{"segment":[{"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Di\u1ec7n t\u00edch h\u00ecnh v\u00e0nh kh\u0103n gi\u1edbi h\u1ea1n b\u1edfi hai \u0111\u01b0\u1eddng tr\u00f2n $(O; 10cm)$ v\u00e0 $(O; 6cm)$ l\u00e0: ","select":["A. $64\\pi \\,\\left( c{{m}^{2}} \\right)$","B. $60\\pi \\,\\left( c{{m}^{2}} \\right)$","C. $72\\pi \\,\\left( c{{m}^{2}} \\right)$","D. $80\\pi \\,\\left( c{{m}^{2}} \\right)$"],"explain":" <span class='basic_left'><center> <img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai21/lv3/img\/h939_K1.1.png' \/><\/center> <br\/> Di\u1ec7n t\u00edch h\u00ecnh tr\u00f2n $(O; 10cm)$ l\u00e0: ${{S}_{1}}=\\pi {{R}^{2}}=\\pi {{.10}^{2}}=100\\pi \\,\\left( c{{m}^{2}} \\right)$ <br\/> Di\u1ec7n t\u00edch h\u00ecnh tr\u00f2n $(O; 6cm)$ l\u00e0: ${{S}_{2}}=\\pi R{{'}^{2}}=\\pi {{.6}^{2}}=36\\pi \\,\\left( c{{m}^{2}} \\right)$ <br\/> Di\u1ec7n t\u00edch h\u00ecnh v\u00e0nh kh\u0103n l\u00e0: $S={{S}_{1}}-{{S}_{2}}=100\\pi -36\\pi =64\\pi \\,\\left( c{{m}^{2}} \\right)$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A <\/span> <\/span>","column":2}],"id_ques":1651},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["60"]]],"list":[{"point":10,"width":50,"ques":"T\u1ee9 gi\u00e1c $ABCD$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n c\u00f3 $\\widehat{DAB}={{120}^{o}}$. V\u1eady s\u1ed1 \u0111o g\u00f3c $BCD$ l\u00e0 _input_ $^o$","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai21/lv3/img\/h939_K1.2.png' \/><\/center> <br\/> Ta c\u00f3: $\\widehat{DAB}+\\widehat{BCD}={{180}^{o}}$ (\u0111\u1ecbnh l\u00ed t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp) <br\/> $\\Rightarrow \\widehat{BCD}={{180}^{o}}-\\widehat{DAB}\\\\ \\hspace{0.5cm}={{180}^{o}}-{{120}^{o}}={{60}^{o}}$ <br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $60$ <\/span><\/span><\/span> "}],"id_ques":1652},{"time":3,"title":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O;R)$ v\u00e0 m\u1ed9t d\u00e2y $AB$, tr\u00ean tia $BA$ l\u1ea5y \u0111i\u1ec3m $C$ n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n. T\u1eeb \u0111i\u1ec3m ch\u00ednh gi\u1eefa $P$ c\u1ee7a cung l\u1edbn $AB$ k\u1ebb \u0111\u01b0\u1eddng k\u00ednh $PQ$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n c\u1eaft d\u00e2y $AB$ t\u1ea1i $D$. Tia $CP$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $I$. C\u00e1c d\u00e2y $AB$ v\u00e0 $QI$ c\u1eaft nhau t\u1ea1i $K$. <br\/> <b> C\u00e2u a: <\/b> Ch\u1ee9ng minh t\u1ee9 gi\u00e1c $PDKI$ n\u1ed9i ti\u1ebfp","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[5],[4],[3],[2],[1]]],"list":[{"point":10,"left":["$\\Rightarrow $ T\u1ee9 gi\u00e1c $PIKD$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n","$\\Rightarrow $ C\u00e1c \u0111i\u1ec3m $I, D$ c\u00f9ng nh\u00ecn $PK$ d\u01b0\u1edbi m\u1ed9t g\u00f3c vu\u00f4ng","M\u00e0 $\\widehat{PIK}={{90}^{o}}$ (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n) "," $\\Rightarrow \\widehat{PDK}={{90}^{o}}$ (\u0111\u1ecbnh l\u00ed \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung)"," Ta c\u00f3 $P$ l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa cung l\u1edbn $AB$"],"top":60,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai21/lv3/img\/h939_K1.3.png' \/><\/center> <br\/> Ta c\u00f3 $P$ l\u00e0 \u0111i\u1ec3m ch\u00ednh gi\u1eefa cung l\u1edbn $AB$ <br\/> $\\Rightarrow \\widehat{PDK}={{90}^{o}}$ (\u0111\u1ecbnh l\u00ed \u0111\u01b0\u1eddng k\u00ednh v\u00e0 d\u00e2y cung) <br\/>M\u00e0 $\\widehat{PIK}={{90}^{o}}$ (g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n) <br\/> $\\Rightarrow $ C\u00e1c \u0111i\u1ec3m $I, D$ c\u00f9ng nh\u00ecn $PK$ d\u01b0\u1edbi m\u1ed9t g\u00f3c vu\u00f4ng <br\/> $\\Rightarrow $ T\u1ee9 gi\u00e1c $PIKD$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n<\/span>"}],"id_ques":1653},{"time":3,"title":"<span class='basic_left'> Cho \u0111\u01b0\u1eddng tr\u00f2n $(O;R)$ v\u00e0 m\u1ed9t d\u00e2y $AB$, tr\u00ean tia $BA$ l\u1ea5y \u0111i\u1ec3m $C$ n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n. T\u1eeb \u0111i\u1ec3m ch\u00ednh gi\u1eefa $P$ c\u1ee7a cung l\u1edbn $AB$ k\u1ebb \u0111\u01b0\u1eddng k\u00ednh $PQ$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n c\u1eaft d\u00e2y $AB$ t\u1ea1i $D$. Tia $CP$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $I$. C\u00e1c d\u00e2y $AB$ v\u00e0 $QI$ c\u1eaft nhau t\u1ea1i $K$. <br\/> <b> C\u00e2u b: <\/b> Ch\u1ee9ng minh $CI.CP = CA.CB$","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[3],[1],[4],[2]]],"list":[{"point":10,"left":["$\\Rightarrow \\Delta CIA\\sim \\Delta CBP\\,\\left( g.g \\right)$ $\\Rightarrow \\dfrac{CI}{CB}=\\dfrac{CA}{CP}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7)","Ta c\u00f3 b\u1ed1n \u0111i\u1ec3m $I, A, B, P$ c\u00f9ng n\u1eb1m tr\u00ean m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n $\\Rightarrow \\widehat{AIC}=\\widehat{PBC}$ (t\u00ednh ch\u1ea5t t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp)"," $\\Rightarrow CI.CP=CA.CB$"," X\u00e9t $\\Delta CIA$ v\u00e0 $\\Delta CBP$ c\u00f3: $\\left\\{ \\begin{align} & \\widehat{AIC}=\\widehat{PBC} \\\\ & \\widehat{C}\\,\\left( \\text{chung} \\right) \\\\ \\end{align} \\right.$"],"top":100,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai21/lv3/img\/h939_K1.4.png' \/><\/center> <br\/> Ta c\u00f3 b\u1ed1n \u0111i\u1ec3m $I, A, B, P$ c\u00f9ng n\u1eb1m tr\u00ean m\u1ed9t \u0111\u01b0\u1eddng tr\u00f2n $(O)$ <br\/> $\\Rightarrow \\widehat{AIC}=\\widehat{PBC}$ (t\u00ednh ch\u1ea5t t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp) <br\/> X\u00e9t $\\Delta CIA$ v\u00e0 $\\Delta CBP$ c\u00f3: <br\/> $\\left\\{ \\begin{align} & \\widehat{AIC}=\\widehat{PBC} \\\\ & \\widehat{C}\\,\\left( \\text{chung} \\right) \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow \\Delta CIA\\sim \\Delta CBP\\,\\left( g.g \\right)$ <br\/> $\\Rightarrow \\dfrac{CI}{CB}=\\dfrac{CA}{CP}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng t\u1ec9 l\u1ec7) <br\/> $\\Rightarrow CI.CP=CA.CB$ <\/span>"}],"id_ques":1654},{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Cho \u0111\u01b0\u1eddng tr\u00f2n $(O)$ \u0111\u01b0\u1eddng