Question 1043743
we are going to determine the equations of AD and BC, this will allow us to determine the altitudes of triangles ABE and CDE
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locate the vertex at the origin (0,0) then the figure is in quadrant 1 (x>0, y>0)
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slope of AD = 6/15 = 2/5 and equation of AD is y = 2x/5
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slope of BC = 12/-15 = -4/5 and equation of BC is y = -4x/5 + 12
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The coordinates for point E is determined by setting the above equations = to each other
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2x/5 = -4x/5 + 12
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2x = -4x + 60
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x = 10 then
y = 2(10)/5 = 4
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point E is located at (10,4)
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we can use the formula for the perpendicular distance from a point to a line
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d = |Am + Bn +C| / square root(A^2 + B^2) where (m,n) is the point (10,4) and
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Ax +By + C = 0
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consider line segment AB, its equation is x = 0
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d for E to AB = 10
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p = (1/2) * 12 * 10 = 60
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consider line segment CD, its equation is x -15 = 0
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d for E to CD = 5
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q = (1/2) * 5 * 6 = 15
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p - q = 60 - 15 = 45
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