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The longer base of an isosceles trapezoid is equal in length to a diagonal of the trapezoid. The shorter base is equal in length to the altitude of the trapezoid. Show that the ratio of the bases is 5:3.

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  • 1 decade ago
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    Let y = the small base and x = the large base.

    Draw the two altitudes from each corner of the small base.

    The large base will then be made up of a section of length y

    and two end sections each of length (x - y)/2.

    Draw a diagonal which is length x. Make this the hypotenuse

    of a triangle which has legs of length y and [y + (x - y)/2].

    Therefore, by Pythagoras,

    x^2 = y^2 + [y + (x - y)/2]^2

    = y^2 + [y + x/2 - y/2]^2

    = y^2 + {x/2 + y/2)^2

    = y^2 + [(x + y)/2]^2

    = y^2 + (x + y)^2 / 4

    Multiply through by 4 :

    4x^2 = 4y^2 + (x + y)^2

    = 4y^2 + x^2 + 2xy + y^2

    Simplify :

    3x^2 = 5y^2 + 2xy

    Divide through by xy :

    3x/y = 5y/x + 2

    3(x/y) = 5(y/x) + 2

    3(x/y) = 5 / (x/y) + 2

    Let x/y = k

    so, 3k = 5/k + 2

    Multiply through by k :

    3k^2 = 5 + 2k

    Rearrange :

    3k^2 - 2k - 5 = 0

    Factorise :

    (k + 1)(3k - 5) = 0

    Thus, k = -1 or 5/3

    That is, x/y = -1 or 5/3

    But x/y cannot be negative, so x/y = 5/3, or x:y = 5:3.

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