Question

Q) Prove that the perimeter of a quadrilateral is greater than twice of any one diagonal.

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Answer 1

Answer:

Lets take  a quadrilateral ABCD and the diagonal AC. 

To prove: AB + BC + CD + DA > 2 AC

we know that the sum of 2 sides of a triangle are greater than the third side. 

In triangle ABC,

AB + BC > AC    - (1)

IN triangle ADC, 

CD + DA > AC    - (2)

On adding (1) and (2), we get

AB + BC + CD + DA > AC + AC

Therefore, AB + BC + CD + DA > 2 AC

 

Answer 2

Answer:

For any quadrilateral ABCD, we can easily prove that

AB + BC + CD + DA + AC + BD.......... (1)

Now ,three cases arise. Either

AC + BD + or AC> BD or AC< BD

If AC=BD, then the result follows from (1).

if AC> BD =>AC + BD >2BDand then the result follows from (1).

If BD>AC=> AC+ BD =>AC + BD>2AC and then result follow from one (1).