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To prove: AB+ BC+CD+DA< 2 (AC+BD)

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To prove: AB+ BC+CD+DA< 2 (AC+BD)

Posted by Abhijit Kishan 2 years, 7 months ago

CBSE > Class 07 > Mathematics

1 answers

Preeti Dabral 2 years, 7 months ago

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

Therefore,

In Δ AOB, AB < OA + OB ……….(i)

In Δ BOC, BC < OB + OC ……….(ii)

In Δ COD, CD < OC + OD ……….(iii)

In Δ AOD, DA < OD + OA ……….(iv)

⇒ AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD

⇒ AB + BC + CD + DA < 2[(AO + OC) + (DO + OB)]

⇒ AB + BC + CD + DA < 2(AC + BD)

Hence, it is proved.

0Thank You

ANSWER

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Posted by Shree Vanshika Reddy Reddy 1 month, 1 week ago

CBSE > Class 07 > Mathematics