{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho hai h\u00ecnh tr\u1ee5. H\u00ecnh tr\u1ee5 th\u1ee9 nh\u1ea5t c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y b\u1eb1ng n\u1eeda b\u00e1n k\u00ednh \u0111\u00e1y c\u1ee7a h\u00ecnh tr\u1ee5 th\u1ee9 hai v\u00e0 c\u00f3 chi\u1ec1u cao g\u1ea5p b\u1ed1n l\u1ea7n chi\u1ec1u cao c\u1ee7a h\u00ecnh tr\u1ee5 th\u1ee9 hai. T\u1ec9 s\u1ed1 c\u00e1c th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh tr\u1ee5 th\u1ee9 nh\u1ea5t v\u00e0 h\u00ecnh tr\u1ee5 th\u1ee9 hai b\u1eb1ng: ","select":["A. $1$ ","B. $2$ ","C. $\\dfrac{1}{2}$ ","D. $\\dfrac{3}{2}$ "],"hint":"\u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c t\u00ednh th\u1ec3 t\u00edch $V=\\pi {{R}^{2}}h$ \u0111\u1ec3 l\u1eadp t\u1ec9 s\u1ed1 ","explain":"<span class='basic_left'>Gi\u1ea3 s\u1eed h\u00ecnh tr\u1ee5 th\u1ee9 nh\u1ea5t c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 $R$ v\u00e0 chi\u1ec1u cao l\u00e0 $h.$ Th\u1ec3 t\u00edch h\u00ecnh tr\u1ee5 th\u1ee9 nh\u1ea5t l\u00e0: ${{V}_{1}}=\\pi {{R}^{2}}h$ <br\/>V\u00ec h\u00ecnh tr\u1ee5 th\u1ee9 nh\u1ea5t c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y b\u1eb1ng n\u1eeda b\u00e1n k\u00ednh \u0111\u00e1y c\u1ee7a h\u00ecnh tr\u1ee5 th\u1ee9 hai v\u00e0 c\u00f3 chi\u1ec1u cao g\u1ea5p b\u1ed1n l\u1ea7n chi\u1ec1u cao c\u1ee7a h\u00ecnh tr\u1ee5 th\u1ee9 hai n\u00ean h\u00ecnh tr\u1ee5 th\u1ee9 hai c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 $2R$ v\u00e0 chi\u1ec1u cao l\u00e0 $\\dfrac{h}{4}.$<br\/> Th\u1ec3 t\u00edch h\u00ecnh tr\u1ee5 th\u1ee9 hai l\u00e0:<br\/>$\\begin{align} & {{V}_{2}}=\\pi .{{\\left( 2R \\right)}^{2}}.\\dfrac{h}{4}=\\pi {{R}^{2}}h={{V}_{1}} \\\\ & \\Rightarrow \\dfrac{{{V}_{1}}}{{{V}_{2}}}=1 \\\\ \\end{align}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span> <\/span>","column":2}]}],"id_ques":1681},{"time":24,"part":[{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["1","3"]],"list":[{"point":10,"img":"","ques":"Ph\u1ea7n m\u1eb7t ph\u1eb3ng \u0111i qua tr\u1ee5c $OO'$ n\u1eb1m trong h\u00ecnh tr\u1ee5 l\u00e0 m\u1ed9t h\u00ecnh ch\u1eef nh\u1eadt c\u00f3 chi\u1ec1u d\u00e0i $3$$cm,$ chi\u1ec1u r\u1ed9ng $2$ $cm.$ T\u00ednh di\u1ec7n t\u00edch xung quanh v\u00e0 th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh tr\u1ee5 \u0111\u00f3.","hint":"Ta c\u1ea7n x\u00e9t hai tr\u01b0\u1eddng h\u1ee3p: H\u00ecnh tr\u1ee5 c\u00f3 chi\u1ec1u cao b\u1eb1ng chi\u1ec1u d\u00e0i h\u00ecnh ch\u1eef nh\u1eadt v\u00e0 h\u00ecnh tr\u1ee5 c\u00f3 chi\u1ec1u cao b\u1eb1ng chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt","column":1,"number_true":2,"select":["A. $S_{xq}\\approx 18,85\\, (cm^2)$ v\u00e0 $V\\approx 9,42 \\,(cm^3)$","B. $S_{xq}\\approx 18,85\\, (cm^2)$ v\u00e0 $V\\approx 10,08 \\,(cm^3)$","C. $S_{xq}\\approx 18,85\\, (cm^2)$ v\u00e0 $V\\approx 14,14 \\,(cm^3)$","D. $S_{xq}\\approx 15,25\\, (cm^2)$ v\u00e0 $V\\approx 9,42 \\,(cm^3)$"],"explain":"<span class='basic_left'><b>Tr\u01b0\u1eddng h\u1ee3p 1: <\/b> Chi\u1ec1u cao c\u1ee7a h\u00ecnh tr\u1ee5 b\u1eb1ng $3$<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai22/lv3/img\/H941_K1c.png' \/><\/center><br\/>Ta c\u00f3 \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y b\u1eb1ng chi\u1ec1u r\u1ed9ng h\u00ecnh ch\u1eef nh\u1eadt v\u00e0 b\u1eb1ng $2$$cm.$ Suy ra b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 $1$ $cm$<br\/>Di\u1ec7n t\u00edch xung quanh h\u00ecnh tr\u1ee5 l\u00e0 $S=2\\pi Rh=2\\pi .1.3=6\\pi \\approx 18,85\\left( c{{m}^{2}} \\right)$ <br\/>Th\u1ec3 t\u00edch h\u00ecnh tr\u1ee5 l\u00e0 $V=\\pi {{R}^{2}}h=\\pi .1.3=3\\pi \\approx 9,42\\left( c{{m}^{3}} \\right)$ <br\/><b>Tr\u01b0\u1eddng h\u1ee3p 2: <\/b> Chi\u1ec1u cao c\u1ee7a h\u00ecnh tr\u1ee5 l\u00e0 $2$ $cm$<br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai22/lv3/img\/H941_K2.png' \/><\/center><br\/>Ta c\u00f3 \u0111\u01b0\u1eddng k\u00ednh \u0111\u00e1y b\u1eb1ng $3$ $cm.$ Suy ra b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 $1,5$ $cm$<br\/>Di\u1ec7n t\u00edch xung quanh h\u00ecnh tr\u1ee5 l\u00e0 $S=2\\pi Rh=2\\pi .1,5.2=6\\pi \\approx 18,85\\left( c{{m}^{2}} \\right)$ <br\/>Th\u1ec3 t\u00edch h\u00ecnh tr\u1ee5 l\u00e0 $V=\\pi {{R}^{2}}h=\\pi .1,{{5}^{2}}.2=4,5\\pi \\approx 14,14\\left( c{{m}^{3}} \\right)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 A v\u00e0 C.<\/span><br\/><span class='basic_green'>L\u01b0u \u00fd:<\/span> C\u1ea7n x\u00e9t \u0111\u1ee7 hai tr\u01b0\u01a1ng h\u1ee3p c\u1ee7a b\u00e0i to\u00e1n. Trong hai tr\u01b0\u1eddng h\u1ee3p, di\u1ec7n t\u00edch xung quanh c\u1ee7a h\u00ecnh tr\u1ee5 \u0111\u1ec1u b\u1eb1ng nhau nh\u01b0ng th\u1ec3 t\u00edch kh\u00e1c nhau. Th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh tr\u1ee5 s\u1ebd l\u1edbn h\u01a1n n\u1ebfu ta l\u1ea5y chi\u1ec1u r\u1ed9ng c\u1ee7a m\u1eb7t c\u1eaft l\u00e0m chi\u1ec1u cao.<\/span>"}]}],"id_ques":1682},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<span class='basic_left'>M\u1ed9t c\u1ed1c th\u1ee7y tinh h\u00ecnh tr\u1ee5 \u0111\u01b0\u1ee3c \u0111\u1ed5 \u0111\u1ea7y s\u1eefa. C\u00f3 th\u1ec3 r\u00f3t ra \u0111\u00fang m\u1ed9t n\u1eeda s\u1ed1 s\u1eefa trong c\u1ed1c m\u00e0 kh\u00f4ng c\u1ea7n s\u1eed d\u1ee5ng c\u00e1c d\u1ee5ng c\u1ee5 \u0111o hay kh\u00f4ng? ","select":["A. C\u00f3 th\u1ec3 \u0111\u01b0\u1ee3c","B. Kh\u00f4ng th\u1ec3 \u0111\u01b0\u1ee3c"],"hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai22/lv3/img\/H941_K3a.png' \/><\/center><br\/>C\u00f3 th\u1ec3 \u0111\u01b0\u1ee3c (nh\u01b0 h\u00ecnh v\u1ebd). Ta nghi\u00eang c\u1ed1c h\u00ecnh tr\u1ee5 v\u00e0 nh\u1eb9 nh\u00e0ng \u0111\u1ed5 ra cho \u0111\u1ebfn khi n\u01b0\u1edbc trong c\u1ed1c v\u1eeba ch\u1ea1m v\u00e0o \u0111\u00e1y. Khi \u0111\u00f3 trong c\u1ed1c c\u00f2n l\u1ea1i \u0111\u00fang m\u1ed9t n\u1eeda s\u1ed1 s\u1eefa c\u00f3 l\u00fac \u0111\u1ea7u<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span> <\/span>","column":2}]}],"id_ques":1683},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["26"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'>M\u1ed9t h\u00ecnh tr\u1ee5 c\u00f3 chu vi \u0111\u00e1y b\u1eb1ng $48$ $cm,$ chi\u1ec1u cao b\u1eb1ng $10$ $cm.$ $ABCD$ l\u00e0 m\u1eb7t c\u1eaft \u0111i qua tr\u1ee5c c\u1ee7a h\u00ecnh tr\u1ee5. T\u00ednh \u0111\u1ed9 d\u00e0i ng\u1eafn nh\u1ea5t tr\u00ean m\u1eb7t h\u00ecnh tr\u1ee5 m\u00e0 m\u1ed9t con ki\u1ebfn ph\u1ea3i b\u00f2 t\u1eeb $A$ \u0111\u1ebfn $C$<br\/><b>\u0110\u00e1p s\u1ed1: <\/b> _input_$(cm)$<img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai22/lv3/img\/H941_K4.png' \/> <\/span>","hint":"Ta c\u1eaft m\u1eb7t xung quanh c\u1ee7a h\u00ecnh tr\u1ee5 theo c\u00e1c \u0111\u01b0\u1eddng sinh $AD,$ $BC$ r\u1ed3i tr\u1ea3i ph\u1eb3ng m\u1ed9t n\u1eeda m\u1eb7t xung quanh c\u1ee7a h\u00ecnh tr\u1ee5, ta \u0111\u01b0\u1ee3c h\u00ecnh ch\u1eef nh\u1eadt $ABCD.$ T\u1eeb \u0111\u00f3 ta t\u00ecm \u0111\u1ed9 d\u00e0i ng\u1eafn nh\u1ea5t \u0111\u1ec3 \u0111i t\u1eeb $A$ \u0111\u1ebfn $C$ tr\u00ean h\u00ecnh khai tri\u1ec3n n\u00e0y.","explain":"<span class='basic_left'>Ta c\u1eaft m\u1eb7t xung quanh c\u1ee7a h\u00ecnh tr\u1ee5 theo c\u00e1c \u0111\u01b0\u1eddng sinh $AD,$ $BC$ r\u1ed3i tr\u1ea3i ph\u1eb3ng m\u1ed9t n\u1eeda m\u1eb7t xung quanh c\u1ee7a h\u00ecnh tr\u1ee5, ta \u0111\u01b0\u1ee3c h\u00ecnh ch\u1eef nh\u1eadt $ABCD$ c\u00f3 $BC =10$$cm,$ $AB$ b\u1eb1ng n\u1eeda chu vi \u0111\u00e1y c\u1ee7a h\u00ecnh tr\u1ee5, t\u1ee9c b\u1eb1ng $48:2=24$ $(cm)$<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai22/lv3/img\/H941_K7.png' \/><\/center><br\/>\u0110\u1ed9 d\u00e0i ng\u1eafn nh\u1ea5t \u0111\u1ec3 \u0111i t\u1eeb $A$ \u0111\u1ebfn $C$ tr\u00ean m\u1eb7t h\u00ecnh tr\u1ee5 b\u1eb1ng \u0111\u1ed9 d\u00e0i $AC$ tr\u00ean h\u00ecnh khai tri\u1ec3n.<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago v\u00e0o tam gi\u00e1c vu\u00f4ng $ABC,$ ta c\u00f3:<br\/>$\\begin{align} & A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}={{24}^{2}}+{{10}^{2}}=676 \\\\ & \\Rightarrow AC=\\sqrt{676}=26\\left( cm \\right) \\\\ \\end{align}$ <br\/>V\u1eady \u0111\u1ed9 d\u00e0i ng\u1eafn nh\u1ea5t tr\u00ean m\u1eb7t h\u00ecnh tr\u1ee5 m\u00e0 m\u1ed9t con ki\u1ebfn ph\u1ea3i b\u1ecf t\u1eeb $A$ \u0111\u1ebfn $C$ l\u00e0 $26$ $cm$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $26.$<\/span><\/span>"}]}],"id_ques":1684},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u ''$>,=,<$'' th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["="],[">"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Cho h\u00ecnh ch\u1eef nh\u1eadt $ABCD$ c\u00f3 $AB=a,$ $BC=b,$ trong \u0111\u00f3 $a>b.$ Quay h\u00ecnh ch\u1eef nh\u1eadt \u0111\u00f3 m\u1ed9t v\u00f2ng quanh c\u1ea1nh $BC,$ ta \u0111\u01b0\u1ee3c m\u1ed9t h\u00ecnh tr\u1ee5 c\u00f3 di\u1ec7n t\u00edch xung quanh b\u1eb1ng $S_1$ v\u00e0 th\u1ec3 t\u00edch $V_1.$ Quay h\u00ecnh ch\u1eef nh\u1eadt \u0111\u00f3 m\u1ed9t v\u00f2ng quanh c\u1ea1nh $AB,$ ta \u0111\u01b0\u1ee3c m\u1ed9t h\u00ecnh tr\u1ee5 c\u00f3 di\u1ec7n t\u00edch xung quanh b\u1eb1ng $S_2$ v\u00e0 th\u1ec3 t\u00edch b\u1eb1ng $V_2.$So s\u00e1nh $S_1$ v\u00e0 $S_2;$ $V_1$ v\u00e0 $V_2$<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $S_1$ _input_ $S_2;$ $V_1$ _input_ $V_2$<\/span>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: V\u1ebd h\u00ecnh cho t\u1eebng tr\u01b0\u1eddng h\u1ee3p<br\/>B\u01b0\u1edbc 2: T\u00ednh $S_1;$ $V_1;$ $S_2;$$V_2$ b\u1eb1ng c\u00e1ch \u00e1p d\u1ee5ng c\u00f4ng th\u1ee9c ${{S}_{xq}}=2\\pi Rh;V=\\pi {{R}^{2}}h$ <br\/>B\u01b0\u1edbc 3: So s\u00e1nh $S_1$ v\u00e0 $S_2.$ \u0110\u1ec3 so s\u00e1nh $V_1$ v\u00e0 $V_2,$ ta so s\u00e1nh $\\dfrac{{{V}_{1}}}{{{V}_{2}}}$ v\u1edbi $1$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai22/lv3/img\/H941_K5.png' \/>$\\hspace{2cm}$<img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai22/lv3/img\/H941_K6.png' \/><br\/><br\/>\u1ede h\u00ecnh th\u1ee9 nh\u1ea5t, ta c\u00f3:<br\/>${{S}_{1}}=2\\pi ab$ <br\/>${{V}_{1}}=\\pi {{a}^{2}}b$ <br\/>\u1ede h\u00ecnh th\u1ee9 hai, ta c\u00f3:<br\/> ${{S}_{2}}=2\\pi ba$<br\/> ${{V}_{1}}=\\pi {{b}^{2}}a$ <br\/>Suy ra <br\/>${{S}_{1}}={{S}_{2}}$ <br\/>$\\dfrac{{{V}_{1}}}{{{V}_{2}}}=\\dfrac{\\pi {{a}^{2}}b}{\\pi {{b}^{2}}a}=\\dfrac{a}{b}>1$ (do $a>b$)<br\/>V\u1eady ${{V}_{1}}>{{V}_{2}}$<br\/> <span class='basic_pink'>V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $=$ v\u00e0 $>$<\/span><\/span>"}]}],"id_ques":1685},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2800"],["490"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>T\u00ednh th\u1ec3 t\u00edch c\u1ee7a v\u1eadt th\u1ec3 h\u00ecnh h\u1ecdc sau v\u1edbi c\u00e1c k\u00edch th\u01b0\u1edbc \u0111\u00e3 cho tr\u00ean h\u00ecnh