Rhombus ABCD has perimeter 148, and one of its diagonals has length \(28\). what is the area of ABCD?
The side of the rhombus = 148 / 4 = 37
Using the Pythagorean Theorem, half the length of the other diagonal = sqrt [ 37^2 - (28/2)^2 ] =
sqrt [ 37^2 - 14^2 ] = sqrt [ 1173 ]
So the length of the other diagonal is twice this = 2sqrt [ 1173 ]
And the area = (1/2) (product of the diagonals ) = (1/2) ( 28) *2 sqrt (1173) = = 28sqrt [ 1173 ] units^2