The two diagonals PR and QS of a trapezoid PQRS intersect at the point (1, 0) and PS is parallel to QR. All the four vertices of the trapezoid lie on the parabola y² = 4x and PR = QS = 100/9.

Determine the area of the trapezoid PQRS.

With brute-force and trial-and-error, with my knowledge of isosceles trapezoids, Pythagorean theorem, line equations, etc., I was was able to deduce the following:

100/9 = SQRT((X₂ - X₁)² + (Y₂ + Y₁)²)
Y₁ = SQRT(4 * X₁) 
Y₂ = SQRT(4 * X₂) 


Y₁ = SQRT(4 * 1/9) = 2/3
Y₂ = SQRT(4 * 9) = 6

(X₁, Y₁) = (1/9, 2/3)
(X₂, Y₂) = (  9 ,   6 )

100/9 = SQRT((9 - 1/9)² + (6 + 2/3)²)

           (X₂,Y₂)
          /| 
         / |     
(X₁,Y₁) /  |      
       |   |  
    b₂ |-h-| b₁
        \  |      
         \ |
          \| 

b₁ = 2*6 = 12
b₂ = 2*(2/3) = 4/3
h = (X₂ - X₁) = (9 - 1/9) = 80/9

Area_Trapezoid = h(b₁+b₂)/2
= 80/9*(12 + 4/3) = 1600/27 = 59 5/27

Posted by Dej Mar on 2007-12-03 08:47:37