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by divya23 » Tue Jun 14, 2011 1:23 am
a thin piece of wire length = 40 is cut into 2 from one is made circle of radius r and from oder square..find the total area of both circle and square so formed.

[spoiler]ans = pir^2+(10-1/2pir)^2[/spoiler]
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by GMATGuruNY » Tue Jun 14, 2011 1:44 am
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. πr²
B. πr² +10
C. πr² + 1/4 π²r²
D. πr² + (40-2πr)²
E. πr² + (10 - (1/2)πr)²
Plug in r = 4.
Area of circle = πr² = π4² = 16π ≈ 48
Circumference of circle = 2πr = 8π ≈ 24.
Perimeter of square = remaining wire = 40-24 = 16.
Side of square = 16/4 = 4.
Area of square = 4² = 16.
Area of circle + area of square ≈ 48+16 ≈ 64. This is our target.

Now we plug r=4 into the answers to see which yields our target of 64.

Only answer choice E works:
πr² + (10 - 1/2πr)² ≈ 3(4²) + (10 - (1/2)*3*4)² = 48+16 = 64.

The correct answer is E.
Last edited by GMATGuruNY on Tue Jun 14, 2011 12:27 pm, edited 1 time in total.
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by [email protected] » Tue Jun 14, 2011 4:28 am
But gmatguru, the answer varies with the value of r. It does not remain 64
eg: I tried taking various values of circumference and hence radius and with every answer i found a different total area of the circle + the square.....
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by [email protected] » Tue Jun 14, 2011 4:44 am
No gmatguru i got it i was misinterpreted the e] equation..
Thnx..
but dont u feel its too lengthy
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by finites » Tue Jun 14, 2011 11:40 am
I got the answer as pi*r^2 + (10 -(pi*r/2))^2
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by Brent@GMATPrepNow » Tue Jun 14, 2011 12:23 pm
For this type of question where there are variables in the answer choices, you typically have two options:
1) Plug in some values and check the answer choices
2) Apply your algebra skills to create the required expression

Mitch already demonstrated option 1, so I'll do option 2.

Since r is the radius of the circle, the area of the circle will be (pi)r^2.

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)r^2 + [10 - (pi)r/2]^2, which is the same as E

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Last edited by Brent@GMATPrepNow on Thu Jan 24, 2013 6:38 am, edited 1 time in total.
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by mmenifi » Wed Jun 15, 2011 5:24 pm
I am interested to see how you would solve it the algebraic approach.


GMATGuruNY wrote:
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. πr²
B. πr² +10
C. πr² + 1/4 π²r²
D. πr² + (40-2πr)²
E. πr² + (10 - (1/2)πr)²
Plug in r = 4.
Area of circle = πr² = π4² = 16π ≈ 48
Circumference of circle = 2πr = 8π ≈ 24.
Perimeter of square = remaining wire = 40-24 = 16.
Side of square = 16/4 = 4.
Area of square = 4² = 16.
Area of circle + area of square ≈ 48+16 ≈ 64. This is our target.

Now we plug r=4 into the answers to see which yields our target of 64.

Only answer choice E works:
πr² + (10 - 1/2πr)² ≈ 3(4²) + (10 - (1/2)*3*4)² = 48+16 = 64.

The correct answer is E.
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