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A trapezium with vertices A (t, 0) B (t+2, 3t+2) and D (t, 2t+1) where t>0 has an...

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chisko12 | Student, Grade 11 | Honors

Posted May 7, 2011 at 4:12 PM via web

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A trapezium with vertices A (t, 0) B (t+2, 3t+2) and D (t, 2t+1) where t>0 has an area of 18 sq units. Find the value of t.

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giorgiana1976 | College Teacher | Valedictorian

Posted May 7, 2011 at 5:26 PM (Answer #1)

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We'll write the equation of parallel sides:

(xB-xA)/(x - xA) = (yB-yA)/(y - yA)

(t+2-t)/(x-t) = (3t+2)/y

2/(x-t) = (3t+2)/y

2y = (3t+2)(x-t)

y = 3tx/2 - 3t^2/2 + x - t

y = (3t/2 + 1)*x - (3t^2/2 + t)

(xD-xC)/(x - xC) = (yD-yC)/(y - yC)

(t-xC)/(x - xC) = (2t+1-yC)/(y - yC)

(2t+1-yC)(x - xC) = (t-xC)(y - yC)

2tx + x - 2txC - xC - x*yC + xC*yC = ty - t*yC - y*xC + xC*yC

ty - t*yC - y*xC = 2tx + x - 2txC - xC - x*yC

y*(t - xC) = x*(2t + 1 - yC) + t*yC - 2txC - xC

y = x*(2t + 1 - yC)/(t - xC) + (t*yC - 2txC - xC)/(t - xC)

Since AB is parallel to CD, we'll get:

(3t/2 + 1)*(t - xC)/(2t + 1 - yC) = - (3t^2/2 + t)*(t - xC)/(t*yC - 2txC - xC)

(3t/2 + 1)/(2t + 1 - yC) = - (3t^2/2 + t)/(t*yC - 2txC - xC)

-(2t + 1 - yC)(3t^2/2 + t)=(3t/2 + 1)(t*yC - 2txC - xC)

-3t^3 -2t^2 - 3t^2/2- t+ 3t^2*yC/2 + yC*t=3t^2*yC/2 - 3t^2*yC - 3t*xC/2 + t*yC - 2txC - xC

-3t^3 -2t^2 - 3t^2/2- t + yC*t = - 3t^2*yC - 3t*xC/2 + t*yC - 2txC - xC

-3t^3 -2t^2 - 3t^2/2 - t*(1 + yC) = - 3t^2*yC + t*(- 3xC/2 + yC - 2xC) - xC

Comparing, we'll get:

-3yC = -2 - 3/2

yC = 2/3 + 1/2

yC = 7/6

1 + 7/6 = - 3xC/2 + 7/6 - 2xC

xc*(3/2 + 2) = -1

xC = -4/7

Since we know the coordinates of the point C, we can determine the area of trapezoid using Heron's formula:

S = (a+b)*sqrt[(p-b)(p-a)(p-b-c)(p-b-d)]/(b-a)

a,b,c,d are the lengths of the sides AB,BC,CD,AD.

p = (a+b+c+d)/2

The length of a side is:

d = sqrt [(x2 - x1)^2 + (y2 - y1)^2]

AB = sqrt(2^2 + 9t^2 + 6t + 4)

AB = sqrt(9t^2 + 6t+8)

18 = (a+b)*sqrt[(p-b)(p-a)(p-b-c)(p-b-d)]/(b-a)

Replacing the lengths of the sides in the relation above, we'll get t.

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