__Category__: Solid Geometry, Plane Geometry"Published in Newark, California, USA"

In the figure is shown a rectangular parallelepiped whose dimensions are 2, 4, 6. Points A, B, C, E, F, and L are each at the midpoint of an edge. Find the area of each of the sections ABEF, ABC, and MNL.

Photo by Math Principles in Everyday Life |

__Solution__:

If points A, B, C, E, F, and L are the midpoint of the edges of a rectangular parallelepiped, then we can label further the figure above as follows

Photo by Math Principles in Everyday Life |

By Pythagorean Theorem, the length of AB is

By Pythagorean Theorem, the length of BC is

By Pythagorean Theorem, the length of AC is

Parallelogram ABEF is a rectangle because AB ║ EF , BE ║ AF, AB ≅ EF, and BE ≅ AF. If AB = 3.1623, then it follows that EF = 3.1623.

By Pythagorean Theorem, the length of MN is

By Pythagorean Theorem, the length of NL is

By Pythagorean Theorem, the length of ML is

We notice that ∆MNL is an isosceles triangle because MN

≅ ML = 6.3246.

Therefore, the area of rectangle ABEF is

For ∆ABC, the semi-perimeter is

Therefore, the area of ∆ABC by using Heron's Formula is

For ∆MNL, the semi-perimeter is

Therefore, the area of ∆MNL by using Heron's Formula is