whats the percentage probability that a point chosen randomly from the interior a rectangle is closer to the rectangle's center than to its vertices?
Answers
Answer:
80 percent the correct answer
Given : a rectangle & its center
To find : probability that a point chosen randomly from the interior a rectangle is closer to the rectangle's center than to it's vertices
Solution:
Let say one rectangle is of size
2a & 2b units
Four vertex are
(0,0) , ( 2a , 0) , (2a , 2b ), ( 0 , 2b)
Center would be ( a , b)
Break rectangle into 4 rectangles
Probability into each rectangle would be equal
one rectangle would be
(0 , 0) , ( a , 0) , ( a , b) , ( 0 , b)
so now Two vertex are from which distance has to be checked
are ( 0 , 0 ) & ( a , b)
opposite vertex of a rectangle
which has equal probability of being closer to any of the vertex
Same will happen for all 4 rectangles
Hence 50 % is the probability that a point chosen randomly from the interior a rectangle is closer to the rectangle's center than to it's vertices
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