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 \(AB + BC + CA>2AD\).
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

Hint:

Remember: The perimeter of a triangle is always greater than the sum of its medians, but less than twice the sum of its medians. This range often helps in geometry logic questions.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Both statements are related to the triangle inequality theorem, which states that the sum of any two sides of a triangle is always greater than the third side.
2 Step 2: Detailed Explanation:
Statement I: In any triangle with sides \(a, b, c\) and medians \(m_a, m_b, m_c\), it is a proven geometric property that:
\[ \frac{3}{4}(\text{Perimeter})<\text{Sum of Medians}<\text{Perimeter} \] Since the sum of medians is always less than the perimeter, Statement I is true.
Statement II: Let \(\triangle ABC\) be a triangle with \(D\) as a point on side \(BC\).
In \(\triangle ABD\): \(AB + BD>AD\) (by triangle inequality)
In \(\triangle ACD\): \(AC + CD>AD\) (by triangle inequality)
Adding the two inequalities:
\[ AB + BD + AC + CD>AD + AD \] \[ AB + (BD + CD) + AC>2AD \] Since \(BD + CD = BC\), we get:
\[ AB + BC + AC>2AD \] This shows the perimeter is greater than twice any line segment from a vertex to the opposite side. Thus, Statement II is true.
Step 3: Final Answer:
Both statements are true.