# Rectangle

In rectangle with sides, 6 and 3 mark the diagonal. What is the probability that a randomly selected point within the rectangle is closer to the diagonal than to any side of the rectangle?

Result

p =  42.7 %

#### Solution:

$a = 6 \ \\ b = 3 \ \\ c = \sqrt{ a^2+b^2 } = \sqrt{ 6^2+3^2 } = 3 \ \sqrt{ 5 } \doteq 6.7082 \ \\ \ \\ S_{ 1 } = \dfrac{ a \cdot \ b }{ 2 } = \dfrac{ 6 \cdot \ 3 }{ 2 } = 9 \ \\ \ \\ r = \dfrac{ 2 \cdot \ S_{ 1 } }{ a+b+c } = \dfrac{ 2 \cdot \ 9 }{ 6+3+6.7082 } \doteq 1.1459 \ \\ S_{ 2 } = \dfrac{ r \cdot \ c }{ 2 } = \dfrac{ 1.1459 \cdot \ 6.7082 }{ 2 } \doteq 3.8435 \ \\ \ \\ p = 100 \cdot \ \dfrac{ S_{ 2 } }{ S_{ 1 } } = 100 \cdot \ \dfrac{ 3.8435 }{ 9 } \doteq 42.7051 = 42.7 \%$

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