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.
To appease my conventionality I turned the parabola upright to y=x^2/4, rather than x=y^2/4.
The diagonals must have some absolute value of slope, say a. One has positive and the other negative, slope. Let's take the one with positive slope, y = ax+1.
Let p be the x value (remembering it's an upright parabola, not sideways as in the statement) of the right end of the lower (shorter) of the parallel sides of the trapezoid, and q be the x value of the right end of the higher (longer) parallel side.
Then q^2/4 = aq+1 and p^2/4 = 1-ap. The lower y value is then 1-ap and the higher is 1+aq.
The following program uses the quadratic formula to solve for p and q, and then y1 and y2 for a given value of a (the slope of the diagonal), and then iteratively solves so that the diagonal is 100/9:
stp = 1
bgn = 1: fin = 5
FOR a = bgn TO fin STEP stp
p = (-4 * a + SQR(16 * a * a + 16)) / 2
q = (4 * a + SQR(16 * a * a + 16)) / 2
PRINT USING "####.#######"; a; p; q;
y1 = 1 - a * p
y2 = 1 + a * q
d = SQR((y1 - y2) ^ 2 + (p + q) ^ 2)
PRINT USING "####.#######"; y1; y2; d
IF d > 100 / 9 THEN EXIT FOR
prevA = a
rsp$ = INKEY$
IF rsp$ = CHR$(27) THEN EXIT FOR
stp = stp / 4
bgn = prevA
fin = a
LOOP UNTIL rsp$ = CHR$(27)
area = (y2 - y1) * (p + q)
The solution found is:
slope of len of
diag p = x1 q = x2 y1 y2 diag
1.3333333 0.6666667 6.0000000 0.1111111 9.0000000 11.1111111
area ~= 59.25925925925926
as one base is twice p, the other base is twice q and the height is y2-y1.
The area looks like 59 + 259/999 = 59 + 7/27 = 1600/27.
This is a correction from my first post, as I had not solved the quadratic correctly.
Edited on December 2, 2007, 2:03 pm
Posted by Charlie
on 2007-12-02 13:51:13