k\u00ednh $AB$. V\u1ebd d\u00e2y $CD\/\/ AB$ th\u00ec: ","select":["A. $AC = BD$","B. $AD=BC$ ","C. $ABDC$ l\u00e0 h\u00ecnh thang c\u00e2n","D. T\u1ea5t c\u1ea3 \u0111\u1ec1u \u0111\u00fang"],"explain":" <span class='basic_left'><center> <img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai21/lv3/img\/h939_K1.5.png' \/><\/center> <br\/> Ta c\u00f3: $AB\/\/CD\\Rightarrow \\overset\\frown{AC}=\\overset\\frown{BD}$ (cung b\u1ecb ch\u1eafn gi\u1eefa hai d\u00e2y song song) <br\/> $\\Rightarrow AC = BD$ (\u0111\u1ecbnh l\u00ed li\u00ean h\u1ec7 gi\u1eefa cung v\u00e0 d\u00e2y) <br\/> $\\Rightarrow $ \u0110\u00e1p \u00e1n A \u0111\u00fang <br\/> Ta c\u00f3: $\\overset\\frown{AC}=\\overset\\frown{BD}$ (ch\u1ee9ng minh tr\u00ean) <br\/> $\\Rightarrow \\overset\\frown{AC} + \\overset\\frown{CD}=\\overset\\frown{BD}+ \\overset\\frown{CD}$ hay $ \\overset\\frown{AD}=\\overset\\frown{BC}$ <br\/> $\\Rightarrow \\widehat{CAB} = \\widehat{DBA}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn hai cung b\u1eb1ng nhau) <br\/> T\u1ee9 gi\u00e1c $ABDC$ c\u00f3: $\\left\\{ \\begin{align} & AB\/\/CD \\\\ & \\widehat{CAB} = \\widehat{DBA} \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow ABDC$ l\u00e0 h\u00ecnh thang c\u00e2n <br\/> $\\Rightarrow $ \u0110\u00e1p \u00e1n C \u0111\u00fang <br\/> $ABDC$ l\u00e0 h\u00ecnh thang c\u00e2n $\\Rightarrow AD=BC$ (t\u00ednh ch\u1ea5t h\u00ecnh thang c\u00e2n) <br\/> $\\Rightarrow $ \u0110\u00e1p \u00e1n B \u0111\u00fang <br\/> Suy ra t\u1ea5t c\u1ea3 c\u00e1c \u0111\u00e1p \u00e1n \u0111\u1ec1u \u0111\u00fang <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D <\/span> <\/span>","column":2}],"id_ques":1655},{"time":3,"title":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$. L\u1ea5y \u0111i\u1ec3m $E$ n\u1eb1m tr\u00ean c\u1ea1nh $AB$ v\u00e0 v\u1ebd \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh $EB$. \u0110\u01b0\u1eddng th\u1eb3ng $CE$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $M, AM$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $N$. <br\/> <b> C\u00e2u a: <\/b> Ch\u1ee9ng minh tia $BA$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $CBN$","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[3],[1],[5],[2],[4]]],"list":[{"point":10,"left":[" $\\Rightarrow \\widehat{NMB}=\\widehat{BCA}$ (t\u00ednh ch\u1ea5t t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp) <br\/> M\u1eb7t kh\u00e1c $\\widehat{NMB}=\\widehat{NEB}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung $NB$)"," Ta c\u00f3: $\\left\\{ \\begin{align} & \\widehat{BME}=\\widehat{BNE}={{90}^{o}}\\,\\left( \\text{g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n} \\right) \\\\ & \\widehat{BAC}={{90}^{o}} \\,\\left( \\text{gi\u1ea3 thi\u1ebft} \\right) \\\\ \\end{align} \\right.