BTGmoderatorDC wrote:Eight points are equally spaced on a circle. If 3 of the 8 points are to be selected at random, what is the probability that a triangle having the 3 points chosen as vertices will be a right triangle?
A) 1/14
B) 1/7
C) 3/14
D) 3/7
E) 6/7
TOUGH question!!!
Key property: An inscribed angle that contains (aka "holds") the DIAMETER will be a 90-degree angle.
For example, let's draw one of the diameters...
The inscribed angle that contains (aka "holds") the DIAMETER will be a 90-degree angle.
Likewise, this inscribed angle also contains (aka "holds") the DIAMETER, so it will also be a 90-degree angle.
So, for the ONE PARTICULAR diameter (shown below)...
...we can see that, if we make any of the 6 points the 3rd vertex of the triangle, we will get a right triangle.
This means that, FOR EACH diameter in our circle, there are
6 points that will create a right triangle.
Since there are
4 diagonals altogether....
....we know that the TOTAL number of right triangles possible = (
4)(
6) =
24
---------------------------------------
Now we need to determine how many different triangles can be created by selecting 3 of the 8 points.
Since the order in which we select the 3 points does not matter, we can use COMBINATIONS.
We can select 3 points from 8 points in 8C3 ways
8C3 = (8)(7)(6)/(3)(2)(1) =
56
-----------------------------------------
So, P(we get a right triangle) = (
total number of RIGHT triangles possible)/(
total number of triangles possible)
=
24/
56
= 3/7
Answer: D
