If ABC is an equilateral triangle, and BC=4*3^(1/2), what is the appro

Since \(ABC\) is an equilateral triangle, we know the following angles are 60° each.
Also, let's let n = the length of each side of the square



Since BWX is also an equilateral triangle, we know that all 3 sides have length n:



Since \(BC=4\sqrt{3}\), and since \(BX = n\), we know that side \(XC=4\sqrt{3}-n\)



At this point, we can see that triangle XYC is a special 30-60-90 right triangle.

When we compare ∆XYC with the base 30-60-90 triangle, we can compare corresponding sides to create the following equation: (4√3 - n)/2 = n/√3
Cross multiply to get: (√3)(4√3 - n)= (2)(n)
Simplify to get: 12 - (√3)n = 2n
Add (√3)n to both sides to get: 12 = 2n + (√3)n
Factor right side to get: 12 = n(2 + √3)
Divide both sides by (2 + √3) to get: n = 12/(2 + √3)

PRO TIP #1: By test day, all students should have the following approximations memorized:
√2 ≈ 1.4
√3 ≈ 1.7
√5 ≈ 2.2
So, 12/(2 + √3) ≈ 12/(2 + 1.7) ≈ 12/3.7

PRO TIP #2: We need not calculate the actual value of 12/3.7
Instead, notice that 12/3 = 4 and 12/4 = 3
Since 3.7 is BETWEEN 3 and 4, we now that 12/3.7 must be between 3 and 4
In other words, 12/3.7 = 3.something.

Answer: C

Cheers,
Brent

Height of Equilateral Triangle is √3/2 * (length of the side); the height of Triangle ABC = 6.
Let, the side of the square WXYZ be "n", then the side of the triangle BWX will also be "n".
Height of Triangle BWX is √3/2 * (n).
Height of ABC = Height of BWX + Height of Square
6=√3/2 * (n)+n
or, n= 12/(√3+2)
or, n = 12/3.7≈3.2

Answer: C

Hi GMATPrepNow,
Can we not assume x to be the midpoint of side BC?

Good idea, but x is not the midpoint of side BC.
Now that we know n ≈ 3.2, we can test that theory by plugging that n-value into our diagram.
When we do so, we see that BX ≠ XC

Cheers,
Brent

We can start out by setting one side of the square as \(x\) and attempt to fill out the rest of the lengths in terms of \(x\). Notice the top triangle is also equilateral so \(BX = WX = x\). Using the right triangle XYC and \(XY = x\), we get \(XC = 2 \sqrt{3}/3 * x\) by the ratio \(\frac{XY}{XC} = \sqrt{3} / 2\)

Therefore we have \(BC = BX + XC = x + 2\sqrt{3}/3* x = 4\sqrt{3}\). We may substitute in \(\sqrt{3} = 1.7\) to help us, \(BC = 6.8 = x + 1.13x = 2.13x\)
So x will be a little bigger then 3, we choose C.