Rhombus ABCD$ has perimeter 148, and one of its diagonals has length \(20\). How long is the other diagonal?
First, you seperate the rhombus into 4 congruent right triangles by drawing the diagonals. In one of these triangles, you know the hypotenuse is 148/4=37 units, and one of the legs is 10 units. Applying the Pythagorean theroem, 10^2+x^2=37^2, where x is the missing leg. Now you can subtract 10^2 on both sides to get x^2=1269. Take the square root of both sides to get x=sqrt1269=3sqrt141, so the answer of the diagonal is 2sqrt1296 as this x is only half of the diagonal, so the answer is 6sqrt141.