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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 $A$?

 

 Aug 13, 2023
 #1
avatar+128732 
+1

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

 

cool cool cool

 Aug 13, 2023

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