Let B = (0,0, 0) Let C = (0, 5, 0) Let P = ( 3, 0 ,0) Let D = (5,5, 0) Let Q = ( 5,,0 , 1)
Equation of CP y = (-5/3)x + 5 → [ y - 5] / (-5/3) = x
Equation of AD = x = 5
Y Intersection of CP, AD [ y - 5 ] / (-5/3) = 5 → y - 5 = 5 (-5/3) → y = -25/3 + 5 = -10/3
So "S" = (5, -10/3, 0 )
So AS = 10/3
Now, the plane apex intersects AD at S.....thus, SRD forms a triangle such that triangle SRD ≈ SQA
And QA/AS ≈ RD / DS
1 / ( 10/3) = RD/ (5 + 10/3)
3/10 = RD / ( 25/3)
(3/10) (25/3) = RD = 25 / 10 = 2.5
Check ...distance from S to R ≈ 8.7 = the hypotenuse of SRA
And DR^2 + SD^2 = sqrt [ (5 + 10/3)^2 + 2.5^2 ] ≈ 8.7= SR
