Question

11.Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2 (BD + AC)

12. Show that in a quadrilateral ABCD,AB + BC + CD + DA > AC + BD

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Answer 1

Answer:

Explanation:

11) 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.

12) Let ABCD be a quadrilateral with AC and BD be its diagonals

In triangle ABC

AB+BC>AC----(1)

In triangle ACD

CD+AD>AC--------(2)

Adding the equation (1)and (2) we get,

(AB+BC+CD+AD)>AC+AC

⇒(AB+BC+CD+AD)>2AC

Similarly,

It can be proved at (AB+BC+CD+AD)>2BD by taking the triangle ABD and BCD.

Now, adding two results,we get

2(AB+BC+CD+AD)>2(AC+BD)

⇒AB+BC+CD+AD>AC+BD

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