Question

Triangle abc is equilateral with side length of 3. a point x is randomly chosen within triangle abc. what is the probability that x is no more than 1 unit away from vertex a?

Answer 1

To find the probability that point x is no more than 1 unit away from vertex A in an equilateral triangle ABC with side length 3, we need to consider the area of the region within 1 unit of vertex A. The probability is 4/9.

Step-by-step explanation:

To find the probability that point x is no more than 1 unit away from vertex A, we need to consider the area of the region within 1 unit of vertex A.

The equilateral triangle ABC has side length 3, so its height can be found using the formula h = √3/2 * s, where s is the length of one side. The height of triangle ABC is h = √3/2 * 3 = √3 * 3/2 = (3√3)/2 units.

The area of an equilateral triangle is given by the formula A = (√3/4) * s^2, where s is the length of one side. The area of triangle ABC is A = (√3/4) * 3^2 = (√3/4) * 9 = (9√3)/4 square units.

Therefore, the probability that point x is no more than 1 unit away from vertex A is the ratio of the area of the region within 1 unit of vertex A to the area of triangle ABC, which is (1 unit)/(9√3/4 square units) = (4/9√3) = (4√3)/(9√3) = 4/9.

Answer 2

First we calculate for the area of the triangle:

A = s^2 * sqrt(3) / 4

A = 3^2 * sqrt(3) / 4

A = 3.9

Then we calculate the area of a sector where from 1 side to the other side where distance or r = 1.

Asector = (θ/360) π r^2

where theta is angle of corner a. In equilateral triangles, all angles are equal and is equivalent to θ = 60°, therefore:

Asector = (60/360) π (1)^2

Asector = 0.52

The probability that x is no more than 1 unit from vertex a is therefore:

P = Asector / A

P = 0.52 / 3.9

P = 0.13 = 13%