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

 Nov 17, 2019
 #1
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Let  P   =  (3, b)

Let  A = (0, 0)

Let C =  ( 3, 6)

 

And we can draw a circle with center (3, b)   and radius PA  that will pass through points A, B and C

 

So....since radius PA  = radius PC, we have that

 

(0 - 3)^2  + ( 0 - b)^2  =  (3 - 3)^2  + ( 6 - b)^2

 

9   +  b^2  =  b^2 - 12b + 36

 

12b  =  27

 

b =  27 / 12   =  9 / 4     = height of triangle APB

 

So  P  = (3, 9/4)

 

And the area of triangle APB =   (1/2) BC * (9/4)  = (1/2) (6) (9/4)  = 27/4 =  6.75 units^2

 

See the following image  :

 

 

 

cool cool cool

 Nov 17, 2019

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