In the diagram, triangle ABE, triangle BCE and triangle CDE are right-angled, with angle AEB = angle BEC = angle CED = 60 degrees, and AE=24.
Find the area of quadrilateral ABCD.
Ps:This question is different it is asking for the area!!!
Area
= (24^2 + 12^2 + 6^2)/2 * (sin 60 cos 60)
= sqrt(107163)/2
= 189sqrt(3) / 2 unit^2
(1/2*BE*AB)+(1/2*CE*BC)+(1/2*ED*CD), and the answer is thus \((\frac{1}{2}*12*12\sqrt{3})+(\frac{1}{2}*6*6\sqrt{3})+(\frac{1}{2}*3*3\sqrt{3})=\boxed{\frac{189\sqrt{3}}{2}}.\)
