# Three circles with centers A, B and C have respective

Three circles with centers A, B and C have respective radii 40, 30 and 20 in. and are tangent to each other externally. Find the perimeter of the curvilinear triangle formed by the three circles.
Select one:
a) 94 in
b) 88 in
c) 127 in
d) 134 in
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Marley Dupont
Step 1
$P{R}^{2}={70}^{2}$
$R{Q}^{2}={50}^{2}$
$P{Q}^{2}={60}^{2}$
By cosine law
$\mathrm{cos}R=\frac{P{R}^{2}+R{Q}^{2}-P{Q}^{2}}{2\left(PR\right)\left(RQ\right)}$
$\mathrm{cos}R=\frac{{\left(70\right)}^{2}+{\left(50\right)}^{2}-{\left(60\right)}^{2}}{2\left(70\right)\left(50\right)}$
$\mathrm{cos}R=\frac{3800}{7000}$
$R={\mathrm{cos}}^{-1}\left(\frac{38}{70}\right)\approx {57.12}^{\circ }$
Step 2
By sinc low
$\frac{\mathrm{sin}R}{PQ}=\frac{\mathrm{sin}Q}{PR}$
$\frac{\mathrm{sin}57×{12}^{\circ }}{60}=\frac{\mathrm{sin}Q}{70}$
$\frac{70}{60}{\mathrm{sin}57.12}^{\circ }=\mathrm{sin}Q$
$Q={\mathrm{sin}}^{-1}\left(\frac{7}{6}\mathrm{sin}\left({57.12}^{\circ }\right)\right)\approx {78.46}^{\circ }$
By angle Sum property
$m\mathrm{\angle }P+mlQ+mlR={180}^{\circ }$
$m\mathrm{\angle }P-{180}^{\circ }-{78.46}^{\circ }-{57.12}^{\circ }$
$m\mathrm{\angle }P={44.42}^{\circ }$
Step 3
Length of $arc=\frac{Q}{360}×2\pi r$
For arc DF radius $=20$
and $m\mathrm{\angle }Q={78.46}^{\circ }$
Length of arc DF $=\frac{{78.46}^{\circ }}{360}×2\pi \left(20\right)$
$=27.39\in$
For arc DE $⇒$ radius $=40$
$m\mathrm{\angle }P={44.42}^{\circ }$
Length of arc DE $=\frac{{44.42}^{\circ }}{360}×2\pi \left(40\right)$
$=31.01\in$
Step 4
For arc EF
radius $\left(r\right)=30$
$m\mathrm{\angle }R={57.12}^{\circ }$
Length of arc EF $=\frac{57.12}{360}×2\pi \left(30\right)$
$=29.91\in$
Total perimeter of curvi linear triangle