Answer:
13.43%
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
The probability that a point selected a random in triangle has distance from D to 1 or less is given by the area of the sector divided by the area of triangle
step 1
Find the area of sector
[tex]A_1=\frac{60}{360}\pi r^{2}[/tex]
we have
[tex]r=1\ unit[/tex]
substitute
[tex]A_1=\frac{60}{360}\pi (1)^{2}[/tex]
[tex]A_1=\frac{\pi}{6}\ unit^{2}[/tex]
step 2
Find the area of triangle
Applying the law of sines
The area of an equilateral triangle is equal
[tex]A_2=\frac{1}{2}b^2sin(60^o)[/tex]
we have
[tex]b=3\ units\\\\sin(60^o)=\frac{\sqrt{3}}{2}[/tex]
[tex]A_2=\frac{1}{2}(3)^2(\frac{\sqrt{3}}{2})[/tex]
[tex]A_2=9\frac{\sqrt{3}}{4}\ units^2[/tex]
step 3
Find the probability
[tex]P=\frac{A_1}{A_2}[/tex]
substitute
[tex]P=\frac{\pi}{6}:9\frac{\sqrt{3}}{4}[/tex]
[tex]P=\frac{4\pi}{54\sqrt{3}}[/tex]
assume
[tex]\pi =3.14[/tex]
[tex]P=\frac{4(3.14)}{54\sqrt{3}}=0.1343[/tex]
Convert to percentage
[tex]P=0.1343*100=13.43\%[/tex]
