Rectangle

In a rectangle with sides, 8 and 9 mark the diagonal. What is the probability that a randomly selected point within the rectangle is closer to the diagonal than any side of the rectangle?

Correct answer:

p =  41.5 %

Step-by-step explanation:

$$\begin{gathered} a=8 \\ b=9 \\ {c}^2=a^2+b^2 \\ c=\sqrt{a^2+b^2}=\sqrt{8^2+9^2}=\sqrt{145}\doteq 12.0416 \\ {S}_1=\frac{a\cdot b}{2}=\frac{8\cdot 9}{2}=36 \\ r=\frac{2\cdot S_1}{a+b+c}=\frac{2\cdot 36}{8+9+12.0416}\doteq 2.4792 \\ {S}_2=\frac{r\cdot c}{2}=\frac{2.4792\cdot 12.0416}{2}\doteq 14.9268 \\ p=100\cdot \frac{S_2}{S_1}=100\cdot \frac{14.9268}{36}=41.5\% \end{gathered}$$

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