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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 10, 2024

Best Answer 

 #1
avatar+1252 
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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! :)

 Jun 10, 2024
 #1
avatar+1252 
+1
Best Answer

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! :)

NotThatSmart Jun 10, 2024

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