Area of Inscribed Circle
Date: 12/01/98 at 13:24:52 From: Anna Subject: Trigonometry The question is: Find the area of the circle inscribed in a triangle a, b, c. The hint is: Draw lines to the center of the circle and use Heron's law. I would appreciate it if you could help me! Anna
Date: 12/01/98 at 15:41:37 From: Doctor Floor Subject: Re: Trigonometry Hi Anna, Thanks for your question! Let us consider triangle ABC with sidelengths BC = a, AC = b, AB = c and center of the inscribed circle (incenter) I. Drop perpendicular altitudes from I to Ia, Ib and Ic on the sides of ABC. The lengths of these altitudes are equal to the radius r of the incircle. This radius we will have to calculate in order to find the area of the inscribed circle (= incircle). C / \ / \ / \ Ib Ia / \ / I \ / \ A-------Ic------B First, we know that area(ABC) = area(AIB) + area(BIC) + area(CIA). We can calculate area(AIB) = 0.5 * AB * IIc = 0.5cr. And in the same way we find area(BIC) = 0.5ar and area(CIA) = 0.5br. The conclusion is that area(ABC) = 0.5(a+b+c)r = sr, where s is the semiperimeter of ABC. Then, from Heron's formula, we also know that: area(ABC) = sqrt(s(s-a)(s-b)(s-c)) Combining the two formulas for the area of ABC we can derive: sr = sqrt(s(s-a)(s-b)(s-c)) r = sqrt(s(s-a)(s-b)(s-c)) / s r = sqrt((s-a)(s-b)(s-c) / s ) And now we can finish off the area of the incircle: area(incircle) = pi * r^2 = pi * sqrt((s-a)(s-b)(s-c) / s )^2 pi(s-a)(s-b)(s-c) = ----------------- s Quite nice! Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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