its diagonal has length x. What is the area of this rectangle.

Homework Equations

i got l + w = 50 (from the perimeter of the rectangle)
and x²= l² + w² (find the diagonal of the right triangle)

i found this on the internet: 2lw= (l+w)² - (l²+w²)
and got 2500 - x² , then they divide it by 2 to get the area of the rectangle.

so basically they put : 2(Area) = (perimeter)² - [this is what i don't get]
and why they square the perimeter?

2 rectangle = 4 triangle = 1 kite ? i don't think there's any connection w/ the kite : A = (1/2)d1*d2

P/s: english is not my language, so if u guys don't understand something just say sth and ill try to explain. sry

btw this is the link of all the questions: http://www.unl.edu/amc/a-activities/a7-problems/practice1012p/html1012p/10pcontest/02c10p.html#10pc#3

 

[Mentallic]

We are told that the perimeter is 100
This is [itex]2(l+w)=100[/itex] since there are 2 lengths and 2 widths in the rectangle.

So we divide by 2 and now have [itex]l+w=50[/itex]

We also have [itex]l^2+w^2=x^2[/itex] by pythagoras' theorem

and [itex]lw=A[/itex] for the Area

Now, the result you saw [itex]2lw=(l+w)^2-(l^2+w^2)[/itex] can easily be proven by expanding [itex](l+w)^2[/itex] and don't think about squaring the perimeter as a geometric thing (don't try picture what shape its creating) just square the number 50 since from before [itex]l+w=50[/itex] by squaring both sides [itex](l+w)^2=2500[/itex]

And the [itex]l^2+w^2[/itex] we already showed is equal to [itex]x^2[/itex].

So substitute everything into the equation:

[tex]2A=(50)^2-(x^2)[/tex]

[tex]A=1250-\frac{x^2}{2}[/tex]

Really, all you needed to do in this problem was find out all the relationships you can to involve x (such as pythagoras' theorem, the area of a rectangle, and the perimeter) and then the tricky bit was to notice that the area [itex]lw[/itex] can be related to the perimeter [itex]l+w[/itex] in such a way that if we square the perimeter, expanding it gives us the other pythagorean relationship [itex]l^2+w^2[/itex] and the area [itex]lw[/itex]. Then if you found the correct formula, all you need to do is substitute all your relationships into it.

 

[HallsofIvy]

[itex](l+ w)^2= l^2+ 2lw+ w^2[/itex]. l+ w= 50 and [itex]l^2+ w^2= x^2[/itex] so (50)^2= x^2+ 2A That gives the same formula Mentallic gave.

 

[LeadyKnight]

thank you guys, it really does help me out.