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A thin piece of wire 40 meters long is cut into two pieces.

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by AAPL » Thu Oct 04, 2018 7:03 am

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GMAT Prep

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 square regions in terms of R?
$$A.\ \pi R^2$$
$$B.\ \pi R^2+10$$
$$C.\ \pi R^2+\frac{1}{4}\pi^2R^2$$
$$D.\ \pi R^2+\left(40-2\pi R\right)^2$$
$$E.\ \pi R^2+\left(10-\frac{1}{2}\pi R\right)^2$$
OA E.
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by GMATGuruNY » Thu Oct 04, 2018 7:06 am
AAPL wrote:GMAT Prep

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 square regions in terms of R?
$$A.\ \pi R^2$$
$$B.\ \pi R^2+10$$
$$C.\ \pi R^2+\frac{1}{4}\pi^2R^2$$
$$D.\ \pi R^2+\left(40-2\pi R\right)^2$$
$$E.\ \pi R^2+\left(10-\frac{1}{2}\pi R\right)^2$$
Since the answer choices are in terms of a variable -- the value of r -- we can PLUG IN.

Let the ENTIRE WIRE be used to form the square.
Then:
Perimeter of the square = 40.
Side = 10.
Area = 100.

Circle area + square area = 0 + 100 = 100. This is our target.
Since the circle has no area, r=0.
Now we plug r=0 into the answers to see which yields our target of 100.
Only E works:
πr² + (10 - (1/2)πr)² = π0² + (10 - (1/2)π0)² = 0 + 10² = 100.

The correct answer is E.
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by Brent@GMATPrepNow » Thu Oct 04, 2018 7:09 am
AAPL wrote:GMAT Prep

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 square regions in terms of R?
$$A.\ \pi R^2$$
$$B.\ \pi R^2+10$$
$$C.\ \pi R^2+\frac{1}{4}\pi^2R^2$$
$$D.\ \pi R^2+\left(40-2\pi R\right)^2$$
$$E.\ \pi R^2+\left(10-\frac{1}{2}\pi R\right)^2$$
OA E.
One approach is to plug in a value for r and see what the output should be.

Let's say r = 0. That is, the radius of the circle = 0
This means, we use the entire 40-meter length of wire to create the square.
So, the 4 sides of this square will have length 10, which means the area = 100

So, when r = 0, the total area = 100

We'll now plug r = 0 into the 5 answer choices and see which one yields an output of 100

A) (π)(0²) = 0 NOPE
B) (π)(0²) + 10 = 10 NOPE
C) (π)(0²) + 1/4([π]² * 0²) = 0 NOPE
D) (π)(0²) + (40 - 2[π]0)² = 1600 NOPE
E) (π)(0²) + (10 - 1/2[π](0))² = 100 PERFECT!

Answer: E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Sun Oct 07, 2018 6:30 pm
AAPL wrote:GMAT Prep

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 square regions in terms of R?
$$A.\ \pi R^2$$
$$B.\ \pi R^2+10$$
$$C.\ \pi R^2+\frac{1}{4}\pi^2R^2$$
$$D.\ \pi R^2+\left(40-2\pi R\right)^2$$
$$E.\ \pi R^2+\left(10-\frac{1}{2}\pi R\right)^2$$
OA E.
We are given that 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.

Since the circumference of a circle with radius R is 2Ï€R, the length of wire used to form the circle is 2Ï€R. Thus, we have (40 - 2Ï€R) left over to form the square. In other words, the perimeter of the square is (40 - 2Ï€R). However, since we need to calculate the total area of the circular and the square regions, we need to determine the side of the square in terms of R. Since the perimeter of the square is (40 - 2Ï€R), the side of the square is:

side = (40 - 2Ï€R)/4

side = 10 - (1/2)Ï€R

Now we can determine the areas of the circle and the square.

Area of circle = πR2

Area of square = side^2 = (10 - (1/2)Ï€R)^2

Thus, the combined area of the circle and square is πR2 + (10 - (1/2)πR)2.

Answer: E

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