Question 1021093
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So if a rhombus has a diagonal of 14 and 2x+7y-3, how would you find the perimeter? 
With the diagonal of 2x+7y-3, one side of the diagonal is 3x and the other side is 5y-1. 
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What should I do: write my (the correct) solution, or comment yours (incorrect)?



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<U>Comment from student</U>: Please write the correct solution. Thank you.
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OK.


<pre>
The diagonals divide the rhombus in four congruent right-angled triangles.
They have the legs of half the diagonals, i.e. {{{7/2}}} units long and {{{(2x+7y-3)/2}}} units long.

According to the Pythagorean theorem, the side of the rhombus has the length

{{{sqrt((7/2)^2 + ((2x+7y-3)/2)^2)}}}.

The perimeter of the rhombus is four times this length, i.e.

{{{4*sqrt((7/2)^2 + ((2x+7y-3)/2)^2)}}}.
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Comment from student #2: I'm sorry but I don't understand still. On my worksheet it says that 
the diagonal that is 2x+7y-3, is split into two and one side of it is 3x and the other is 5y-1.
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OK.

<pre>
I see that we do not understand each other.


Then please 

1. Disregard and ignore what I responded to you.


2. Resubmit your assignment, word-by-word, to the forum and somebody will help you.

   Do not add any additional word from yourself to the assignment formulation.
   Only the assignment formulation as is.
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