Class 9 - Ganita Manjari Chapter 9 - Part 14
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Class 9 Ganita Manjari Chapter 9: Propositions and their Converses
Part 14
MASTERY ASSESSMENT II Synthesizing Geometry, Algebra, and Logical Rules EQUILATERAL AREA TRAP Area = 10 Area = 10 ? Does equal area force equal sides in an equilateral triangle?
[cite: 3, 6] The Synthesis of Rules
In Mastery Assessment II, we are breaking down the walls between geometry and algebra. [cite: 6] We saw earlier that triangles with the same area are not necessarily congruent because you can stretch the base and shrink the height. [cite: 3]

But what if we combine that geometric rule with an algebraic constraint? What if we force the triangle to be equilateral? Suddenly, the flexible base and height are locked together by an algebraic equation. Will the false converse suddenly become true? It is time to chain our logical rules together!
Chained Logic Rules Combining propositions changes their truth values! Base Rule (False Converse): If \(Area(A) = Area(B)\), then \(A\) and \(B\) are congruent. (False, because a short/wide triangle can equal a tall/thin one). Chained Rule (True Converse!): If \(Area(A) = Area(B)\) AND both are Equilateral, then they are congruent. (True, because Equilateral locks the shape! Only one size fits the area).
1 Equilateral Area Logic
In Example 4, we learned that the converse "If two triangles have the same area, then they are congruent" is False. [cite: 3]

However, a student modifies the statement to say: "If two equilateral triangles have the same area, then they are congruent."

Evaluate this new proposition. Is it true or false? Provide a rigorous mathematical justification for your answer using the properties of equilateral triangles.
Concept:
1
A standard triangle's area depends on two independent variables: base and height. [cite: 3]
2
An equilateral triangle is completely locked by a single variable: its side length. The area of an equilateral triangle is exactly calculated as \( (\frac{\sqrt{3}}{4}) \times side^2 \). Therefore, a specific area can only ever be produced by one specific side length!
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Hint:
1
If two equilateral triangles have the exact same area, can one have a side of 4 and the other have a side of 6? No, because the larger side would create a larger area.
2
If their side lengths must be identical, what does SSS rule tell us about them?
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Mistakes to Avoid:
!
Applying the general rule blindly: Just because the general area converse is false doesn't mean all specific versions are false. By adding the constraint "equilateral," the student effectively removed the variables that allowed for a counterexample!
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Solution:
The new proposition is definitively True.

Justification:
In a standard triangle, area is flexible because you can stretch the base while shrinking the height to maintain the same value. [cite: 3]
However, an equilateral triangle has a fixed geometric ratio. Its area is determined entirely by a single variable (its side length, \(s\)), using the formula \(Area = (\frac{\sqrt{3}}{4})s^2\).
If two equilateral triangles have the exact same area, their side lengths \(s\) must be mathematically identical.
Because all three sides of the first triangle match all three sides of the second triangle, they are congruent by the SSS criterion. The new condition successfully eliminates all possible counterexamples!
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2 Chained Algebraic Roots
Consider the chained algebraic proposition:

"If \(x^3 = y^3\), then \(x^2 = y^2\)." [cite: 6]

Is this chained proposition true or false for all real numbers? Justify your logic step-by-step.
Concept:
1
This problem chains two concepts together. First, we must resolve the "If" condition. Does $x^3 = y^3$ force a specific relationship between $x$ and $y$? Yes, it forces $x = y$ because cubes preserve their negative/positive signs. [cite: 6]
2
If we know $x = y$ is absolutely true, can we then safely square both sides to reach the "Then" conclusion? Yes!
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Hint:
1
Let's test negative numbers. If \(x^3 = -27\) and \(y^3 = -27\), then both \(x\) and \(y\) must be -3. [cite: 6]
2
If \(x\) and \(y\) are both exactly -3, what happens when you square them? Does \((-3)^2 = (-3)^2\)? Yes!
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Mistakes to Avoid:
!
Confusing the direction of the chain: If the statement was reversed ("If $x^2 = y^2$, then $x^3 = y^3$"), it would be FALSE! (Because $2^2 = (-2)^2$, but $8 \neq -8$). However, moving from cube to square works perfectly.
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Solution:
The chained proposition is definitively True. [cite: 6]

