2007-01-08 07:01:20 · 14 answers · asked by armywife327 2 in Pregnancy & Parenting ➔ Baby Names
Skylar is an American name. It's a form of the name Schuyler and it means "scholar"
2007-01-08 07:04:47 · answer #1 · answered by CelebrateMeHome 6 · 0⤊ 1⤋
Skylar is a surname meaning 'Scholar'.
There are different meanings for names all over the place because so many websites and cheap name books publish false information. You can't just give a name any meaning you want because legit names come from words and are translated from different cultures. I recommend the website www.behindthename.com to anyone looking for reliable naming histories and meanings.
Good luck to you!
2007-01-08 16:31:05 · answer #2 · answered by Verity 3 · 0⤊ 0⤋
Name: Skylar
Gender: (female)
Origin: American English
Meaning: Eternal life, strength, love and beauty.
2007-01-08 15:13:41 · answer #3 · answered by mistymae33 1 · 0⤊ 0⤋
Skylar is a variation of the name Schuyler, meaning "scholar" and is of Dutch origin.
2007-01-08 16:26:10 · answer #4 · answered by LindaLou 7 · 0⤊ 0⤋
Here you go...Skylar (Dutch) protective; sheltering.
You can check out a baby book at your local library and check many or even online. I hope this helps, have a great week!
2007-01-08 15:08:33 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Skylar...American English. Meaning...Eternal life, strength, love and beauty.
2007-01-08 15:10:03 · answer #6 · answered by criley27 4 · 0⤊ 0⤋
lol it depends on where you look, I found Shelter, scholar, love and peace. If you look at a Christian based place you will get the last three. If you look in a regular old baby name book you get the word Shelter and comoforter. Hope that helps.
2007-01-08 16:32:03 · answer #7 · answered by trhwsh 5 · 0⤊ 0⤋
Scholar
2007-01-08 15:25:06 · answer #8 · answered by Anonymous · 0⤊ 0⤋
From the Dutch surname Schuyler which means "scholar".
2007-01-08 15:05:45 · answer #9 · answered by Anonymous · 0⤊ 0⤋
From Dutch...variation of Schylar...means "to shelter"
2007-01-08 15:11:20 · answer #10 · answered by alex f 2 · 0⤊ 0⤋