# Finding Area of a Triangle without Trignometric ratios.

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/

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can you use known triangle side lengths ratios? like a 30-60-90 triangle? –  Joseph Skelton Jun 17 '12 at 22:21
The area of a triangle is defined as $A=\frac{1}{2}ab\sin(C)$, where $a, b$ are sides and $C$ is the angle between them. –  Shaktal Jun 17 '12 at 22:21
Thats what is in the figure. You cant disect it from the middle because its not equilateral and you cant apply phythagoras here cause u dont know if its a right angle triangle.. –  MistyD Jun 17 '12 at 22:22
@Shaktal Thats interesting.. Any link on where i could read more about that –  MistyD Jun 17 '12 at 22:22
@Каднон Мстакуи You have recently made plenty of suggested edits‌​, mostly to old posts. It is better to avoid bumping too many old posts in short time, see meta: When a low-rep user suggests many edits to old posts?, Editing Binge Etiquette, How much bumping is too much?. (cont...) –  Martin Sleziak Dec 7 '12 at 6:04

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

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I agree but then again this method is a bit risky to use since it requires us to assume that the other angles are 60 and 90. So ill always be doubtful of the answer –  MistyD Jun 17 '12 at 22:30
i'll upload an image of what i'm talking about, as soon as i figure out how to do that... –  Joseph Skelton Jun 17 '12 at 22:32
there you go, i put in a link, hope that clears up what i'm talking about. –  Joseph Skelton Jun 17 '12 at 22:41
I did a little calculation and i dont think that the triangle could qualify for a 30-60-90 triangle. Since a 30-60-90 triangles has sides of x,x.root3 and 2x respectively. Pleases correct me if i am wrong –  MistyD Jun 17 '12 at 22:55
because i'm silly and mislabel things. The important side is correct though. The height is 3. the other side should be $3\sqrt{3}$ Should be fixed now. –  Joseph Skelton Jun 17 '12 at 23:00

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.

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How did you get HA=3 from my understanding you split the triangle into two right angle triangle and applied phythagoras ? –  MistyD Jun 17 '12 at 22:41
@MistyD No. Let $G$ be a point on the line passing through $A$ and $H$ such that $|GH| = |AH|$ and $|AG| = 2|GH|$. Then $\triangle AGC$ is equilateral, so $|AG| = |AC| = 6$ and $|AH| = 3$. This the same solution as Joseph's in different wording, he was just faster ;-) –  dtldarek Jun 17 '12 at 22:44

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

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If I can read correctly, the OP asked for an answer without the trigonometric ratios... –  dtldarek Jun 17 '12 at 22:37
@dtldarek I interpreted that as him wishing to stay clear of right-triangle specific trigonometric ratios: $\sin(\theta)=\frac{o}{h}$ for instance, rather than trigonometric functions in general. –  Shaktal Jun 17 '12 at 22:38