Question

The area (in sq units) of the triangle whose sides are $6$, $5$, $\sqrt{13}$, is

• A. $5\sqrt{2}$
• B. $9$
• C. $6\sqrt{2}$
• D. $11$

Answer 1

The correct option is B

$9$

Explanation for correct answer:

Determining the area of the triangle:

Given sides of a triangle are $6$, $5$, and .$\sqrt{13}$.

Let $a=6$, $b=5$ and $c=\sqrt{13}$

We know that,

$$\begin{gathered} \cos \left(C\right)=\frac{a^2+b^2-c^2}{2ab} \\ =\frac{6^2+5^2-{\left(\sqrt{13}\right)}^2}{2\times 5\times 6} \\ =\frac{36+25-13}{60} \\ =\frac{48}{60} \\ =\frac{4}{5} \end{gathered}$$

Now,

$$\begin{gathered} {\sin }^2\left(C\right)+{\cos }^2\left(C\right)=1 \\ \Rightarrow \sin \left(C\right)=\sqrt{1-{\cos }^2\left(C\right)} \\ =\sqrt{1-{\left(\frac{4}{5}\right)}^2} \\ =\sqrt{1-\left(\frac{16}{25}\right)} \\ =\sqrt{\frac{9}{25}} \\ =\frac{3}{5} \end{gathered}$$

We know the area of the triangle is given as;

$$\begin{gathered} \mathrm{Area}=\frac{1}{2}\times a\times b\times \sin \left(C\right) \\ =\frac{1}{2}\times 6\times 5\times \frac{3}{5} \\ =9 \end{gathered}$$

Therefore the area of the triangle is equal to $9\text{square units}$

Hence, the correct answer is option (B).