The question is as follows:
Both the slant height and the base diameter of a cone are 12 inches. What is distace between two opposite points on base circle of cone, if it is required that the path must lie on the lateral surface of the cone?
I am not sure of where to start with this problem, therefore, any help with how to start this will be very much appreciated. Thank you in advance.
On
I came upon this math question while looking for something else. The question is a complicated way (perhaps meant to test one’s understanding of the terms) of addressing the very fundamental geometry of the Pythagorean theory: A squared plus B squared equals C squared-the sides A and B being the horizontal and vertical sides of a right triangle, and side C being the diagonal of the triangle. Never mind the cone, what we’re looking at is a simple right triangle. And since the question tells us the height and width are both 12”-actually we don’t even need to know the height in this case- the answer is double the base width, or 24”. It’s a very simple question.
The answer is best seen on a development.
Semi-vertcal angle is $\sin^{-1}\frac12 = 30^{\circ}$
On development angle subtended at cone apex is
$$ \frac12* 360^{\circ}=180^{\circ}$$
If $l=2r = 12 inches, \, $say, cone develpment is a semi-circle of r= 6 units radius.
Minimum (geodesic) distance is shown by red line $$ = r \sqrt 2 \, or \, 6\sqrt 2 $$