# Find the new lengths of a right triangle if hypotenuse shifts to a distance W

Given that sides $$AC$$ and $$A'C'$$ are parallel and lengths $$AB$$ and $$BC$$ are known. Also the distance between $$AC$$ and $$A'C'$$ is known and is $$w$$, What I would like to know is the lengths $$AA'$$ and $$CC'$$ and also the difference between AC and A'C'

Appreciate any pointers. Thank you for your help.

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Kumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
• Also we need to find the difference AC - A'C' – Kumar Jul 12 at 13:52
• this might help – Lozenges Jul 12 at 14:22

Let's assume $$AB=a$$, $$BC=b$$, $$A'B=a'$$, $$BC'=b'$$ and $$A'D=C'D'=w$$.

Clearly triangle $$ABC$$ and $$ADA'$$ are similar. Therefore,

$$\dfrac{AA'}{AC}=\dfrac{A'D}{BC}$$

Or

$$\dfrac{AA'}{\sqrt{a^2+b^2}}=\dfrac{w}{b}$$

$$AA'=\dfrac{w}{b}\sqrt{a^2+b^2}$$

Similarly,

Triangle $$BCA$$ and $$D'CC'$$ are similar. Therefore,

$$\dfrac{CC'}{AC}=\dfrac{C'D'}{AB}$$

Or

$$\dfrac{CC'}{\sqrt{a^2+b^2}}=\dfrac{w}{a}$$

$$CC'=\dfrac{w}{a}\sqrt{a^2+b^2}$$

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Akash Karnatak is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
• Thank you for your help. So then the difference between AC and A'C' (which I had asked in comments) should be : w(a/b+b/a) is that right? – Kumar Jul 12 at 16:58
• @Kumar If you mean $AC-A'C'$ then it's correct – Akash Karnatak Jul 13 at 2:16

Draw a perpendicular from $$A'$$ to $$AC$$. Let $$X$$ be the point of intersection of the perpendicular and $$AC$$. $$A'AX$$ is a right triangle, similar to $$ABC$$. $$A'X=W$$. From similarity of triangles, $$AX:W=AB:BC$$ or $$AX=\frac{AB \cdot W}{BC}$$. Now use Pythagorean theorem to find $$AA'$$. Similarly, construct triangle $$CC'Y$$ and obtain that $$CY=\frac{BC \cdot W}{AB}$$.