Friday, November 6, 2015
Unizor - Geometry3D - Cones - Problems 2
Unizor - Creative Minds through Art of Mathematics - Math4Teens
Notes to a video lecture on http://www.unizor.com
Right Circular Cones - Problems 2
In this course we will be dealing only with right circular cones and will call them simply cones.
A doctor has recommended for a sick girl to take some liquid medicine a couple of times a day by 15-20 grams each time.
Mother was going to use a small conical glass that could hold exactly 20 gram of liquid for this. She filled it up, but a girl said that she remembered the taste of this medicine and did not like it.
Then the mother suggested a compromise: "How about you drink only half a glass?"
The girl agreed, drank some medicine and left some in a glass. What she left took exactly half the height of a glass.
How much medicine did a girl drink?
17.5 grams - as doctor ordered.
A semicircle of a radius R is rolled into a cone.
What is a volume of this cone?
V = πR³√3/24
A right circular cylinder of a radius r is inscribed into a cone of a radius R (greater than r) and an altitude H such that the lower base of a cylinder lies in the same plane as the base of a cone with the centers of these bases coinciding, the cylinder height ends at an intersection of a side of a cylinder with a side of a cone.
What is a volume of the cylinder?
V = πr²(R−r)H/R
Given a cone with a radius of a base R and height H.
Determine a central angle of a circular sector obtained if we cut a side of a cone along a generatrix and roll it out on a plane (in radians).
φ = 2πR/√R²+H²
Consider a right circular cylinder. Two congruent cones are inscribed into it. One cone has its base coinciding with the bottom base of a cylinder and has its apex at the center of an upper base of a cylinder. Another cone has its base coinciding with the upper base of a cylinder and has its apex at the center of a bottom base of a cylinder.
Find the ratio of a volume of the common part of these two cones (part of their intersection) to a volume of one of them.