# 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?

Correct result:

p =  42.7 %

#### Solution:

$a=6 \ \\ b=3 \ \\ c=\sqrt{ a^2+b^2 }=\sqrt{ 6^2+3^2 } \doteq 3 \ \sqrt{ 5 } \doteq 6.7082 \ \\ \ \\ S_{1}=\dfrac{ a \cdot \ b }{ 2 }=\dfrac{ 6 \cdot \ 3 }{ 2 }=9 \ \\ \ \\ r=\dfrac{ 2 \cdot \ S_{1} }{ a+b+c }=\dfrac{ 2 \cdot \ 9 }{ 6+3+6.7082 } \doteq 1.1459 \ \\ S_{2}=\dfrac{ r \cdot \ c }{ 2 }=\dfrac{ 1.1459 \cdot \ 6.7082 }{ 2 } \doteq 3.8435 \ \\ \ \\ p=100 \cdot \ \dfrac{ S_{2} }{ S_{1} }=100 \cdot \ \dfrac{ 3.8435 }{ 9 }=42.7 \%$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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