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# help

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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.

Jun 22, 2019
edited by Logic  Jun 22, 2019

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Area

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

= sqrt(107163)/2

= 189sqrt(3) / 2 unit^2

Jun 22, 2019
edited by MaxWong  Jun 22, 2019
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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.

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Jun 22, 2019
edited by tertre  Jun 22, 2019