Question 646150
What is the length of the diagonals of a rhombus with side = 5cm and area = 24cm^2?
<pre>
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locate(-1.5,0,x), locate(1.5,0,x), locate(.1,2,y), locate(.1,-1.5,y),
locate(-1.8,2.2,5),locate(1.7,2.2,5),locate(-1.8,-1.9,5),locate(1.7,-1.9,5),
line(-3,0,0,-4), line(0,-4,3,0), line(3,0,0,4), line(0,4,-3,0) )}}}

The rhombus is made of 4 right triangles. The base of each of those
right triangles is x and the height of each one is y.

The area of each of those 4 right triangles is

        A = {{{1/2}}}·b·h
        A = {{{1/2}}}·x·y

So the area of the whole rhombus is 4 times that.

Area of rhombus = 4·{{{1/2}}}·x·y = 2xy

We are given that the area of the rhombus is 24 cm², so

     2xy = 24

By the Pythagorean theorem on each right triangle:

  x² + y² = 5²
  x² + y² = 25

So we have this system:

  x² + y² = 25
      2xy = 24

Add equals to equals:

x² + 2xy + y² = 49

Factor the left side as a perfect square:

     (x + Y)² = 49

Use the principle of square roots

       x + y = ±7  but we can ignore the negative sign.

       x + y = 7
           y = 7 - x

Substitute in 2xy = 24

           2xy = 24

       2x(7-x) = 24

Divide both sides by 2

        x(7-x) = 12
         7x-x² = 12
     -x²+7x-12 = 0

Multiply both sides by -1

      x²-7x+12 = 0

Factor the left side

    (x-4)(x-3) = 0

  x-4 = 0;  x-3 = 0
    x = 4;    x = 3

Substitute each in y = 7 - x

    y = 3;    y = 4

That looks like two different solutions but actually
they are both the same.  One diagonal is 2x = 2(4) = 8 cm
The other diagonal is 2y = 2(3) = 6 cm.

Edwin</pre>