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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.
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| 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
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| 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
