Stuck on the sides of a triangle problem

  • Context: MHB 
  • Thread starter Thread starter tony700
  • Start date Start date
  • Tags Tags
    Stuck Triangle
Click For Summary
SUMMARY

The discussion revolves around solving for the sides 'a' and 'b' of a triangle given its height and base, specifically using the Pythagorean theorem and the Geometric Mean Theorem. The equations presented are $$464^2 + e^2 = a^2$$ and $$464^2 + (1218 - e)^2 = b^2$$, alongside the equation $$464^2 = e \cdot (1218 - e)$$. The problem involves three unknowns: 'a', 'b', and 'e', with the goal of determining their values and confirming if the angle between 'a' and 'b' is 90 degrees.

PREREQUISITES
  • Understanding of the Pythagorean theorem
  • Familiarity with the Geometric Mean Theorem
  • Basic knowledge of right triangles
  • Ability to solve systems of equations
NEXT STEPS
  • Study the application of the Pythagorean theorem in triangle problems
  • Explore the Geometric Mean Theorem in more depth
  • Learn methods for solving systems of nonlinear equations
  • Investigate properties of right triangles and their angles
USEFUL FOR

Mathematicians, geometry students, and anyone involved in solving triangle-related problems will benefit from this discussion.

tony700
Messages
5
Reaction score
0
I have figured out the triangle's height and base, but I need to figure out sides a and b. I have tried Pythagorean theorem and similar triangle ratios, but it is not working out. Please help. See picture below. Thank you.View attachment 6489
 

Attachments

  • problem2.jpg
    problem2.jpg
    11.5 KB · Views: 88
Mathematics news on Phys.org
We have that two right triangles and from the Pythagorean Theorem for those two we get:
$$464^2+e^2=a^2$$ and $$464^2+(1218-e)^2=b^2$$

From the Geometric mean theorem we have that $$464^2=e\cdot (1218-e)$$

Now we have three unknown variables, $a,b,e$, and three equations. So, we can find the values for $a,b,e$.
 
Is the angle contained by $a$ and $b$ equal to $90^\circ$? For that matter, what is the complete problem?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 59 ·
2
Replies
59
Views
78K
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K