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If I could only remember my algebra! So, I have a right triangle and I know the hypotenuse (28) and I know that the legs are the same length, so I have:

$A^2 + B^2 = C^2$

Where: $A=B$ and $C=28$

Pedestrian, I know, but how do I solve for $A$?

Many thanks!

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If $A = B$ then $A^2 = B^2$, hence $A^2 + B^2 = 2 A^2 = 28$, meaning $A = \pm 2\sqrt{7}$ – Pragabhava Oct 11 '12 at 0:28
@Pragabhava: You didn't square the $C$. – Asaf Karagila Oct 11 '12 at 0:32
@AsafKaragila You are correct my friend! Also, I didn't divided by $\sqrt{2}$. Shame on me :) – Pragabhava Oct 11 '12 at 0:49
@Pragabhava: And $A$ negative may be a difficult triangle to draw. Or should we draw the legs in red and the hypotenuse in black? – André Nicolas Oct 11 '12 at 1:04
@AndréNicolas With our pseudo-Riemannian ruler and pseudo-Riemannian compass? – Pragabhava Oct 11 '12 at 1:37
up vote 2 down vote accepted

Try this:

$A^2 + B^2 = C^2$

Since A = B and C = 28, then:

$A^2 + A^2 = 28^2$

$2A^2 = 28^2$

$A^2 = \frac{784}{2}$

$A = \sqrt{\frac{784}{2}}$

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Perfect! Thank you for working it out fully! A ~= 19.79898987... for the lazy ;) – Campbeln Oct 11 '12 at 0:39
@Campbeln Not a problem! If you found my answer helpful, could you accept the answer, please? – Oliver Spryn Oct 11 '12 at 0:43

Put the numbers in the formula: $$2A^2=28^2=784$$

Now you have that $A^2=392$. You should know it from here.

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Correct, thank you for sharing! And thank you for catching the missed square in the comment above! – Campbeln Oct 11 '12 at 0:44

I know the question already received good answers, but I want just to comment something: try working first with letters and only in the end substitute the numbers. This means that you would do:

$A^2 + A^2 = C^2 \Longrightarrow 2A^2=C^2$

Hence $A=\pm \dfrac{\strut C}{\strut \sqrt{2}}=\pm \dfrac{\strut C\sqrt{2}}{2}$

Doing so you just need to put the value for $C$ and you get $A=\pm 14\sqrt{2}$ which is the answer. I just wanted to comment that because doing things this way you get practice to work with more general equations and on the other side you reduce the number of numerical calculations needed.

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