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In a right trapezoid, the diagonals are perpendicular.  The parallel sides have lengths 4 and 9.  Find the area of the trapezoid.

 

 May 8, 2020
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Call the height of the trapezoid "x"   --->   AD  =  x

Since CD = 4, triangle(CDA) is a right triangle with CA  =  sqrt(42 + x2)  =  sqrt(16 + x2)

Since AB = 9, triangle(BAD) is a right triangle with BD  = sqrt(92 + x2)  =  sqrt(81 + x2)

 

The area of the trapezoid will be the area of triangle(CDA) + area triangle(CAB).

Let CA be the base of each of these triangles.

Since the diagonals are perpendicular, the sum of the heights of these triangles is BD.

 

Area of trapezoid(ABCD)  =  Area of triangle(CDA) + area triangle(CAB)  =  ½·sqrt(16 + x2)·sqrt(81 + x2).

 

Also, Area of trapezoid(ABC)  =  ½·x·(4 + 9)  =  (13/2)·x

 

Setting these two area equal to each other:  (13/2)·x  =  ½·sqrt(16 + x2)·sqrt(81 + x2).

--->      13x  =  sqrt(16 + x2)·sqrt(81 + x2).

--->   169x2  =  (16 + x2)·(81 + x2).

--->   169x2  =  1296 + 97x2 + x4

--->           0  =  x4 - 72x2 + 1296

--->           0  =  (x2 - 36)2

--->           0  =  x2 - 26

--->           x  =  6

 

Area  =  (13/2)·x  =  (13/2)·6  =  39

 May 8, 2020

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