Cho hình chóp \(S.ABC\)có thể tích \(70a^3\). Gọi M, N là accs điểm trên SB, SC sao cho \(SM=\frac{2}{3}SB, SN=\frac{4}{5}SC\). Thể tích khối chóp \(S.AMN\) bằng
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGtb % Gaamytaiabg2da9maalaaabaGaaGOmaaqaaiaaiodaaaGaam4uaiaa % dkeacqGHshI3daWcaaqaaiaadofacaWGcbaabaGaam4uaiaad2eaaa % Gaeyypa0ZaaSaaaeaacaaIZaaabaGaaGOmaaaaaeaacaWGtbGaamOt % aiabg2da9maalaaabaGaaGinaaqaaiaaiwdaaaGaam4uaiaadoeacq % GHshI3daWcaaqaaiaadofacaWGdbaabaGaam4uaiaad6eaaaGaeyyp % a0ZaaSaaaeaacaaI1aaabaGaaGinaaaaaaaa!5240! \begin{array}{l} SM = \frac{2}{3}SB \Rightarrow \frac{{SB}}{{SM}} = \frac{3}{2}\\ SN = \frac{4}{5}SC \Rightarrow \frac{{SC}}{{SN}} = \frac{5}{4} \end{array}\)
Ta có:
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaWcaa % qaaiaadAfadaWgaaWcbaGaam4uaiaac6cacaWGbbGaamOqaiaadoea % aeqaaaGcbaGaamOvamaaBaaaleaacaWGtbGaaiOlaiaadgeacaWGnb % GaamOtaaqabaaaaOGaeyypa0ZaaSaaaeaacaWGtbGaamyqaaqaaiaa % dofacaWGbbaaaiaac6cadaWcaaqaaiaadofacaWGcbaabaGaam4uai % aad2eaaaGaaiOlamaalaaabaGaam4uaiaadoeaaeaacaWGtbGaamOt % aaaacqGH9aqpcaaIXaGaaiOlamaalaaabaGaaG4maaqaaiaaikdaaa % GaaiOlamaalaaabaGaaGynaaqaaiaaisdaaaGaeyypa0ZaaSaaaeaa % caaIXaGaaGynaaqaaiaaiIdaaaaabaGaeyO0H4TaamOvamaaBaaale % aacaWGtbGaaiOlaiaadgeacaWGnbGaamOtaaqabaGccqGH9aqpcaWG % wbWaaSbaaSqaaiaadofacaGGUaGaamyqaiaadkeacaWGdbaabeaaki % aacQdadaWcaaqaaiaaigdacaaI1aaabaGaaGioaaaacqGH9aqpcaaI % 3aGaaGimaiaadggadaahaaWcbeqaaiaaiodaaaGccaGG6aWaaSaaae % aacaaIXaGaaGynaaqaaiaaiIdaaaGaeyypa0ZaaSaaaeaacaaIXaGa % aGymaiaaikdacaWGHbWaaWbaaSqabeaacaaIZaaaaaGcbaGaaG4maa % aaaaaa!739E! \begin{array}{l} \frac{{{V_{S.ABC}}}}{{{V_{S.AMN}}}} = \frac{{SA}}{{SA}}.\frac{{SB}}{{SM}}.\frac{{SC}}{{SN}} = 1.\frac{3}{2}.\frac{5}{4} = \frac{{15}}{8}\\ \Rightarrow {V_{S.AMN}} = {V_{S.ABC}}:\frac{{15}}{8} = 70{a^3}:\frac{{15}}{8} = \frac{{112{a^3}}}{3} \end{array}\)
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
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