+207 Let ABC be a triangle, and let its angle bisectors be AD, BE, and CF which intersect at I. If DI=3, BD=4 and BI=6 then compute the area of triangle BID.
+1908 We can use a really handy formula to help solve this question.
First, we have to calculate the semiperimeter of the triangle, which is half the sum of the three side.
We get \(\frac{3+4+6}{2}=\frac{13}{2}\)
That may seem useless, but it will come in handy very soon.
Now, we apply Heron's Formula, which states that the area of a triangle is equal to \(\sqrt{s(s-a)(s-b)(s-c)}\). Subbing everything in, We get
\(Area = \sqrt{\frac{13}{2}(\frac{13}{2}-3)(\frac{13}{2}-4)(\frac{13}{2}-6)}\)
\(Area = \sqrt{\frac{455}{16}}=\frac{\sqrt{455}}{4}\)
So our final answer is \(\frac{\sqrt{455}}{4}\)
Thanks! :)
