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

Guest Aug 3, 2020

#1**+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)·x^{2}

Area( triangle(DCQ) ) = ½·2·sqrt(3)·y·y = sqrt(3)·y^{2}

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)·x^{2} + sqrt(3)·y^{2} + sqrt(3)·xy + sqrt(3)·xy

= sqrt(3)·( x^{2} + 2xy + y^{2} ) = 36·sqrt(3)

---> x^{2} + 2xy + y^{2} = 36

---> (x + y)^{2} = 36

---> x + y = 6

Each diagonal is 2x + 2y = 12.

geno3141 Aug 3, 2020