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29 December, 03:26

Let $AB = 6$, $BC = 8$, and $AC = 10$. What is the area of the circumcircle of $/triangle ABC$ minus the area of the incircle of $/triangle ABC$?

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  1. 29 December, 03:33
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    The problem is illustrated by the figure shown below.

    It is clear that ΔABC is a right triangle because

    (a) it is inscribed in a circle,

    (b) Its sides satisfy the Pythagorean theorem because

    6² + 8² = 36 + 64 = 100, and 10² = 100.

    Therefore, AC is the diameter of the circumscribing circle, and the radius is

    r = 5.

    The area of the circumscribing circle is

    A₁ = π (5²) = 25π

    The area of the triangle is

    A₂ = (1/2) 8*6 = 24

    The required area, of the circumcircle minus that of the triangle (shaded area) is

    A₁ - A₂ = 25π - 24 = 54.54 square units.

    Answer:

    54.54 square units.
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