Can you find the area

Can you find the area of polygon ABCDE given that BC is 10 and from point E we draw a perpendicular Suppose point p lies on bc then in aepb these three angles are each 90 degrees That means this angle will also be 90 degrees so it will form a rectangle Since a e is 4 b p will also be 4 and since a b is 8 e p will also be 8 We find that ep is 8 and bp is 4 and now if we connect ec then cp it becomes bc minus bp and bc is 10 minus bp which is 4 that will be 6 cp is 6 and angle epc It will be e p squared minus b c squared and e c squared e p is 8 squared minus b c is 6 squared and e c squared is 64 minus 36 and e c squared is 100 which means e c is 10 EC is 10 and now the angle ECD is ECD ED is 10 which means It is an isosceles triangle so if we drop a perpendicular from point E to CD Suppose this point is Q then it will bisect the base and since CD is 4 CQ and DQ will each be 2 divided by 2 which will be 4 divided by 2 and that is 2 we have c d is 4 and from point e if we make A perpendicular from point c to line d suppose this point is q then cq will be 2 and dq will be 2 and the square of the inter angle e c q e c s is equal to e q s squared minus c q s squared and ec is 10s² which is eqs² minus cq is 2s² and 100 which is eqs² minus 4 times e q s squared it becomes 100 minus 4 times e q s squared which is 96 so e q equals root 96 which is 4 times root 6 equals 4 times root 6 and now the area of AEPB It will be e times e p and it is 4 times 8 which will be 32 this area is 32 the area of e p c The area is half the base times the height where the base is pc and the height is ep and half of pc is 6 EP is 8 which will make the area 24 this area is 24 and the area of triangle ECD will be 1 over 2 times base CD and the height is equal to 4 and since 1 over 2 times c d is 4 and e q is 4 root 6 that will be 8 root 6 s area is 8√6 and the area of ABCDE will be 32 minus 24 minus 8√6 which equals 56 minus 8 times the square root of 6