Unknown lengths of triangle, known area and internal angles

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Homework Help Overview

The discussion revolves around determining the lengths of the sides of a right-angled triangle when the area and the ratio of the two non-hypotenuse sides are known. Participants explore the relationship between the area and the sides of the triangle, as well as the implications of the given ratio.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the area and the sides of the triangle, questioning how to express the lengths in terms of the given area and ratio. Some mention the formula for the area and attempt to manipulate it to find the side lengths.

Discussion Status

The discussion is active, with participants offering insights into the mathematical relationships involved. Some have provided algebraic manipulations that could lead to finding the lengths, while others express uncertainty about the process. There is no explicit consensus yet, but productive lines of reasoning are being explored.

Contextual Notes

Participants note the challenge of working with the ratio of the sides and the area without knowing the actual lengths. There is also mention of the time constraints affecting the discussion.

tim9000
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Homework Statement



I know the area of a right angled triangle, I also know the ratio of the two non-hypotenuse sides.
Is there anyway of finding the lengths?

Thanks!

Homework Equations



I don't think there are any like Area = 1/2abSinC

The Attempt at a Solution



I obviously know that tan of the ratio will give one angle, so I know the internal angles.

buggered if I know, it's 2:45am here.
 
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tim9000 said:

Homework Statement



I know the area of a right angled triangle, I also know the ratio of the two non-hypotenuse sides.
Is there anyway of finding the lengths?

If the sides are [itex]a[/itex] and [itex]b[/itex] then you know that the area of the triangle is [itex]\frac12 ab[/itex].
 
If you call the two legs a and b then the area is (1/2)ab= A so ab= 2A.
The ratio of the two sides is a/b= r so that a= br. Replace a in the first equation by that: ab= (br)b= rb^2= A so b^2= A/r. Solve for b, then solve for a.
 
pasmith said:
If the sides are [itex]a[/itex] and [itex]b[/itex] then you know that the area of the triangle is [itex]\frac12 ab[/itex].

I DON'T know the sides, just the ratio of the sides (the gradient of the hypotenuse)
 
HallsofIvy said:
If you call the two legs a and b then the area is (1/2)ab= A so ab= 2A.
The ratio of the two sides is a/b= r so that a= br. Replace a in the first equation by that: ab= (br)b= rb^2= A so b^2= A/r. Solve for b, then solve for a.

Wow, that's brilliant, thanks very much!
 

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