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A square with sides 6 inches is shown. If P is a point such that the segment \(\overline{PA}\), \(\overline{PB}\), \(\overline{PC}\) are equal in length, and segment \(\overline{PC}\) is perpendicular to segment \(\overline{FD}\), what is the area, in square inches, of triangle APB?

 

If you could figure it out that would be great! 

THX in Advance!!

 May 17, 2020
 #1
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ContinueAP down to line AB. Call this point of intersection "Q".

 

Let  AP  =  BP  =  CP  =  x.

Then PQ  =  6 - x.

 

Triangle(APQ) is a right triangle with  AP  =  x,  AQ  =  3,  and  PQ  =  6 - x.

 

By the Pythagorean Theorem:  x2  =  32  +  (6 - x)2

                                                  x2  =  9  +  (36 - 12x + x2)

                                                   0  =  9  +  36  - 12x

                                                12x  =  45

                                                    x  =  3.75  =  AP  =  BP  =  CP

 

PQ  =  CQ - CP

PQ  =  6 - 3.75  =  2.25

 

Area(triangle(APB)  =  ½·AB·PQ  =  ½·6·2.25  =  6.75 sq in. 

 May 17, 2020
 #2
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oh that explains the steps! I was just assuming that lines AP, BP, and CP were all 3. 

Thx!  

Guest May 17, 2020

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