作DE⊥AC于点E∵∠DAB=∠ACB=90°∴∠DAE=∠B∵AB=AD,∠AED=∠ACB∴△ADE≌△ABC设BC=a,则AC=DE=4a,CE=BC=a,CE=3a根据勾股定理CD=5a∴y=1/2*AC*DE+1/2AC*BC=8a²+2a²=10a²∵5a=x∴a=x/5∴y=10(x/5)
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