Indians’ Kluber allows Angels 3 hits, no runs


The Associated Press



The Indians' Corey Kluber got his first shutout of the season Saturday against the Los Angeles Angels in Cleveland. (AP photo)

The Indians' Corey Kluber got his first shutout of the season Saturday against the Los Angeles Angels in Cleveland. (AP photo)


CLEVELAND — Corey Kluber pitched a three-hitter for his first shutout this season and new arrival Leonys Martin homered again for Cleveland, leading the Indians to a 3-0 win over the Los Angeles Angels on Saturday night.

Kluber (14-6) matched zeros for five innings with Felix Pena (1-3), who carried a no-hitter into the sixth inning before Martin hit a leadoff homer. Martin has two homers in just three games since coming to Cleveland in a trade with Detroit.

It was the seventh career shutout for Kluber, and the performance eased any concerns the Indians had about the two-time Cy Young Award winner. Kluber had lost two of his three previous starts and he received an injection in his right knee before the All-Star break. He walked one, hit a batter and struck out seven.

The AL Central-leading Indians only got four hits but notched their 60th win.

The Angels were again without All-Star outfielder Mike Trout, who missed his third straight game with a sore right wrist.

The Indians' Corey Kluber got his first shutout of the season Saturday against the Los Angeles Angels in Cleveland. (AP photo)
https://www.limaohio.com/wp-content/uploads/sites/54/2018/08/web1_onlineindians-1.jpgThe Indians' Corey Kluber got his first shutout of the season Saturday against the Los Angeles Angels in Cleveland. (AP photo)

The Associated Press

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