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Sixth grade math, solve the fourth and fifth questions, tell me which answer I will give, thank you, master.
It's a little difficult for sixth graders!

S A-S B = 16

Then, the area of ABCD is added to S A and S B at the same time, and their difference remains unchanged.

(area of S A+trapezoidal ABCD)-(area of S B trapezoidal ABCD) = 16.

Namely: the area of ABCD-the area of -ABE = 16.

Suppose: CE is X.

(8+x)x6\2-8x6= 16

(8+x)x3-48= 16

x=38\3

CE is 38\3.

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