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(As you edited your question, I will edit my answer.)

Area 

= (24^2 + 12^2 + 6^2)/2 * (sin 60 cos 60)

= sqrt(107163)/2

= 189sqrt(3) / 2 unit^2

30-60-90 triangles override this problem. If AE=24, then AB=$\frac{24}{2}\sqrt{3}=12\sqrt{3}.$ BE is equal to twelve inches, and this is opposite to the ninety(90) degrees angle in triangle BCE. And, CE=6 inches, while BC is equal to $6\sqrt{3}$ inches. Thus, opposite to the ninety-degrees in the triangle CDE, means that ED=three inches, and CD equals $3\sqrt{3}$ inches. Thus, the perimeter of quadrilateral ABCD is $12\sqrt{3}+6\sqrt{3}+3\sqrt{3}+24+3=21\sqrt{3}+27$ inches.