Shown below is rectangle EFGH. Its diagonals meet at Y. Let X be the foot of in altitude is dropped from E to FH. If EX = 24 and GY = \(26\), find the perimeter of rectangle .
+130071 The diagonals bisect each other.....therefore EY = GY = 26 and EG = 52 = HF
And because triangle EYX is right then XY = sqrt ( EY^2 - EX^2) = sqrt (26^2 - 24^2) = sqrt 100 = 10
But the diagonals of a rectangle are congruent so EY = FY = GY = HY
And HX = HY - XY = 26 - 10 = 16
And since triangle EHX is right, then EH = sqrt ( HX^2 + EX^2) = sqrt ( 16^2 + 24^2 ) = sqrt ( 832)
And triangle HEF is also right so EF =sqrt ( HF^2 - EH^2 ) = sqrt ( 52^2 - 832) = sqrt (1872)
So.....the perimeter of EFGH = 2 [ EH + EF ] = 2 [ sqrt 832 + sqrt 1872 ] ≈ 144.22
