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The rhombus (AFCE) is inscribed in rectangle (ABCD). The length of

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The rhombus (AFCE) is inscribed in rectangle (ABCD). The length of  [#permalink]

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New post 31 Jan 2016, 15:43
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Would anyone please be so kind to help me with this question? I have no idea how to tackle it. I literally spent hours trying to solve the question.

The rhombus (AFCE) is inscribed in rectangle (ABCD). The length of BF = DE.
If the rectangle has a width of 20 yards and a length of 25 yards, what would be the total length (the perimeter) of a fence along the sides defined by AFCE?

A. 80 yards
B. 82 yards
C. 84 yards
D. 85 yards
E. 90 yards

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Re: The rhombus (AFCE) is inscribed in rectangle (ABCD). The length of  [#permalink]

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New post 02 Feb 2016, 05:23
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saiesta wrote:
Would anyone please be so kind to help me with this question? I have no idea how to tackle it. I literally spent hours trying to solve the question.

The rhombus (AFCE) is inscribed in rectangle (ABCD). The length of BF = DE.
If the rectangle has a width of 20 yards and a length of 25 yards, what would be the total length (the perimeter) of a fence along the sides defined by AFCE?

A. 80 yards
B. 82 yards
C. 84 yards
D. 85 yards
E. 90 yards


Hi,
the method explained above is the correct method..
But another method which you could use is the POE..

now the perimeter of rectangle is 20+25+20+25=90..
so perimeter of rhombus has to be less than 90.. elimiate E..

Also perimeter of the rhombus has to be greater than 20+20+20+20=80, as the side of rhombus is the hypotenuse , which has to be more than the side of the same triangle, 20
eliminate A..

lets see the remaining choices..
A rhombus has all equal sides, so each side = perimeter/4
B. 82 yards... side= 20.5
C. 84 yards..side=21
D. 85 yards..side=21.25

now this side = hypotenuse osf a right angle triangle whose one side is 20..
lets see if hyp=20.5 and one side = 20, what is the second side...
\(\sqrt{20.5^2-20^2}\)=\(\sqrt{(20.5-20)(20.5+20)}\)
...\(\sqrt{.5*40.5}\)=\(\sqrt{.5*.5*81}\)=4.5..

this exactly fits in as other side of rhombus..
25=4.5=20.5..
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The rhombus (AFCE) is inscribed in rectangle (ABCD). The length of  [#permalink]

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New post 31 Jan 2016, 18:15
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The definition of a rhombus states that every side is of equal length. This definition is the key to solving this question.

1) We state that the hypotenuse of either side triangle is x. For this solution, I will be working with the left triangle.
2) We state that the bottom and top sides of the rhombus are also x, due to the definition of a rhombus.
3) We will define the bottom of the left triangle as y. That is, for the rectangle bottom itself, we can define it as 25 = y + x, where x is a side for the rhombus and y is the bottom part of the left triangle.
4) From (1), we know that x = \(\sqrt{20^2 + y^2}\).
5) From (3), we know x = 25 - y.
6) From (4) and (5), we have the equation \(\sqrt{20^2 + y^2} = 25 - y\)
7) Solving for y, we get 4.5.
8) 25 - 4.5 = 20.5 = x
9) Since all sides of the rhombus are the same, 20.5 * 4 = 41 * 2 = 82.

The answer is B, 82
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Re: The rhombus (AFCE) is inscribed in rectangle (ABCD). The length of  [#permalink]

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New post 02 Feb 2016, 03:22
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Beixi88 wrote:
The definition of a rhombus states that every side is of equal length. This definition is the key to solving this question.

1) We state that the hypotenuse of either side triangle is x. For this solution, I will be working with the left triangle.
2) We state that the bottom and top sides of the rhombus are also x, due to the definition of a rhombus.
3) We will define the bottom of the left triangle as y. That is, for the rectangle bottom itself, we can define it as 25 = y + x, where x is a side for the rhombus and y is the bottom part of the left triangle.
4) From (1), we know that x = \(\sqrt{20^2 + y^2}\).
5) From (3), we know x = 25 - y.
6) From (4) and (5), we have the equation \(\sqrt{20^2 + y^2} = 25 - y\)
7) Solving for y, we get 4.5.
8) 25 - 4.5 = 20.5 = x
9) Since all sides of the rhombus are the same, 20.5 * 4 = 41 * 2 = 82.

The answer is B, 82

Thank you very much. I am forever grateful. :-D
+1 kudos to you!
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The rhombus (AFCE) is inscribed in rectangle (ABCD). The length of  [#permalink]

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New post 02 Feb 2016, 05:39
4
saiesta wrote:
Would anyone please be so kind to help me with this question? I have no idea how to tackle it. I literally spent hours trying to solve the question.

The rhombus (AFCE) is inscribed in rectangle (ABCD). The length of BF = DE.
If the rectangle has a width of 20 yards and a length of 25 yards, what would be the total length (the perimeter) of a fence along the sides defined by AFCE?

A. 80 yards
B. 82 yards
C. 84 yards
D. 85 yards
E. 90 yards


You can solve this question by assuming DE = x units. As rhombus has all 4 sides equal ---> AE=EC=CF=FA = 25-x (as side CD=25).

Consider right angled triangle ADE, right angled at D,

\((AD)^2+(DE)^2=(AE)^2\) ----> \(20^2+x^2=(25-x)^2\) ---> \(400+x^2=625+x^2-50x\) ---> \(50x=225\) ---> \(x = 4.5\) units.

Thus the sides of the rhombus = 20.5 units each ---> Perimeter = 20.5*4 = 82 units.

Hence B is the correct answer.

Hope this helps.
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Re: The rhombus (AFCE) is inscribed in rectangle (ABCD). The length of  [#permalink]

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New post 02 Feb 2016, 10:38
Thank you all! The explanations have been of much help.
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Re: The rhombus (AFCE) is inscribed in rectangle (ABCD). The length of  [#permalink]

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New post 23 Dec 2017, 23:57
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Let the sides of rhombus be x and the base of triangle formed i.e DE = FB = y
x+y = 25

Applying Pythagoras theorem in ADE.
x^2 = 20^2 + y^2
(x-y)(x+y) = 400
We know that x + y = 25.
x-y = 16 and hence solving it we can get x= 20.5.
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Re: The rhombus (AFCE) is inscribed in rectangle (ABCD). The length of  [#permalink]

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