![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/7af40ad162d9f2d33fa1aa5eaaec8a136227cccb?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
解:延长BM交AC于点D,再延长BD至E,使DE=DM,连接CE,
∵M是△ABC的重心,
∴AD=CD,MD=
BM,
∵∠ADM=∠CDE(对顶角相等),DE=DM,
∴△AMD≌△CDE(SAS),
∴AM=EC=3,
∵DE=DM,MD=
BM,
∴BM=EM=4,
在△CME中,CM=5,ME=4,EC=3,根据勾股定理可得△CME为直角三角形,
S
△CME=
×3×4=6,
由以上可证得S
△AMC=S
△CME
∵M是△ABC的重心,
∴S
△ABC=3S
△AMC=18.
故答案填:18.