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 Prove that in a quadrilateral the sum of all sides is greater than the sum of it's diagonals .

oh sorry!

the correct solution is -

consider a quadrilateral ABCD where AC and BD are the diagonals

AB+BCis greater than AC .......1 (SUM OF TWO SIDES IS GREATER THAN THE THIRD SIDE)

AD+DCis greater thanAC..........2

AB+ADis greater thanBD ..........3

DC+BCis greater thanBD..........4

ADDING 1,2,3,4

AB+BC+AD+DC+AB+AD+DC+BCis grater than AC+AC+BD+BD

2(AB+BC+CD+DA)is greater than 2(AC+BD)

AB+BC+CD+DA is greater than AC+BC

hence proved

  • 52
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eazy question

take a quadrilateral abcd.

in triangle abc:

ab+bc is greater than ac(coz the sum of two sides of a triangle is greater than the third side)-1

similarly, cd+ad is greater than bd. -2

adding 1 and 2:

ab+bc+cd+ad is greater than cd+ad

hence proved

  • 6

excuse me homicidalmaniac. CD,AD,BD are not forming a triangle. so,your solution is wrong.

the correct solution is -

consider a quadrilateral ABCD where AC and BD are the diagonals

AB+BC AC .......1 (SUM OF TWO SIDES IS GREATER THAN THE THIRD SIDE)

AD+DCAC..........2

AB+ADBD ..........3

DC+BCBD..........4

ADDING 1,2,3,4

AB+BC+AD+DC+AB+AD+DC+BC AC+AC+BD+BD

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

AB+BC+CD+DA AC+BC

hence proved

  • 10
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