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Think of an equilateral triangle in which the length of a side is s, the perimeter (distance around the triangle) is p, and the area is A. Find the missing parts. s=? p=? area= 9 sqaureroot of 3

Mathematics
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ill try:)
thank you soo much!! i really need someones help! :)

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Other answers:

ok so.. the formula of the area of an equilateral area is
http://www.mathwords.com/a/a_assets/area%20equilateral%20triangle%20formula.gif
but how do i get s? if i have to divide it? Im so confused.
:/ tryna help but i really dont know:/ im soo sorry
\[A=9 \sqrt{3}=\frac{1}{2} b h\] So we have that the base, b, of the triangle is s b=s So we have \[9 \sqrt{3}=\frac{1}{2}s h\] So this means we have \[(2) 9 \sqrt{3} =\frac{1}{2}sh (2)\] \[18 \sqrt{3} =sh\] So is there a way to write h in terms of s Well remember we have an equilateral triangle with each side s and h, the height of the equilateral can be found in terms of s by using the Pythagorean Theorem |dw:1337557593984:dw|
no look at one of those triangles inside the equilateral |dw:1337557653191:dw|
Write h in terms of s
Use the Pythagorean thm to do so
now not no*
so 18 square root of 3 using pythagorean thm we get the S?
@seashell thank you anyways! :) <3
I want you to write h in terms of s Use the Pythagorean thm
lol:) how sweet!
Look at the triangle I drew
And use the Pythagorean thm to write h in terms of s
a^2 + b^2 = C^2 so H will eaqul 54?
|dw:1337558023888:dw| \[(h)^2+(\frac{s}{2})^2=s^2\]
Solve for h
is H equal 13.5
Solve what I have up there for h please \[h^2+\frac{s^2}{4}=s^2\] this right here can you do that?
im really sorry Im trying to understand it, but i get confused. Im trying my best. Sorry :/
@myininaya will H equal 4?
\[\text{ subtract } \frac{s^2}{4} \text{ on both sides of } h^2+\frac{s^2}{4}=s^2\]
\[h^2=s^2-\frac{s^2}{4}\] Combine the fractions on the right hand side of the equation \[h^2=\frac{4s^2}{4}-\frac{s^2}{4} \] \[h^2=\frac{3s^2}{4}\] Take square root of both sides So we have \[h=\sqrt{\frac{3s^2}4}\] \[h=\frac{\sqrt{3s^2}}{\sqrt{4}}\] \[h=\frac{\sqrt{3} \sqrt{s^2}}{2}\] \[h=\frac{\sqrt{3}s}{2}\]
I wrote h in terms of s
Now we can go back to solving for s Recall we had the equation \[18 \sqrt{3}=sh\] Now we just wrote h in terms of s So we have \[18 \sqrt{3}=s \frac{\sqrt{3} s}{2}\] Can you solve this for s ?
\[18 \sqrt{3} = s s \frac{\sqrt{3}}{2}\] \[18 \sqrt{3} = \frac{\sqrt{3}}{2} s^2\] There that might look a bit better to you Try solving this for s

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