A circle is given with an inscribed equilateral triangle and each side of the triangle having a measurement of 24 cm. What is the probability of selecting a point at random inside the circular region, but not inside the triangle, assuming that the point cannot lie outside the circular region. Leave your answer in exact form.
2007-08-23 09:30:12 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
That is all the info in the problem. There is no more details to the question. :[
2007-08-23 09:47:47 · update #1
We draw a radius r to the height of the triangle, and the resultant inside right triangle has an hypothenuse of r, one leg of 12, and one of [12sqrt(3) - r]. This leads to r = 24/sqrt(3). Then inscribed triangle's area = 12*12sqrt(3) = 144 sqrt(3).
Questioned region = pi * 576/3 = 192pi - 144sqrt(3), and prob = [192pi - 144sqrt(3)]/192pi
2007-08-23 10:07:56 · answer #1 · answered by John V 6 · 0⤊ 0⤋
The radius of the circle is (24/2)(2/√3) = 24/√3
The area of the circle is π(24/√3)^2 = 192π
The area of the triangle is (12)(24√3)/2 = 144√3
The probability of selecting a point inside the circle but outside the triangle is
(192π - 144√3)/192π = 1 - (3√3)/(4π)
2007-08-23 10:09:47 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋
how big is the circle compared to the triangle
2007-08-23 09:40:11 · answer #3 · answered by Xggnome 2 · 0⤊ 0⤋
you are able to take 5 out of 5m + 15 as a element 5m + 15 ------------ ..m + 3 ...5(m + 3) = ------------ .....m + 3 you are able to now cancel the m + 3 from the perfect and the backside 5(m + 3) ------------ ..m + 3 = 5 so the respond is a
2016-10-03 03:32:31 · answer #4 · answered by ? 4 · 0⤊ 0⤋
need more info.... how large is the cirlce?
2007-08-23 09:38:25 · answer #5 · answered by hunter_o_redheads 3 · 0⤊ 0⤋