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

enter image description here

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
up vote 4 down vote accepted

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 $\frac{b\cdot h}{2}=\frac{8\cdot3}{2}=12$


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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.

enter image description here

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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:


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

In this case, you can do the following:


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

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