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# In the circle above, chord AC = 85. What is the area of the circle?

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Re: In the circle above, chord AC = 85. What is the area of the circle? [#permalink]
Bunuel wrote:

In the circle above, chord AC = 85. What is the area of the circle?

(1) Angle ABC = 90°
(2) AB = 51 and BC = 68

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

If we know chord AB were a diameter, then we would divide by 2 to find the radius. Once we know the radius, we could find the area. Furthermore, we would know that AB were a diameter if the angle at B were a right angle, because an inscribed angle of 90° always intersects a semicircular arc.

Statement #1 simply tells us that the angle at B is 90°, so we are done! This statement, alone and by itself, is sufficient.

Statement #2 gives us the other two sides of triangle ABC, we now we know we have a {51, 68, 85} triangle. If these three sides satisfy the Pythagorean Theorem, then the triangle would be a right triangle, and the angle at B would be 90°. Well, it would be HUGE mistake to square these numbers as is and try to do anything with the results! Instead, can we factor out a common factor? Well, 51 = 3*17. Also, notice that 85 = 5*17. This makes us suspicious, and it turns out, 68 = 4*17. This is simply the good old 3-4-5 triangle, with everything multiplied by 17. Of course, this is a right triangle, so the angle at B is 90°. This statement, alone and by itself, is sufficient.

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Re: In the circle above, chord AC = 85. What is the area of the circle? [#permalink]
Let's say that 2) gave that AB=51 and BC=69 (not 68). In this case, it would be an inscribed triangle but not an inscribed right triangle. Would you still be able to find the area of the ellipse with these 3 pieces of data? Is the resulting ellipse a circle?

Just wondering in case the GMAT decided to take this question a step further and test something more complex.
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Re: In the circle above, chord AC = 85. What is the area of the circle? [#permalink]
Another way to see why statement 2 is sufficient is to consider the following property of triangles:

For any given triangle there is exactly one possible circle into which that triangle can be inscribed (the triangle's 'curcumcircle').

So, since statement 2 completely defines a triangle, we know that there is only one possible circle, with some specific area, that will pass through every vertex. Sufficient.

brooklyndude, the same reasoning can be applied to see that your hypothetical statement would also be sufficient to find the area of a circumscribed circle (not necessarily some other ellipse), though the actual calculation would involve trigonometry that the GMAT doesn't test. (it's also important to reiterate that the GMAT couldn't include your hypothetical statement along with the first given statement, since DS statements never contradict each other)
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Re: In the circle above, chord AC = 85. What is the area of the circle? [#permalink]
Bunuel wrote:
Bunuel wrote:

In the circle above, chord AC = 85. What is the area of the circle?

(1) Angle ABC = 90°
(2) AB = 51 and BC = 68

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

If we know chord AB were a diameter, then we would divide by 2 to find the radius. Once we know the radius, we could find the area. Furthermore, we would know that AB were a diameter if the angle at B were a right angle, because an inscribed angle of 90° always intersects a semicircular arc.

Statement #1 simply tells us that the angle at B is 90°, so we are done! This statement, alone and by itself, is sufficient.

Statement #2 gives us the other two sides of triangle ABC, we now we know we have a {51, 68, 85} triangle. If these three sides satisfy the Pythagorean Theorem, then the triangle would be a right triangle, and the angle at B would be 90°. Well, it would be HUGE mistake to square these numbers as is and try to do anything with the results! Instead, can we factor out a common factor? Well, 51 = 3*17. Also, notice that 85 = 5*17. This makes us suspicious, and it turns out, 68 = 4*17. This is simply the good old 3-4-5 triangle, with everything multiplied by 17. Of course, this is a right triangle, so the angle at B is 90°. This statement, alone and by itself, is sufficient.

Hi Bunuel,

Ac or AB is the diameter of the circle?

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Re: In the circle above, chord AC = 85. What is the area of the circle? [#permalink]
zanaik89 wrote:
Bunuel wrote:
Bunuel wrote:

In the circle above, chord AC = 85. What is the area of the circle?

(1) Angle ABC = 90°
(2) AB = 51 and BC = 68

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

If we know chord AB were a diameter, then we would divide by 2 to find the radius. Once we know the radius, we could find the area. Furthermore, we would know that AB were a diameter if the angle at B were a right angle, because an inscribed angle of 90° always intersects a semicircular arc.

Statement #1 simply tells us that the angle at B is 90°, so we are done! This statement, alone and by itself, is sufficient.

Statement #2 gives us the other two sides of triangle ABC, we now we know we have a {51, 68, 85} triangle. If these three sides satisfy the Pythagorean Theorem, then the triangle would be a right triangle, and the angle at B would be 90°. Well, it would be HUGE mistake to square these numbers as is and try to do anything with the results! Instead, can we factor out a common factor? Well, 51 = 3*17. Also, notice that 85 = 5*17. This makes us suspicious, and it turns out, 68 = 4*17. This is simply the good old 3-4-5 triangle, with everything multiplied by 17. Of course, this is a right triangle, so the angle at B is 90°. This statement, alone and by itself, is sufficient.

Hi Bunuel,

Ac or AB is the diameter of the circle?