Question

please solve q 6 plz.......​

Answer 1

☆☞Diagram☜☆

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$\sf{\underline{\boxed{\large{\blue{\mathsf{Solution}}}}}}$

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

• AM is the median of triangle ABC

_______________________

$\sf{\bold{\green{\underline{\underline{To\:Prove}}}}}$

• AB + BC + CA > 2AM

______________________

$\sf{\bold{\green{\underline{\underline{proof}}}}}$

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In triangle ABM

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AB + BM > AM ( sum of 2 sides of triangle is greater than the 3rd one ) ---- ( i )

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In triangle AMC

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AC + CM > AM ( Sum of 2 sides of triangle is greater than the 3rd one ) ---- ( ii )

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• Adding eq ( i ) and ( ii )

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AB + BM + CM + AC > AM + AM

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AB + ( BM + CM ) + AC > 2AM ( BM + CM = BC )

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AB + BC + AC > 2AM

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀......Hence Proved

Answer 2

$\displaystyle\large\underline{\sf\red{Given}}$

✭ AM is a median of the triangle ABC

$\displaystyle\large\underline{\sf\blue{To \ Prove}}$

◈ AB+BC+CA > 2 AM

$\displaystyle\large\underline{\sf\gray{Solution}}$

Actually in simple words they are asking us to prove that the sum of all the sides of a triangle is greater than twice the Median!!

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$\underline{\bigstar\:\textsf{According to the given Question :}}$

So here we may use the concept that,

»» Sum of any two sides of a triangle is greater than the third side

So on solving we see that,

In ∆ABM

⪼ $\displaystyle\sf AB + BM > AM \:\:\: -eq(1)$

In ∆AMC

⪼ $\displaystyle\sf CA + MC > AM \:\:\: -eq(2)$

Adding eq(1) & eq(2)

➳ $\displaystyle\sf AB+BM+CA+MC > AM+AM$

➳ $\displaystyle\sf AB+BM+CA+MC > 2AM \:\:\: -eq(3)$

From the diagram we may observe that,

➝ $\displaystyle\sf BC = BM+MC$

Substituting this in eq(3) we get that,

›› $\displaystyle\sf AB+CA+(BM+MC) > 2AM$

›› $\displaystyle\sf \pink{AB+BC+CA > 2AM}$

$\displaystyle\sf Hence \ Proved!!$

$\displaystyle\sf \star\: Diagram\:\star$

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