An irregular hexagon DEFMNL is drawn as follows We start with a right triangle abc draw sqaures on the legs and hypotenuse and then join in sequence the free vertices of the squares. If the hypotenuse of the triangle ABC has the length 13 and the triangle ABC has the area 30 find the area of the hexagon defmnl

Prove (a) cosh2(x) − sinh2(x) = 1 and (b) 1 − tanh 2(x) = sech 2(x). solution (a) cosh2(x) − sinh2(x) = ex + e−x 2 2 − 2 = e2x + 2 + e−2x 4 − = 4 = . (b) we start with the identity proved in part (a): cosh2(x) − sinh2(x) = 1. if we divide both sides by cosh2(x), we get 1 − sinh2(x) cosh2(x) = 1 or 1 − tanh 2(x) = .