Why is is this answer incorrect? [Geometry]

[uranium_235]

Why is this answer incorrect? [Geometry]

---edited
I see where I went wrong, but with my current level of mathematical knowledge, I am at a dead end:


I do not know how do draw this question in ASCII, so I will describe it. There is an equilateral triangle with a square inscribed within it. The square has sides of 6 (cm), it asks how long each side of the triangle is. (NOTE: the top of the square fits perfectly to form another triangle, and the two sides of the square form 2 right triangles)

Now, in the top triangle, since it is as well equilateral, and the base (the top of the square) is 6, so the other sides must be 6.

-let y represent the hypotenuse of the two right triangles formed by the square and the larger triangle.
-let n represent the base of the two right triangles (the side other than that formed by the 6 cm side of the square)
-let x represent the side lengths of the larger equilateral triangle which the square is in.
(it would help to draw this while following along)

AXIOMS:
- $$x = 6 + y$$
- $$n = ( x - 6 ) / 2$$
therefore $$n = [ ( y + 6 ) - 6 ] / 2 = y / 2$$
Now, with this all settled...

$$y^2 = 6^2 + n^2$$

$$y^2 = 36 + (y/2)^2$$

$$y^2 = 36 + (y^2 /4)$$

$$y = \sqrt{36+(y^2/4)}$$

$$y = \sqrt{ \frac {36} {1} + \frac {y^2} {4} }$$

$$y = \sqrt{ \frac {144 + y^2} {4} }$$

$$y = \frac { \sqrt{144+y^2} } {2}$$

Is it possible to derive a value of $$y$$ from this?

 

[faust9]

Go back up to $$y^2=36+\frac{y^2}{4}$$ retry from here.

 

[Tide]

Yes, you can find y from that equation but I think a simpler way would be to recognize that if you take the two triangle are 30-60-right triangles (i.e. if you put them together then they form a equilateral triangle!). The altitude of the triangles is the same as the side of the square. Pythagoras will give you the side of the triangles in short order.

 

[uranium_235]

Go back up to $$y^2=36+\frac{y^2}{4}$$ retry from here.

Thanks; I always miss out on the simple details in math.