v\u1ebd <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai22/lv3/img\/H941_K9.png' \/><\/center><br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $V =$_input_$+$ _input_$\\pi$$(cm^3)$<\/span>","hint":"","explain":"<span class='basic_left'>Ta c\u00f3: <br\/>- Ph\u1ea7n d\u01b0\u1edbi c\u1ee7a h\u00ecnh v\u1ebd l\u00e0 h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt v\u1edbi \u0111\u00e1y c\u00f3 k\u00edch th\u01b0\u1edbc $20$$cm,$ $14$ $cm$ v\u00e0 chi\u1ec1u cao l\u00e0 $10$ $cm.$<br\/>Th\u1ec3 t\u00edch h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt l\u00e0: ${{V}_{1}}=14.20.10=2800\\left( c{{m}^{3}} \\right)$<br\/> - Ph\u1ea7n tr\u00ean l\u00e0 n\u1eeda h\u00ecnh tr\u1ee5 v\u1edbi b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 $\\dfrac{14}{2}=7\\left( cm \\right)$ v\u00e0 chi\u1ec1u cao l\u00e0 $20$ $cm.$<br\/> Th\u1ec3 t\u00edch n\u1eeda h\u00ecnh tr\u1ee5 l\u00e0 ${{V}_{2}}=\\dfrac{\\pi R^2h}{2}=\\dfrac{\\pi .7^2.20}{2}=490\\pi \\left( c{{m}^{3}} \\right)$<br\/> Th\u1ec3 t\u00edch c\u1ee7a v\u1eadt th\u1ec3 l\u00e0:<br\/>$V={{V}_{1}}+{{V}_{2}}=2800+490\\pi \\left( c{{m}^{3}} \\right)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $2800$ v\u00e0 $490.$ <\/span><\/span>"}]}],"id_ques":1686},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["140"],["140"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>H\u00e3y t\u00ednh th\u1ec3 t\u00edch v\u00e0 di\u1ec7n t\u00edch b\u1ec1 m\u1eb7t m\u1ed9t chi ti\u1ebft m\u00e1y theo k\u00edch th\u01b0\u1edbc \u0111\u00e3 cho tr\u00ean h\u00ecnh v\u1ebd sau:<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai22/lv3/img\/H941_K10.png' \/><\/center><br\/><b>\u0110\u00e1p s\u1ed1: <\/b><br\/>(a) Th\u1ec3 t\u00edch chi ti\u1ebft m\u00e1y l\u00e0: _input_$\\pi$$(cm^3)$ <br\/>(b) Di\u1ec7n t\u00edch b\u1ec1 m\u1eb7t c\u1ee7a chi ti\u1ebft m\u00e1y l\u00e0 _input_ $\\pi$$(cm^2)$<\/span>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn: <\/span><br\/>B\u01b0\u1edbc 1: Ph\u00e2n t\u00edch b\u00e0i to\u00e1n: Chi ti\u1ebft m\u00e1y g\u1ed3m hai h\u00ecnh tr\u1ee5 c\u00f3 c\u00e1c k\u00edch th\u01b0\u1edbc th\u1ebf n\u00e0o?<br\/>B\u01b0\u1edbc 2: T\u00ednh th\u1ec3 t\u00edch chi ti\u1ebft: B\u1eb1ng t\u1ed5ng th\u1ec3 t\u00edch c\u1ee7a hai h\u00ecnh tr\u1ee5<br\/>B\u01b0\u1edbc 3: T\u00ednh di\u1ec7n t\u00edch b\u1ec1 m\u1eb7t: B\u1eb1ng t\u1ed5ng di\u1ec7n t\u00edch xung quanh v\u00e0 di\u1ec7n t\u00edch c\u00e1c m\u1eb7t ngo\u00e0i c\u00f2n l\u1ea1i (g\u1ed3m \u0111\u00e1y tr\u00ean, ph\u1ea7n \u0111\u00e1y d\u01b0\u1edbi ph\u00eda ngo\u00e0i c\u1ee7a h\u00ecnh tr\u1ee5 tr\u00ean v\u00e0 \u0111\u00e1y d\u01b0\u1edbi c\u1ee7a c\u1ee7a h\u00ecnh tr\u1ee5 d\u01b0\u1edbi)<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Chi ti\u1ebft m\u00e1y g\u1ed3m hai h\u00ecnh tr\u1ee5<br\/>+ H\u00ecnh tr\u1ee5 tr\u00ean c\u00f3 b\u00e1n k\u00ecnh \u0111\u00e1y l\u00e0 $6cm, $ chi\u1ec1u cao $3 cm$<br\/>+ H\u00ecnh tr\u1ee5 d\u01b0\u1edbi c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 $2cm$ v\u00e0 chi\u1ec1u cao $8cm$<br\/> (a) Th\u1ec3 t\u00edch c\u1ee7a chi ti\u1ebft m\u00e1y l\u00e0:<br\/>$V=\\pi {{.6}^{2}}.3+\\pi {{.2}^{2}}.8=140\\pi \\left( c{{m}^{3}} \\right)$ <br\/>(b) Di\u1ec7n t\u00edch xung quanh c\u1ee7a hai h\u00ecnh tr\u1ee5 l\u00e0:<br\/>${{S}_{1}}=2\\pi .6.3+2\\pi .2.8=68\\pi \\left( c{{m}^{2}} \\right)$ <br\/>T\u1ed5ng di\u1ec7n t\u00edch c\u00e1c m\u1eb7t ngo\u00e0i c\u00f2n l\u1ea1i l\u00e0:<br\/>${{S}_{2}}=\\pi {{.6}^{2}}+\\pi \\left( {{6}^{2}}-{{2}^{2}} \\right)+\\pi {{.2}^{2}}=72\\pi \\left( c{{m}^{2}} \\right)$ <br\/>Di\u1ec7n t\u00edch b\u1ec1 m\u1eb7t c\u1ee7a chi ti\u1ebft m\u00e1y l\u00e0: $S={{S}_{1}}+{{S}_{2}}=68\\pi +72\\pi =140\\pi \\left( c{{m}^{2}} \\right)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u1ea7n l\u01b0\u1ee3t l\u00e0 $140$ v\u00e0 $140$ <\/span><\/span>"}]}],"id_ques":1687},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<span class='basic_left'>M\u1ed9t b\u0103ng gi\u1ea5y d\u1ea3i \u0111\u01b0\u1ee3c cu\u1ed9n ch\u1eb7t l\u1ea1i $60$ v\u00f2ng l\u00e0m th\u00e0nh m\u1ed9t cu\u1ed9n gi\u1ea5y h\u00ecnh tr\u1ee5 r\u1ed7ng. Bi\u1ebft \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n trong c\u00f9ng b\u1eb1ng $2$$cm,$ \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ngo\u00e0i c\u00f9ng b\u1eb1ng $6$$cm.$ T\u00ednh chi\u1ec1u d\u00e0i c\u1ee7a b\u0103ng gi\u1ea5y.<\/span>","select":["A. $747,7$ $(cm)$ ","B. $856,4$ $(cm)$ ","C. $647,6$ $(cm)$ ","D. $912,3$ $(cm)$ "],"hint":"Chi\u1ec1u d\u00e0i c\u1ee7a b\u0103ng gi\u1ea5y b\u1eb1ng t\u1ed5ng \u0111\u1ed9 d\u00e0i c\u1ee7a $60$ v\u00f2ng tr\u00f2n. Ta c\u1ea7n x\u00e1c \u0111\u1ecbnh \u0111\u1ed9 d\u00e0y c\u1ee7a b\u0103ng gi\u1ea5y, t\u1eeb \u0111\u00f3 bi\u1ec3u di\u1ec5n b\u00e1n k\u00ednh c\u1ee7a c\u00e1c v\u00f2ng tr\u00f2n b\u00ean trong theo \u0111\u1ed9 d\u00e0y, b\u00e1n k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n trong c\u00f9ng v\u00e0 b\u00e1n k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ngo\u00e0i c\u00f9ng.","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai22/lv3/img\/H941_K11.png' \/><\/center><br\/><br\/>G\u1ecdi l l\u00e0 chi\u1ec1u d\u00e0i c\u1ee7a b\u0103ng gi\u1ea5y, chi\u1ec1u d\u00e0i \u0111\u00f3 b\u1eb1ng t\u1ed5ng \u0111\u1ed9 d\u00e0i c\u1ee7a $60$ \u0111\u01b0\u1eddng tr\u00f2n c\u00f3 b\u00e1n k\u00ednh ${{r}_{1}},{{r}_{2}},...,{{r}_{60}}$, trong \u0111\u00f3 ${{r}_{1}}=1cm,{{r}_{60}}=3cm$ <br\/>Ta c\u00f3:<br\/>$l=2\\pi \\left( {{r}_{1}}+{{r}_{2}}+...