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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. 

 Feb 7, 2024
 #1
avatar+394 
+1

Here is a solution that doesn't require any complex geometry. We see that we know all 3 side lengths of triangle BID. We know a formula that can tell us the area directly knowing 3 sides of the triangle, Heron's formula.

First we calculte the semi-perimeter \(\frac{3+4+6}{2}=\frac{13}{2}\)

Then we apply Heron's formula. \(Area = \sqrt{\frac{13}{2}(\frac{13}{2}-3)(\frac{13}{2}-4)(\frac{13}{2}-6)}\). Simplifying, we get \(Area = \sqrt{\frac{455}{16}}=\frac{\sqrt{455}}{4}\) 

So, the Area is sqrt(455)/4 which is around 5.3327.

 Feb 9, 2024
 #2
avatar+1624 
+2

Nice! Heron's Formula is a very important formula: The area of a triangle with side lengths \(a\), \(b\), and \(c\), with semi-perimeter = \(a + b + c\over{2}\) is:

\(\sqrt{s(s-a)(s-b)(s-c)}\)

This formula will come in handy, especially if you are doing competition math. wink

proyaop  Feb 9, 2024

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