In a quadrilateral ABCD AB + BC + CD + DA > 2BD

A B + B C + C D + D A < 2 A C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A B + B C + C D + D A < 2 B D

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A B + B C + C D + D A > 2 A C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A B + B C + C D + D A > 2 B D

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B $A B + B C + C D + D A > 2 B D$
In $△ A B D$ , we have sum of two sides > third
side
$=> A B + A D > B D$ -- (1)
Also, in $△ B C D$ , we have sum of two sides > third side
$=> B C + C D > B D$ ---- (2)
Adding equations $1, 2$ we get
$=> A B + B C + C D + A D > 2 B D$

Suggest Corrections

Similar questions

A B C D

A B + B C + C D + D A > 2 B D

(A B + B C + C D + D A) < 2 (B D + A C)

A B C D

A B + B C + C D + D A > 2 B D

A B C D

A B + B C + C D + D A > 2 A C

Join BYJU'S Learning Program