Geometry Class 4 by Rakesh Yadav Sir |CGL CHSL,CPO 2023 | Geometry #rakeshsir #geometry

Summary

Rakesh Yadav teaches triangle properties through ruler-and-compass construction, worked examples, and exam-style problems. The central topic is the triangle inequality and its use in deciding whether a triangle is possible, bounding an unknown side, counting possible integer side lengths, and solving perimeter problems.

1. Constructing a triangle and the triangle inequality

To construct a triangle with sides 5 cm, 6 cm, and 7 cm, draw a 7 cm base. From its endpoints, draw arcs with radii 5 cm and 6 cm. Their intersection gives the third vertex.

A triangle is possible only if the sum of any two sides is greater than the third side:

[ a+b>c,\qquad a+c>b,\qquad b+c>a ]

For example, sides 2, 3, and 7 cannot form a triangle because the two shorter sides do not add up to more than the longest side.

An equivalent rule is that the difference between two sides must be less than the third side. For sides (a), (b), and (c):

[ |a-b|<c<a+b ]

Both inequalities are strict.

2. Finding an unknown side

For two known sides (a) and (b), the unknown third side (x) must satisfy:

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

Examples from the lesson:

• With sides 3 and 8: (5<x<11).
• With sides 5 and 7: (2<x<12).
• With sides 6 and 10: (4<x<16). If (x) must be an integer, its minimum is 5.
• With sides 12 and 5: (7<x<17). The integer possibilities are 8 through 16.

The teacher emphasizes the difference between greater than and greater than or equal to. Equality would make the points lie in a straight line rather than form a triangle.

3. Counting integer third-side values

When two integer side lengths are given, let the smaller side be (a). The number of possible integer values for the third side is:

[ 2a-1 ]

This follows because the third side must be greater than the difference between the two sides and less than their sum.

Examples:

• Given sides 2 and 3, the third side can be 2, 3, or 4: 3 triangles.
• Given sides 3 and 5, it can be 3, 4, 5, 6, or 7: 5 triangles.
• Given sides 7 and 5, the possible integer third sides run from 3 to 11: 9 triangles.

This shortcut assumes the third side is an integer. If decimal lengths are allowed, infinitely many values are possible.

4. Summing perimeters across possible triangles

When two sides are fixed, their combined length is the same in every triangle. To find the sum of the perimeters:

1. Count the possible integer third sides.
2. Multiply the sum of the two fixed sides by that count.
3. Add the sum of all possible third-side values.

Examples:

• For fixed sides 2 and 3, the possible third sides are 2, 3, and 4. The fixed sides contribute (5\times3=15), and the third sides sum to 9, giving a total perimeter sum of 24.
• For fixed sides 3 and 5, there are five triangles. The fixed sides contribute (8\times5=40), and the third sides sum to (3+4+5+6+7=25), giving a total of 65.

5. Summing consecutive integers

The teacher reviews the formula:

[ 1+2+\cdots+n=\frac{n(n+1)}2 ]

To find the sum over a range, subtract the sum up to the number immediately before the range from the sum up to its last number. This is useful for quickly adding possible third-side values when calculating total perimeters.

6. Problems involving factors

For questions relating two integer side lengths by their product:

1. List the distinct factor pairs.
2. For each pair, treat the two numbers as fixed sides and count possible integer third sides using (2\times(\text{smaller side})-1).
3. Add the counts.

For example, if the product of two side lengths is 4, the distinct pairs are ((1,4)) and ((2,2)). They allow 1 and 3 possible integer third sides, respectively, for a total of 4 triangles.

For a product of 12, the factor pairs are ((1,12)), ((2,6)), and ((3,4)). They allow 1, 3, and 5 possible integer third sides, respectively, for a total of 9 possibilities.

7. Classifying triangles using side squares

For a triangle whose largest side is (c), compare (a^2+b^2) with (c^2):

• If (a^2+b^2=c^2), the triangle is right-angled.
• If (a^2+b^2>c^2), the triangle is acute-angled.
• If (a^2+b^2<c^2), the triangle is obtuse-angled.

Examples:

• For sides 12, 17, and (k), the lesson considers (k) as the largest side, opposite the obtuse angle. Since (12^2+17^2=433), (k^2) must exceed 433, so integer (k) must be at least 21. The triangle inequality also requires (k<29). Thus, (k) can be 21 through 28: 8 values.
• For sides 9, 21, and (x), the worked answer assumes 21 is the largest side. The triangle inequality requires (12<x<30), while the obtuse-angle condition requires (9^2+x^2<21^2), or (x^2<360). The integer possibilities are 13 through 18: 6 values.

8. Class participation and practice announcement

The teacher encourages students to comment on whether they understand the explanations and where the pace may need adjustment. He also announces a daily maths test of 25 questions at 9 p.m. through a testing app, with solutions to be shared afterward.

Speakers and sources

• Rakesh Yadav Sir — the instructor and main speaker.
• Students and viewers — represented through comments and responses addressed by the instructor; no separate student speaker is clearly identifiable in the subtitles.