Cho hình chóp S.ABC có đáy là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam % yqaiaadkeacaWGdbaaaa!39AE! \Delta ABC\) vuông cân ở B, \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaado % eacqGH9aqpcaWGHbWaaOaaaeaacaaIYaaaleqaaOGaaiilaiaaykW7 % aaa!3C89! AC = a\sqrt 2 ,\,\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaadg % eacqGHLkIxdaqadaqaaiaadgeacaWGcbGaam4qaaGaayjkaiaawMca % aiaacYcaaaa!3DD0! SA \bot \left( {ABC} \right),\) SA = a. Gọi G là trọng tâm của \(\Delta SBC\) , \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiaadc % hadaqadaqaaiabeg7aHbGaayjkaiaawMcaaaaa!3B02! mp\left( \alpha \right)\) đi qua AG và song song với BC chia khối chóp thành hai phần. Gọi V là thể tích của khối đa diện không chứa đỉnh S. Tính V
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Lời giải:
Báo saiTrong mặt phẳng (SBC) . Qua G kẻ đường thẳng song song với BC và lần lượt cắt SC, SB tại E, F . Khi đó ta được khối đa diện không chứa đỉnh S là ABCEF
Ta có G là trọng tâm của \(\Delta SBC\) nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGwbWaaSbaaSqaaiaadofacaGGUaGaaeyqaiaabAeacaWGfbaabeaa % aOqaaiaadAfadaWgaaWcbaGaam4uaiaac6cacaWGbbGaamOqaiaado % eaaeqaaaaakiabg2da9maalaaabaGaam4uaiaadgeaaeaacaWGtbGa % amyqaaaacaGGUaWaaSaaaeaacaWGtbGaamOraaqaaiaadofacaWGcb % aaaiaac6cadaWcaaqaaiaadofacaWGfbaabaGaam4uaiaadoeaaaGa % eyypa0ZaaSaaaeaacaaIYaaabaGaaG4maaaacaGGUaWaaSaaaeaaca % aIYaaabaGaaG4maaaacqGH9aqpdaWcaaqaaiaaisdaaeaacaaI5aaa % aiaac6caaaa!5452! \frac{{{V_{S.{\rm{AF}}E}}}}{{{V_{S.ABC}}}} = \frac{{SA}}{{SA}}.\frac{{SF}}{{SB}}.\frac{{SE}}{{SC}} = \frac{2}{3}.\frac{2}{3} = \frac{4}{9}.\)
Do đó: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGtbGaaiOlaiaabgeacaqGgbGaamyraaqabaGccqGH9aqp % daWcaaqaaiaaisdaaeaacaaI5aaaaiaac6cacaWGwbWaaSbaaSqaai % aadofacaGGUaGaamyqaiaadkeacaWGdbaabeaakiabgkDiElaadAfa % daWgaaWcbaGaamyqaiaadkeacaWGdbGaamyraiaadAeaaeqaaOGaey % ypa0JaamOvamaaBaaaleaadaWgaaadbaGaam4uaiaac6cacaWGbbGa % amOqaiaadoeaaeqaaaWcbeaakiabgkHiTmaalaaabaGaaGinaaqaai % aaiMdaaaGaaiOlaiaadAfadaWgaaWcbaGaam4uaiaac6cacaWGbbGa % amOqaiaadoeaaeqaaOGaeyypa0ZaaSaaaeaacaaI1aaabaGaaGyoaa % aacaGGUaGaamOvamaaBaaaleaacaWGtbGaaiOlaiaadgeacaWGcbGa % am4qaaqabaGccaGGUaaaaa!61B0! {V_{S.{\rm{AF}}E}} = \frac{4}{9}.{V_{S.ABC}} \Rightarrow {V_{ABCEF}} = {V_{_{S.ABC}}} - \frac{4}{9}.{V_{S.ABC}} = \frac{5}{9}.{V_{S.ABC}}.\)
Vì tam giác \(\Delta ABC\) vuông cân ở B, \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaado % eacqGH9aqpcaWGHbWaaOaaaeaacaaIYaaaleqaaaaa!3A44! AC = a\sqrt 2 \) nên AB = BC = a
Mặt khác \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGtbGaaiOlaiaadgeacaWGcbGaam4qaaqabaGccqGH9aqp % daWcaaqaaiaaigdaaeaacaaIZaaaamaalaaabaGaaGymaaqaaiaaik % daaaGaamyyaiaac6cacaWGHbGaaiOlaiaadggacqGH9aqpdaWcaaqa % aiaadggadaahaaWcbeqaaiaaiodaaaaakeaacaaI2aaaaiaac6caaa % a!4770! {V_{S.ABC}} = \frac{1}{3}\frac{1}{2}a.a.a = \frac{{{a^3}}}{6}.\) Suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGbbGaamOqaiaadoeacaWGfbGaamOraaqabaGccqGH9aqp % daWcaaqaaiaaiwdaaeaacaaI5aaaaiaac6cadaWcaaqaaiaadggada % ahaaWcbeqaaiaaiodaaaaakeaacaaI2aaaaiabg2da9maalaaabaGa % aGynaiaadggadaahaaWcbeqaaiaaiodaaaaakeaacaaI1aGaaGinaa % aaaaa!460E! {V_{ABCEF}} = \frac{5}{9}.\frac{{{a^3}}}{6} = \frac{{5{a^3}}}{{54}}\) .