Q.451. Challenging area problem of geometry.

Question 451.
Equilateral triangle PQR is on square ABCD, as at the picture:
For given areas of right triangles (22 and 23) find unknown trapezoidal area, exact value.

Answer 451.
Let ∠ DRP = x
=> ∠ CRQ = 120° - x
Let a = length of the side of the equilateral triangle
=> (1/2) acosx * asinx = 22
=> a^2/4 sin2x = 22 ... ( 1 )
Similarly,
a^2/4 sin(240° - 2x) = 23 ... ( 2 )

Taking ratio ( 2 ) to ( 1 ),
sin(240° - 2x) / sin2x = 23/22
=> 22 * [- (√3/2) cos2x + (1/2) sin2x] = 23 * sin2x
=> - 11√3 cos2x = 12sin2x
=> tan2x = - (11√3)/12
=> sin2x = (11√3)/13√3 = 11/13
and cos2x = sin2x/tan2x = - 12/(13√3)

Plugging sin2x in ( 1 ),
a^2 = 88 / (11/13) = 104

Area of triangle = (√3/4) a^2 = 26√3

Length of side of the square
= acosx + acos(120° - x)
= acosx - (a/2)cosx + (a√3/2) sinx
= (a/2) cosx + (a√3/2) sinx

Area of the square
= [(a/2) cosx + (a√3/2) sinx]^2
= a^2 [(1/4) cos^2 x + (3/4) sin^2 x + √3 sinx cosx]
= (104) * [(1/8)(1 + cos2x) + (3/8)(1 - cos2x) + (√3/2) sin2x]
= (104) * [(1/2) - (1/4) cos2x + (√3/2) sin2x]
= (104) * [(1/2) + (1/4) (12/13√3) + (√3/2) * (11/13)]
= 52 + 30√3

=> area of the trapezium
= area of the square - area of equilateral triangle - areas of the two right triangles
= 52 + 30√3 - 22 - 23 - 26√3
= 7 + 4√3.

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