Question

Question No: 7
the
percentage
probability that
point chosen
randomly from the
interior a rectangle is
closer
the
rectangle's center
than to it's vertices?
5
75
50
25
67​

Answer 1

the answer to your question for BNAT is 67%

Answer 2

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  due to symmetry

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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