Cho hình hộp chữ nhật ABCD.A'B'C'D' có đáy ABCD là hình vuông cạnh \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyamaaka % aabaGaaGOmaaWcbeaaaaa!37B1! a\sqrt 2 \) , AA'=2a. Tính khoảng cách giữa hai đường thẳng BD và CD' .
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Lời giải:
Báo saiGọi O, O' lần lượt là tâm của hai mặt đáy.Khi đó tứ giác COO'C' là hình bình hành và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4qayaafa % Gabm4tayaafaGaeyypa0ZaaSaaaeaacaWGbbGaam4qaaqaaiaaikda % aaGaeyypa0Jaamyyaaaa!3CF4! C'O' = \frac{{AC}}{2} = a\)
Do \(BD//B'D'\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % OqaiaadseacaaMe8Uaae4laiaab+cacaaMe8+aaeWaaeaacaWGdbGa % bmOqayaafaGabmirayaafaaacaGLOaGaayzkaaaaaa!4257! \Rightarrow BD\;{\rm{//}}\;\left( {CB'D'} \right)\) nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamOqaiaadseacaGG7aGaam4qaiqadseagaqbaaGaayjkaiaa % wMcaaiabg2da9iaadsgadaqadaqaaiaad+eacaGG7aWaaeWaaeaaca % WGdbGabmOqayaafaGabmirayaafaaacaGLOaGaayzkaaaacaGLOaGa % ayzkaaGaeyypa0JaamizamaabmaabaGabm4qayaafaGaai4oamaabm % aabaGaam4qaiqadkeagaqbaiqadseagaqbaaGaayjkaiaawMcaaaGa % ayjkaiaawMcaaaaa!4E5A! d\left( {BD;CD'} \right) = d\left( {O;\left( {CB'D'} \right)} \right) = d\left( {C';\left( {CB'D'} \right)} \right)\) .
Ta có :\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiqadkeagaqbaiqadseagaqbaiabgwQiEjqadgeagaqbaiqadoea % gaqbaaqaaiqadkeagaqbaiqadseagaqbaiabgwQiEjaadoeaceWGdb % GbauaaaaGaay5EaaGaeyO0H4TabmOqayaafaGabmirayaafaGaeyyP % I41aaeWaaeaacaWGdbGaam4taiqad+eagaqbaiqadoeagaqbaaGaay % jkaiaawMcaaaaa!4B98! \left\{ \begin{array}{l} B'D' \bot A'C'\\ B'D' \bot CC' \end{array} \right. \Rightarrow B'D' \bot \left( {COO'C'} \right)\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H49aae % WaaeaacaWGdbGabmOqayaafaGabmirayaafaaacaGLOaGaayzkaaGa % eyyPI41aaeWaaeaacaWGdbGaam4taiqad+eagaqbaiqadoeagaqbaa % GaayjkaiaawMcaaaaa!42D4! \Rightarrow \left( {CB'D'} \right) \bot \left( {COO'C'} \right)$\)
Lại có : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGdbGabmOqayaafaGabmirayaafaaacaGLOaGaayzkaaGaeyykIC8a % aeWaaeaacaWGdbGaam4taiqad+eagaqbaiqadoeagaqbaaGaayjkai % aawMcaaiabg2da9iaadoeaceWGpbGbauaaaaa!4312! \left( {CB'D'} \right) \cap \left( {COO'C'} \right) = CO'\)
Trong \(\Delta CC'O'\) hạ \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4qayaafa % GaamisaiabgwQiEjaadoeaceWGpbGbauaacqGHshI3ceWGdbGbauaa % caWGibGaeyyPI41aaeWaaeaacaWGdbGabmOqayaafaGabmirayaafa % aacaGLOaGaayzkaaaaaa!4496! C'H \bot CO' \Rightarrow C'H \bot \left( {CB'D'} \right)\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % izamaabmaabaGaamOqaiaadseacaGG7aGaaGPaVlaadoeaceWGebGb % auaaaiaawIcacaGLPaaacqGH9aqpceWGdbGbauaacaWGibaaaa!42E1! \Rightarrow d\left( {BD;\,CD'} \right) = C'H\)
Khi đó : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGabm4qayaafaGaamisamaaCaaaleqabaGaaGOmaaaaaaGc % cqGH9aqpdaWcaaqaaiaaigdaaeaacaWGdbGabm4qayaafaWaaWbaaS % qabeaacaaIYaaaaaaakiabgUcaRmaalaaabaGaaGymaaqaaiqadoea % gaqbaiqad+eagaqbamaaCaaaleqabaGaaGOmaaaaaaGccqGH9aqpda % WcaaqaaiaaigdaaeaadaqadaqaaiaaikdacaWGHbaacaGLOaGaayzk % aaWaaWbaaSqabeaacaaIYaaaaaaakiabgUcaRmaalaaabaGaaGymaa % qaaiaadggadaahaaWcbeqaaiaaikdaaaaaaOGaeyypa0ZaaSaaaeaa % caaI1aaabaGaaGinaiaadggadaahaaWcbeqaaiaaikdaaaaaaaaa!4FDE! \frac{1}{{C'{H^2}}} = \frac{1}{{C{{C'}^2}}} + \frac{1}{{C'{{O'}^2}}} = \frac{1}{{{{\left( {2a} \right)}^2}}} + \frac{1}{{{a^2}}} = \frac{5}{{4{a^2}}}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Tabm % 4qayaafaGaamisaiabg2da9maalaaabaGaaGOmamaakaaabaGaaGyn % aaWcbeaakiaadggaaeaacaaI1aaaaaaa!3E4D! \Rightarrow C'H = \frac{{2\sqrt 5 a}}{5}\)
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
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