Question

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

[ Figure in the attachment ]

Note :- Wrong or invalid answer will be reported .
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Answer 1

✴Question:-

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

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✴Solution:-

→Given:-

• ABCD is a quadrilateral.

→To Find:-

• Is AB + BC + CD + DA < 2(AC + BD) ?

→Concept:-

• Simple Geometry Problem.

→Proof:-

⇒In ∆AOB,

Sum of two sides > Third side

= AO+BO > AB ...(1)

➢ Similarly,

⇒In triangle ∆BOC,

=BO+CO > BC ...(2)

⇒In triangle ∆COD,

=CO+DO > CD ...(3)

⇒In triangle ∆AOD

=DO+AO > DA ...(4)

➢Adding (1),(2),(3),(4)

=AO+ BO + BO + CO + CO + DO + DO + AO > AB+BC+CD+DA

=2AO + 2BO + 2CO + 2DO > AB + BC + CD + DA

=2(AO + CO) + 2(BO + DO) > AB + BC + CD + DA

=2AC + 2BD > AB + BC + CD + DA

=2(AC + BD) > AB+ BC + CD + DA

=AB + BC + CD + DA < 2(AC + BD)

Hence Proved,

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Answer 2

Answer:

Step-by-step explanation: