Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
BM=MC
=>\(S_{AMB}=S_{AMC};S_{OMB}=S_{OMC}\)
=>\(S_{AMB}-S_{OMB}=S_{AMC}-S_{OMC}\)
=>\(S_{AOB}=S_{AOC}\)
Ta có: AN=2NC
=>\(S_{BNA}=2\times S_{BNC};S_{ONA}=2\times S_{ONC}\)
=>\(S_{BNA}-S_{ONA}=2\times\left(S_{BNC}-S_{ONC}\right)\)
=>\(S_{BOA}=2\times S_{BOC}\)
=>\(S_{AOC}=2\times S_{BOC}\)
=>\(S_{COA}=2\times S_{COB}\)
Ta có; F nằm giữa A và B
=>\(\frac{S_{CFA}}{S_{CFB}}=\frac{FA}{FB};\frac{S_{OFA}}{S_{OFB}}=\frac{FA}{FB}\)
=>\(\frac{S_{CFA}-S_{OFA}}{S_{CFB}-S_{OFB}}=\frac{FA}{FB}\)
=>\(\frac{S_{COA}}{S_{COB}}=\frac{FA}{FB}\)
=>\(\frac{FA}{FB}=2\)
=>\(\frac{AF}{AB}=\frac23\)
Ta có: BM=2MC
=>\(S_{AMB}=2\times S_{AMC};S_{OMB}=2\times S_{OMC}\)
=>\(S_{AMB}-S_{OMB}=2\times\left(S_{AOC}-S_{MOC}\right)\)
=>\(S_{AOB}=2\times S_{AOC}\)
=>\(S_{AOC}=\frac{40}{2}=20\left(\operatorname{cm}^2\right)\)
Ta có: CN=3NA
=>\(S_{BNC}=3\times S_{BNA};S_{ONC}=3\times S_{ONA}\)
=>\(S_{BNC}-S_{ONC}=3\times\left(S_{BNA}-S_{NOA}\right)\)
=>\(S_{BOC}=3\times S_{BOA}=3\times40=120\left(\operatorname{cm}^2\right)\)
Diện tích tam giác ABC là:
\(S_{ABC}=S_{OAB}+S_{OAC}+S_{OBC}\)
\(=20+40+120=60+120=180\left(\operatorname{cm}^2\right)\)
Ta có: BM=2MC
=>\(S_{AMB}=2\times S_{AMC};S_{OMB}=2\times S_{OMC}\)
=>\(S_{AMB}-S_{OMB}=2\times\left(S_{AMC}-S_{OMC}\right)\)
=>\(S_{AOB}=2\times S_{AOC}\)
=>\(S_{AOC}=\frac12\times S_{AOB}\)
Ta có: AN=2NC
=>\(S_{BNA}=2\times S_{BNC};S_{ONA}=2\times S_{ONC}\)
=>\(S_{BNA}-S_{ONA}=2\times\left(S_{BNC}-S_{ONC}\right)\)
=>\(S_{BOA}=2\times S_{BOC}\)
=>\(S_{BOC}=\frac12\times S_{AOB}\)
Ta có: \(S_{AOC}+S_{BOC}+S_{AOB}=S_{ABC}\)
=>\(S_{AOB}+\frac12\times S_{AOB}+\frac12\times S_{AOB}=S_{ABC}=840\)
=>\(2\times S_{AOB}=840\)
=>\(S_{AOB}=420\)