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This isosceles trapezoid has an area of 36*sqrt(3)​, with diagonals intersecting at a 120∘ angle. How long is each diagonal?

 

 Aug 3, 2020
 #1
avatar+21953 
+2

Label the trapezoid in this manner:

   A  is the lower-left corner point

   B  is the lower-right corner point

   C  is the upper-right corner point

   D  is the upper-left corner point.

   Q is the point of intersection of the diagonals.

 

Construct PQR with Q the midpoint of CD, R with midpoint of AB. Since the trapezoid is isosceles, line segment

PR will pass through Q and PR will be perpendicular to both AB and CD.

The plan is to find the areas of each of the four triangles.

 

Triangle(ARQ) is a 30-60-90 right triangle.

Call QR = x   --->   QA = 2x   --->   AR = sqrt(3)·x   --->   AB  =  2·sqrt(3)·x.

 

Triangle(DPQ) is a 30-60-90 right triangle.

Call PQ = y   --->   QD = 2y   --->   DP = sqrt(3)·y   --->   DC = 2·sqrt(3)·y.

 

Using the formula:  Area  =  ½·base·height:

Area( triangle(ABQ) )  =  ½·2·sqrt(3)·x·x  =  sqrt(3)·x2

Area( triangle(DCQ) )  =  ½·2·sqrt(3)·y·y  =  sqrt(3)·y2

 

Using the formual:  Area  =  ½·a·b·sin(C):

Area( triangle(ADQ) )  =  ½·2x·2y·sin(60)  =  2xy·sqrt(3)/2  =  sqrt(3)·xy

Area( triangle(BCQ) )  =  ½·2x·2y·sin(60)  =  2xy·sqrt(3)/2  =  sqrt(3)·xy

 

Total area of the trapezoid  =  

=   Area( triangle(ABQ) ) + Area( triangle(ABQ) ) + Area( triangle(ADQ) ) + Area( triangle(BCQ) )

=  sqrt(3)·x2 + sqrt(3)·y2 + sqrt(3)·xy + sqrt(3)·xy  

=  sqrt(3)·( x2 + 2xy + y2 )  =  36·sqrt(3)

 

--->   x2 + 2xy + y2  =  36

--->            (x + y)2  =  36

--->                x + y  =  6

 

Each diagonal is  2x + 2y  =  12.

 Aug 3, 2020
 #2
avatar+1326 
+1

Great job, geno!!!

Dragan  Aug 4, 2020

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