Question

ABCD is a Quadrilateral. Is AB + BC + CD + DA > 2 (AC + BD) ?

Answer 1

Since, the sum of length of any two sides in a triangle should be greater than the Length of third side.


Therefore, In ΔAOB,
$\: \: \: \: \bold{AB < OA + OB}$ ..........(i)

In ΔBOC,
$\: \: \: \: \bold{BC < OB + OC}$ ..........(ii)

In ΔCOD,
$\: \: \: \: \bold{CD < OC + OD}$ ..........(iii)

In ΔAOD,
$\: \: \: \: \bold{DA < OD + OA}$ ..........(iv)


$\text{\underline{Adding equations (i), (ii), (iii) and (iv) we get:}}$


$\textbf{AB + BC + CD + DA < OA + OB + OB +OC + OC + OD + OD + OA}$


$⇒\textbf{AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD}$


$⇒\textbf{AB + BC + CD + DA < 2[(AO + OC) + (DO +OB)]}$


$\bold{\red{\fbox{AB + BC + CD + DA < 2(AC + BD)}}}$


Hence, it is proved ✔✅


[ Based on NCERT solutions ]

Answer 2

Answer:

Refer to the attachment above. ⬆️

hope it helps!!