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A rancher uses 64 feet of fencing to create a rectangular horse corral

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A rancher uses 64 feet of fencing to create a rectangular horse corral  [#permalink]

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New post 05 Mar 2020, 06:54
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A rancher uses 64 feet of fencing to create a rectangular horse corral. If the ratio of the corral’s length to width is 3:1, which of the following most closely approximates the minimum length of additional fencing needed to divide the rectangular corral into three triangular corrals, one of which is exactly twice the area of the other two?

(A) 24 feet
(B) 29 feet
(C) 36 feet
(D) 41 feet
(E) 48 feet

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A rancher uses 64 feet of fencing to create a rectangular horse corral  [#permalink]

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New post 13 Mar 2020, 03:06
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My approach (May be something more efficient? But this got me the right answer.)

The rectangle clearly has length 3x and width x, with perimeter 64, so x=8. (3x+3x+x+x=8x | 8x=64, x=8)

To create the triangles --> imagine the rectangle is being cut in half lengthwise. each of those subsets is being cut in half at the diagonal. The two diagonal cuts are where the fence goes, creating one large triangle and two small triangles.

Calculate the hypotenuse, or one of the diagonal cuts, with a triangle: side 1 = 1.5x, side 2 = x | side1=12, side2=8.
\(8^2+12^2=y^2\) --> \(64+144=y^2\) --> \(208=y^2\) --> \(y=4 \sqrt{13}\)

two diagonals , so the extra fencing is \(2*4\sqrt{13}\)
The only close answer is 29.
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Re: A rancher uses 64 feet of fencing to create a rectangular horse corral  [#permalink]

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New post 02 Apr 2020, 04:15
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SajjadAhmad wrote:
A rancher uses 64 feet of fencing to create a rectangular horse corral. If the ratio of the corral’s length to width is 3:1, which of the following most closely approximates the minimum length of additional fencing needed to divide the rectangular corral into three triangular corrals, one of which is exactly twice the area of the other two?

(A) 24 feet
(B) 29 feet
(C) 36 feet
(D) 41 feet
(E) 48 feet


Solution:


Let x and 3x be the width and length of the corral.
Image
    • Perimeter = 64 feet
      o \(2(x+3x) = 64\)
      o \(8x = 64\)
      o \(x = 8\) feet.
         Length = \(3*8 = 24\) feet.
         Width = 8 feet.
Let’s divide the length of the corral in equal half and join AE and DE.
    o Length of extra fence is AE and DE
      o CE = DE [triangle ADE is congruent to triangle BCE]
         Extra fence required = \(CE + DE = 2*DE\)
      o ADE is a right-angle triangle.
      o Applying Pythagoras theorem,
         \(DE^2 = AD^2 + AE^2\)
         \(DE^2 = 8^2 + 12^2\)
         \(DE^2 = 64 + 144 = 208\)
         \(DE = 4\sqrt{13}\)
      o Extra fence required = \(8*\sqrt{13}\)
         Now, we know that,
         \(3< \sqrt{13} < 4\)
         \(8*3 < 8*\sqrt{13}<8*4\)
         \(24 < 8*\sqrt{13}<32\)
      o Required length is nearest to 29 feet.
Hence, the correct answer is Option B.
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Re: A rancher uses 64 feet of fencing to create a rectangular horse corral   [#permalink] 02 Apr 2020, 04:15

A rancher uses 64 feet of fencing to create a rectangular horse corral

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