Cone Problems

Problem 1: Find a formula for the area of the lateral surface of the cone shown on the left above.

Solution to Problem 1:

• When cut open along the slant height H and flattened, the cone becomes a sector of a circle as shown in the figure above right. Therefore the area of the lateral surface of the cone is equal to the area of the sector shown on the right. The length S of the arc of the sector is the perimeter of the base of the cone and is given by:
  
  S = 2 * Pi * r
  
  
• Also the relationship between S, the central angle t and H the radius of the sector is:
  
  S = t * H
  
  
• Equating the right sides of S = 2 * Pi * r and S = t * H, and solve for t to obtain:
  
  t = 2 * Pi * r / H
  
  
• The area A of the sector is given by:
  
  A = (1 / 2) * t * H ² = (1 / 2) * (2 * Pi * r / H) * H ²
  
  
• Simplify to obtain the area of the sector which is also the lateral surface of the cone:
  
  A = Pi * r * H
  
  
• We now return to the cone and consider triangle OCM which a right triangle; use Pythagora's theorem:
  
  H = sqrt [ h ² + r ² ]
  
  
• Substitute the above expression for H into the formula for A to obtain an expression in terms of r the radius of the base and h the height of the cone.
  
  A = Pi * r * sqrt [ h ² + r ² ]
  
  

Problem 2:A tool is made up of a cone on top of a cylinder (see figure below). The cylinder has a height h of 15 cm and a radius of 5 cm. The volume of the cone is 100 Pi cm ². O is the vertex of the cone, AB is the diameter of the base of the cone and C its center. Points O, A, B and C are in the same plane.

1 - Approximate angle AOB.

2 - Calculate the area of the lateral surface of the tool.

Solution to Problem 2:

• Note that the radius of the cylinder and the radius of the base of the cone have the same size. We first use the formula of the volume of the cone to find its height H:
  
  (1 / 3) * Pi * 5 ² * H = 100 Pi
  
  
• Solve for H :
  
  H = 12 cm
  
  
• Since OC is orthogonal to the base, triangle AOC is a right triangle. tan(angle AOC) is given by:
  
  tan(angle AOC) = r / H = 5 / 12
  
  angle AOC = arctan(5 / 12)
  
  
• Because of the symmetry of the tool, the size of angle AOB is twice the size of angle AOC, hence
  
  size of angle AOB = 2 * arctan (5 / 12) = 45.2 degrees (approximated to 1 decimal place).
  
  
• The area lateral surface of the cone and that of the cylinder are added to obtain the total area A of the surface.
  
  A = Pi * r * sqrt [ H ² + r ² ] + 2 * Pi * r * h
  
  = Pi * 5 * sqrt [ 12 ² + 5 ² ] + 2 * Pi * 5 * 15
  
  = 675.4 cm ² (approximated to 1 decimal place).

More references on geometry.