## anonymous 4 years ago 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

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1. anonymous

2. anonymous

ill try:)

3. anonymous

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

4. anonymous

ok so.. the formula of the area of an equilateral area is

5. anonymous
6. anonymous

but how do i get s? if i have to divide it? Im so confused.

7. anonymous

:/ tryna help but i really dont know:/ im soo sorry

8. myininaya

$A=9 \sqrt{3}=\frac{1}{2} b h$ 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|

9. myininaya

no look at one of those triangles inside the equilateral |dw:1337557653191:dw|

10. myininaya

Write h in terms of s

11. myininaya

Use the Pythagorean thm to do so

12. myininaya

now not no*

13. anonymous

so 18 square root of 3 using pythagorean thm we get the S?

14. anonymous

@seashell thank you anyways! :) <3

15. myininaya

I want you to write h in terms of s Use the Pythagorean thm

16. anonymous

lol:) how sweet!

17. myininaya

Look at the triangle I drew

18. myininaya

And use the Pythagorean thm to write h in terms of s

19. anonymous

a^2 + b^2 = C^2 so H will eaqul 54?

20. myininaya

|dw:1337558023888:dw| $(h)^2+(\frac{s}{2})^2=s^2$

21. myininaya

Solve for h

22. anonymous

is H equal 13.5

23. myininaya

Solve what I have up there for h please $h^2+\frac{s^2}{4}=s^2$ this right here can you do that?

24. anonymous

im really sorry Im trying to understand it, but i get confused. Im trying my best. Sorry :/

25. anonymous

@myininaya will H equal 4?

26. myininaya

$\text{ subtract } \frac{s^2}{4} \text{ on both sides of } h^2+\frac{s^2}{4}=s^2$

27. myininaya

$h^2=s^2-\frac{s^2}{4}$ 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 $h=\sqrt{\frac{3s^2}4}$ $h=\frac{\sqrt{3s^2}}{\sqrt{4}}$ $h=\frac{\sqrt{3} \sqrt{s^2}}{2}$ $h=\frac{\sqrt{3}s}{2}$

28. myininaya

I wrote h in terms of s

29. myininaya

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 ?

30. myininaya

$18 \sqrt{3} = s s \frac{\sqrt{3}}{2}$ $18 \sqrt{3} = \frac{\sqrt{3}}{2} s^2$ There that might look a bit better to you Try solving this for s