ABCD is a square with AC = 49.5 cm. P is a point inside ABCD such that PB = PC, and the area of triangle PCB is one third of the area of ABCD. What is the length, in cm., of PA? Round your answer off to the nearest integer.
+128408 If AC = 49.5 cm......the the side of the square is 49.5 / sqrt 2 cm ≈ 35 cm
And the area of the square = 35^2 ≈ 1225 cm^2
1/3 of this is ≈ 408.33 cm^2
Since PB = PC......then triangle PBC is isoceles and its height can be found as
408.33 = (1/2) BC * height
816.66 = 35 * height
816.66 / 35 = height ≈ 23.33 cm
Call the point P , ( 35 - 23.33 , 17.5) = ( 11.67, 17.5)
Let A = (0, 35)
So PA = sqrt [ ( 11.67 - 0 )^2 + ( 35 -17.5)^2 ] ≈ 21 cm
