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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Our percentage calculator will help you quickly calculate various typical tasks with percentages. Pythagorean theorem is the base for the right triangle calculator. See also our trigonometric triangle calculator. Would you like to compute count of combinations?

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