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Let the diagonals be x and y:-

If: x+y = 24.5 then y = 24.5-x

If: 0.5xy = 73.7 then 0.5x(24.5-x) = 73.5

So: 24.5x -x^2 -147 = 0

Solving the quadratic equation: x = 14 or x = 10.5

The rhombus will then consist of 4 right angle triangles of base 5.25 and height 7

Using Pythagoras: 5.25^2+7^2 = 76.5625 and its square root is 8.75

Therefore the perimeter of the rhombus: 4*8.75 = 35 cm

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8y ago
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8y ago

Suppose the diagonals are of length 2a and 2bthen sum of diagonals = 2*a + 2*b = 24.5 = 49/2 => a + b = 49/4

Also, area = 2*a*b = 73.5 = 147/2


Suppose the side of the rhombus is c. Then

c^2 = a^2 + b^2 = (a + b)^2 - 2*a*b = (49/4)^2 - 147/2 = 1225/16

=> c = 35/4


Then, perimeter = 4*c = (35/4)*4 = 35 cm


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Q: What is the perimeter of a rhombus when its diagonals add up to 24.5 cm and has an area of 73.5 square cm showing work?
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