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?

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%

answer: a. 74

step-by-step explanation:

it shows divided evenly, so add the 37+37 = 74

add the table and i will answer correctly

step-by-step explanation: