Cho hình nón tròn xoay có chiều cao h = 20(cm) , bán kính đáy r = 25(cm) . Một thiết diện đi qua đỉnh của hình nón có khoảng cách từ tâm đáy đến mặt phẳng chứa thiết diện là 12(cm) . Tính diện tích của thiết diện đó.

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9iaaiwdacaaIWaGaaGimaiaaykW7daqadaqaaiaabogacaqGTbWa % aWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzkaaGaaiOlaaaa!4094! S = 500\,\left( {{\rm{c}}{{\rm{m}}^2}} \right).\)

B. \(S = 300\,\left( {{\rm{c}}{{\rm{m}}^2}} \right).\)

C. \(S = 400\,\left( {{\rm{c}}{{\rm{m}}^2}} \right).\)

D. \(S = 406\,\left( {{\rm{c}}{{\rm{m}}^2}} \right).\)

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Chủ đề: Đề thi THPT QG

Môn: Toán

Lời giải:

Báo sai

Theo bài ra ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaad+ % eacqGH9aqpcaWGYbGaeyypa0JaaGOmaiaaiwdacaGG7aaaaa!3CCB! AO = r = 25;OK=12\); \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaad+ % eacqGH9aqpcaWGObGaeyypa0JaaGOmaiaaicdacaGG7aGaaGPaVdaa % !3E59! SO = h = 20;\)(Hình vẽ).

Lại có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaam4taiaadUeadaahaaWcbeqaaiaaikdaaaaaaOGaeyyp % a0ZaaSaaaeaacaaIXaaabaGaam4taiaadMeadaahaaWcbeqaaiaaik % daaaaaaOGaey4kaSYaaSaaaeaacaaIXaaabaGaam4taiaadofadaah % aaWcbeqaaiaaikdaaaaaaOGaeyO0H4Taam4taiaadMeacqGH9aqpca % aIXaGaaGynaiaaykW7daqadaqaaiaabogacaqGTbaacaGLOaGaayzk % aaaaaa!4D71! \frac{1}{{O{K^2}}} = \frac{1}{{O{I^2}}} + \frac{1}{{O{S^2}}} \Rightarrow OI = 15\,\left( {{\rm{cm}}} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadk % eacqGH9aqpcaaIYaGaamyqaiaadMeacqGH9aqpdaGcaaqaaiaaikda % caaI1aWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGymaiaaiwdada % ahaaWcbeqaaiaaikdaaaaabeaakiabg2da9iaaisdacaaIWaGaaGPa % VpaabmaabaGaae4yaiaab2gaaiaawIcacaGLPaaacaGG7aGaaGjbVl % aadofacaWGjbGaeyypa0ZaaOaaaeaacaWGtbGaam4tamaaCaaaleqa % baGaaGOmaaaakiabgUcaRiaad+eacaWGjbWaaWbaaSqabeaacaaIYa % aaaaqabaGccqGH9aqpcaaIYaGaaGynaiaaykW7daqadaqaaiaaboga % caqGTbaacaGLOaGaayzkaaGaeyO0H4Taam4uamaaBaaaleaacqqHuo % arcaWGtbGaamyqaiaadkeaaeqaaOGaeyypa0ZaaSaaaeaacaaIXaaa % baGaaGOmaaaacaGGUaGaaGOmaiaaiwdacaGGUaGaaGinaiaaicdacq % GH9aqpcaaI1aGaaGimaiaaicdacaaMc8+aaeWaaeaacaqGJbGaaeyB % amaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawMcaaiaac6caaaa!7387! AB = 2AI = \sqrt {{{25}^2} - {{15}^2}} = 40\,\left( {{\rm{cm}}} \right);\;SI = \sqrt {S{O^2} + O{I^2}} = 25\,\left( {{\rm{cm}}} \right) \Rightarrow {S_{\Delta SAB}} = \frac{1}{2}.25.40 = 500\,\left( {{\rm{c}}{{\rm{m}}^2}} \right).\)

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