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
Answers
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:
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.