We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
117
2
avatar+1040 

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
 #1
avatar+7711 
+1

(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

 Jun 22, 2019
edited by MaxWong  Jun 22, 2019
 #2
avatar+4322 
+1

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.

.
 Jun 22, 2019
edited by tertre  Jun 22, 2019

6 Online Users