In the diagram to the right, what is the value of x?

How do you go from
\(a^2 + (2a - \sqrt{3}a)^2 = \sqrt{2}^2\).

to
\(a^2 = \frac{1}{4 - 2\sqrt{3}} = \frac{1}{(\sqrt{3} - 1)^2}\)

\(a^2 + (2a - \sqrt{3}a)^2 = \sqrt{2}^2\)
\(a^2 + a^2(2 - \sqrt{3})^2 = 2\)
\(a^2(1 + (2 - \sqrt{3})^2) = 2\)
\(a^2(1 + 4 + 3 - 4\sqrt{3}) = 2\) (Using \((a + b)^2 = a^2 + b^2 + 2ab\))
\(a^2 = 2/(8 - 4\sqrt{3})\)
\(a^2 = 1/(4 - 2\sqrt{3})\)

Now, \((4 - 2\sqrt{3})\) = \((1 + 3 - 2\sqrt{3})\) = \((1^2 + \sqrt{3}^2 - 2*1*\sqrt{3})\)
= \((\sqrt{3} - 1)^2\)

Therefore, \(a^2 = 1/(\sqrt{3} - 1)^2\)

We can also use the law of cosines to arrive at the solution quickly:
a^2 = b^2 + c^2 - 2bc(cos A) [here a,b,and c are the sides of the triangle and A is the angle opposite a]

Applying it to this question:
=> {sqrt(2)}^2 = x^2 + x^2 - 2x^2 (cos 30)
=> 2 = 2x^2 - 2x^2 (sqrt(3)/2)
=> 2 = 2x^2 - sqrt(3)*x^2
=> x^2 = 2/(2 -sqrt(3))= 2(2 + sqrt(3))/(4-3)
=> x^2 = 4 + sqrt(3) = 1 + 3 + 2*sqrt(3) = (1+sqrt(3))^2
=> x = 1+ sqrt(3)

The MGMAT book also talks about estimating in a problem like this. Assume the diagram is drawn with the altitude added as above to make a 30:60:90 triangle:

√2 ≈ 1.4 (a bit more than) Now, obviously the perpendicular line is a bit less than this. However, if we assume that its length is also 1.4 then we can use the 30:60:90 triangle ratios right away and eliminate some answers.

If 1.4 = s (angle with 30) then x = 2s, or 2.8 So we know our maximum possible value is a bit less than this. Right away we can eliminate C, D and E.

C) 2*1.4 = 2.8
D) 1.4 + 1.7 = 3.1
E) 2*1.7 = 3.4

We're left with A and B.
A) 1+1.4 = 2.4
B) 1+1.7 = 2.7

Here I would just guess and move on, we assumed the line is just a bit less than 1.4 so B) is closest. You can also note that the answers vary by .3 except for B and C which can potentially be a hint (or not...)

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