Trapezoid ABCD has base AB = 20 units and base CD = 30 units. Diagonals AC and BD intersect at X. If the area of trapezoid ABCD is 300 square units, what is the area of triangle BXC?
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We can find the height of the ttrapezoid thusly
300 = h ( 30 + 20) /2
300 = h * 25
!2 = h
Let A = (5,12) B = (25,12) C = (30,0) and D = (0,0)
The area of triangle DBC = (1/2) CD * 12 = (1/2) (30) * 12 = 180 units^2 (1)
The equation of line segment BD is
y =(12/25)x
The y coordinate of X is the height of triangle CXD and can be found as
y = (12/25) (15) = 3/5 * 12 = 7.2
So....the area of triangle CXD = (1/2)(CD)(7.2) = (1/2)(30)(7.2) = 108 units^2 (2)
So...the area of triangle BXC = (1) - (2) = 180 - 108 = 72 units^2
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