Question

Let $DEF$ be an equilateral triangle with side length $3.$ At random, a point $G$ is chosen inside the triangle. Compute the probability that the length $DG$ is less than or equal to $1.$

Answer 1

$|\Omega|=(\text{the area of the triangle})=\dfrac{a^2\sqrt3}{4}=\dfrac{3^2\sqrt3}{4}=\dfrac{9\sqrt3}{4}\\|A|=(\text{the area of the sector})=\dfrac{\alpha\pi r^2}{360}=\dfrac{60\pi \cdot 1^2}{360}=\dfrac{\pi}{6}\\\\\\P(A)=\dfrac{\dfrac{\pi}{6}}{\dfrac{9\sqrt3}{4}}\\\\P(A)=\dfrac{\pi}{6}\cdot\dfrac{4}{9\sqrt3}\\\\P(A)=\dfrac{2\pi}{27\sqrt3}\\\\P(A)=\dfrac{2\pi\sqrt3}{27\cdot3}\\\\P(A)=\dfrac{2\pi\sqrt3}{81}\approx13.4\%$

Answer 2

Answer:

13.44%

Step-by-step explanation:

For DG to have length of 1 or less, point G must be contained in a sector of a circle with center at point D, radius of 1, and a central angle of 60°.

The area of that sector is

$A_s = \dfrac{n}{360^\circ}\pi r^2$

$A_s = \dfrac{60^\circ}{360^\circ} \times 3.14159 \times 1^2$

$A_s = 0.5254$

The area of the triangle is

$A_t = \dfrac{1}{2}ef \sin D$

$A_t = \dfrac{1}{2}\times 3 \times 3 \sin 60^\circ$

$A_t = 3.8971$

The probability is the area of the sector divided by the area of the triangle.

$p = \dfrac{A_s}{A_t} = \dfrac{0.5254}{3.8971} = 0.1344$