# perimeter very easy

• Feb 24th 2008, 02:24 AM
kri999
perimeter very easy
There's a square drawn here and it's diagonal is shown as being 6sqrt3 cm long
and i have to find the perimeter..how do i do this?
• Feb 24th 2008, 02:28 AM
Jhevon
Quote:

Originally Posted by kri999
There's a square drawn here and it's diagonal is shown as being 6sqrt3 cm long
and i have to find the perimeter..how do i do this?

let the side length be $x$.

then by Pythagoras' theorem, $x^2 + x^2 = (6 \sqrt{3})^2 = 108$

solve for $x$. then the perimeter is given by $4x$
• Feb 24th 2008, 02:35 AM
kri999
oh yes sorry i forgot to mention that i have to use the rule about 45 45 90 triangles with this problem
• Feb 24th 2008, 02:38 AM
Jhevon
Quote:

Originally Posted by kri999
oh yes sorry i forgot to mention that i have to use the rule about 45 45 90 triangles with this problem

why? that would only make your life harder. then you'd have to use sine or cosine or tangent or something

if you insist, you could do $\sin 45 = \frac x{6 \sqrt{3}}$ and solve for x from that
• Feb 24th 2008, 02:49 AM
kri999
Yeah sorry the whole chapter is on sines and cosines and special triangles, so they insist on making life harder. haha..they gaave the answer as 12 sqrt 6

That's all i don't understand
• Feb 24th 2008, 03:38 AM
mr fantastic
Quote:

Originally Posted by kri999
Yeah sorry the whole chapter is on sines and cosines and special triangles, so they insist on making life harder. haha..they gaave the answer as 12 sqrt 6

That's all i don't understand

Well $x = \frac{6 \sqrt{3}}{\sqrt{2}} = \frac{6 \sqrt{3} \sqrt{2}}{\sqrt{2} \sqrt{2}} = \frac{6 \sqrt{6}}{2} = 3 \sqrt{6}$.

And the perimeter is equal to 4x.

Therefore ......
• Feb 24th 2008, 11:50 AM
Jhevon
Quote:

Originally Posted by kri999
Yeah sorry the whole chapter is on sines and cosines and special triangles, so they insist on making life harder. haha..they gaave the answer as 12 sqrt 6

That's all i don't understand

it would be good if you told us your difficulty. if you have trouble setting up the sine, cosine and tangent trig ratios, say so, and we can address it. if your problem was the algebra, then Mr. Fantastic already addressed that. do you understand the problem now? or is there something else?