+817 Help this is a problem
Point $P$ is chosen at random inside triangle $ABC$. If $AB = \sqrt{2}$, $BC = 1$, and $CA = 1$, then what is the probability that $P$ is closer to $C$ than it is to $A$?
+129881 See the following :
When P falls below the line segment ED, it will be closer to C than to A because any triangle ADC will be isosceles with AD = CD
The area of ECBD is a trapezoid with bases of 1 and 1/2 and height 1/2 so its area= (1/2) (1/2) ( 1 + 1/2) =
(1/4)(3/2) = 3/8
And the total area of triangle ABC = (1/2)AC * BC = (1/2) (1) (1) = 1/2
So....the probability that P is closer to C than A is (3/8) / (1/2) = 6/8 = 3/4
