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Some of pyhtagron triplets we need to keep it in mind. Like {( 2,3,5) , ( 5,12,13) ,( 7, 24,25), ( 11, 60,61). ( 9, 40, 41) is also an Pythagron triplet always appears on the GMAT probs.

So now we know the triangle is an right angle triangle. The circle circumscribes the triangle.

The circumraduis of the circle that circumscribes the right angle triangle = hypotanse / 2 = 41 / 2 = 20.5

Thanks Asimov for correcting me. Here is a consolidated list of all the common Pythogron triplets. (3,4,5) (5,12,13)(7,24,25)(8,15,17)(9,40,41)(11,60,61)(12,35,37)(16,60,65)(20,21,29)

That all. I just put them together.incase u might need them.

whats the best way to find the diameter of a circle with an inscribed triangle if you are given an arc? so lets say an inscribed triangle with an arc of 24 that is approximately 3/4ths of the circumference. any suggestions?

If I understood you correctly,you mean to say one side of the inscribed triangle creates an arc equal to 24 which is approx 3/4 of the circumference.

If the above understanding is correct, the solution is pretty simple . circumference = 24 * 4/3 = 32 2*pi*r=32 d*pi=32 d=32/pi

dakhan wrote:

whats the best way to find the diameter of a circle with an inscribed triangle if you are given an arc? so lets say an inscribed triangle with an arc of 24 that is approximately 3/4ths of the circumference. any suggestions?

I think it is not necessary to know that this triangle is a right triangle.

My 10sec approach. Any sides of a circumscribed triangle is equal or less than a diameter of the circle that circumscribes the triangle. In our case we have a long and thin triangle, so the answer will be 41/2 or something a bit larger. Let's look our options... only B fits. _________________

I think it is not necessary to know that this triangle is a right triangle.

My 10sec approach. Any sides of a circumscribed triangle is equal or less than a diameter of the circle that circumscribes the triangle. In our case we have a long and thin triangle, so the answer will be 41/2 or something a bit larger. Let's look our options... only B fits.

Awesome. Well, thats how I thought. But something came in my mind. How can we consider the largest side as the diameter of the circle without knowing that the triangle is a right angled one? What I mean to say is, if we do consider the largest side as a diameter 41, in this case, isn't it possible that the other two points can lay either within or outside the circle but not on the circle. I dont know how correct I am. But the answer would definately help.

You are awesome with Quant, thats for sure. _________________

GMAT offended me. Now, its my turn! Will do anything for Kudos! Please feel free to give one.

...How can we consider the largest side as the diameter of the circle without knowing that the triangle is a right angled one? What I mean to say is, if we do consider the largest side as a diameter 41, in this case, isn't it possible that the other two points can lay either within or outside the circle but not on the circle. I dont know how correct I am. But the answer would definately help.

When I said "41/2 or a bit larger" I missed to mention one thing. There are my thoughts in detail.

First of all, diameter cannot be smaller than 41/2, otherwise it would be impossible to circumscribe the triangle by the such small circle. So, we have two options: 41 (B radius = 20.5) or 90 (E radius = 45). 90 is more than twice greater than the biggest side of triangle. It would be possible for triangle with one angle >>90 (for example, 40, 48.5, 9). In our case, we have one small angle (between 41 and 40) and two angle near 90. So only B is an option. It is fast method to figure out an answer with 98% accuracy (let's say a fast guessing technique). But if you have time, it is always possible to spend remaining 2 mins on getting 100% accuracy. You may ask me what if E would be 21? Yeah, this approach doesn't work but for this particular problem it helps save at least 1:30 for next hard question. So, If GMAT wants to trick us, why we cannot trick it? _________________

...How can we consider the largest side as the diameter of the circle without knowing that the triangle is a right angled one? What I mean to say is, if we do consider the largest side as a diameter 41, in this case, isn't it possible that the other two points can lay either within or outside the circle but not on the circle. I dont know how correct I am. But the answer would definately help.

When I said "41/2 or a bit larger" I missed to mention one thing. There are my thoughts in detail.

First of all, diameter cannot be smaller than 41/2, otherwise it would be impossible to circumscribe the triangle by the such small circle. So, we have two options: 41 (B radius = 20.5) or 90 (E radius = 45). 90 is more than twice greater than the biggest side of triangle. It would be possible for triangle with one angle >>90 (for example, 40, 48.5, 9). In our case, we have one small angle (between 41 and 40) and two angle near 90. So only B is an option. It is fast method to figure out an answer with 98% accuracy (let's say a fast guessing technique). But if you have time, it is always possible to spend remaining 2 mins on getting 100% accuracy. You may ask me what if E would be 21? Yeah, this approach doesn't work but for this particular problem it helps save at least 1:30 for next hard question. So, If GMAT wants to trick us, why we cannot trick it?

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

What is the measure of the radius of the circle that circumscribes a triangle whose sides measure 9, 40 and 41? (A) 6 (B) 4 (C) 24.5 (D) 20.5 (E) 12.5

First of all we can notice that a triangle whose sides measure 9, 40 and 41 is a right triangle because 9^2 + 40^2 = 41^2.

Next, a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle (the reverse is also true: if the diameter of a circle is also the triangle’s side, then that triangle is a right triangle).

Thus the diameter of the circle is the hypotenuse of the triangle --> diameter = hypotenuse = 41 --> radius = 41/2 = 20.5.

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