Point P is selected at random from the interior of the pentagon with vertices \(A=(0,2), B= (4,0), C = (2\pi +1, 0), D=(2\pi +1,4), and E=(0,4).\)What is the probability that angle APB is obtuse? Express your answer as a common fraction.
+118723 I did this the hard way before I realised that it is really quite easy. (This is the easy version)
Any triangle in a semicircle is a right triangle. In this question the diameter will be AB.
The third point of the triangle is P
So if P is one the circumference then angle APB will be a rtight angle
If the point P lies inside the semicircle then angle APB will be obtuse and
if the point P lies outside the semicircle then the triangle APB will be accute.
A(0,2) B(4,0)
\(AB=\sqrt{4^2+(-2)^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}\\ \mbox{So the radius of the circle is }\sqrt{5}\\ \mbox{The centre of the circle is (2,1)}\)
The area of this semicircle is
\(area=\frac{1}{2}*\pi*(\sqrt{5})^2\\ area=\frac{1}{2}*\pi*5\\ area=\frac{5\pi}{2}\;\;units ^2\\ \)
Any point P inside this semicircle will meet the criterion.
That is APB will be obtuse.
Now the area of the pentagon
\(Area\;of\; pentagon=4(2\pi+1)-\frac{1}{2}*2*4\\ Area\;of\; pentagon=8\pi+4-4\\ Area\;of\; pentagon=8\pi\\~\\ \mbox{P(APB is obtuse)}=\frac{5\pi}{2}\div8\pi=\frac{5\pi}{2*8\pi}=\frac{5}{16}\)
Here is a diagram to help you understand.
+130516 Very nice, Melody........the easiest way of doing things is always the best !!!!!
+118723 Thanks Chris, it was a neat question.
I had not considered before that all the triangles inside a semicircle with the diameter as one side would be obtuse. It makes sense but I had to go through the question the long way before I really understood. So this question helped me hone a bit of geometry knowledge which was great. :)
