This isosceles trapezoid has an area of 36*sqrt(3), with diagonals intersecting at a 120∘ angle. How long is each diagonal?
+23252 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.
