Three circles with centers A, B and C have respective

Question

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

Answer 1

Step 1

P R^{2} = 70^{2}

R Q^{2} = 50^{2}

P Q^{2} = 60^{2}

By cosine law

\cos R = \frac{P R^{2} + R Q^{2} - P Q^{2}}{2 (P R) (R Q)}

\cos R = \frac{{(70)}^{2} + {(50)}^{2} - {(60)}^{2}}{2 (70) (50)}

\cos R = \frac{3800}{7000}

R = \cos^{- 1} (\frac{38}{70}) \approx {57.12}^{\circ}

Step 2
By sinc low

\frac{\sin R}{P Q} = \frac{\sin Q}{P R}

\frac{\sin 57 \times 12^{\circ}}{60} = \frac{\sin Q}{70}

\frac{70}{60} {\sin 57.12}^{\circ} = \sin Q

Q = \sin^{- 1} (\frac{7}{6} \sin ({57.12}^{\circ})) \approx {78.46}^{\circ}

By angle Sum property

m ∠ P + m l Q + m l R = 180^{\circ}

m ∠ P - 180^{\circ} - {78.46}^{\circ} - {57.12}^{\circ}

m ∠ P = {44.42}^{\circ}

Step 3
Length of

a r c = \frac{Q}{360} \times 2 π r

For arc DF radius

= 20

and

m ∠ Q = {78.46}^{\circ}

Length of arc DF

= \frac{{78.46}^{\circ}}{360} \times 2 π (20)

= 27.39 \in

For arc DE

\Rightarrow

radius

= 40

m ∠ P = {44.42}^{\circ}

Length of arc DE

= \frac{{44.42}^{\circ}}{360} \times 2 π (40)

= 31.01 \in

Step 4
For arc EF
radius

(r) = 30

m ∠ R = {57.12}^{\circ}

Length of arc EF

= \frac{57.12}{360} \times 2 π (30)

= 29.91 \in

Total perimeter of curvi linear triangle

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