Question

prove that any three sides of a quardilateral are together greater than the fourth side​

Answer 1

Answer:

Join any one of the diagonal, say BD.

We know that, in a triangle, sum of two sides is greater than the third side.

Now, in triangle ABD, we have:

AB + AD > BD ... (1)

In triangle BDC, we have:

BD + CD > BC ... (2)

Adding (1) and (2), we get,

AB + AD + BD + CD > BD + BC

AB + AD + CD > BC (Subtracting BD from both sides)

Thus, we have proved that sum of three sides is greater than the fourth side.

Answer 2

Answer:

Consider a quadrilateral ABCD

Consider ΔACD

By triangle inequality

AC<AD+CD⋯(1)

Consider ΔACB

By triangle inequality

AC<AB+CB⋯(2)

Consider ΔBCD

By triangle inequality

BD<BC+CD⋯(3)

Consider ΔADB

By triangle inequality

BD<AD+AB⋯(4)

Adding all the equations

2(AB+BC+CD+AD)>2(AC+BD)

AB+BC+CD+DA>AC+BD

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