Area of interior triangle of pyramid normal to a side length

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Homework Help Overview

The discussion revolves around calculating the area of a triangle formed within a pyramid, specifically focusing on the triangle defined by points on the pyramid's vertex and base. The pyramid has an equilateral triangular base, and the problem involves understanding the geometric relationships and angles involved in this configuration.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the pyramid's geometry and the area of triangle AzB, questioning the implications of taper angles and the positioning of points within the pyramid. There is also a discussion about the consistency of geometric conditions presented in the problem.

Discussion Status

The discussion is ongoing, with participants providing insights and clarifications regarding the geometric setup. Some guidance has been offered, but there remains a lack of consensus on certain aspects of the problem, particularly concerning the relationships between the points and the area calculations.

Contextual Notes

Participants are navigating through potential inconsistencies in the geometric definitions and relationships, particularly regarding the positioning of points z and O, as well as the implications of the taper angle on the area calculations.

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Homework Statement
Given a pyramid who's base is an equilateral triangle with corner taper ##\psi##, what is the cross-sectional area of a plane that is normal to one of the interior corners (not normal to the altitude of the pyramid) ##z## distance from the corner taper?
Relevant Equations
A = 1/2 bh for a triangle
This isn't homework, but I figured it's fine if I make it a HW problem and post here (if not, please let me know).

Let ##z^*=0## be the vertex of the pyramid, and let ##z^*## run the altitude. It's easy to show the area of the base normal to the altitude is ##A = 4 \left.z^*\right.^2 \tan(30^\circ)\tan(\psi)^2##. However, if we now let ##z## define the length along one of the corner vertices, I'm not sure how to calculate the area of the plane within the pyramid normal to ##z##.

If we lay the pyramid on a side so that ##z## is going in the flat direction, I believe the taper angle would be ##2\psi## from ##z=0##, making me think the height of the interior triangle would be ##z\tan(2\psi)##, but I'm having issues thinking about that interior angle that used to be ##30^\circ##. Any help?

If this is confusing, let me pose the problem differently: given a pyramid with an equilateral triangular base ##ABC## and vertex ##V##, define ##O## to be the point on base ABC such that VO is the altitude (height) of the pyramid. ##\angle CVO= \psi##, the taper angle of the pyramid. Now let the point ##z## lie on ##\overleftrightarrow{VC}## such that ##\angle AzV = \angle BzV = 90^\circ##. What is the area of triangle ##AzB##?
 
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joshmccraney said:
What is the area of triangle AzB?
Say it S
\frac{S\cdot VC}{3}=V
where V is volume of the pyramid.
 
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joshmccraney said:
Now let the point z lie on VC↔ such that ∠AzV=∠BzV=90∘. What is the area of triangle AzB?
Then z and O are the same point and the area of triangle AzB is 1/3 the area of triangle ABC.
these two conditions are inconsistent with the geometry of the problem you first stated
 
I think I've been consistent with the notation, but in case not, here's a picture.
IMG_3761.jpg

Cool, so ##ABC \cdot VO = S\cdot Vz + S\cdot zC = S\cdot (Vz+zC) = S \cdot VC ##, which is three times the total volume. Thanks so much @anuttarasammyak !
 
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