2008 Australian Open – Women's singles explained

See main article: 2008 Australian Open.

Maria Sharapova defeated Ana Ivanovic in the final, 7–5, 6–3 to win the women's singles tennis title at the 2008 Australian Open. It was her third major singles title, and she became the first Russian to win the title. She did not lose a set during the tournament or face a tiebreak in any set. It was Ivanovic's second runner-up finish in as many major finals, though she would win the French Open a few months later.

Serena Williams was the defending champion, but was defeated in the quarterfinals by Jelena Janković.

Justine Henin, whose consecutive streak of 33 match wins dated back at the 2007 Rogers Cup, lost in the quarterfinals to Sharapova.

Agnieszka Radwańska became the first Polish player to reach the quarterfinals in the Open era, the first player was to do so since Jadwiga Jędrzejowska in the 1939 Wimbledon Championships.

This was the first Australian Open appearance for future champion Caroline Wozniacki, who lost to Ivanovic in the fourth round.

Qualifying

See main article: 2008 Australian Open – Women's singles qualifying.

Draw

Top half

Section 4

Bottom half

Section 8

Championship match statistics

Category Sharapova Ivanovic
1st serve %27/50 (54%) 42/70 (60%)
1st serve points won24 of 27 = 89% 26 of 42 = 62%
2nd serve points won16 of 23 = 70% 14 of 28 = 50%
Total service points won40 of 50 = 80.00% 40 of 70 = 57.14%
Aces1 3
Double faults 3 4
Winners16 14
Unforced errors15 33
Net points won 7 of 10 = 70% 4 of 11 = 36%
Break points converted 4 of 9 = 44% 1 of 2 = 50%
Return points won 30 of 70 = 43% 10 of 50 = 20%
Total points won bgcolor=98FB98970 50
Source

External links