+1887 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.
+1804 Let's use Heron's Formula to solve this problem.
First, let's note what Heron's Formula is. The formula states that \(\text{Area of Triangle} = \sqrt{s(s-a)(s-b)(s-c)}\) for a triangle with sides \(a,b,c\) and semiperimeter of \(\frac{a+b+c}{2}\). We can use this formula to find the area of BID.
First, let's find the semiperimeter. Plugging in values, we get \(\frac{3+4+6}{2}=\frac{13}{2}\) as semiperimeter.
Now, we can go forward and find the area. We get
\(Area = \sqrt{\frac{13}{2}(\frac{13}{2}-3)(\frac{13}{2}-4)(\frac{13}{2}-6)}\)
\(Area = \sqrt{\frac{455}{16}}=\frac{\sqrt{455}}{4}\)
This rounds to approximately 5.3327.
Thanks! :)
