Math, asked by VIVEKKUMARMMEHTA, 1 year ago

prove that the perimeter of a triangle is greater than the sum of its three medians

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Answered by asif3049

Let ABC be the triangle and D. E and F are midpoints of BC, CA and AB respectively.Recall that the sum of two sides of a triangle is greater than twice the median bisecting the third side, Hence in ΔABD, AD is a median ⇒ AB + AC > 2(AD)Similarly, we get BC + AC > 2CF BC + AB > 2BE On adding the above inequations, we get (AB + AC) + (BC + AC) + (BC + AB )> 2AD + 2CD + 2BE 2(AB + BC + AC) > 2(AD + BE + CF) ∴ AB + BC + AC > AD + BE + CF Hence, we can say that the perimeter of a triangle is greater than the sum of the medians.

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asif3049: The best example is that of an equilateral triangle. If the sides are 2 units each, the medians are 2*sin 60 = 1.732050808 units each. The three sides are 3*2 = 6 units while the three medians are 3*1.732050808 = 5.196152423 units. Thus the perimeter is more than the sum of the lengths of the medians.

It clearly shows that the sum of the three medians is less than the perimeter of the EAT. This holds true for all triangles.

VIVEKKUMARMMEHTA: sorry, i want in solution

asif3049: i will im tryping wait.. i will give you the solution..

asif3049: Consider ABC is a triangle and AL, BM and CN are the altitudes.

asif3049: to prove : AL+BM+CN

asif3049: We know that peependicular AL drawn from the point A to the line BC is shorter than the line segment AB drawn from the point A to the line BC

asif3049: ALBMCN on adding equation 1,2,3 AL+BM+CN

asif3049: look again the graph I sent you then my solution.. I hope this will help you..

VIVEKKUMARMMEHTA: now it is okay

VIVEKKUMARMMEHTA: and thank you

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