Question

Question no.6 plz...​

Answer 1

We will prove this by using a property of a triangle i.e. Sum of two sides of a triangle is always greater than its third side.

So,
Given :- ABCD is a quadrilateral.

To prove :- AB+ BC + CD + DA < 2(AC + BD)

Proof:- In triangle AOB,

=> AO + BO > AB ---1 { sum of two sides of a triangle is always greater than its third side.}

Similarly,

In triangle AOD,

=> AO + DO > DA ---------- 2

In Triangle BOC,

=> BO + CO > BC ------- 3

In triangle COD,

=> CO + DO > CD -------- 4

Adding all four equations :-

=> AO + BO + AO + DO + BO + CO + CO + DO > AB + BC + CD + DA

=> AO + AO + BO + BO + CO + CO + DO + DO > AB + BC + CD + DA

=> 2( AO + CO + BO + DO) > AB + BC + CD + DA

=> 2 ( AC + BD) > AB + BC + CD + DA

OR,

$\bold{ AB+ BC+ CD+ DA< 2( AC+ BD)}$

$\bold{ Hence \:\:Proved}$

Answer 2

We will prove this by using a property of a triangle i.e. Sum of two sides of a triangle is always greater than its third side.

So,

Given :- ABCD is a quadrilateral.

To prove :- AB+ BC + CD + DA < 2(AC + BD)

Proof:- In triangle AOB,

=> AO + BO > AB ---1 { sum of two sides of a triangle is always greater than its third side.}

Similarly,

In triangle AOD,

=> AO + DO > DA ---------- 2

In Triangle BOC,

=> BO + CO > BC ------- 3

In triangle COD,

=> CO + DO > CD -------- 4

Adding all four equations :-

=> AO + BO + AO + DO + BO + CO + CO + DO > AB + BC + CD + DA

=> AO + AO + BO + BO + CO + CO + DO + DO > AB + BC + CD + DA

=> 2( AO + CO + BO + DO) > AB + BC + CD + DA

=> 2 ( AC + BD) > AB + BC + CD + DA

OR,

\bold{ AB+ BC+ CD+ DA< 2( AC+ BD)}AB+BC+CD+DA<2(AC+BD)

\bold{ Hence \:\:Proved}HenceProved