Question

4. In the given figure, ABCD is a quadrilateral in which AC and BD are its diagonals.
Show that AB + BC + CD + DA > AC + BD

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Answer 1

Given :-

ABCD is a quadrilateral .

AC and BD are its diagonals .

Solution :-

ABCD is a quadrilateral and its diagonal are AC and BD

As we know that , Sum of two sides of triangle is greater than Third side .

Now considering that ,

ABC , BCD , ADC and BAD are triangles .

Therefore ,

AB + BC > AC ......( 1 )

BC + CD > BD .......( 2 )

AD + CD > AC .........( 3 )

AB + AD > BD .........( 4 )

Now , Adding 1 , 2 , 3 and 4 equation ,

AB + BC + BC + CD + AD + CD + AB + AD > AC + BD + AC + BD

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

Therefore ,

AB + BC + CD + AD > AC + BD

Hence , Proved

Answer 2

Answer:

ABCD is a quadrilateral and AC, and BD are the diagonals.

Sum of the two sides of a triangle is greater than the third side.

So, considering the triangle ABC, BCD, CAD and BAD, we get

AB + BC > AC

CD + AD > AC

AB + AD > BD

BC + CD > BD

Adding all the above equations,

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

⇒ 2(AB + BC + CA + AD) > 2(AC + BD)

⇒ (AB + BC + CA + AD) > (AC + BD)

HENCE, PROVED

Hope this is helpful for you.