Pythagorean formula question of sorts with B related to length of A

**freedda** · November 2nd 2012, 06:02 PM

I am a non-math person with a math question. Suppose you have an Pythagorean formula, but there is a known relationship in length between A and B. So:

Given that A = A and B = (0.75A) . . . your formula is:

A(squared) + 0.75A(squared) = 361 or 19(squared)

Can you solve this, since you now only have one unknown, A? and if so, could you show me the steps to get there? Thanks, D

**MarkFL** · November 2nd 2012, 06:36 PM

$A^2+\left(\frac{3}{4}A \right)^2=19^2$

$A^2\left(1+\frac{9}{16} \right)=19^2$

$\frac{25}{16}A^2=19^2$

$A^2=\left(\frac{4}{5}\cdot19 \right)^2=\left(\frac{76}{5} \right)^2$

Assuming $0<A$ we have:

$A=\frac{76}{5}$

**freedda** · November 2nd 2012, 06:49 PM

Thanks Mark. I don't follow how you got from the first line (which is just restating my question) to the second, which includes (1 + 9/16). Where the heck did the 1 come from? But actually, the final equation works and can be used with other numbers, so understanding step two is only a curiousity for me now. David

Originally Posted by MarkFL2

$A^2+\left(\frac{3}{4}A \right)^2=19^2$

$A^2\left(1+\frac{9}{16} \right)=19^2$

$\frac{25}{16}A^2=19^2$

$A^2=\left(\frac{4}{5}\cdot19 \right)^2=\left(\frac{76}{5} \right)^2$

Assuming $0<A$ we have:

$A=\frac{76}{5}$

**MarkFL** · November 2nd 2012, 07:02 PM

Both terms on the left side have $A^2$ as a factor, perhaps if you put this in between the first and second lines:

$A^2\cdot1+A^2\cdot\frac{9}{16}=19^2$

It is more clear now that when we factor out the $A^2$ we get the second line of my first post.

**freedda** · November 3rd 2012, 06:53 AM

Mark, I don't really get it, but that's okay, since it works.

If I was to use another fraction to multiply A by, for example B = 5/8A, I assume that I could still use your formula, only the second step would be:

Asq (1 + 25/64) = C sq

Is that correct? I get this by squaring each number of the fraction 5/8 which gives me the 25/64. I assume that's what you do to get to the fraction in the second step, but that's only a guess. Thanks again, David.

**MarkFL** · November 3rd 2012, 11:15 AM

Yes, that's correct.

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