Area from Medians length

Area of a triangle = $\large \dfrac 4 3 *$ (area of triangle formed by the three medians.)
If u,v,w are the lengths of the medians, the area is also = $\large \dfrac{ 1}{ 3} *\sqrt{2(u^2v^2+v^2w^2+w^2u^2) -(u^4+v^4+w^4)}.$

#Geometry

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

http://jwilson.coe.uga.edu/emt725/Medians.Triangle/Area.Medians.Tri.html. That's for extra information, I found it wonderful

Sathvik Acharya - 4 years, 2 months ago

" $\color{#D61F06}{Forbidden\\ \\~~~\\ You~ don't~ have ~permission~ to ~access \\ /emt725/Medians.Triangle/ ~on~ this~ server. }$

I was not allowed ! Any other way to see what you suggest ? Thanks.

Niranjan Khanderia - 4 years, 2 months ago

sir maybe you could just go to the site of University of Georgia and then look for this page on area of triangle given median lengths or you could just Google the above topic and I believe there is another webpage on the same topic which is very informative and there are other forms of the formula as well. thank you

Sathvik Acharya - 4 years, 2 months ago

Thank you.

Niranjan Khanderia - 4 years, 2 months ago

Cannot understand what you nean by 4/3

Aman Real - 6 years, 1 month ago

One and one third.

Niranjan Khanderia - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$