If we draw a vertical height from A down to BC, that divides the triangle into a 45-45-90 triangle on the left, and a 30-60-90 triangle on the right. The hypotenuse of the 45-45-90 is AC, so is of length 1, and the height we've drawn is thus of length 1/√2 = √2/2 (since the sides of a 45-45-90 are in a 1 to 1 to √2 ratio). This is opposite the 30 degree angle in the 30-60-90 triangle, so it is the shortest side of that triangle, and the longest side AB is twice as long (since in a 30-60-90 triangle, sides are in a 1 to √3 to 2 ratio), so the length of AB is 2(√2/2) = √2.

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What is the length of line segment \(AB\) in the figure above?

A. \(\frac{\sqrt{2}}{2}\)

B. \(\frac{\sqrt{3}}{2}\)

C. \(\sqrt{2}\)

D. \(\frac{3}{2}\)

E. \(\sqrt{3}\)

OA C

If we drop a perpendicular from vertex A to side BC, that is, if we draw the height, we will divide the triangle into two special right triangles: a 45-45-90 triangle on the left and a 30-60-90 triangle on the right. Let's call this height AD; that is, D is a point on BC such that AD is perpendicular to BC.

We know that the side ratio of a 45-45-90 triangle is x : x : x√2. Since AC = 1, we see that if we let AD = x, then x√2 = 1. So x = 1/√2 = √2/2 = AD.

We also know that the side ratio of a 30-60-90 triangle is x : x√3 : 2x. Since AD = √2/2, we see that AB must be twice as much. So AB = 2(√2/2) = √2.

Answer: C