Question

Given below are two statements
Statement I - The perimeter of a triangle is greater than the sum of its three medians
Statement II - In any triangle ABC,if D is any point on BC, then AB+BC+CA > 2 AD
In the light of the above statements,choose the correct answer from the options given below:

• A. Both Statement I and Statement II are true
• B. Both Statement I and Statement II are false
• C. Statement I is true but Statement II is false
• D. Statement I is false but Statement II is true

Answer 1

The Correct Option is A

To evaluate the statements given in the question, let's examine each one: 

1. Statement I: The perimeter of a triangle is greater than the sum of its three medians.
   • A triangle's perimeter is the sum of its three sides. Let the sides be \(a\), \(b\), and \(c\). Thus, the perimeter is \(a + b + c\).
   • A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. A triangle has three medians.
   • It is a known mathematical theorem that the sum of the medians of a triangle is always less than the perimeter of the triangle. Hence, \(a + b + c \gt m_a + m_b + m_c\), where \(m_a, m_b,\) and \(m_c\) are the medians.
2. Statement II: In any triangle ABC, if D is any point on BC, then \(AB + BC + CA \gt 2 \cdot AD\).
   • This statement is a consequence of the triangle inequality theorem, which states that the sum of the lengths of any two sides of a triangle is greater than the length of the remaining side.
   • Considering segments created by point D on side BC, \(AB + BD \gt AD\) and \(CA + DC \gt AD\).
   • Adding these inequalities gives \((AB + BD) + (CA + DC) \gt 2 \cdot AD\) which simplifies to \(AB + BC + CA \gt 2 \cdot AD\).

Considering both statements and the logical explanation behind them, the correct answer is: Both Statement I and Statement II are true.

Answer 2

Statement I is true because the perimeter of any triangle is indeed greater than the sum of the lengths of its medians. This is a well-established property in geometry, and it holds for all types of triangles. 
Statement II is also true. It is known as a triangle inequality for a point inside the triangle: if D is any point on side BC, then the sum of the distances from A to D, B to D, and C to D is always greater than the length of any side of the triangle. This inequality holds true for all triangles and is one of the fundamental results in geometry