ABCD is a square with AB = 25. P is a point within ABCD such that PA = 24 and PB = 7. What is the value of PD^2?
+128475 Let A =(0,0) B =(25,0) C =(25,25) D =(0,25)
APB is a 7-24-25 right triangle
Its area is (1/2) ( 7) (24) = 84
Also...its area = (1/2) AB * height
84 = (1/2) 25 * height
168/25 = height = y coordinate of P = 6.72
And using the equation of a circle x^2 + y^2 = 24^2...we can find the x coordinate as
x^2 + (6.72)^2 = 24^2
x= sqrt [ 24^2 - ( 6.72)^2 ] = 23.04
So DP^2 =
( 23.04)^2 + ( 25 - 6.72)^2 = 865
Also.....note something interesting....
AP^2 + CP^2 = BP^2 + DP^2
24^2 + CP^2 = 7^2 + 865
576 + CP^2 = 914
CP^2 = 914 - 576 = 338
