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.

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 √2

*R*.

So √2

*R* =

*R* + 1

√2

*R* –

*R* = 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 = 2π

*R* = 2π × 2.4 = 4.8π

The closest answer choice is D.

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**bad one.. GMAT does not make you approximate.**