find the area of the triangle with sides 5, 6 and root 13 units
 

Dear Student,

To find the area of a triangle when the three sides are given, we use the Heron's formula.

Let the triangle be ABC and a = 5, b=6 and c = $\sqrt{13}$

Now, Heron's formula is given as:

$$\begin{gathered} Area = \sqrt{S(S-a)(S-b)(S-c)} \\ where S= \frac{a+b+c}{2}, and a,b,c are the sides of the triangle. \end{gathered}$$
$$\begin{gathered} So, \\ S=\frac{5+6+\sqrt{13}}{2} \\ So, \\ Area = \sqrt{\frac{5+6+\sqrt{13}}{2}(\frac{5+6+\sqrt{13}}{2}-5)(\frac{5+6+\sqrt{13}}{2}-6)(\frac{5+6+\sqrt{13}}{2}-\sqrt{13})} \\ = \sqrt{\frac{11+\sqrt{13}}{2}\left(\frac{11+\sqrt{13}-10}{2}\right)\left(\frac{11+\sqrt{13}-12}{2}\right)\left(\frac{11+\sqrt{13}-2\sqrt{13}}{2}\right)} \\ =\sqrt{\frac{11+\sqrt{13}}{2}\left(\frac{1+\sqrt{13}}{2}\right)\left(\frac{-1+\sqrt{13}}{2}\right)\left(\frac{11-\sqrt{13}}{2}\right)} \\ =\sqrt{\frac{11+\sqrt{13}}{2}\left(\frac{1+\sqrt{13}}{2}\right)\left(\frac{-11+\sqrt{13}+11\sqrt{13}-13}{4}\right)} \\ =\sqrt{\frac{11+\sqrt{13}}{2}\left(\frac{1+\sqrt{13}}{2}\right)\left(\frac{-24+12\sqrt{13}}{4}\right)} \\ =\sqrt{\frac{11+\sqrt{13}}{2}\left(\frac{-24+12\sqrt{13}-24\sqrt{13}+156}{8}\right)}=\sqrt{\frac{11+\sqrt{13}}{2}\left(\frac{132-12\sqrt{13}}{8}\right)} \\ =\sqrt{\frac{1452-132\sqrt{13}+132\sqrt{13}-156}{16}}=\sqrt{\frac{1296}{16}}=\sqrt{81} \\ =9 \end{gathered}$$

Thus, the correct answer is 9 units.

Hope it helps.

Regards