SOLUTION: The diagonals of a rhombus differ by 4. If the perimeter is 40, find its area.

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Question 1012047: The diagonals of a rhombus differ by 4. If the perimeter is 40, find its area.
Answer by ValorousDawn(53) About Me  (Show Source):
You can put this solution on YOUR website!
Lets say the shorter diagonal has length 2x, and the longer one is of length 2x+4. The center to each vertex is therefore x and x+2. Using the Pythagorean theorem, the length of each side of the diagonal is sqrt%28x%5E2%2B%28x%2B2%29%5E2%29
Which equates to sqrt%282x%5E2%2B4x%2B4%29
There are four of these lengths, and the sum of those four is the perimeter-40. Thus, we have 4sqrt%282x%5E2%2B4x%2B4%29=40
sqrt%282x%5E2%2B4x%2B4%29=10
2x%5E2%2B4x%2B4=100
2x%5E2%2B4x-96=0
x%5E2%2B2x-48=0
%28x%2B8%29%28x-6%29=0
The negative answer doesn't work, because negative distance doesn't exist. Our answer is therefore x=6.
The diagonals are thus 12, and 16. The area of a rhombus given by it's diagonals are d_1%2Ad_2%2F2 so the area is 12%2A16%2F2

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