Is the perimeter of a right triangle equal to its area? (Part 2)

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SUMMARY

The discussion centers on proving that the perimeter of a right triangle is numerically equal to its area using specific leg lengths and hypotenuse values. Given the legs u = [2(m + n)]/n and v = 4m/(m - n), and the hypotenuse w = [2(m^2 + n^2)/(m - n)n, the equation (1/2)(uv) = u + v + w is established. Participants confirm that multiplying u and v by (1/2) and summing u, v, and w will demonstrate this equality. The exercise effectively illustrates the relationship between the perimeter and area of a right triangle.

PREREQUISITES
  • Understanding of right triangle properties
  • Familiarity with algebraic manipulation
  • Knowledge of the Pythagorean theorem
  • Basic grasp of geometric formulas for area and perimeter
NEXT STEPS
  • Explore geometric proofs involving right triangles
  • Study the derivation of the Pythagorean theorem
  • Investigate the relationship between area and perimeter in various geometric shapes
  • Learn about algebraic expressions and their simplifications
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Mathematicians, geometry students, educators, and anyone interested in the properties of right triangles and their geometric relationships.

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A right triangle is given. One leg is u units and the other leg is v units. The hypotenuse is given to be w units.

If u = [2(m + n)]/n, v = 4m/(m - n), and
w = [2(m^2 + n^2)/(m - n)n, show that

(1/2)(uv) = u + v + w

I must multiply u times v times (1/2), right? I then must add u + v + w. The right side must equal the left, right?

This exercise will show that the perimeter is numerically equal to the area.
 
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RTCNTC said:
I must multiply u times v times (1/2), right? I then must add u + v + w. The right side must equal the left, right?
Yes.
 
Evgeny.Makarov said:
Yes.

Cool.
 

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