Question 57348: There is a circle inside a triangle the bottom of the triangle is 240 kilometers, one side is 330 kilometers, and the other side is 270 kilometers. The circle is the largest possible circle that can fit inside the triangle. What is the area in square kilometers, rounding to the nearest hundred.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Inscribed Circle
From this we see that the intersection of any two angle bisectors is the center if the inscribed circle. It follows that all three internal angle bisectors intersect at one point, which is the center of the inscribed circle, or "incircle". The perpendicular distance from point I to any of the sides is the radius, r, of the incircle. The area, K, of triangle ABC is the sum of the areas of AIB, BIC, and CIA. If we label the lengths of the sides of triangle ABC in traditional form, with side a opposite vertex A, b opposite B, and c opposite C, then the areas of these three small triangles are cr/2, ar/2, and br/2 because the height of each small triangle is the radius of the incircle. The sum of these areas is
K = (a+b+c)(r)/2, or
K = sr. Also,
r = K/s
where s = (a+b+c)/2 is the semiperimeter of triangle ABC.
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I don't know what methods you have for solving your problem.
I found the above on-line at MathWorld Inscribed Circle.
You can find the area of your triangle (K) = (1/2)base*height
height=sqrt(330^2-120^2)
base = 240
So, K= 0.5*240*307.4=47187.21
s=(240+330+270)/2=420
Then r=K/s=112.35
Area of the circle =(pi)r^2
=(pi)(112.35)^2=39655.18 sq. km
Cheers,
Stan H.
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