Question

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

Answer 1

Step-by-step explanation:

We know that the length of the diagonals will be more than the side.

Lets suppose the quadrilateral is a rectangle, with sides 6 and 4.

The length of a diagonal is:

√(6*6)+(4*4)

√36+16

√52

7.21(rounded off to 3 sig. fig.)

So, 6+4+6+4<2(7.21+7.21)

20< 2*14.42

20<28.84, which is true.

:)