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Right Triangle Inscribed in a Circle Question

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Right Triangle Inscribed in a Circle Question

by morganmthomson7 » Sun Apr 15, 2018 10:45 am
Hi there,

I just had a quick question if anyone can help explain this to me!

The preceding diagram shows a figure composed of a right triangle ABC adjacent to semicircle Arc AC. If AB = OB = x, what is the perimeter of the figure?
Initially, since it is a right triangle, I would have assumed that length BC (hypotenuse of right triangle in the given figure) was x root 3, because the other sides of the right triangle are x and 2x (typically what I would associate as a 30-60-90 triangle). But in the solution, they use pythagorean theorem to solve for the length of BC instead. Why would assuming this be wrong in this case?

Thanks so much for the help!

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by [email protected] » Mon Apr 16, 2018 11:01 am
Hi morganmthomson7,

To start, when posting questions, you should post the ENTIRE prompt - including any pictures and the 5 answer choices. Without that drawing, it's unclear exactly what the shapes look like. If you post the full prompt, then I'll be happy to help you work through it.

In addition, with a 30/60/90 right triangle, the side across from the 90-degree angle is the 2X side. If the two 'legs' of the right triangle are X and 2X, then the hypotenuse is (root5X), so we do NOT have a 30/60/90 right triangle.

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TTT

by KCTester » Wed Apr 25, 2018 11:48 am
As the poster above me said, the first thing to note is that the ratio of sides in a 30-60-90 right triangle is 1 - root(3) - 2. And 2 is the longest side, which is the hypotenuse.

Another thing to note is that, for a right triangle inscribed in a circle, the hypotenuse is going to be the diameter of the circle. Chances are, that is also a useful connection to make on the way to an efficient solution.
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