What is the perimeter of a rhombus whose diagonals add up to 41 cm and has an area of 210 square cm showing all work with final answer?

Answer 1

Double the area then find two number that have a sum of 41 and a product of 420:-

If: x+y = 41

Then: y = 41-x

If: xy = 420

Then: x(41-x) = 420

So: 41x -x^2 - 420 = 0

Solving the above quadratic equation: x = 21 or x = 20 meaning y is 20

The rhombus will then have 4 right angle triangles with sides of 10.5 and 10

Using Pythagoras' theorem: 10.5^2 + 10^2 = 210.25 and its square root is 14.5

Therefore the perimeter is: 4*14.5 = 58 cm

Check: 0.5*21*20 = 210 square cm

Answer 2

Suppose the diagonals are A and B where A < Bthen area = AB = 210

Also A + B = 41 => B = 41 - A

Substituting in the area: A*(41 - A) = 210

Or A^2 - 41A + 210 = 0

Solving the quadratic, A = 6 and B = 35.

Now, the semi-diagonals and a side of the rhombus form a right angled triangle.

So, Side^2 = (6/2)^2 + (35/2)2 = 315.25

=> side = sqrt(315.25) = 17.7553 cm (approx).

and so perimeter = 4*side = 71.0211 cm.