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 Isosceles Trapezoid (Posted on 2006-01-02)
Let ABCD be an isosceles trapezoid with AB parallel to CD, angle ADC equal to angle BCD and AD = DC = CB. Let M be the midpoint of BC and N be the midpoint of CD.

If AB=MN=1 then how long is CD?

 See The Solution Submitted by Brian Smith No Rating

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 Solution | Comment 1 of 3
` `
`Let x = |BM| = |CM| = |CN|, y = |BN|,     p = angle CNM, and t = angle BNM.`
`From triangle CNM:`
`    2*x = 1/cos(p)                                    (1)`
`From triangle ABN:`
`    2*y = 1/cos(p+t)                                  (2)`
`From triangle BNM:`
`    x/sin(t) = 1/sin(p-t)                             (3)`
`    y/sin(p) = 1/sin(p-t)                             (4)`
`Combining (1) and (3) we get`
`    sin(p-t) = 2*cos(p)*sin(t)`
`             or`
`    tan(t) = tan(p)/3                                 (5)`
`Combining (2) and (4) we get `
`    sin(p-t) = 2*cos(p+t)*sin(p)     `
`             or`
`    tan(p) - tan(t) = 2*sin(p)*(1 - tan(p)*tan(t))    (6)`
`Combining (5) and (6) we get`
`    4*cos(p)^2 - cos(p) - 1 = 0`
`             or`
`              1 + sqrt(17)    cos(p) = --------------                   8`
`Therefore,`
`                        8    |CD| = 2*x = -------------- ~= 1.56155                  1 + sqrt(17)`
` `

 Posted by Bractals on 2006-01-02 12:25:56

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