Question

Circle O is inscribed in ABC. What is the perimeter of ABC

Answer 1

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 $P$ of triangle ABC will be twice the sum of the labeled edges:

$P=2(12\,\mathrm{cm}+16\,\mathrm{cm}+6\,\mathrm{cm})=72\,\mathrm{cm}$

Answer 2

Answer:

68 cm

Step-by-step explanation:

When you have a circle, and an external point $P$, the two segments $PM$ and $PN$ that are tangent to the circle are congruent always. (See Figure 1).

In this case, we can apply the same theorem three times in you figure! (See figure 2)

$BD = BC =16$

$AD = FA =12$

$CF = CE =6$

Then, we can find the size of the sides:

$AB = AD + BD = 12 + 16=28$

$BC = BE + EC = 16 + 6=22$

$CA = CF + FA = 6 + 12=18$

Finally, we can find the perimeter that is the sum of the three sides:

$Perimeter = AB + BC + CA = 28 + 22 + 18 = 68$