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This was quite a challenge - more difficult having just definitions and no wordplay. It got easier once a few answers could be fitted into the grid. The definition to form the decagon held me up as I was looking for ‘t??e rather than w??d, but I got there in the end. Some of the t???s were unfamiliar. Let’s hope that next week will be easier as I must get going with Christmas preparations.

Finally got a full grid, but surely the longer answer in 10d (5) should have an e and not an o for the second letter. Otherwise the longer word in 11a (8) is meaningless

Finally got a full grid, but surely the longer answer in 10d (5) should have an e and not an o for the second letter. Otherwise the longer word in 11a (8) is meaningless

Casanova, the 5 letter word in 10d and the 8 letter word in 11a both have the second letter e and that is the cell where they intersect

It looks as if Casanova may have missed the removal of a word in 10d. Easily done.

Thank you quisling and rad. Yes I omitted to leave out the extra word in 10d

Only got to this today after busy weekend. I found the grid fill tedious but the end game was fun. I have no problem with the preamble. All objects completely within the definition wherever you draw the decagon and all in distinct columns/rows.
IT is highly unlikely that all these objects would be found in the same defined area in nature but it is Christmas so will forgive that.

Not my idea of fun, but a fine construction. I have 6 of the 9 things (I had 7 until ule failed the preamble test) - being 2 across and 4 down. I'm assuming that the lower 5 highlighted letters does not do double duty as one of the 9. Would someone confirm that, and solve for x and y in the statement "there are x across things and y down ones"? (I'm also assuming that they don't overlap.)

Cockle, I have five Across and 4 Down 'things'. One of the former IS the lower highlighted thing.

Thanks, Malone ... but that would seem to imply that the lower highlighted element (as one of the 9) shares two letters with two down 9s ... and that would suggest that the two across members of the 9 in Row 7 share two letters also ... and that would imply that two of the 9s are the same ... and that would imply that there are 10 9s, because there's another duplicate in Column 2. Or am I barking up the wrong ... oops!