Circle o is inscribed in abc. what is the perimeter of abc

Since O is inscribed in triangle ABC, you know that AB, AC, and BC are all tangent to the circle. The tangent segments theorem asserts that AD is congruent to A F; BD is congruent to BE; and CF is congruent to CE. (NOTE: E is the point where the circle touches BC - I've taken the liberty of labeling it as such.)

Then the perimeter of triangle ABC will be twice the sum of the labeled edges:

answer: dead jah

step-by-step explanation:

bullet + jah= dead jah

geometry: triangles

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call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

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