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 Post subject: GMAT GeometryPosted: Wed Jul 27, 2011 4:31 am

Joined: Sun May 30, 2010 3:15 am
Posts: 424

In the figure above (not drawn to scale), the triangle ABC is inscribed in the circle with the center O and the diameter AB. AC is 4 inches shorter than AB. The ratio of the angle COB’s measure to the measure of angle OBC is 2:1. What is the length of OB?
A. 2√2
B. 2√3
C. 4
D. 4 + 2√2
E. 4 + 2√3

(D) Let us use the fact that the ratio of the angle COB’s measure to the measure of the angle OBC is 2 to 1. Denote the measure of the angle OBC as x degrees. Then, the measure of the angle COB is 2x degrees. OB = OC, as the both are radii. Therefore the triangle BOC is isosceles and angle OCB = angle OBC = x degrees. Since the sum of all the angle measures in any triangle is 180 degrees, we can write the equation:
x + x + 2x = 180
4x = 180
x = 45

So, the measure of the angle BOC is 2x = 90 degrees. The angles AOC and COB are contiguous so angle AOC = 180° – 90° = 90°. Also AO = OC as radii. Therefore, triangle AOC is an isosceles right triangle (90°-45°-45°). The sides of such triangle have 1 : 1 : √2 ratio.

The question statement defines AC = AB – 4. Since AB is a diameter and AO is a radius AC = 2AO – 4. When we plug it in AO : AC = 1 : √2 we get AO : (2AO – 4) = 1 : √2 .
√2AO = 2AO – 4
4 = 2AO – √2AO
4 = (2 – √2)AO
AO = 4/(2 – √2)
AO = 4(2 + √2) / ((2 – √2)(2 + √2))
AO = (8 + 4√2) / 2
AO = 4 + 2√2

The correct answer is D.

If you don't remember the property of a 90°-45°-45° triangle, there is an alternative ending of the solution. Let us use the fact that AC is 4 inches shorter than AB. Denote radii by y inches. So, we can deduce three things:
1. AO = OB = OC = y
2. AB = 2y
3. AC = 2y – 4

Then we use the Pythagorean Theorem for triangle AOC and get the following equation:
(2y – 4)² = y² + y²
4y² – 16y + 16 = 2y²
2y² – 16y + 16 = 0
y² – 8y + 8 = 0
y = 4 + 2√2 and y = 4 – 2√2 are two solutions to this equation. But, the length of AC (2y – 4) must be a positive number, so y = 4 + 2√2 is the only option.
The correct answer is D.
----------
since the triangle AOC is right isoceles, can we simply use the Pythagorean Theorem 1:1:√2?
1:√2 = AO:4, AO = 4√2 /2 = OB

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 Post subject: Re: GMAT GeometryPosted: Wed Jul 27, 2011 5:48 am

Joined: Sun May 30, 2010 2:23 am
Posts: 498
Quote:
since the triangle AOC is right isoceles, can we simply use Pythagorean Theorem 1:1:√2?
1:√2 = AO:4, AO = 4√2 /2 = OB
The property of the sides AO:OC:AC = 1:1:√2 is true. But we do NOT know the length of either side. In your reasoning you assume AC = 4 inches. But the question statement does NOT provide us such value.

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 Post subject: Re: GMAT GeometryPosted: Wed Jul 27, 2011 6:06 am

Joined: Sun May 30, 2010 3:15 am
Posts: 424
I believe (and hope) that the step below (1 to 2) is not usually if ever tested in the exam. Could anyone please confirm that there it is not worth memorizing the long solution for a quadratic equation? (as opposed to solution factoring, which is indeed frequently tested)
1) y² – 8y + 8 = 0
2) y = 4 + 2√2 and y = 4 – 2√2 are both solutions to this equation

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 Post subject: Re: GMAT GeometryPosted: Wed Jul 27, 2011 6:29 am

Joined: Sun May 30, 2010 2:23 am
Posts: 498
Yes, indeed, it is used very rarely. And there are very few chances that the same quadratic equation will be very hard to solve using factoring. However knowing the formula will let you solve the equation even if you have hard time factoring it. In other words the good thing about this method is that it works for ANY quadratic equation. You also will know right away whether it is solvable or NOT.

I also would like to note, that this formula is provided in the GMAT math review. So it is assumed you should know it.

So to summarize my idea. There is a very low chance that a quadratic equation in the test can NOT be solved by factoring, but I advise to know the formula, because it is universal and can be used in many situations. Sometimes the formula works faster or can be used if you are stuck on the question.

P.S. For example, in this case the provided solution is universal and will work in many similar situations. But, as you can see in the posts above, there is an alternative solution, which can be used in this particular case without need to solve a quadratic equation. You'll be able to use it if you remember the property of a 45⁰-45⁰-90⁰ triangle. But what if NOT? Universal method can be applied.

A shortcut will save you time, but if you don't see it, a general method will solve a question.

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