Originally Posted by

**Unknown008** If you take only quadrilaterals with two common adjacent sides, you have 5x10 = 50 quadrilaterals. (5 possible quadrilaterals for each of the 10 pair of adjacent sides)

Now, taking non adjacent sides, it will be a problem similar to:

How many different pair of letters you can make from a through j, without repeat and order is not important, ie ab and ba are the same, provided two adjacent letters cannot be paired, including a and j.

You have the sides:

ad

ae

af

ag

ah

be

bf

bg

bh

bi

cf

cg

ch

ci

cj

dg

dh

di

da (but you already have ad!, so this is rejected. I won't put other repeats)

eh

ei

ej

fi

fj

gj

For a total of (24 + 50) possible quadrilaterals.