+{{r}_{60}} \\right)$ <br\/>\u0110\u1ed9 d\u00e0y c\u1ee7a $60$ v\u00f2ng gi\u1ea5y l\u00e0 $d={{r}_{60}}-{{r}_{1}}=3-1=2$<br\/>Suy ra \u0111\u1ed9 d\u00e0y c\u1ee7a t\u1edd gi\u1ea5y l\u00e0 $\\dfrac{2}{60}$$(cm)$<br\/>Do \u0111\u00f3:<br\/>$\\begin{align} & {{r}_{2}}={{r}_{1}}+\\dfrac{2}{60} \\\\ & {{r}_{3}}={{r}_{2}}+\\dfrac{2}{60}={{r}_{1}}+2.\\dfrac{2}{60} \\\\ & .... \\\\ & {{r}_{60}}={{r}_{59}}+\\dfrac{2}{60}={{r}_{1}}+59.\\dfrac{2}{60} \\\\ \\end{align}$ <br\/>V\u1eady chi\u1ec1u d\u00e0i c\u1ee7a b\u0103ng gi\u1ea5y l\u00e0:<br\/>$\\begin{align} & l=2\\pi \\left[ {{r}_{1}}+\\left( {{r}_{1}}+\\dfrac{2}{60} \\right)+\\left( {{r}_{1}}+2.\\dfrac{2}{60} \\right)+...+\\left( {{r}_{1}}+59.\\dfrac{2}{60} \\right) \\right] \\\\ & \\,\\,\\,=2\\pi \\left[ 60{{r}_{1}}+\\dfrac{2}{60}\\left( 1+2+3+...+59 \\right) \\right] \\\\ & \\,\\,\\,=2\\pi \\left[ 60.1+\\dfrac{2}{60}\\dfrac{\\left( 59+1 \\right)59}{2} \\right]\\\\ & \\,\\,\\,=747,7\\left( cm \\right) \\\\ \\end{align}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":1688},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<span class='basic_left'>M\u1ed9t h\u1ed9p thu\u1ed1c h\u00ecnh tr\u1ee5 \u0111\u01b0\u1ee3c \u0111\u1eb7t kh\u00edt trong m\u1ed9t h\u1ed9p gi\u1ea5y h\u00ecnh ch\u1eef nh\u1eadt. Th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh tr\u1ee5 b\u1eb1ng bao nhi\u00eau ph\u1ea7n tr\u0103m th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh h\u1ed9p ch\u1eef nh\u1eadt? <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai22/lv3/img\/H941_K12.png' \/><\/center><\/span>","select":["A. $87,54\\% $ ","B. $78,54\\%$ ","C. $88,52\\%$","D. $75,45\\%$"],"hint":"V\u00ec h\u1ed9p thu\u1ed1c \u0111\u1eb7t kh\u00edt trong h\u1ed9p gi\u1ea5y n\u00ean \u0111\u00e1y h\u1ed9p gi\u1ea5y c\u00f3 chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng b\u1eb1ng \u0111\u01b0\u1eddng k\u00ednh c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n \u0111\u00e1y c\u1ee7a h\u1ed9p thu\u1ed1c","explain":"<span class='basic_left'>Gi\u1ea3 s\u1eed h\u00ecnh tr\u1ee5 c\u00f3 b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 $R$ v\u00e0 chi\u1ec1u cao l\u00e0 $h.$ <br\/>Khi \u0111\u00f3, th\u1ec3 t\u00edch c\u1ee7a h\u1ed9p gi\u1ea5y l\u00e0: ${{V}_{1}}=\\pi {{R}^{2}}h$ (\u0111\u01a1n v\u1ecb th\u1ec3 t\u00edch)<br\/>V\u00ec h\u1ed9p thu\u1ed1c \u0111\u1eb7t kh\u00edt trong h\u1ed9p gi\u1ea5y n\u00ean \u0111\u00e1y h\u1ed9p gi\u1ea5y c\u00f3 chi\u1ec1u d\u00e0i v\u00e0 chi\u1ec1u r\u1ed9ng l\u00e0 $2R$ v\u00e0 chi\u1ec1u cao l\u00e0 $h.