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ABCD is a square. Given PA = 1; PB=3 and PD=7​, find the area of square ABCD.

 

 Jun 23, 2020
 #1
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This is what I did:

 

1)  I looked at triangle(BAP) and used the Law of Cosines to find angle(BAP).

     I set the length of AB = x.

     BP2  =  AP2 + AB2 - 2·AP·AB·cos( BAP )

        32  =  12 + x2 - 2·1·x·cos( BAP )

          9  =  1 + x2 - 2x·cos( BAP )

         ( 8 - x2 ) / ( -2x )  =  cos( BAP )

          cos( BAP ) =  ( x2 - 8 ) / (2x )

 

2)  Then, I looked at triangle(DAP) and used the Law of Cosines to find angle(DAP).

     Since it's a square, AD = x.

     DP2  =  AP2 + AD2 - 2·AP·AD·cos( DAP )

        72  =  12 + x2 - 2·1·x·cos( DAP )

         49  =  1 + x2 - 2x·cos( DAP )

         ( 48 - x2 ) / ( -2x )  =  cos( DAP )

          cos( DAP ) =  ( x2 - 48 ) / (2x )

 

3)  Since angle(DAP) = 90o - angle(BAP)   --->   cos( DAP )  =  sin( BAP )

                      --->   sin( BAP )  =  ( x2 - 48 ) / (2x )

 

4)  Since  sin2( BAP ) + cos2( BAP )  =  1

                [ ( x2 - 48 ) / (2x ) ]2  +  [ ( x2 - 8 ) / (2x ) ]2  =  1

           [ ( x4 - 96x2 + 2304 ) / ( 4x2 ) ]  +  [ ( x4 - 16x2 + 64 ) / ( 4x2 ) ]  =  1

 

multiplying both sides by 4x2:

            ( x4 - 96x2 + 2304 )  +  ( x4 - 16x2 + 64 )  =  4x2

                 2x4 - 116x2  + 2368  =  0

 

But this doesn't have any real solutions?

 

Where did I mess up?

 Jun 24, 2020
 #2
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ABCD is a rectangle. Given PA = 1; PB=3 and PD=7​, find the area of rectangle ABCD.

 

Let angle BAP be 45º

 

Draw the altitudes from point P to point M on side AB, and to point N on side AD.

 

MP = NP = sqrt( 0.5) = 0.707106781

 

MB = sqrt ( 32 - 0.5 ) = 2.915475947

 

ND = sqrt ( 72 - 0.5 ) = 6.964194139

 

Area  = ( AN + ND )*( AM + MB )  smiley

 Jun 25, 2020

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