Question

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

Answer 1

Diagram :

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Given :

• ∆ABC is a traingle.

To Proof :

• The peremeter of a triangle is greater than the sum of its three medians.

Proof :

Let us assume that, AD, BE and CF be the three medians of a ∆ABC.

As we know that

The sum of any two sides of a triangle is greater than twice the median drawn to the third side.

=> AB + AC > 2AD .....(i)

=> AB + BC .....(ii)

=> BC + AC > 2CF ....(iii)

Adding, the corresponding sides of (i), (ii) and (ii),

=> 2(AB + BC + AC) > 2(AD+ BE + CF)

=> (AB+ BC + AC) > (AD + BE + CF).

Hence, the perimeter of a triangle is greater than the sum of its three medians.

Hence Proved !

Answer 2

$\huge{\boxed{\rm{\red{Question}}}}$

Prove that the peremeter of a triangle is greater than the sum of its three medians.

$\huge{\boxed{\rm{\red{Answer}}}}$

${\bigstar}\large{\boxed{\sf{\pink{Given \: that}}}}$

• ABC is a triangle ∆.

${\bigstar}\large{\boxed{\sf{\pink{To \: prove}}}}$

• Thee perimeter of a triangle is greater than the sum of its three medians.

${\bigstar}\large{\boxed{\sf{\pink{As \: we \: know \: that}}}}$

• The sum of any two sides of the triangle ∆ is greater than twice the median drawn to the third side.

${\bigstar}\large{\boxed{\sf{\pink{Solution}}}}$

• $\textsf{AB + AC > 2AD}$
• $\textsf{AB + BC}$
• $\textsf{BC + AC > 2CF}$

Adding the corresponding sides 1 , 2 and 3.

• (AB + BC + AC) > 2(AD + BE + CF)
• (AB + BC + AC) > (AD + BE + CF)

Hence, the peremeter of a triangle is greater than the sum of its three medians

Hence, proved.

@Itzbeautyqueen23

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