+1839 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.
+1926 While it may not seem the most important, let's calculate the semiperimeter. We get \(\frac{3+4+6}{2}=\frac{13}{2}\)
The reaon why we must find the semiperimeter is because we can use Heron's Formula.
It states that the area is \(\sqrt{s(s-a)(s-b)(s-c)}\) where s is the semiperimeter and a,b, and c are the sidelengths.
Applying this, 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 about \(5.3327\)
Thanks! :)
+1926 While it may not seem the most important, let's calculate the semiperimeter. We get \(\frac{3+4+6}{2}=\frac{13}{2}\)
The reaon why we must find the semiperimeter is because we can use Heron's Formula.
It states that the area is \(\sqrt{s(s-a)(s-b)(s-c)}\) where s is the semiperimeter and a,b, and c are the sidelengths.
Applying this, 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 about \(5.3327\)
Thanks! :)
