Consider the following statements : 1. The sum of any two sides of a triangle is less than twice the median drawn to the third side. 2. The perimeter of a triangle is greater than the sum of the three medians. Which of the above statements is/are correct?
2 only
Analyzing Triangle Median Properties
This solution examines two statements concerning the properties of medians in a triangle. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side.
Statement 1: Sum of Two Sides vs. Twice the Median
The first statement claims: "The sum of any two sides of a triangle is less than twice the median drawn to the third side." Let's analyze this using triangle ABC, with sides \(a\), \(b\), \(c\), and let \(m_c\) be the median drawn to side \(c\) (the side AB). Let D be the midpoint of AB.
Mathematical Derivation for Statement 1
Consider extending the median CD to a point E such that CD = DE. This means \(CE = 2 \times m_c\). The quadrilateral ACBE has diagonals AB and CE that bisect each other at D. Therefore, ACBE is a parallelogram.
In a parallelogram, opposite sides are equal in length. So, \(AE = BC = a\) and \(BE = AC = b\).
Now, consider the triangle BCE. The lengths of its sides are \(a\), \(b\), and \(2m_c\). According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
Applying the triangle inequality to triangle BCE:
- \(BC + BE > CE \implies a + b > 2m_c\)
- \(BC + CE > BE \implies a + 2m_c > b\)
- \(BE + CE > BC \implies b + 2m_c > a\)
The key inequality here is \(a + b > 2m_c\). This means the sum of the two sides (\(a\) and \(b\)) is actually greater than twice the median (\(m_c\)) drawn to the third side (\(c\)).
The statement claims \(a + b < 2m_c\). Since our derivation shows \(a + b > 2m_c\), Statement 1 is incorrect.
Statement 2: Perimeter vs. Sum of Medians
The second statement claims: "The perimeter of a triangle is greater than the sum of the three medians." The perimeter (\(P\)) is \(a + b + c\), and the sum of the medians is \(m_a + m_b + m_c\). We need to verify if \(P > m_a + m_b + m_c\).
Mathematical Derivation for Statement 2
We can use the triangle inequality applied to smaller triangles formed within the main triangle, or specifically relating sides to medians. Consider the medians \(m_a\), \(m_b\), \(m_c\) drawn to sides \(a\), \(b\), \(c\) respectively.
It is a known property derived from the triangle inequality that:
- The median \(m_a\) is less than half the sum of the other two sides (\(b\) and \(c\)). That is, \(m_a < \frac{b+c}{2}\).
- Similarly, \(m_b < \frac{a+c}{2}\).
- And, \(m_c < \frac{a+b}{2}\).
Let's add these three inequalities:
\(m_a + m_b + m_c < \frac{b+c}{2} + \frac{a+c}{2} + \frac{a+b}{2}\)
Combine the terms on the right side:
\(m_a + m_b + m_c < \frac{(b+c) + (a+c) + (a+b)}{2}\)
\(m_a + m_b + m_c < \frac{2a + 2b + 2c}{2}\)
\(m_a + m_b + m_c < a + b + c\)
This inequality shows that the sum of the lengths of the three medians (\(m_a + m_b + m_c\)) is strictly less than the perimeter of the triangle (\(a+b+c\)).
Therefore, Statement 2, "The perimeter of a triangle is greater than the sum of the three medians," is correct.
Conclusion
Based on the analysis:
- Statement 1 is incorrect because the sum of two sides is greater than, not less than, twice the median to the third side.
- Statement 2 is correct because the perimeter is always greater than the sum of the three medians.
Thus, only Statement 2 is correct.
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