Wire split into circle and square

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Wire split into circle and square

by josh80 » Wed Dec 11, 2013 4:24 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 square regions in terms of r?

A. πr^2
B. πr^2 + 10
C. πr^2 + (1/4)π^2r^2
D. πr^2 + (40-2πr)^2
E. πr^2 + [10-(1/2)πr]^2

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by Brent@GMATPrepNow » Wed Dec 11, 2013 4:27 pm
A thin piece of 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 term of r?

A)(pi)r²
B)(pi)r² + 10
C)(pi)r² + 1/4([pi]² * r²)
D)(pi)r² + (40 - 2[pi] * r)²
E)(pi)r² + (10 - 1/2[pi] * r)²
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) (pi)(0²) = 0 NOPE
B) (pi)(0²) + 10 = 10 NOPE
C) (pi)(0²) + 1/4([pi]² * 0²) = 0 NOPE
D) (pi)(0²) + (40 - 2[pi]0)² = 1600 NOPE
E) (pi)(0²) + (10 - 1/2[pi](0))² = 100 PERFECT!

Answer: E

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by Brent@GMATPrepNow » Wed Dec 11, 2013 4:28 pm
A thin piece of 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 term of r?

A)(pi)r²
B)(pi)r² + 10
C)(pi)r² + 1/4([pi]² * r²)
D)(pi)r² + (40 - 2[pi] * r)²
E)(pi)r² + (10 - 1/2[pi] * r)²
Here's an algebraic approach:

Since r is the radius of the circle, the area of the circle will be (pi)r²

If r is the radius of the circle, the length of wire used for this circle will equal its circumference which is 2(pi)r

So, the length of wire to be used for the square must equal 40 - 2(pi)r

In other words, the perimeter of the square will be 40 - 2(pi)r

Since squares have 4 equal sides, the length of each side of the square will be [40 - 2(pi)r]/4, which simplifies to be 10 - (pi)r/2

If each side of the square has length 10 - (pi)r/2, the area of the square will be [10 - (pi)r/2]²

So, the total area will equal (pi)r² + [10 - (pi)r/2]², which is the same as E

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by Amrabdelnaby » Tue Nov 24, 2015 5:11 pm
Hi Brent,

I had a very different approach and it led me to very different answer as well.

Please tell me what i did wrong.

since the wire is 40 i wrote 2 pi r = 40
r = 20/pi
the area is equal to pi r^2
so pi(20/pi)^2 = 400/pi

how does this answer relates to the answer choices!
Brent@GMATPrepNow wrote:
A thin piece of 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 term of r?

A)(pi)r²
B)(pi)r² + 10
C)(pi)r² + 1/4([pi]² * r²)
D)(pi)r² + (40 - 2[pi] * r)²
E)(pi)r² + (10 - 1/2[pi] * r)²
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) (pi)(0²) = 0 NOPE
B) (pi)(0²) + 10 = 10 NOPE
C) (pi)(0²) + 1/4([pi]² * 0²) = 0 NOPE
D) (pi)(0²) + (40 - 2[pi]0)² = 1600 NOPE
E) (pi)(0²) + (10 - 1/2[pi](0))² = 100 PERFECT!

Answer: E

Cheers,
Brent

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by [email protected] » Wed Nov 25, 2015 1:46 pm
Hi josh80,

TESTing VALUES (the approach that Brent used) is perfect for this type of question. In the event that you get 'stuck' on a prompt that is complex though, you have to be ready to take a guess and move on. Here's how you could increase your chances of selecting the correct answer...

The question asks for the TOTAL area of the circular and square regions. We're clearly going to be adding two numbers together, so we'd likely need an answer that fits that format. Eliminate A (it's just the area of a circle).

Since we don't know what the area of the square is (and since it will change depending on how much of the wire we put into the square's perimeter), it seems unlikely that the area would be exactly 10. Eliminate B.

You could then guess from the remaining 3 options.

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by Brent@GMATPrepNow » Wed Nov 25, 2015 1:59 pm
Amrabdelnaby wrote:Hi Brent,

I had a very different approach and it led me to very different answer as well.

Please tell me what i did wrong.

since the wire is 40 i wrote 2 pi r = 40
r = 20/pi
the area is equal to pi r^2
so pi(20/pi)^2 = 400/pi

how does this answer relates to the answer choices!
You've created a LOT of work for yourself, but you can still go this route.
Your solution makes the area of the square ZERO and uses all of the wire to make a circle. Fine.
So, when r = 20/pi, the total area of the circle and square = 400/pi + 0 = 400/pi

At this point, you must take each answer choice and replace r with 20/pi, and see which one yields an output of 400/pi

If you try it, you'll find that E yields an output of 400/pi

In my solution, I replaced r with 0, which made the calculations MUCH EASIER

Cheers,
Brent
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