$ "," $\\Rightarrow \\widehat{NBE} =\\widehat{ABC}$ <br\/> $\\Rightarrow$ Tia $BA$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $CBN$ ","$\\Rightarrow $ T\u1ee9 gi\u00e1c $AMBC$ c\u00f3 c\u00e1c \u0111i\u1ec3m $A, M$ c\u00f9ng nh\u00ecn $BC$ d\u01b0\u1edbi m\u1ed9t g\u00f3c vu\u00f4ng n\u00ean n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n","$\\Rightarrow \\widehat{BCA}=\\widehat{NEB}=\\widehat{NMB}$ <br\/> L\u1ea1i c\u00f3: $ \\widehat{NBE}+\\widehat{NEB}=\\widehat{ABC}+\\widehat{ACB}={{90}^{o}}$ (t\u1ed5ng hai g\u00f3c nh\u1ecdn trong tam gi\u00e1c vu\u00f4ng)"],"top":100,"hint":"Ch\u1ee9ng minh qua t\u00ednh ch\u1ea5t b\u1eafc c\u1ea7u $\\widehat{NMB}=\\widehat{NEB}=\\widehat{BCA}\\Rightarrow \\widehat{NBE}+\\widehat{NEB}=\\widehat{ABC}+\\widehat{ACB}={{90}^{o}}$","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai21/lv3/img\/h939_K1.6.png' \/><\/center> <br\/> Ta c\u00f3: $\\left\\{ \\begin{align} & \\widehat{BME}=\\widehat{BNE}={{90}^{o}}\\,\\left( \\text{g\u00f3c n\u1ed9i ti\u1ebfp ch\u1eafn n\u1eeda \u0111\u01b0\u1eddng tr\u00f2n} \\right) \\\\ & \\widehat{BAC}={{90}^{o}} \\,\\left( \\text{gi\u1ea3 thi\u1ebft} \\right) \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow $ T\u1ee9 gi\u00e1c $AMBC$ c\u00f3 c\u00e1c \u0111i\u1ec3m $A, M$ c\u00f9ng nh\u00ecn $BC$ d\u01b0\u1edbi m\u1ed9t g\u00f3c vu\u00f4ng n\u00ean n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n <br\/> $\\Rightarrow \\widehat{NMB}=\\widehat{BCA}$ (t\u00ednh ch\u1ea5t t\u1ee9 gi\u00e1c n\u1ed9i ti\u1ebfp) <br\/> M\u1eb7t kh\u00e1c $\\widehat{NMB}=\\widehat{NEB}$ (hai g\u00f3c n\u1ed9i ti\u1ebfp c\u00f9ng ch\u1eafn cung $NB$) <br\/> $\\Rightarrow \\widehat{BCA}=\\widehat{NEB}=\\widehat{NMB}$ <br\/> L\u1ea1i c\u00f3: $ \\widehat{NBE}+\\widehat{NEB}=\\widehat{ABC}+\\widehat{ACB}={{90}^{o}}$ (t\u1ed5ng hai g\u00f3c nh\u1ecdn trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow \\widehat{NBE} =\\widehat{ABC}$ <br\/> $\\Rightarrow$ Tia $BA$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $CBN$<\/span>"}],"id_ques":1656},{"time":3,"title":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$. L\u1ea5y \u0111i\u1ec3m $E$ n\u1eb1m tr\u00ean c\u1ea1nh $AB$ v\u00e0 v\u1ebd \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh $EB$. \u0110\u01b0\u1eddng th\u1eb3ng $CE$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $M, AM$ c\u1eaft \u0111\u01b0\u1eddng tr\u00f2n t\u1ea1i $N$. <br\/> <b> C\u00e2u b: <\/b> G\u1ecdi $K$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AC$ v\u00e0 $BM$. Ch\u1ee9ng minh $KE\\bot BC$","title_trans":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh","temp":"sequence","correct":[[[4],[3],[1],[2]]],"list":[{"point":10,"left":[" $\\Rightarrow KE\\bot BC$"," M\u00e0 $AB\\cap MC=E\\Rightarrow E$ l\u00e0 tr\u1ef1c t\u00e2m tam gi\u00e1c $BKC$ "," X\u00e9t $\\Delta BKC$ c\u00f3:","$\\left\\{ \\begin{aligned} & \\widehat{BAC}={{90}^{o}} \\\\ & \\widehat{BME}={{90}^{o}} \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & AB\\bot KC \\\\ & MC\\bot BK \\\\ \\end{aligned} \\right.$"],"top":80,"explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai21/lv3/img\/h939_K1.7.png' \/><\/center> <br\/> X\u00e9t $\\Delta BKC$ c\u00f3: <br\/> $\\left\\{ \\begin{aligned} & \\widehat{BAC}={{90}^{o}} \\\\ & \\widehat{BME}={{90}^{o}} \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & AB\\bot KC \\\\ & MC\\bot BK \\\\ \\end{aligned} \\right.