Step-by-step Justification:
1. The Initial Condition: We start with the assumption that \(x^3 = y^3\).
2. The Root Logic: As we learned earlier, because odd powers preserve their signs, a real number has only one unique real cube root. Therefore, if \(x^3 = y^3\), it is a mathematical guarantee that \(x = y\). [cite: 6]
3. The Final Operation: Now that we have established that \(x\) and \(y\) are the exact same number, we apply the multiplication property of equality. Multiplying a number by itself (squaring) on both sides will always yield the same result.
Therefore, since \(x = y\), it mathematically follows that \(x^2 = y^2\). The logic chain is unbroken.
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Objective 1: Re-evaluating the Square Trap
If we restrict our rule to only positive integers, does the converse of "If \(x = y\), then \(x^2 = y^2\)" become true? [cite: 6]
Concept:
1
A counterexample relies on exploiting a loophole in the proposition's domain. In the case of $x^2 = y^2$, the loophole was the existence of negative roots. [cite: 6]
2
If a new rule artificially restricts the domain to remove that loophole, the proposition's truth value will change! [cite: 6]
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Hint:
1
The original counterexample was \(x = 3\) and \(y = -3\).
2
If you are no longer allowed to use \(-3\), can you find any two different positive numbers that square to the same result? (e.g., Does \(4^2\) equal \(5^2\)? No.)
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Mistakes to Avoid:
!
Clinging to old answers: You memorized that the converse for squares is False. But mathematics requires constant re-evaluation when the conditions change. The restriction to positive integers changes everything!
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Solution:
The converse statement is "If \(x^2 = y^2\), then \(x = y\)." [cite: 6]
Originally, this was false because \((-3)^2 = 9\) and \((3)^2 = 9\), but \(-3 \neq 3\).
If we restrict the domain to only positive integers, we eliminate the ability to use negative numbers as counterexamples.
In the positive integer domain, every perfect square has exactly one positive principal root. Therefore, if their squares are equal, the base numbers must be identically equal. The converse becomes True.
Correct Option: (B)
Objective 2: The Altitude Center
In Question 3, we defined the incentre as the intersection of angle bisectors. What geometric center is formed if we instead drop altitudes (perpendicular heights) from all three vertices to the opposite sides? [cite: 6]
Concept:
1
A triangle has four primary "centers" based on different sets of intersecting lines: Medians (Centroid), Angle Bisectors (Incentre), Perpendicular Bisectors (Circumcentre), and Altitudes (Orthocentre). [cite: 6]
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Hint:
1
"Ortho" comes from the Greek word for "straight" or "right angle". Which set of lines explicitly drops down at a right angle (perpendicular) from the vertex? The altitudes!
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Mistakes to Avoid:
!
Confusing altitudes with perpendicular bisectors: An altitude drops from a vertex and hits the opposite side at 90 degrees. A perpendicular bisector hits the side at 90 degrees, but it must hit exactly in the middle, and it doesn't necessarily pass through the opposite vertex!
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Solution:
The textbook defines the incentre as the intersection of the angle bisectors. [cite: 6]
By definition in geometry, the intersection of the three altitudes (the lines drawn from a vertex perpendicular to the opposite side) is called the Orthocentre.
Correct Option: (D)
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Extra Practice 1
Let's synthesize Divisibility and Prime Factor counting!

A student makes the following claim: "If a number is perfectly divisible by exactly two different prime numbers, then it must have exactly 4 factors." [cite: 6]

Evaluate this proposition. Is it true or false? If false, provide the smallest possible counterexample.
Concept:
1
The proposition says "divisible by exactly two different primes". This means the prime factorization contains ONLY two unique primes (like 2 and 3). [cite: 6]
2
However, it does NOT limit the exponents of those primes! What if one of the primes is squared?
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Hint:
1
Let the primes be 2 and 3.
2
If the number is \(2 \times 3 = 6\), it has exactly 4 factors (1, 2, 3, 6). The rule looks true!
3
But what if the number is \(2^2 \times 3 = 12\)? It is still ONLY divisible by two different primes (2 and 3). How many factors does 12 have?
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Mistakes to Avoid:
!
Assuming the base implies a power of 1: The statement in Question 11 explicitly said "a product of two unequal prime numbers" ($p \times q$). This new statement only says "divisible by...". Being divisible by a prime does not restrict the number from containing multiple copies of that prime! [cite: 6]
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Solution:
The proposition is False. [cite: 6]

Reasoning: The "If" condition states the number is divisible by exactly two unique prime numbers. However, it places no limits on the exponents of those prime factors.
Counterexample: Consider the number 12.
Its prime factorization is \(2^2 \times 3\). It is perfectly divisible by exactly two different prime numbers (2 and 3), completely satisfying the "If" condition.
However, let's list the factors of 12: 1, 2, 3, 4, 6, 12.
It has exactly 6 factors, not 4. The proposition fails!
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