Ex.6.4 Q3 The-Triangle-and-its-Properties Solutions-NCERT Maths Class 7

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Question

\(AM\) is a median of a triangle \(ABC.\) Is \(AB + BC + CA > 2 AM?\) (Consider the sides of triangles \(\Delta{ABM}\) and \(\Delta{AMC}\).)

Text Solution

What is known?

\(ABC\) is a triangle and \(AM\) is a median of triangle \(ABC\).

What is unknown?

Is \(AB + BC + CA > 2\, AM?\)

Reasoning:

In this question it is asked if \(AB + BC + CA > 2 AM\) or not. This question is also based on the property that the sum of lengths of two sides of a triangle is always greater than the third side. In such kind of problems, you just visually identify the triangle \(ABC\) and \(AM\) is the median which further divides triangle \(ABC\) into two more triangles i.e. triangle \(ABM\) and \(AMC.\) Now consider any two sides of first triangle \(ABM\) and use the above property then consider the any two sides of another triangle \(AMC\) and use the above property. Now, add \(\rm{}L.H.S. \)and \(\rm{}R.H.S.\) of both the triangles.

Steps:

In triangle \(ABM,\)

       \(AB + BM > AM{\rm{ }} \qquad  \ldots .{\rm{ }}\left( 1 \right)\)

In triangle \(AMC,\)

       \(AC + MC > AM{\rm{ }} \qquad  \ldots .{\rm{ }}\left( 2 \right)\)

Adding equation \(\rm{}(1)\) and \(\rm{}(2)\) we get,

\[\begin{align}&AB \!+\! BM \!+\! AC \!+\! MC \!>\! AM \!+\! AM\\&AB \!+\! AC \!+\! BM \!+\! MC \!>\! 2AM\\&AB \!+\! AC \!+\! BC \!>\! 2AM\end{align}\]

Hence, it is true

Useful Tip:

Whenever you encounter problems of this kind, it is best to think of the property based on sum of lengths of any two sides of a triangle is always greater than the third side

  
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