Answer:
The probability is about 60%
Step-by-step explanation:
Given a circle is inscribed in an equilateral triangle.
A point in the figure is selected at random then we have to find its probability.
Let the radius of circle is r and side of equilateral triangle is a
OD=r
As centroid of the triangle divides the median into 2:1
∴ AO=2r
In ΔABD,
[tex]AB^2=BD^2+AD^2[/tex]
[tex]a^2=(\frac{a}{2})^2+(3r)^2[/tex]
[tex]a^2-\frac{a^2}{4}=9r^2[/tex]
[tex]\frac{3a^2}{4}=9r^2[/tex]
[tex]a^2=12r^2[/tex]
[tex]\text{Area of circle=}\pi r^2[/tex]
[tex]\text{Area of triangle=}\frac{1}{2}\times a\times 3r=\frac{1}{2}\times \sqrt{12}r\times 3r[/tex]
[tex]=3\sqrt3 r^2[/tex]
[tex]Probability=\frac{\text{area of circle}}{\text{area of triangle}}=\frac{\pi r^2}{3\sqrt3 r^2}=0.604599788078\sim 0.605[/tex]
In percentage:
[tex] 0.605\times 100=60.5[/tex]
which is about 60%
Hence, the correct option is A.
