Right Angle Triangle Geometry Lecture Notes

Jul 16, 2024

Geometry Concepts - Right Angle Triangles and Related Calculations

Introduction

  • Instructor: Ravi Prakash
  • This is part of the sixth lecture on geometry focusing on right angle triangles and related calculations such as in-radius, circumradius, and Pythagorean triplets.

Right Angle Triangle Basics

  • Given: Triangle with sides 15, 20, and 25. AD is perpendicular to the side 25.
  • Objective: Find the sum of in-radii of triangles ABD and ADS.

Steps to Solve

  1. Identify Triangle Type: Sides 15, 20, 25 form a Pythagorean triplet (3, 4, 5 scaled by 5) -> Right angle triangle.
  2. Hypotenuse: 25 (longest side, opposite the right angle).
  3. Area Calculation: Use Area = 1/2 * base * height to find the height (AD).
    • Area: 1/2 * 15 * 20 = 150
    • Using area to find height: 1/2 * 25 * height = 150 -> Height (AD) = 12
  4. Finding BD Using Pythagoras: BD^2 = 15^2 - 12^2 -> BD = 9
  5. In-radii Calculations:
    • Triangle ABD (9, 12, 15): In-radius = (semi-perimeter - hypotenuse) = 3
    • Triangle ADS (12, 16, 20): In-radius = 4
    • Sum of in-radii = 3 + 4 = 7

Additional Problems and Concepts

Problem 1: Distance Between In-Centers of Triangles within a Rectangle

  • Rectangle ABCD: Sides 9 and 12.
  • Objective: Find the distance between the in-centers of triangle ABC and triangle ADC.
  • Concept:
    • Explain why circles (incenters) do not coincide/touch in a rectangle unlike a square.
    • Use Pythagorean theorem to find the distance.
    • Key Steps: Calculate in-radii -> Draw perpendiculars -> Apply Pythagorean theorem.
    • Result: Distance = $3 \sqrt{5}$

Problem 2: Area and Perimeter of Right Angle Triangle

  • Given: Area = 80, Perimeter = 80, find the hypotenuse.
  • Using formulas:
    • In-radius: Semi-perimeter - hypotenuse
    • Solve: Perimeter (40) - hypotenuse -> Hypotenuse = 38

Problem 3: Triangle with Given Circumradius and In-radius

  • Given: Circumradius = 18, In-radius = 8
  • Objective: Find the area of the triangle.
  • Steps: Use R = 1/2 hypotenuse, R = 18 -> Hypotenuse = 36
    • Parameter and corresponding area: $8 \times 44 = 352$

Problem 4: Solving for C in a Specific Triangle Condition

  • Given: 2a + 7c = 9b, a = 12.
  • Objective: Find hypotenuse (C).
  • Methods: Identify triplet satisfying the given conditions using derived ratio: $(a - b) / (b - c) = 7 / 2$.
  • Solution: Apply common triplet (8, 15, 17) scaled appropriately -> C = 25.5.

Key Geometry Formulas and Theorems

  • Pythagorean Theorem: $a^2 + b^2 = c^2$
  • Area Calculation: Area = 1/2 * base * height
  • In-radius Formula: In-radius (R) = Area / Semi-perimeter
  • Circumradius in Right Angle Triangle: R = Hypotenuse / 2
  • Pythagorean Triplets: Common triplets like (3, 4, 5), (5, 12, 13), etc.

Conclusion

  • Detailed problem solving applying right-angle triangle properties, in-radius, circumradius, and Pythagorean identities.
  • Thorough understanding and application of basic geometric theorems help solve complex geometry problems efficiently.

Next Session

  • Further concepts will be discussed in the following lectures.