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In right trapezoid MNPQ

MQ

NP
,
MN

NP
, the diagonal is perpendicular to the leg and has 60ft length. A△MPQ = 2.25· A△MPN. Find perimeter of the trapezoid.

User ESI
by
7.1k points

1 Answer

4 votes

Answer:

about 252.78 ft

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)

__

PQ = 60tan(α)

area MPQ = (1/2)(MP)(PQ) = 1800tan(α)

__

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)

__

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.

_____

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

perimeter = (6√10)(7+√6+√15)

In right trapezoid MNPQ MQ ∥ NP , MN ⊥ NP , the diagonal is perpendicular to the leg-example-1
User Pscl
by
6.8k points
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