# Thread: Is this the correct area of this triangle?

1. ## Is this the correct area of this triangle?

TRIANGLE: http://img172.imageshack.us/img172/1344/trig.jpg

I worked out that it would be 127 cm2

2. Originally Posted by olivia59
TRIANGLE: http://img172.imageshack.us/img172/1344/trig.jpg

I worked out that it would be 138 cm2
No.

For any triangle, it's area can be found using the formula

$A = \frac{1}{2}\,ab\,\sin{C}$, where $C$ is the angle in between sides $a$ and $b$.

So $A = \frac{1}{2}\cdot 12\,\textrm{m}\cdot 30\,\textrm{m} \cdot \sin{45^{\circ}}$

$= 180\,\textrm{m}^2\cdot \frac{\sqrt{2}}{2}$

$= 90\sqrt{2}\,\textrm{m}^2$.

3. But i thought the rule was 1/2 * bc sin A

4. Originally Posted by olivia59
But i thought the rule was 1/2 * bc sin A
That's exactly the same rule. I've just chosen two different sides of the triangle.

Point is, you need two sides of the triangle and the angle between them.

BTW I always like to keep the solutions exact where possible, hence the square root value rather than the decimal.

5. ok, but when i enter 1/2 * 12 * 30 * sin 45 on the calculator it gives me 127....is that not right? i dont understand where you get the 90 from?

6. I'm not sure why you reposted this question when I already answered it in your other thread: http://www.mathhelpforum.com/math-he...-triangle.html. Did you not see it?

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7. Originally Posted by olivia59
ok, but when i enter 1/2 * 12 * 30 * sin 45 on the calculator it gives me 127....is that not right? i dont understand where you get the 90 from?
Like I said, I've kept the answer exact.

The answer is not $90$.

The answer is $90 \times \sqrt{2}$.

Notice that when you put it into the calculator you get a decimal answer. This is NOT exact, it is a decimal approximation to $90\times\sqrt{2}$.

You have done the right thing but have gotten the decimal answer rather than the exact answer.

8. Originally Posted by olivia59
ok, but when i enter 1/2 * 12 * 30 * sin 45 on the calculator it gives me 127....is that not right? i dont understand where you get the 90 from?
$\frac{1}{2} \cdot 12 \cdot 30 \cdot \sin 45^{\circ}$
\begin{aligned}
&= \frac{1}{2} \cdot 12 \cdot 30 \cdot \frac{\sqrt 2}{2} \\
&= 90\sqrt {2}
\end{aligned}

Do you see where the 90 comes from now?

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EDIT: beaten to it by Prove It...

9. oh i see now, thanks for your help.

(not sure how to reply on forums, havent really used them before, dont know whether im sposed to click post reply or quote)

10. Originally Posted by olivia59
oh i see now, thanks for your help.

(not sure how to reply on forums, havent really used them before, dont know whether im sposed to click post reply or quote)
Either way is fine.

On this forum though, it is proper etiquette to thank the people who have helped you by clicking on the "Thanks" link in their posts.