Cho hình chóp S.ABCD có đáy ABCD là hình vuông tâm O cạnh a, SO vuông góc với mặt phẳng (ABCD) và SO = a. Khoảng cách giữa SC và AB bằng
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Lời giải:
Báo saiGọi M,N lần lượt là trung điểm của các cạnh AB; CDCD; H là hình chiếu vuông góc của O trên SN
Vì AB// CD nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamyqaiaadkeacaGGSaGaai4uaiaacoeaaiaawIcacaGLPaaa % cqGH9aqpcaWGKbWaaeWaaeaacaWGbbGaamOqaiaacYcacaGGOaGaam % 4uaiaadoeacaWGebGaaiykaaGaayjkaiaawMcaaiabg2da9iaadsga % daqadaqaaiaad2eacaGGSaGaaiikaiaadofacaWGdbGaamiraiaacM % caaiaawIcacaGLPaaacqGH9aqpcaaIYaGaamizamaabmaabaGaam4t % aiaacYcacaGGOaGaam4uaiaadoeacaWGebGaaiykaaGaayjkaiaawM % caaaaa!57EE! d\left( {AB,SC} \right) = d\left( {AB,(SCD)} \right) = d\left( {M,(SCD)} \right) = 2d\left( {O,(SCD)} \right)\)
Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadoeacaWGebGaeyyPI4Laam4uaiaad+eaaeaacaWGdbGaamir % aiabgwQiEjaad+eacaWGobaaaiaawUhaaiabgkDiElaadoeacaWGeb % GaeyyPI4LaaiikaiaadofacaWGpbGaamOtaiaacMcacqGHshI3caWG % dbGaamiraiabgwQiEjaad+eacaWGibaaaa!51A2! \left\{ \begin{array}{l} CD \bot SO\\ CD \bot ON \end{array} \right. \Rightarrow CD \bot (SON) \Rightarrow CD \bot OH\)
Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadoeacaWGebGaeyyPI4Laam4taiaadIeaaeaacaWGpbGaamis % aiabgwQiEjaadofacaWGobaaaiaawUhaaiabgkDiElaad+eacaWGib % GaeyyPI4LaaiikaiaadofacaWGdbGaamiraiaacMcacqGHshI3caWG % KbWaaeWaaeaacaWGpbGaai4oaiaacIcacaWGtbGaam4qaiaadseaca % GGPaaacaGLOaGaayzkaaGaeyypa0Jaam4taiaadIeacaGGUaaaaa!57E2! \left\{ \begin{array}{l} CD \bot OH\\ OH \bot SN \end{array} \right. \Rightarrow OH \bot (SCD) \Rightarrow d\left( {O;(SCD)} \right) = OH.\)
Tam giác SON vuông tại O nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaam4taiaadIeadaahaaWcbeqaaiaaikdaaaaaaOGaeyyp % a0ZaaSaaaeaacaaIXaaabaGaam4taiaad6eadaahaaWcbeqaaiaaik % daaaaaaOGaey4kaSYaaSaaaeaacaaIXaaabaGaam4taiaadofadaah % aaWcbeqaaiaaikdaaaaaaOGaeyypa0ZaaSaaaeaacaaIXaaabaWaaS % aaaeaacaWGHbWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGinaaaaaaGa % ey4kaSYaaSaaaeaacaaIXaaabaGaamyyamaaCaaaleqabaGaaGOmaa % aaaaGccqGH9aqpdaWcaaqaaiaaiwdaaeaacaWGHbWaaWbaaSqabeaa % caaIYaaaaaaakiabgkDiElaad+eacaWGibGaeyypa0ZaaSaaaeaaca % WGHbaabaWaaOaaaeaacaaI1aaaleqaaaaaaaa!5489! \frac{1}{{O{H^2}}} = \frac{1}{{O{N^2}}} + \frac{1}{{O{S^2}}} = \frac{1}{{\frac{{{a^2}}}{4}}} + \frac{1}{{{a^2}}} = \frac{5}{{{a^2}}} \Rightarrow OH = \frac{a}{{\sqrt 5 }}\)
Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamyqaiaadkeacaGGSaGaai4uaiaacoeaaiaawIcacaGLPaaa % cqGH9aqpcaaIYaGaam4taiaadIeacqGH9aqpdaWcaaqaaiaaikdaca % WGHbWaaOaaaeaacaaI1aaaleqaaaGcbaGaaGynaaaaaaa!43FF! d\left( {AB,SC} \right) = 2OH = \frac{{2a\sqrt 5 }}{5}\)
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
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