Question

[Precalculus] Find the area of this triangle:

Answer 1

check the picture below.

notice, by drawing the altitude from C, we end up with a 30-60-90 and a 45-45-90 triangle, thus we use the 30-60-90 and 45-45-90 rules as you see there.

$\bf \textit{area of a triangle}\\\\ A=\cfrac{1}{2}bh~~ \begin{cases} b=\stackrel{w}{6\sqrt{3}}+\stackrel{y}{6}\\ h=6 \end{cases}\implies A=\cfrac{1}{2}(6\sqrt{3}+6)(6) \\\\\\ A=18\sqrt{3}+18\implies A=18(1+\sqrt{3})$

Answer 2

Look at the picture.

In a 30-60-90 (45-45-90) triangle, you can find the measure of any of the three sides (picture 2).

$CD=\dfrac{1}{2}AC\to CD=\dfrac{1}{2}\cdot12=6$

$AD=CD\sqrt3\to AD=6\sqrt3$

$DB=CD\to DB=6$

$AB=AD+DB\to AB=6\sqrt3+6$

The formula of the area of a triangle ABC:

$A_\triangle=\dfrac{1}{2}(AB)(CD)$

substitute:

$A_\triangle=\dfrac{1}{2}\cdot(6\sqrt3+6)(6)=3(6\sqrt3+6)=18+18\sqrt3=18(1+\sqrt3)$

Answer: $\boxed{C.\ 18(1+\sqrt3)}$