Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
AB//CD
=>\(\frac{OA}{OC}=\frac{OB}{OD}=\frac{AB}{CD}=\frac46=\frac23\)
Ta có: \(\frac{OA}{OC}=\frac23\)
=>\(\frac{S_{BOA}}{S_{BOC}}=\frac23\)
=>\(\frac{10}{S_{BOC}}=\frac23=\frac{10}{15}\)
=>\(S_{BOC}=15\left(\operatorname{cm}^2\right)\)
Ta có: \(\frac{OB}{OD}=\frac23\)
=>\(\frac{S_{AOB}}{S_{AOD}}=\frac23\)
=>\(\frac{10}{S_{AOD}}=\frac23=\frac{10}{15}\)
=>\(S_{AOD}=15\left(\operatorname{cm}^2\right)\)
TA có: \(\frac{OA}{OC}=\frac23\)
=>\(\frac{S_{AOD}}{S_{DOC}}=\frac23\)
=>\(S_{DOC}=15\cdot\frac32=22,5\left(\operatorname{cm}^2\right)\)
\(S_{ABCD}=S_{OAB}+S_{OAD}+S_{OBC}+S_{OCD}\)
\(=10+15+15+22,5=40+22,5=62,5\left(\operatorname{cm}^2\right)\)
\(ABssCD\Rightarrow\dfrac{AB}{CD}=\dfrac{OB}{OD}=\dfrac{OA}{OC}=\dfrac{2}{3}\)
a)\(S_{AOD}=\dfrac{1}{2}OA.OD.sinAOB\)
\(S_{BOC}=\dfrac{1}{2}OB.OC.sinBOC\)
\(\Rightarrow\dfrac{S_{AOD}}{S_{BOC}}=\dfrac{OA.OD}{OB.OC}\) vì \(\widehat{AOD}=\widehat{BOC}\Rightarrow sinAOD=sinBOC\)
\(\Leftrightarrow\dfrac{S_{AOD}}{S_{BOC}}=\dfrac{2}{3}.\dfrac{3}{2}=1\)
b) vì \(ABssCD\Rightarrow\dfrac{OH}{OK}=\dfrac{2}{3}\Rightarrow\dfrac{OH}{HK}=\dfrac{2}{5}\)
\(S_{AOB}=\dfrac{1}{2}.OH.AB\\ S_{ABCD}=\dfrac{1}{2}\left(AB+CD\right).HK=\dfrac{1}{2}\left(AB+\dfrac{3}{2}AB\right).HK=\dfrac{1}{2}.\dfrac{5}{2}AB.HK\)
\(\Rightarrow\dfrac{S_{AOB}}{S_{ABCD}}=\dfrac{\dfrac{1}{2}OH.AB}{\dfrac{1}{2}HK.\dfrac{5}{2}AB}=\dfrac{2}{5}.\dfrac{1}{\dfrac{5}{2}}=\dfrac{4}{25}\)
\(\Rightarrow S_{ABCD}=\dfrac{4}{\dfrac{4}{25}}=25\)
Ta kí hiệu S (MNP) là diện tích tam giác MNP