Geometry Class 5 by Rakesh Yadav Sir |CGL CHSL,CPO 2023 | Geometry #rakeshsir #geometry

Main Ideas

The lesson develops a method for counting integer side lengths that produce acute, right, or obtuse triangles. The key is to combine the triangle inequality with a comparison of the squares of the sides.

1. Classify a Triangle Using Its Longest Side

Let (L) be the longest side, and let the other sides be (a) and (b).

• Acute triangle: (a^2+b^2>L^2)
• Right triangle: (a^2+b^2=L^2)
• Obtuse triangle: (a^2+b^2<L^2)

The side opposite the largest angle is the longest side. For an obtuse triangle, the square of that side is greater than the sum of the squares of the other two sides.

2. Apply the Triangle Inequality

For three side lengths (a), (b), and (x), the third side must satisfy

[ |a-b|<x<a+b. ]

When the sides must be integers, use these strict bounds to list the possible values of (x). If a question also specifies the longest side, enforce that condition as well.

For example, with sides (9), (21), and (x), if (21) is the longest side and the triangle is obtuse:

• The triangle inequality requires (12<x<30).
• The obtuse condition requires (21^2>9^2+x^2), so (x^2<360), and therefore (x\le18).
• Thus, (x) can be (13,14,15,16,17,) or (18): six possible triangles.

To count consecutive integers, subtract the lower endpoint from the upper endpoint and add one, after accounting for strict inequalities.

3. Split into Cases When the Longest Side Is Not Specified

If it is unclear whether the unknown side (x) is the longest, consider each possible longest side separately. Apply both the triangle inequality and the appropriate square comparison in each case, then combine the valid values without counting any value twice.

Examples include:

• Sides (8,15,x), obtuse:

  • If (x) is longest, (x) can be (18) through (22): five possibilities.
  • If (15) is longest, (x) can be (8) through (12): five possibilities.
  • Total: 10 possibilities.

• Sides (12,18,x), obtuse:

  • If (18) is longest, (x) can be (7) through (13): seven possibilities.
  • If (x) is longest, (x) can be (22) through (29): eight possibilities.
  • Total: 15 possibilities.

These case-based questions take longer, but careful separation of cases makes them manageable.

4. Pay Attention to “Scalene” and Equality Cases

A scalene triangle has three different side lengths. If the question asks for a scalene triangle, exclude any value of (x) that equals one of the fixed sides.

For the acute-triangle example with sides (14), (22), and (x), where (22) is the longest side:

• The square comparison requires (x^2>22^2-14^2=288), so integer (x\ge17).
• The scalene condition excludes (x=22).
• The valid values are (17,18,19,20,21), whose average is 19.

The instructor initially mentions a different average, then corrects the count after accounting for the scalene requirement.

Equality cases also need care when dividing possibilities according to which side is longest. A triangle such as (15,17,17) should be counted once, not omitted or counted twice.

5. Combine Cases for Acute Triangles When Possible

For an acute triangle, the sum of the squares of the two smaller sides must exceed the square of the longest side. When checking which side is longest, the resulting ranges may fit together into one continuous interval.

For example, with sides (15), (17), and (x), the valid integer values of (x) run from (9) through (22), giving 14 possibilities.

The general approach is to establish the triangle bounds first, then check the acute condition according to whichever side is longest.

6. Use Factor Pairs When a Product of Two Sides Is Given

If the problem states that the product of two integer sides is fixed, list the factor pairs first. For each pair:

1. Treat the third side as an integer variable.
2. Apply the triangle inequality to find possible third sides.
3. Check the square rule to identify acute, right, or obtuse triangles.

For example, when two sides have product (5), the only positive-integer factor pair is (1,5). The triangle inequality forces the third side to be (5), giving the triangle (1,5,5).

7. Check Integer Pythagorean Triples for Right Triangles

If two sides have a specified product, factor the product and test the resulting pairs against the Pythagorean condition. For a right triangle, the square of the longest side must equal the sum of the squares of the other two sides.

For a product of (12), the factor pairs are ((1,12)), ((2,6)), and ((3,4)). Only ((3,4)) can be completed to an integer right triangle, using the familiar (3)-(4)-(5) triple. A later product example illustrates checking factor pairs and finding no valid integer right triangle.

Overall Method

• Identify whether the question specifies the longest side.
• Use (|a-b|<x<a+b) to establish the possible side lengths.
• Apply the square comparison for acute, right, or obtuse classification.
• If the longest side is unknown, split into cases and combine the results carefully.
• Check restrictions such as integer side lengths or scalene triangles.
• Count integer values only after applying all conditions.

Speakers or Sources Featured

• Rakesh Yadav Sir — the instructor and sole identifiable speaker in the subtitles.
• The instructor refers to student comments and questions, but no other speaker is independently identifiable.