Basic geometry problem with triangle

  • Context: High School 
  • Thread starter Thread starter Alex126
  • Start date Start date
  • Tags Tags
    Geometry Triangle
Click For Summary

Discussion Overview

The discussion revolves around solving a geometry problem involving a right triangle with a hypotenuse of 5 and an area of 6. Participants explore various mathematical approaches to find the lengths of the legs of the triangle, particularly focusing on the use of the Pythagorean theorem and algebraic manipulation.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Technical explanation

Main Points Raised

  • One participant proposes letting one leg be represented as x and expresses the other leg in terms of x using the Pythagorean theorem.
  • Another participant suggests using the equations a² + b² = 25 and ab/2 = 6 to derive the lengths of the legs.
  • Some participants discuss the possibility of squaring the equation to form a quadratic equation in x², indicating that this approach may be more straightforward.
  • A later reply clarifies that the equation can be transformed into a quartic equation in x, but it is also a quadratic equation in x², which can be solved using standard methods.
  • Participants express uncertainty about the steps involved in squaring the equation and manipulating it to find the solution.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to solve the problem, with multiple approaches being discussed and some uncertainty remaining about the algebraic steps required.

Contextual Notes

There are unresolved mathematical steps regarding the manipulation of the equations, and participants express varying levels of familiarity with the techniques involved in solving quadratic equations.

Alex126
Messages
84
Reaction score
5
Given a right triangle, where the hypotenuse measures 5, and the area measures 6, calculate the measurements of the legs.

This is a problem I thought of, and I was wondering how to mathematically solve it with an equation.

I tried calling one leg x.
So the other leg, because of Pitagora's theorem, is: √(52 - x2)

The area is equal to the product of the legs divided by two, so:

6 = (x * √(52 - x2))/2
12 = x * √(52 - x2)

Problem is, I don't know how to solve the equation beyond the first basic step lol

Btw, I know it can be solved with Euclide's second theorem, but I was wondering how to solve it with Pitagora, if it's possible to solve that equation I wrote. So basically I'm asking how to solve that equation.
 
Last edited:
Mathematics news on Phys.org
Well, you know a2 + b2 = 25 and ab/2 = 6. So you can work out a2 + 2ab + b2 and a2 - 2ab + b2; hence a + b and a - b, hence a and b.
 
  • Like
Likes   Reactions: Alex126 and mfb
Alex126 said:
Btw, I know it can be solved with Euclide's second theorem, but I was wondering how to solve it with Pitagora, if it's possible to solve that equation I wrote. So basically I'm asking how to solve that equation.
You can solve the last equation by squaring it, afterwards it is a quadratic equation in x2.
The approach described by mjc123 is easier, however.
 
mjc123 said:
Well, you know a2 + b2 = 25 and ab/2 = 6. So you can work out a2 + 2ab + b2 and a2 - 2ab + b2; hence a + b and a - b, hence a and b.
That's actually nice lol Thanks for the input.
mfb said:
You can solve the last equation by squaring it, afterwards it is a quadratic equation in x2.
I think that's what I was looking forward to doing, but I don't know how that would work. So, could you elaborate on the passages required? I only know about the property where you can sum/subtract, multiply/divide something left and right in an equation, but I don't know how to proceed with "squaring".
 
If c=d, then c2=d2. If you know that both c and d are positive (which is the case here), the reverse direction is true as well.
 
Wouldn't I get 144 = x2 * (25-x2) from 12 = x * √(52 - x2) then?
It would continue as 144 = 25 x2 - x4, which would be a grade-four equation.
 
It is a quartic equation in x, but a quadratic equation in x2.

144 = 25 (x2) - (x2)2

You can solve for x2 with the usual formula for quadratic equations. Afterwards you can take the square root to find x.
 
  • Like
Likes   Reactions: Alex126
Aaaah I see. Neat, thanks a lot :D
 

Similar threads

  • · Replies 38 ·
2
Replies
38
Views
6K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K