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Find the area of parallelogram $ABCD.$

 

 Dec 30, 2023
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We   find AD as follows

 

BD^2 = AD^2  + AB^2 - 2(AD * AB) cos (60°)

 

6^2 = AD^2 + 5^2 - 2 (AD * 5) (1/2)

 

36 -25 = AD^2 -5AD

 

11 = AD^2 - 5AD

 

AD^2 - 5AD  =  11

 

AD^2 - 5AD  + 25/4  = 11 + 25/4

 

(AD - 5/2)^2  =  [ 69 / 4]

 

AD - 5/2  =  sqrt (69) / 2

 

AD  = [ 5 + sqrt (69) ] / 2

 

The height =  (5/2)sqrt (3) =  5sqrt (3)/2 

 

Area  =  AD * height  =      [5sqrt (3)] / 2  *   [ 5 + sqrt (69)] / 2  =   25sqrt(3) / 4  + 5sqrt [ 207] / 4  =

 

 

25sqrt (3) / 4 + 15sqrt (23) / 4 =

 

(5/4) [ 5sqrt (3)  + 3 sqrt (23) ] ≈  28.8

 

cool cool cool

 Dec 30, 2023

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