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avatar+111 

In quadrilateral \(BCED\) we have \(BD = 11\)\(BC = 9\), and \(CE = 2 \). Sides \(\overline{BD}\) and \(\overline{CE}\) are extended past \(B\) and \(C\), respectively, to meet at point \(A\). If \(AC = 28\) and \(AB = 24\), then what is \(DE\)?

 

Any help would be appriciated. Thank you so much!

 May 28, 2020
 #1
avatar+21953 
+1

Notice that you have two triangles, triangle(ABC) and triangle(ADE).

 

You can use the Law of Cosines on each of these triangles.

 

First:  triangleABC):  BC2  =  AB2 + AC2 - 2·AB·AC·cos( angle(A) )

                                    92  =  242 + 282 - 2·24·28·cos( angle(A) )

   Solve this for angle(A).

 

then:  triangle(ADE):  DE2  =  AD2 + AE2 - 2·AD·AE·cos( angle(A) )

                                   DE2  =  352 + 302 - 2·35·30·cos( angle(A) )                 (use the value of angle(A) from above)

 

     Solve this for DE.

 May 28, 2020

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