Question

Let triangle $DEF$ be equilateral, where the side length is $3.$ A point $G$ is chosen at random inside the triangle. Find the probability that the length $DG$ is at most $1.$

Answer 1

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

$A_1=\frac{60}{360}\pi r^{2}$

we have

$r=1\ unit$

substitute

$A_1=\frac{60}{360}\pi (1)^{2}$

$A_1=\frac{\pi}{6}\ unit^{2}$

step 2

Find the area of triangle

Applying the law of sines

The area of an equilateral triangle is equal

$A_2=\frac{1}{2}b^2sin(60^o)$

we have

$b=3\ units\\\\sin(60^o)=\frac{\sqrt{3}}{2}$

$A_2=\frac{1}{2}(3)^2(\frac{\sqrt{3}}{2})$

$A_2=9\frac{\sqrt{3}}{4}\ units^2$

step 3

Find the probability

$P=\frac{A_1}{A_2}$

substitute

$P=\frac{\pi}{6}:9\frac{\sqrt{3}}{4}$

$P=\frac{4\pi}{54\sqrt{3}}$

assume

$\pi =3.14$

$P=\frac{4(3.14)}{54\sqrt{3}}=0.1343$

Convert to percentage

$P=0.1343*100=13.43\%$