Spaceland Prep Strategy Question #3

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Spaceland Prep Strategy Question #3

by spaceland prep » Mon Mar 07, 2011 8:56 am
Q. A rectangle with sides w and h has a circle with radius 2 cut out of it. If the missing circle is tangent to three of the rectangles edges and the ratio of w to h is 2:1, what is the area of the remaining pieces of the rectangle?

8 - 4*pi
8 - 2*pi
32 - 4*pi
24
32 - 2*pi

Think about how one would go about solving this question and how that corresponds to the way the answer choices are structured. Why does the correct answer have to be one of the "integer minus pi-term" choices? Why must the pi-term be 4*pi? What is the only logical answer?

Solution

[spoiler]The question describes a rectangle with a circle cut out. So the area will be found by subtracting the area of the circle from the area of the rectangle. This is represented by the "integer minus pi-term" answer choices, since a circle with an integer radius will have an area that is a factor of pi.

Since the circle is tangent to three of the rectangle's edges it must be completely within the rectangle, which means its entire area will be subtracted from the area of the rectangle. Because it is given that the circle's radius is 2, its area is 4*pi.

Looking at the two answer choices with 4*pi terms, only one is logically valid. If 4*pi were subtracted from 8, the result would be negative, an impossibility for a geometric area on the GMAT. Thus, the only logical answer is 32 - 4*pi.

When the GMAT uses answer choices that contain multiple term expressions, often the opportunity to reverse-engineer the answers is present. This phenomena extends beyond geometry questions and occasionally can be worked the other way. Check out PS Q. 145 in the Quantitative Review, 2nd Ed. How can the expression of the area of the shaded region be used to answer the question?[/spoiler]
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by srcc25anu » Mon Mar 07, 2011 10:19 am
Radius of circle = 2 and sides of rectangle are in the ratio 2:1. If radius is 2, diameter = 4.
Lets take width (h) = 4 (= diameter) and length of rectangle (w) would be 8 in this case. as l:w = 2:1
Area of rectangle = 4*8 = 32
Area of circle = 4 pi
Area of rem. Part = 32-4pi