$A B C D$ is a quadrilateral.
Is $A B + B C + C D + D A > A C + B D$ ?

True

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False

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Solution

The correct option is A True
Since we know that sum of two sides of a triangle is greater than the third side
so In $Δ A B D$
$A B + A D > B D - (1)$
In $Δ A B C$
$A B + B C > A C - (2)$
In $Δ B C D$
$B C + C D > B D - (3)$
In $Δ A D C$
$A D + D C > A C - (4)$
$2 (A B + B C + C D + D A) > 2 (A C + B D)$
$\Rightarrow A B + B C + C D + D A > A C + B D$

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Similar questions

$A B C D$ is quadrilateral. Is $A B + B C + C D + D A < 2 (A C + B D)$ ?

A B C D

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

A B C D

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

Hello sir or mam I didn't understand the concept of triangle and it's properties.my doubt is abcd is a quadrilateral.is ab+BC+CD+da>AC+bd?

Let ABCD be a quadrilateral with diagonals AC and BD. Prove the following statements (Compare these with the previous problem);

(a) AB + BC + CD > AD;

(b) AB + BC + CD + DA > 2AC;

(d) AB + BC + CD + DA > AC + BD.

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