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 Post subject: GMAT Geometry
PostPosted: Mon Feb 14, 2011 11:18 am 
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Joined: Sun May 30, 2010 3:15 am
Posts: 424
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A circle, with center O, is inscribed in square WXYZ. Point P, as shown above, is where the circle and the line segment ZX intersect. If PX = 1, which of the following is closest to the value of the circumference of the circle?

A. 2π
B. 3π
C. 4π
D. 5π
E. 6π

(D) To solve this problem, we need to draw a triangle in the above figure. Set point B as the midpoint of XY, and then draw the triangle OBX.

Image

Triangle OBX will be an isosceles right triangle, where the two shorter sides, OB and BX, have a length that is equal to the radius of the circle, which we'll call R.

The hypotenuse of this triangle can be measured two ways:
First, as the sum of the segments OP and PX, it has a length of R + 1. Second, as the hypotenuse of an isosceles right triangle, it has a length that is √2 times the length of each of the shorter sides, or √2R.

So √2R = R + 1
√2RR = 1
R(√2 – 1) = 1
R = 1/(√2 – 1) = (√2 + 1) / (2 – 1) = √2 + 1
Since √2 equals approximately 1.4,
R = 1.4 + 1 = 2.4

Now that we know the radius of the circle, we can determine the circumference:
Circumference = 2πR = 2π × 2.4 = 4.8π

The closest answer choice is D.
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bad one.. GMAT does not make you approximate.


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 Post subject: Re: math (test 2, question 1): geometry, number theory.
PostPosted: Mon Feb 14, 2011 11:57 am 
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Joined: Sun May 30, 2010 2:23 am
Posts: 498
The skill of approximation is a very good skill to use in real life as also in GMAT questions. GMAT does have questions in forms "Which of the following is the closest to _____ ?"; "_____ is in between "; etc. Those questions can be solved using approximation. Note, that GMAT also deals with rounded numbers in some questions.

Questions, where you can use approximation are often can be solved in a different way, but in many cases approximation can save you some time. (Note, that the fastest possible solution depends on the question itself as also on what skills you are the most convenient with).

Let me show how this question can be solved without approximation.

When we get the value of the circumference, (2√2 + 2)π, we can compare it to the answer choices.

2π | 3π | 4π | 5π | 6π
(2√2 + 2)π

Divide by π and deduct 2 from all the values:
0 | 1 | 2 | 3 | 4
2√2

All the values are non-negative, so let's square:
0 | 1 | 4 | 9 | 16
8

Therefore 2√2 is closer to 3 than to 2. And (2√2 + 2)π is closer to 5π than any other answer choice.


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