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A Circle is inscribed in a triangle $ABC$, where $AB=10 cm$, $BC=9 cm$ and $AC=7 cm$ . $X$, $Y$, $Z$ are points of contact of the sides $AC$, $BC$ and $AB$ with the circle respectively. $BZ=?$

In the first look, question looked simple to me. But I am not able to get to the answer. I found the In-radius using $rs = \sqrt{s(s-a)(s-b)(s-c)}$. But that is not helping me to find the required length.

In the above formula,
    r= indradius
    s=(a+b+c)/2
    a,b,c -> Lengths of the sides of triangle

I would be glad to see the answer of this question. Any Hints or answers are welcome. Thanks.

Sandy

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    $\begingroup$ The equations are easy to solve, but let's make it nicer. Let $BZ=p$, $CX=q$, $AZ=r$. Then $p+q+r=(10+9+7)/2=s$ (half-perimeter). So $p=s-b=13-7$. Note the appearance of the term $s-b$, which will remind you of Heron's Formula. $\endgroup$ – André Nicolas Jul 21 '12 at 17:24
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It's much easier than you're making it. XC = YC, XA = ZA and BZ = BY, because in each case, you've got two tangents up to the same point.

That gives you enough to solve for all of the lengths. In particular (with all lengths in centimetres) BZ = 10 - AZ = 10 - AX = 3 + XC. But because BY + YC = 9, this gives BZ = 6.

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  • $\begingroup$ Hey thanks. Dint look at the problem that way. $\endgroup$ – mk.. Jul 21 '12 at 11:53

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