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Math Help - trapezium

  1. #1
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    Aug 2009
    Posts
    96

    trapezium

    ABCD is a trapezium with corners A(6,0), B(0,4), C(-2,0) and D(1,-2). The line passing through AB has the equation 2x + 3y - 12 = 0

    a. Find acute angle between AB and the pos. x-axis
    Is there a formula for this?? I can't remember it

    b. Midpoint AB (M)
    that's just [(6+0)/2 , (0+4)/2]
    which is (3,2)
    is that right??

    c. show equation of line CD is 2x + 3y + 4 = 0
    m=-2/3 x=-2 y=0
    using y-y1=m(x-x1)
    y-0=-2/3(x+2)
    3y=-2x-4
    then 2x + 3y + 4 = 0
    yes???

    d. show distance from M to CD is (16sqrt13)/13
    do you just use the perpendicular distance formula??
    so that would be
    |ax1 + by1 + c|/(sqrta^2 + b^2)
    and you get 16/sqrt13
    so you just rationalise and get (16sqrt13)/13

    e. gradient CD = -2/3
    m AB = (0-4)/(6-0)
    = -4/6
    =-2/3
    therefore parallel because m1 = m2

    f.given CD = 5 units, calculate distance AB and find area of trapezium
    dAB = sqrt[(x2-x1)^2 + (y2-y1)^2]
    = sqrt52
    and area = 1/2(a+b)h
    = 1/2 x (5+sqrt52) x [(16sqrt13)/13]
    so you get
    =27.09units^2

    f. M is centre of circle, CD is tangent, what is the equation??
    M(3,2) CD- 2x + 3y + 4 = 0
    is it just
    (x-a)^2 + (y-b)^2 = r^2
    which is (3-2)^2 + (2-3)^2 = ((16sqrt13)/13)^2
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  2. #2
    Junior Member
    Joined
    Nov 2008
    Posts
    26
    Quote Originally Posted by deej813 View Post
    ABCD is a trapezium with corners A(6,0), B(0,4), C(-2,0) and D(1,-2). The line passing through AB has the equation 2x + 3y - 12 = 0

    a. Find acute angle between AB and the pos. x-axis
    Is there a formula for this?? I can't remember it

    b. Midpoint AB (M)
    that's just [(6+0)/2 , (0+4)/2]
    which is (3,2)
    is that right??

    c. show equation of line CD is 2x + 3y + 4 = 0
    m=-2/3 x=-2 y=0
    using y-y1=m(x-x1)
    y-0=-2/3(x+2)
    3y=-2x-4
    then 2x + 3y + 4 = 0
    yes???

    d. show distance from M to CD is (16sqrt13)/13
    do you just use the perpendicular distance formula??
    so that would be
    |ax1 + by1 + c|/(sqrta^2 + b^2)
    and you get 16/sqrt13
    so you just rationalise and get (16sqrt13)/13

    e. gradient CD = -2/3
    m AB = (0-4)/(6-0)
    = -4/6
    =-2/3
    therefore parallel because m1 = m2

    f.given CD = 5 units, calculate distance AB and find area of trapezium
    dAB = sqrt[(x2-x1)^2 + (y2-y1)^2]
    = sqrt52
    and area = 1/2(a+b)h
    = 1/2 x (5+sqrt52) x [(16sqrt13)/13]
    so you get
    =27.09units^2

    f. M is centre of circle, CD is tangent, what is the equation??
    M(3,2) CD- 2x + 3y + 4 = 0
    is it just
    (x-a)^2 + (y-b)^2 = r^2
    which is (3-2)^2 + (2-3)^2 = ((16sqrt13)/13)^2




    a- 2x + 3y - 12 = 0
    to find the acute angel between the line and the positive x axis there is many ways the way i9 remmber is to find the deravative of this equation
    so by arranging this equation y=4-2/3 x
    so dy/dx = -2/3
    so the tan of the angle u want is -2/3 by using calculator
    and i think the angle u want will be 142.56
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