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16 February, 22:45

Three circles with centers A, B and C have respective radii 50, 30 and 20 inches and are tangent to each other externally. Find the area in (in^2) of the curvilinear triangle formed by the three circles

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  1. 16 February, 23:07
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    Refer to the figure below which illustrates the given problem. The yellow

    shaded area is the curvilinear triangle.

    First, we should find angles A, B, and C of triangle ABC.

    a = 50 in, b = 70 in, c = 80 in.

    From the Law of Cosines: c² = a² + b² - 2ab cosC

    cos C = (a² + b² - c²) / (2ab) = (50²+70²-80²) / (2*50*70) = 0.1429

    C = arcos (0.1429) = 81.79°.

    From the Law of Sines,

    sin (B) / b = sin (C) / c

    sin (B) = (b/c) sin (C) = (70/80) * sin (81.79°) = 0.866

    B = arcsin (0.866) = 60°

    Because the sum of angles in a triangle is 180°, therefore

    A = 180 - (60 + 81.79) = 38.21°.

    Calculate the area of ΔABC from the Law of Sines.

    Area = (1/2) ab sinC = (1/2) * 50*70*sin (81.79°) = 1732.065 in²

    Calculate the areas of arc segments created by circle centers A, B, and C.

    Aa = (38.2/360) * π (50²) = 833.39 in²

    Ab = (60/360) * π (30²) = 471.2 in²

    Ac = (81.79/360) * π (20²) = 285.5 in²

    The area of the curvilinear triangle is

    1732.065 - (833.39 + 471.2 + 285.5) = 141.975 in²

    Answer: 142.0 in² (nearest tenth)
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