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divya23: Master | Next Rank: 500 Posts; Posts: 113; Joined: Mon Apr 26, 2010 8:58 amFollowed by: 14 members; Thanked: 2 times

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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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 » 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. 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

Plug in r = 4.
Area of circle = Ï€rÂ² = Ï€4Â² = 16Ï€ â‰ˆ 48
Circumference of circle = 2Ï€r = 8Ï€ â‰ˆ 24.
Perimeter of square = remaining wire = 40-24 = 16.
Side of square = 16/4 = 4.
Area of square = 4Â² = 16.
Area of circle + area of square â‰ˆ 48+16 â‰ˆ 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:
Ï€rÂ² + (10 - 1/2Ï€r)Â² â‰ˆ 3(4Â²) + (10 - (1/2)*3*4)Â² = 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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[email protected]: Legendary Member; Posts: 934; Joined: Tue Nov 09, 2010 5:16 amFollowed by: 14 members; Location: AAMCHI MUMBAI LOCAL; Thanked: 63 times

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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finites: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Thu Jun 02, 2011 7:21 am

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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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

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

Cheers,
Brent

Last edited by Brent@GMATPrepNow on Thu Jan 24, 2013 6:38 am, edited 1 time in total.

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mmenifi: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Thu Oct 29, 2009 9:17 am

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. 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
Plug in r = 4.
Area of circle = Ï€rÂ² = Ï€4Â² = 16Ï€ â‰ˆ 48
Circumference of circle = 2Ï€r = 8Ï€ â‰ˆ 24.
Perimeter of square = remaining wire = 40-24 = 16.
Side of square = 16/4 = 4.
Area of square = 4Â² = 16.
Area of circle + area of square â‰ˆ 48+16 â‰ˆ 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:
Ï€rÂ² + (10 - 1/2Ï€r)Â² â‰ˆ 3(4Â²) + (10 - (1/2)*3*4)Â² = 48+16 = 64.

The correct answer is E.

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