Question

11. ABCD is a quadrilateral. Prove that AB + BC + CD + DA > AC + BD​

Answer 1

Answer:

ABCD IS A QUADRILATERAL, AND AC AND BD ARE THE DIAGONALS.

SUM OF THE TWO SIDES OF TRIANGLE IS GREATER THAN THE THIRD SIDE.

Step-by-step explanation:

SO CONSIDERING THE TRIANGLE ABC, BCD, CAD, BAD. WE GET,

AB + BC >AC

CD +AD > AC

AB +AD> BD

BC+CD> BD

ADDING ALL THE ABOVE EQUATION,

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

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

HOPE THIS MAY HELP YOU.

Answer 2

Given :-

ABCD is a quadrilateral.

To Prove :-

AB + BC + CD + DA > AC + BD

Solution :-

$\sf{Firstly\:Join\:AC\:and\:BD}</p><p>$

Since , The Sum of two Sides of a Triangle is greater than the third Side .

In ∆ ABC :

AB + BC > AC --- (1)

In ∆ ABD :

AD + DC > AC -- (2)

In ∆ ABD :

AB + AD > BD -- (3)

In ∆ BCD :

BC + CD > BD -- (4)

Adding 1 , 2 , 3 and 4 :-

$\sf{AB+BC+AD+DC+AB+AD+BC+ CD>AC+AC+BD+BD}$

$\sf{2AB+2BC+2AD+CD>2AC+2BD}$

$\sf{\cancel{2}(AB+BC+AD+CD)>\cancel{2}(AC+BD)}$

Yes , AB + BC + CD + DA > AC + BD ....

Hence Proved !