How do you find the area of a triangle with side lengths of 8, 10, and 12? Reply within 24 hours after midnight tomorrow, I will go to bed so don't bother.

Anyways, help please!

Printable View

- Jan 25th 2006, 08:52 PMextreme_piArea of triangle with side lengths 8, 10, 12
How do you find the area of a triangle with side lengths of 8, 10, and 12? Reply within 24 hours after midnight tomorrow, I will go to bed so don't bother.

Anyways, help please! - Jan 25th 2006, 09:16 PMCaptainBlackQuote:

Originally Posted by**extreme_pi**

Let $\displaystyle s$ be the semi-perimeter of a triangle with sides $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$, that is:

$\displaystyle s=\frac{a+b+c}{2}$.

Then area of the triangle is:

$\displaystyle \mbox{Area}=\sqrt{s(s-a)(s-b)(s-c)}$.

So in your case $\displaystyle s=\frac{8+10+12}{2}=15$ and so:

$\displaystyle \mbox{Area}=\sqrt{15(15-8)(15-10)(15-12)}\approx140.7$.

RonL

PS

You won't get banned for posting the same**question**multiple times, but the

duplicates will be deleted. This is so that people do not waste their time

answering questions already answered elsewhere.