Cho hình chóp tam giác đều có cạnh đáy bằng a và cạnh bên tạo với đáy một góc \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqOXdOgaaa!37B0! \varphi \) . Thể tích của khối chóp đó bằng
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Lời giải:
Báo sai.png)
Xét hình chóp tam giác đều S.ABC . Gọi M là trung điểm BC , G là trọng tâm của tam giác đều ABC cạnh a thì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaad2 % eacqGH9aqpdaWcaaqaaiaadggadaGcaaqaaiaaiodaaSqabaaakeaa % caaIYaaaaaaa!3B25! AM = \frac{{a\sqrt 3 }}{2}\) ; \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaadE % eacqGHLkIxdaqadaqaaiaadgeacaWGcbGaam4qaaGaayjkaiaawMca % aaaa!3D26! SG \bot \left( {ABC} \right)\), SG là chiều cao của hình chóp nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGtbGaamyqaiaadEeaaiaawkWaaiabg2da9iabeA8aQbaa!3BE2! \widehat {SAG} = \varphi \).
Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacqGHuoarcaWGbbGaamOqaiaadoeaaeqaaOGaeyypa0ZaaSaa % aeaacaaIXaaabaGaaGOmaaaacaWGbbGaamytaiaac6cacaWGcbGaam % 4qaiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaGaeyyXIC9aaSaa % aeaacaWGHbWaaWbaaSqabeaacaaIYaaaaOWaaOaaaeaacaaIZaaale % qaaaGcbaGaaGOmaaaacqGH9aqpdaWcaaqaaiaadggadaahaaWcbeqa % aiaaikdaaaGcdaGcaaqaaiaaiodaaSqabaaakeaacaaI0aaaaaaa!4E10! {S_{\Delta ABC}} = \frac{1}{2}AM.BC = \frac{1}{2} \cdot \frac{{{a^2}\sqrt 3 }}{2} = \frac{{{a^2}\sqrt 3 }}{4}\)
Xét tam giác SAG vuông tại G, \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGtbGaamyqaiaadEeaaiaawkWaaiabg2da9iabeA8aQbaa!3BE2! \widehat {SAG} = \varphi \), \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadE % eacqGH9aqpdaWcaaqaaiaaikdaaeaacaaIZaaaaiaadgeacaWGnbGa % eyypa0ZaaSaaaeaacaWGHbWaaOaaaeaacaaIZaaaleqaaaGcbaGaaG % 4maaaaaaa!3F47! AG = \frac{2}{3}AM = \frac{{a\sqrt 3 }}{3}\):\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaadE % eacqGH9aqpcaWGbbGaam4raiaac6caciGG0bGaaiyyaiaac6gacqaH % gpGAcqGH9aqpdaWcaaqaaiaadggadaGcaaqaaiaaiodaaSqabaGcci % GG0bGaaiyyaiaac6gacqaHgpGAaeaacaaIZaaaaaaa!4798! SG = AG.\tan \varphi = \frac{{a\sqrt 3 \tan \varphi }}{3}\)
Vậy thể tích hình chóp S.ABC: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9maalaaabaGaaGymaaqaaiaaiodaaaGaam4uaiaadEeacaGGUaGa % am4uamaaBaaaleaacqGHuoarcaWGbbGaamOqaiaadoeaaeqaaOGaey % ypa0ZaaSaaaeaacaaIXaaabaGaaG4maaaacqGHflY1daWcaaqaaiaa % dggadaGcaaqaaiaaiodaaSqabaGcciGG0bGaaiyyaiaac6gacqaHgp % GAaeaacaaIZaaaaiabgwSixpaalaaabaGaamyyamaaCaaaleqabaGa % aGOmaaaakmaakaaabaGaaG4maaWcbeaaaOqaaiaaisdaaaGaeyypa0 % ZaaSaaaeaacaWGHbWaaWbaaSqabeaacaaIZaaaaOGaciiDaiaacgga % caGGUbGaeqOXdOgabaGaaGymaiaaikdaaaaaaa!5B3F! V = \frac{1}{3}SG.{S_{\Delta ABC}} = \frac{1}{3} \cdot \frac{{a\sqrt 3 \tan \varphi }}{3} \cdot \frac{{{a^2}\sqrt 3 }}{4} = \frac{{{a^3}\tan \varphi }}{{12}}\) .
