Cho lăng trụ tam giác đều có cạnh đáy bằng a cạnh bên bằng b. Tính thể tích của khối cầu đi qua các đỉnh của lăng trụ.

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaaGymaiaaiIdadaGcaaqaaiaaiodaaSqabaaaaOWaaOaa % aeaadaqadaqaaiaaisdacaWGHbWaaWbaaSqabeaacaaIYaaaaOGaey % 4kaSIaaG4maiaadkgadaahaaWcbeqaaiaaikdaaaaakiaawIcacaGL % PaaadaahaaWcbeqaaiaaiodaaaaabeaakiaac6caaaa!426D! \frac{1}{{18\sqrt 3 }}\sqrt {{{\left( {4{a^2} + 3{b^2}} \right)}^3}} .\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % aHapaCaeaacaaIXaGaaGioamaakaaabaGaaG4maaWcbeaaaaGcdaGc % aaqaamaabmaabaGaaGinaiaadggadaahaaWcbeqaaiaaikdaaaGccq % GHRaWkcaaIZaGaamOyamaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaa % wMcaamaaCaaaleqabaGaaG4maaaaaeqaaOGaaiOlaaaa!436F! \frac{\pi }{{18\sqrt 3 }}\sqrt {{{\left( {4{a^2} + 3{b^2}} \right)}^3}} .\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % aHapaCaeaacaaIXaGaaGioamaakaaabaGaaG4maaWcbeaaaaGcdaGc % aaqaamaabmaabaGaaGinaiaadggadaahaaWcbeqaaiaaikdaaaGccq % GHRaWkcaWGIbWaaWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzkaaWa % aWbaaSqabeaacaaIZaaaaaqabaGccaGGUaaaaa!42B2! \frac{\pi }{{18\sqrt 3 }}\sqrt {{{\left( {4{a^2} + {b^2}} \right)}^3}} \)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % aHapaCaeaacaaIXaGaaGioamaakaaabaGaaGOmaaWcbeaaaaGcdaGc % aaqaamaabmaabaGaaGinaiaadggadaahaaWcbeqaaiaaikdaaaGccq % GHRaWkcaaIZaGaamOyamaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaa % wMcaamaaCaaaleqabaGaaG4maaaaaeqaaOGaaiOlaaaa!436E! \frac{\pi }{{18\sqrt 2 }}\sqrt {{{\left( {4{a^2} + 3{b^2}} \right)}^3}} \)

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Chủ đề: Đề thi THPT QG

Môn: Toán

Lời giải:

Báo sai

Gọi I;I' lần lượt là tâm hai đáy, O là trung điểm của II'. Khi đó ta có O là tâm mặt cầu ngoại tiếp lăng trụ.

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadM % eacqGH9aqpdaWcaaqaaiaadggadaGcaaqaaiaaiodaaSqabaaakeaa % caaIZaaaaiaacYcacaaMc8Uaamysaiaad+eacqGH9aqpdaWcaaqaai % aadkgaaeaacaaIYaaaaaaa!41B9! AI = \frac{{a\sqrt 3 }}{3},\,IO = \frac{b}{2}\) suy ra bán kính mặt cầu ngoại tiếp lăng trụ là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiabg2 % da9maakaaabaWaaSaaaeaacaWGHbWaaWbaaSqabeaacaaIYaaaaaGc % baGaaG4maaaacqGHRaWkdaWcaaqaaiaadkgadaahaaWcbeqaaiaaik % daaaaakeaacaaI0aaaaaWcbeaakiabg2da9maalaaabaGaaGymaaqa % aiaaikdadaGcaaqaaiaaiodaaSqabaaaaOWaaOaaaeaacaaI0aGaam % yyamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiodacaWGIbWaaWba % aSqabeaacaaIYaaaaaqabaaaaa!47AA! R = \sqrt {\frac{{{a^2}}}{3} + \frac{{{b^2}}}{4}} = \frac{1}{{2\sqrt 3 }}\sqrt {4{a^2} + 3{b^2}} \)

Vậy: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaadaqadaqaaiaad+eacaGG7aGaaGPaVlaadkfaaiaawIcacaGL % PaaaaeqaaOGaeyypa0ZaaSaaaeaacaaI0aaabaGaaG4maaaacqaHap % aCcaWGsbWaaWbaaSqabeaacaaIZaaaaOGaeyypa0ZaaSaaaeaacqaH % apaCaeaacaaIXaGaaGioamaakaaabaGaaG4maaWcbeaaaaGcdaGcaa % qaamaabmaabaGaaGinaiaadggadaahaaWcbeqaaiaaikdaaaGccqGH % RaWkcaaIZaGaamOyamaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawM % caamaaCaaaleqabaGaaG4maaaaaeqaaOGaaiOlaaaa!511D! {V_{\left( {O;\,R} \right)}} = \frac{4}{3}\pi {R^3} = \frac{\pi }{{18\sqrt 3 }}\sqrt {{{\left( {4{a^2} + 3{b^2}} \right)}^3}} .\)

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