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.
While it may not seem the most important, let's calculate the semiperimeter. We get 3+4+62=132
The reaon why we must find the semiperimeter is because we can use Heron's Formula.
It states that the area is √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=√132(132−3)(132−4)(132−6)
Area=√45516=√4554
This rounds to about 5.3327
Thanks! :)
While it may not seem the most important, let's calculate the semiperimeter. We get 3+4+62=132
The reaon why we must find the semiperimeter is because we can use Heron's Formula.
It states that the area is √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=√132(132−3)(132−4)(132−6)
Area=√45516=√4554
This rounds to about 5.3327
Thanks! :)