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Triangle 1:
The triangle on top is 30-60-90 triangle
So, we know the ratio of sides of the triangle is \(x:x\sqrt{3}:2x\)

From the triangle 1 on top, we know the side 2x=14, so x=7

Base of triangle 1 = \(7\sqrt{3}\)

Triangle 2:
The triangle in the bottom is 45-45-90 triangle
So, we know the ratio of sides fof the triangle is x:x:\(x\sqrt{2}\)
The hypotenuse of the triangle with side \(x\sqrt{2}\) = \(7\sqrt{3}\)*\(\sqrt{2}\)

The hypotenuse of the triangle 2 = \(7 \sqrt{6}\)

Circle:
The hypotenuse of the triangle is the diameter of the circle.
Diameter = \(7 \sqrt{6}\)
Circumference = d\(\pi\) = \(7 \sqrt{6}*\pi\)

Circumference of semicircle = \( (7 \sqrt{6}*\pi) / 2\)

Ans: C
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Bunuel

What is the circumference of the semicircle in the figure shown?


A. \(14\pi\sqrt{6}\)

B. \(7\pi\sqrt{6}\)

C. \(\frac{7\pi\sqrt{6}}{2}\)

D. \(\frac{7\pi\sqrt{6}}{3}\)

E. \(\frac{7\pi\sqrt{6}}{4}\)


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Attachment:
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Bunuel Please help me out here

Isn't circumference for semi circle = pi*r+2r ???
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Bunuel

What is the circumference of the semicircle in the figure shown?


A. \(14\pi\sqrt{6}\)

B. \(7\pi\sqrt{6}\)

C. \(\frac{7\pi\sqrt{6}}{2}\)

D. \(\frac{7\pi\sqrt{6}}{3}\)

E. \(\frac{7\pi\sqrt{6}}{4}\)


Solution:

Since the top triangle is a 30-60-90 right triangle and its hypotenuse is 14, its long leg (the one that is opposite 60-degree angle) is 14/2 * √3 = 7√3. Since that leg is also a leg of the 45-45-90 right triangle at the bottom, the hypotenuse of the 45-45-90 right triangle is 7√3 * √2 = 7√6. Since that hypotenuse is also the diameter of the semicircle, the length of its semicircular arc is ½ * 7√6 * π = 7π√6/2.

Answer: C
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