Rectangle

In a 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:

$$\begin{gathered} a=6 \\ b=3 \\ c=\sqrt{a^2+b^2}=\sqrt{6^2+3^2}=3\sqrt{5}\doteq 6.7082 \\ {S}_1={\displaystyle \frac{a\cdot b}{2}}={\displaystyle \frac{6\cdot 3}{2}}=9 \\ r={\displaystyle \frac{2\cdot S_1}{a+b+c}}={\displaystyle \frac{2\cdot 9}{6+3+6.7082}}\doteq 1.1459 \\ {S}_2={\displaystyle \frac{r\cdot c}{2}}={\displaystyle \frac{1.1459\cdot 6.7082}{2}}\doteq 3.8435 \\ p=100\cdot {\displaystyle \frac{S_2}{S_1}}=100\cdot {\displaystyle \frac{3.8435}{9}}=42.7\mathrm{\%} \end{gathered}$$

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