rectangle and right angled triangle with equal area and a combined perimeter of 44

Parallel_Platypus

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rectangle and right angled triangle with equal area and a combined perimeter of 44

I believe I could solve the problem through trial and error, but I am curious of the algebraic method of solving the problem. I though a set of simultaneous equations would have worked, but I can't derive enough equations to solve it. This is my only working, as such:
The rectangle has sides a and b. Triangle has sides c, d, and e (where e is equal to the square root of c^2 + d^2)
Therefore- ab=1/2cd
And 2a+2b+c+d+ ((square root of c^2 + d^2) or e)=44

Am I approaching the problem incorrectly, or do I need to correct my notation and derive more equations?
Thx in advance
 
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thankyou

Thanks for the response, so there was no way to work out the answer completely algebraicly. I was considering that the triangle must have had a Pythagorean triad of whole numbers and that would narrow it down somewhat. Anyway, thanks for the in depth reply :)
 
Extension of this question

Hey, another user from a different website suggested that I plug in the Pythagorean triads into the original equations, giving me a solvable simultaneous set. This seemed logical so when I tried it, using the correct triad, 6, 8 and 10. I came to these equations:
xy=24
2x+2y=20
This is my working using the substitution method (the most reliable one I know at year 9 level, although I am open to learning some more advanced ones :D)
Anyway: I made x the subject of the first equation:
x=24/y
I plug this into the second equation:
2(24/y)+2y=20
Now this is where things get sketchy;
48/y+2y=20
I think I should remove the fraction by multiplying?
Giving: 48+2y^2=20y
Now If I did what I thought is logical I would end up with( Sq root of:(48)) +2y=Sq root of: 20y
Most of these are surds I believe.
Keeping it in this form, it would become: Sq:48=Sq:20y - 2y

This is obviously wrong, but when I popped it through algebrator, as always the steps were rediculously overcomplicated, and I was wondering if there was any more logical way to solve it.
I know this is now in the wrong section...
 
Thanks

I guess you didn't read my post where I suggested exactly that.

xy = 24

2x+2y=20
x+y=10
y = 10-x

xy = x(10-x) = 24

10x - x2 = 24

x2 - 10x + 24 = 0

(x-6)(x-4) = 0

x = 6, or x = 4

Not that complicated.
Yeah sorry for that. Thanks for the help :)
 
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