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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.

 Jun 12, 2024
 #1
avatar+1334 
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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! :)

 Jun 12, 2024

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