# question about right triangle

Suppose there is given right triangle($AC=6$) and circle (radius$=5$) which goes through $A$ and $C$ and meets $AB$ leg at the midpoint. Here is picture

We are asked to find length of leg.in my point of view, we can connect center of circle with $A$ and $C$ point,get isosceles triangle we know all sides ,find angle at center, then connect center also to $A$ and $D$ here we know both length (radius) angle will be $180-a$ ($a$ it is which we got by cosine law) calculate $AD$ by the cosine law and got finally $AB$, can you show me shortest way? or am I correct or wrong? Please help!

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What do we know? We know the radius of the circle and one angle. Do we know any lengths or angles? – mixedmath Jul 23 '11 at 10:13
oops sorry AC=6 i forgot – dato datuashvili Jul 23 '11 at 10:16
in fact it is not homework it is just token from national exam tasks – dato datuashvili Jul 23 '11 at 10:21
@user3196: Sorry for being presumptuous. I think the tag is unnecessary in that case. – anon Jul 23 '11 at 10:28
@anon no dont worry please there is not any problem – dato datuashvili Jul 23 '11 at 10:31

Let $S$ be the center of the circle and $M$ be the midpoint of $AC$.

From the right triangle AMS we can get:

$|MS|=\sqrt{5^2-3^2}=4$.

Now we use the right triangle AMD. (This triangle is right since D is the midpoint of AB - have a look at similar triangles ADD' and ACB, where D' is the point of AC such that DD' is perpendicular to AC. You should see that D'=M.)

This right triangle gives us:

$|AD|=\sqrt{3^2+9^2}=3\sqrt{10}$ and $|AB|=2|AD|=6\sqrt{10}$

What is correct English terminology for this: "D' is the point of AC such that DD' is perpendicular to AC"? If I used word by word translation from my language, it would be "D' is the foot of the perpendicular from the point D to the line AC".

If I try to follow your suggestions and compute the angles then I get:

$\sin\alpha=\frac35$ and $\cos\alpha=\frac45$ ($\alpha$ denotes the angle ASM)

Now I digress a little from your suggestion.

$\frac{|AB|}4=5\cos\frac\alpha2=5\sqrt{\frac{1+\cos\alpha}2}=3\sqrt{\frac52}$

$|AB|=4.3\sqrt{\frac52}=6\sqrt{10}$.

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yes absolutely right thanks very much – dato datuashvili Jul 23 '11 at 11:08
I'm no native speaker, but "perpendicular foot" seems to be used in English too. – t.b. Jul 23 '11 at 11:12

$$\left\vert AC\right\vert ^{2}+\left\vert BC\right\vert ^{2}=\left\vert AB\right\vert ^{2}=4\left\vert AD\right\vert ^{2}$$

$$6^{2}+\left\vert BC\right\vert ^{2}=4\left\vert AD\right\vert ^{2}$$

$$h^{2}=5^{2}-3^{2}=16,$$

where $h$ is the distance from the center to $AC$. Hence $h=4$ and $$|AD|^2=\left( 5+h\right) ^{2}+3^{2}=9^{2}+3^{2}=90.$$

Thus

$$\left\vert AD\right\vert =3\sqrt{10}.$$

And $|BC|$ is such that

$$6^{2}+\left\vert BC\right\vert ^{2}=4\cdot 90,$$

$$\left\vert BC\right\vert =18.$$

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thank you very much – dato datuashvili Jul 23 '11 at 11:32
@user3196: You are welcome! – Américo Tavares Jul 23 '11 at 11:39