3 circles of radius 3 cm, 4cm, 5 cm touch each other externally at $A$, $B$, $C$. Tangents drawn at $A$, $B$, $C$ intersect at $P$.
Find $ PA + PB + PC$ .
Thanks.
My thoughts and approach:
Well I made the figure but still could not proceed ahead.
3 circles of radius 3 cm, 4cm, 5 cm touch each other externally at $A$, $B$, $C$. Tangents drawn at $A$, $B$, $C$ intersect at $P$.
Find $ PA + PB + PC$ .
Thanks.
My thoughts and approach:
Well I made the figure but still could not proceed ahead.
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$PA+PB+PC$ is just three times the inradius of a triangle having side lengths $(a,b,c)=(7,8,9)$. Since $$ r(a+b+c) = 2\Delta,$$ we only need to find the area of the triangle thorugh Heron's formula. We have: $$\Delta^2 = p(p-a)(p-b)(p-c) = 12\cdot 3\cdot 4\cdot 5,$$ hence $\Delta=12\sqrt{5}$ and: $$PA+PB+PC = 3r=\frac{6\Delta}{a+b+c} = 3\sqrt{5}.$$