$<br\/> Khi \u0111\u00f3 th\u1ec3 t\u00edch h\u1ed9p gi\u1ea5y l\u00e0: ${{V}_{2}}=2R.2R.h=4{{R}^{2}}.h$ (\u0111\u01a1n v\u1ecb th\u1ec3 t\u00edch)<br\/>Suy ra $\\dfrac{{{V}_{1}}}{{{V}_{2}}}.100\\%=\\dfrac{\\pi {{R}^{2}}h}{4{{R}^{2}}.h}.100\\%=\\dfrac{\\pi }{4}.100 \\% \\approx 78,54\\%$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":1689},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<span class='basic_left'>Trong c\u00e1c h\u00ecnh tr\u1ee5 c\u00f3 c\u00f9ng th\u1ec3 t\u00edch, h\u00ecnh tr\u1ee5 c\u00f3 \u0111\u01b0\u1eddng cao b\u1eb1ng b\u00e1n k\u00ednh \u0111\u00e1y l\u00e0 h\u00ecnh c\u00f3 di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n nh\u1ecf nh\u1ea5t, <b>\u0111\u00fang<\/b> hay <b>sai<\/b>?<\/span>","select":["A. \u0110\u00fang ","B. Sai "],"hint":"$S_{tp}=2\\pi rh+2\\pi {{r}^{2}}=2\\pi \\left( rh+{{r}^{2}} \\right)$ n\u00ean $S_{tp}$ nh\u1ecf nh\u1ea5t khi $rh+{{r}^{2}}$ nh\u1ecf nh\u1ea5t. Ta bi\u1ec3u di\u1ec5n $rh$ qua th\u1ec3 t\u00edch $V$ sau \u0111\u00f3 \u00e1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c C\u00f4 si cho ba s\u1ed1 kh\u00f4ng \u00e2m \u0111\u1ec3 \u0111\u00e1nh gi\u00e1 t\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t.","explain":"<span class='basic_left'>G\u1ecdi $S$ l\u00e0 di\u1ec7n t\u00edch to\u00e0n ph\u1ea7n, $r$ l\u00e0 b\u00e1n k\u00ednh \u0111\u00e1y v\u00e0 $h$ l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a h\u00ecnh tr\u1ee5.<br\/>Ta c\u00f3:<br\/>$S=2\\pi rh+2\\pi {{r}^{2}}=2\\pi \\left( rh+{{r}^{2}} \\right)$<br\/>$S$ nh\u1ecf nh\u1ea5t khi v\u00e0 ch\u1ec9 khi $rh+{{r}^{2}}$ nh\u1ecf nh\u1ea5t.<br\/>G\u1ecdi $V$ l\u00e0 th\u1ec3 t\u00edch c\u1ee7a h\u00ecnh tr\u1ee5, $V$ l\u00e0 h\u1eb1ng s\u1ed1 v\u00e0 $V=\\pi {{r}^{2}}h.$ Suy ra $rh=\\dfrac{V}{\\pi r}$ <br\/>Ta c\u00f3: $rh+{{r}^{2}}=\\dfrac{V}{\\pi r}+{{r}^{2}}=\\dfrac{V}{2\\pi r}+\\dfrac{V}{2\\pi r}+{{r}^{2}}$<br\/>\u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c C\u00f4 si cho 3 s\u1ed1 kh\u00f4ng \u00e2m, ta c\u00f3:<br\/>$rh+{{r}^{2}}\\ge 3\\sqrt[3]{\\dfrac{V}{2\\pi r}.\\dfrac{V}{2\\pi r}.{{r}^{2}}}=3\\sqrt[3]{\\dfrac{{{V}^{2}}}{4{{\\pi }^{2}}}}=$ h\u1eb1ng s\u1ed1.<br\/>D\u1ea5u ''$=$'' x\u1ea3y ra $\\Leftrightarrow \\dfrac{V}{2\\pi r}={{r}^{2}}\\Leftrightarrow \\dfrac{rh}{2}={{r}^{2}}\\Leftrightarrow h=2r$ <br\/>V\u1eady $S$ nh\u1ecf nh\u1ea5t khi $h=2r$ hay chi\u1ec1u cao b\u1eb1ng \u0111\u01b0\u1eddng k\u00ednh.<br\/>V\u1eady kh\u1eb3ng \u0111\u1ecbnh b\u00e0i to\u00e1n l\u00e0 sai.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B.<\/span><\/span>","column":2}]}],"id_ques":1690}],"lesson":{"save":0,"level":3}}