Question

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Answer 1

Correct Question :

Prove that 3(AB + BC + AC) > 2(AD + BE + CF)

Solution:

Rule to be used: The sum of two sides of a ∆ is greater than its third side.

In ∆ABD:

AB + BD > AD ___(1)

In ∆BCE:

BC + CE > BE ___(2)

In ∆ACF:

AC + AF > CF ___(3)

(1)+(2)+(3):-

AB + BC + AC + BD + CE + AF > AD + BE + CF

Since AD, BE, CF are medians,

• BD = BC/2
• CE = AC/2
• AF = AB/2

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

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

=> 3(AB + BC + AC) > 2(AD + BE + CF)

hence proved.

Answer 2

Step-by-step explanation:

hope this helps u.............