GMATPrep 1 Geometry Question

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Mjkourtis: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Tue Feb 07, 2012 3:31 am

by Mjkourtis » Sat Sep 29, 2012 12:28 pm

A thin wire of 40 meters is cut into 2 pieces. One piece is used to form a circle with radius r and the other is used to form a square. No wire is left over. Which represents the total area?
a. Ï€r^2
b. Ï€r^2 +10
c. Ï€r^2 + (1/4)*Ï€^2*r^2
d. Ï€r^2 + (40- 2Ï€r)^2
e. Ï€r^2 + (10-(1/2)*Ï€r)^2

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Source: Beat The GMAT — Problem Solving | Sat Sep 29, 2012 12:28 pm

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Sep 29, 2012 12:35 pm

A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square regions in terms of r?

A. Ï€rÂ²
B. Ï€rÂ² +10
C. Ï€rÂ² + 1/4 Ï€Â²rÂ²
D. Ï€rÂ² + (40-2Ï€r)Â²
E. Ï€rÂ² + (10 - (1/2)Ï€r)Â²

Let r=0, so that the ENTIRE wire is used to form the square.
Perimeter of the square = 40.
Each side of the square = 40/4 = 10.
Area of the square = 10Â² = 100.
Area of the circle + area of the square = 0 + 100 = 100. This is our target.

Now we plug r=0 into the answers to see which yields our target of 100.

A quick scan reveals that only E works:
Ï€rÂ² + (10 - 1/2Ï€r)Â² = 0 + 10Â² = 100.

The correct answer is E.

Last edited by GMATGuruNY on Sat Sep 29, 2012 12:46 pm, edited 1 time in total.

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anuprajan5: Master | Next Rank: 500 Posts; Posts: 279; Joined: Mon Jun 25, 2012 10:56 pmFollowed by: 10 members; Thanked: 60 times

by anuprajan5 » Sat Sep 29, 2012 12:36 pm

We can find the answer as such.

There are 2 parts of the wire.

x and (40-x)

Assume x as the circumference of the circle with radius r.

x = 2pi*r

Since 40-x forms the perimeter of the square, side a = (40-x)/4

a = 10-x/4

Area = a^2 = (10-(x/4))^2

Substituting for x = (10-(2pi*r)/4)^2 = (10-pi*r/2)^2

Total area = Area of circle + Are of square

= pi*r^2 + (10-pi*r/2)^2

Answer E

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