Question

in a trapezium ab is parallel to dc o is the center ao is 3x-19 bo is x-3 do is 3 and Co is x-5

Answer 1

ab is || to dc then
ad = bc
ao×do = bo×co
(3x-19)×3 = (x-3)(x-5)
9x-57=x^2-8x+15
x^2-17x+72
x^2-9x-8x+72
x(x-9)-8(x-9)
(x-9)(x-8)
x= 8,9