## DLS 2 years ago An interesting question! :D ABC is an equilateral triangle such that the vertices B and C lie on two parallel line at a distance of 6.If A lies between the parallel lines at a distance 4 from one of them then the length of a side of the equilateral triangle is=?

1. hartnn

if B and C lie on parallel lines, then BC = side = 6 ...?

2. DLS

units

3. DLS

no unit mentioned if u are asking for one

4. hartnn

i was not asking, i was telling side =6

5. DLS

oh :P

6. DLS

|dw:1356782149154:dw| my diagram

7. hartnn

if ABC is equilateral AB=BC=AC=6

8. DLS

|dw:1356782275365:dw| so cant we do like this.. sin60=h/4

9. hartnn

u wanted to know length of side ... = BC = 6 thats it.

10. DLS

11. DLS

we have to find length

12. hartnn

of ?

13. DLS

side of the eq triangle :/

14. hartnn

maybe the side BC is tilted , then

15. DLS

yes..thats what im saying

16. hartnn

17. DLS

:/

18. hartnn

where u have shown BC =6

19. experimentX

|dw:1356782465079:dw|

20. DLS

why wont that be 60 :o

21. hartnn

|dw:1356782731705:dw|

22. DLS

thats nt touching?

23. hartnn

ofcourse, it is...

24. DLS

then w|dw:1356783287510:dw|

25. hartnn

i have a long way.

26. hartnn

|dw:1356783439425:dw|

27. hartnn

$$a=\sqrt{x^2-4^2} \\ b=\sqrt{x^2-6^2} \\ a+b=\sqrt{x^2-2^2}$$ you can find 'x' from here.

28. hartnn

i hope ther's a shorter way...

29. DLS

x=root 52 or smt?

30. DLS

solution and answer given in my book is DANGEROUS :/

31. hartnn
32. DLS

I have.. $\LARGE 2 \frac{\sqrt{28}}{\sqrt{3}}$

33. hartnn

which is exactly what i got.

34. hartnn

$$\LARGE 2 \frac{\sqrt{28}}{\sqrt{3}}=\LARGE 4\frac{\sqrt{7}}{\sqrt{3}}$$

35. DLS

okay yes

36. hartnn

using $$a=\sqrt{x^2-4^2} \\ b=\sqrt{x^2-6^2} \\ a+b=\sqrt{x^2-2^2}$$

37. DLS

hold on one sec..let me go through it once again

38. DLS

lol that was a nice method :|

39. hartnn

just pythagoras, thrice...

40. DLS

yeah..!