如图，已知△ABC的周长是21，OB，OC分别平分∠ABC和∠ACB，OD⊥BC于D，且OD=4，△ABC的面积是______

过O作OE⊥AB于E，OF⊥AC于F，连接OA，∵OB，OC分别平分∠ABC和∠ACB，OD⊥BC，∴OE=OD，OD=OF，即OE=OF=OD=4，∴△ABC的面积是：S △AOB +S △AOC +S △OBC = 1 2 ×AB×OE+ 1 2 ×AC×OF+ 1 2 ×BC×OD= 1 2 ×4×（AB+AC+BC）= 1 2 ×4×21=42，故答案为：42．