Question

In right trapezoid MNPQ

Answer 1

Answer:

  about 252.78 ft

Step-by-step explanation:

Define angle QMP as α. Then ...

  MN = 60·sin(α)

  NP = 60·cos(α)

  area MPN = (1/2)(MN)(NP) = 1800sin(α)cos(α)

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  PQ = 60tan(α)

  area MPQ = (1/2)(MP)(PQ) = 1800tan(α)

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The ratio of areas is 2.5, so we have ...

  1800tan(α) = 2.5·1800sin(α)cos(α)

  1 = 2.5cos(α)² . . . . . . divide by 1800tan(α)

  cos(α) = √0.4 . . . . . . solve for cos(α)

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Then the perimeter is ...

  Perimeter = MN +NP +PQ +QM = 60sin(α) +60cos(α) +60tan(α) +60/cos(α)

  = 60(sin(α) +cos(α) +tan(α) +sec(α))

  = 60(0.774597 +0.632456 +1.224745 +1.581139)

  = 60(4.212936) = 252.776

The perimeter of the trapezoid is about 252.776 feet.

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With perhaps a little more trouble, you can find the exact value to be ...

  perimeter = (6√10)(7+√6+√15)