We are given the two bases of the trapezium:
[tex] ED = 5,\qquad AB = 9 [/tex]
The formula for the area of the trapezium is
[tex] A = \dfrac{(B+b)h}{2} [/tex]
So, we only need to figure out the length of the height EF.
We know that FA+EF = 11. Also, we're given that the perimeter of ABCF is 28, which means
[tex] 2AB+2FA = 28 \iff 2\cdot 9 + 2FA = 28 \iff 2FA = 10 \iff FA = 5 [/tex]
So, we can deduce
[tex] EF = 11-FA = 11-5=6 [/tex]
And so we're ready to use the solving formula:
[tex] A = \dfrac{(5+9)\cdot 6}{2} = \dfrac{14\cdot 6}{2} = 7\cdot 6 = 42 [/tex]
