help with geometry

Let's use Heron's Formula to solve this problem. 

First, let's note what Heron's Formula is. The formula states that $\text{Area of Triangle} = \sqrt{s(s-a)(s-b)(s-c)}$ for a triangle with sides $a,b,c$ and semiperimeter of $\frac{a+b+c}{2}$. We can use this formula to find the area of BID. 

First, let's find the semiperimeter. Plugging in values, we get $\frac{3+4+6}{2}=\frac{13}{2}$ as semiperimeter. 

Now, we can go forward and find the area. 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 approximately 5.3327. 

Thanks! :)