What is the area of a rhombus of side 13 such that the sum of its two diagonals is 34?
A formula for finding the area of a rhombus, given the two diagonals d1 and d2, is: Area = ½·d1·d2
Since the diagonals of a rhombus are perpendicular, they form four right triangles.
Since the sum of its two diagonals is 34, the sum of one-half of one diagonal (one side of a right triangle)
and one-half of the other diagonal (the other side of a right triangle) is 17.
Let one side be x then the other side is 17 - x.
Using the Pythagorean Theorem: (x)2 + (17 - x)2 = 132
x2 + 289 - 34x + x2 = 169
2x2 - 34x + 120 = 0
x2 - 17x + 60 = 0
(x - 5)(x - 12) = 0
x = 5 or x = 12
So, the diagonals are 10 and 24 and the area is ½·10·24 = 120