Is the area of a parallelogram equal to a/b times sine alpha?

  • Context: MHB 
  • Thread starter Thread starter leprofece
  • Start date Start date
  • Tags Tags
    Area Parallelogram
Click For Summary

Discussion Overview

The discussion revolves around the formula for the area of a parallelogram, specifically whether it can be expressed as \( \frac{a}{b} \sin \alpha \). Participants explore the relationships between the sides and angles of the parallelogram and the implications for calculating its area.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that the area of the parallelogram can be expressed as \( \frac{a}{b} \sin \alpha \), proposing a relationship between two triangles within the figure.
  • Another participant points out that there is a misunderstanding regarding the terms "triangle a" and "triangle B," clarifying that "a" and "b" refer to lengths, not triangles, and emphasizes the need for the actual lengths of the sides to calculate the area.
  • A later reply corrects the initial claim, stating that the area should be \( ab \sin \alpha \), where \( a \) and \( b \) are the sides of the parallelogram, not the perpendicular lengths.
  • One participant mentions a reference from a book that presents the area as \( \frac{a \sin \alpha}{b} \), prompting further confusion about the validity of this expression for area.
  • Another participant challenges the book's expression, asserting that an area cannot be represented as a length divided by a length, reinforcing the requirement for area to be a product of lengths.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correct formula for the area of the parallelogram, with multiple competing views on the relationship between the sides and angles. No consensus is reached on the validity of the proposed expressions.

Contextual Notes

There are unresolved assumptions regarding the definitions of the lengths and angles involved, as well as the context of the figures referenced. The discussion reflects uncertainty about the correct formulation of the area based on the provided diagram.

leprofece
Messages
239
Reaction score
0
View attachment 2089

demonstrate or show that the figure area is = a/b sen alpha

triangle a = triangle B maybe this is the main premise
 

Attachments

  • Scan2.jpg
    Scan2.jpg
    16 KB · Views: 139
Mathematics news on Phys.org
You talk about "triangle a" and "triangle B" but
1) there is no "B" in your picture
2) there are "a" and "b" but they are lengths, not triangles.

You do have two triangles with altitudes of length a and b. The areas of those triangles add to half the area of the parallelogram but you still need to know the lengths of the sides to find the areas of those two triangles.
 
HallsofIvy said:
You talk about "triangle a" and "triangle B" but
1) there is no "B" in your picture
2) there are "a" and "b" but they are lengths, not triangles.

You do have two triangles with altitudes of length a and b. The areas of those triangles add to half the area of the parallelogram but you still need to know the lengths of the sides to find the areas of those two triangles.

B = b type error sorry friends
 
Please help me it is a demostration you have the diagrama oo figure book here
B= b
angle x = angle alpha in the figure
 
leprofece said:
https://www.physicsforums.com/attachments/2089

demonstrate or show that the figure area is = a/b sen alpha

triangle a = triangle B maybe this is the main premise
There must be something wrong here. The area of the parallelogram is $ab\sin\alpha$, where $\alpha$ is as in the diagram, but $a$ and $b$ have to be the sides of the parallelogram, not the perpendicular lengths shown in the diagram. In any case, the answer surely has to be $ab\sin\alpha$, not $a/b\sin\alpha$, which could never represent an area.
 
Sorry In the book is : (a) (sinalpha)/(b)
Do you see now??
This is Nominator:a*sin (alpha)
and denominato;r b
 
Last edited:
leprofece said:
Sorry In the book is : (a) (sinalpha)/(b)
Do you see now??
This is Nominator:a*sin (alpha)
and denominato;r b
No, that cannot be right! An area is always a length times a length. A length divided by a length could never be an area. (Shake)
 

Similar threads

MHB Calculating Perimeter & Area of a Parallelogram & Triangle
  • · Replies 1 ·
Replies
1
Views
1K
MHB [ASK] Find the ratio of the area of triangle BCH and triangle EHD
  • · Replies 4 ·
Replies
4
Views
2K
MHB Prove Equal Area Triangle & Parallelogram, $\angle ADC=90^{\circ}$
  • · Replies 9 ·
Replies
9
Views
4K
MHB Area congruence in two quadrilaterals
  • · Replies 6 ·
Replies
6
Views
2K
MHB Proof of Parallelogram ABCD: Midpoint X & Y Show Area $\frac{1}{4}$
  • · Replies 5 ·
Replies
5
Views
3K
MHB What is the area of a parallelogram without knowing the height?
  • · Replies 11 ·
Replies
11
Views
4K
MHB What is the smallest side length BC of triangle ABC with fixed angle and area?
  • · Replies 2 ·
Replies
2
Views
2K
MHB Find the area of the red region
  • · Replies 2 ·
Replies
2
Views
1K
MHB Proving Nine-Point Circle Theorem w/ Parallelogram & Symmetry
  • · Replies 1 ·
Replies
1
Views
2K
MHB Area of multiple circles inside a rectangle
  • · Replies 3 ·
Replies
3
Views
2K