Finding Area of a Triangle without Trignometric ratios.

Question

Hi I need to figure out the area of the following triangle, without using Trigonometric ratios. Any suggestions on the best approach.

The answer is 12 square units

Edit: I also think that the above triangle can't qualify for a $30-60-90$ triangle since it fails the $x,x.\sqrt3,2x$ rule/

Answer 1

If you use a 30-60-90 triangle with hypotenuse 6, then the height is 3. So the height of this triangle is 3. Thus $\frac{b\cdot h}{2}=\frac{8\cdot3}{2}=12$

triangle

Answer 2

You can find the area of any triangle by applying the following area formula:

$$A=\frac{1}{2}ab\sin(C),$$

Where $a$ and $b$ are sides of the triangle and $C$ is the angle between them.

In this case, you can do the following:

$$\frac{6\times8}{2}\sin(30^{\circ})=12$$

Which is the answer you want.

EDIT: In response to your comment, you can find more about this formula and it's derivation, here

Answer 3

This can be done without using any trigonometric functions. Let $|AC| = 6$, $|BC| = 8$ and $|\angle ACB| = 30^\circ$. Let $H \in BC$ be a point such that $HA$ is the height of the triangle starting at $A$. Then, the $\triangle CAH$ is a half of equilateral triangle and therefore $|HA| = 3$. Using the basic formula for the triangle's area we get $\frac{|CB|\cdot|HA|}{2} = \frac{8\cdot 3}{2} = 12$.

Edit: Considering your last modifications to the question, please take look at the picture below. Let $G$ be a point on the line passing through $A$ and $H$ such that $|GH|=|AH|$ and $|AG|=2|GH|$. Please note that $\triangle AGC$ is now equilateral and thus $\triangle AHC$ is the 30-60-90 triangle.