$ <br\/> M\u00e0 $AB\\cap MC=E\\Rightarrow E$ l\u00e0 tr\u1ef1c t\u00e2m tam gi\u00e1c $BKC$ <br\/> $\\Rightarrow KE\\bot BC$ <\/span>"}],"id_ques":1657},{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"<span class='basic_left'> Cho tam gi\u00e1c c\u00e2n $ABC$, c\u00e1c \u0111\u01b0\u1eddng cao $AG, BE, CF$ c\u1eaft nhau t\u1ea1i $H$. <br\/> <b> C\u00e2u a: <\/b> T\u00e2m $I$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp t\u1ee9 gi\u00e1c $AEHF$ l\u00e0:","select":["A. Tr\u1ecdng t\u00e2m tam gi\u00e1c $AEH$ ","B. Tr\u1ecdng t\u00e2m tam gi\u00e1c $AFH$","C. Trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh $EF$ ","D. Trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh $AH$ "],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai21/lv3/img\/h939_K1.8.png' \/><\/center><br\/> Ta c\u00f3: $\\widehat{AFH}=\\widehat{AEH}={{90}^{o}}$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow$ C\u00e1c \u0111i\u1ec3m $E, F$ c\u00f9ng nh\u00ecn c\u1ea1nh $AH$ d\u01b0\u1edbi m\u1ed9t g\u00f3c vu\u00f4ng <br\/> $\\Rightarrow$ T\u1ee9 gi\u00e1c $AEHF$ n\u1ed9i ti\u1ebfp \u0111\u01b0\u1eddng tr\u00f2n \u0111\u01b0\u1eddng k\u00ednh $AH$ <br\/> $\\Rightarrow$ T\u00e2m \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp t\u1ee9 gi\u00e1c $AEHF$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh $AH$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D <\/span><\/span>","column":2}],"id_ques":1658},{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{10\\pi}{9}$","B. $\\dfrac{8\\pi}{9}$","C. $\\dfrac{11\\pi}{9}$"],"ques":"<span class='basic_left'> Cho tam gi\u00e1c c\u00e2n $ABC$ , c\u00e1c \u0111\u01b0\u1eddng cao $AG, BE, CF$ c\u1eaft nhau t\u1ea1i $H$. <br\/> Cho b\u00e1n k\u00ednh \u0111\u01b0\u1eddng tr\u00f2n $(I)$ l\u00e0 $2cm$, $\\widehat{BAC}={{50}^{o}}$. T\u00ednh \u0111\u1ed9 d\u00e0i cung $\\overset\\frown{FHE}$ c\u1ee7a \u0111\u01b0\u1eddng tr\u00f2n t\u00e2m $(I)$ v\u00e0 di\u1ec7n t\u00edch h\u00ecnh qu\u1ea1t tr\u00f2n $IFH$ <br\/> <b> Di\u1ec7n t\u00edch h\u00ecnh qu\u1ea1t tr\u00f2n $IFH$ l\u00e0 ?$cm^2$ <br\/>","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai21/lv3/img\/h939_K1.9.png' \/><\/center> <br\/> Ta c\u00f3: $AG$ l\u00e0 \u0111\u01b0\u1eddng cao n\u00ean l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow \\widehat{FAH} = \\dfrac{1}{2} \\widehat{BAC} = \\dfrac{50^o}{2}=25^o$ <br\/> <br\/> $\\widehat{FIH}=2\\widehat{FAH}={{2.25}^{o}}={{50}^{o}}$ (h\u1ec7 qu\u1ea3 g\u00f3c n\u1ed9i ti\u1ebfp)<br\/> Di\u1ec7n t\u00edch h\u00ecnh qu\u1ea1t tr\u00f2n $IFH$ l\u00e0: <br\/> ${{S}_{IEH}}=\\dfrac{\\pi .{{R}^{2}}.n}{360}=\\dfrac{\\pi .4.50}{360}=\\dfrac{10\\pi }{9}\\,\\left( c{{m}^{2}} \\right)$<\/span><\/span> "}],"id_ques":1659},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","temp":"fill_the_blank","correct":[[["25"]]],"list":[{"point":10,"width":50,"ques":"Cho h\u00ecnh v\u1ebd: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai21/lv3/img\/h939_K1.10.png' \/><\/center> <br\/> G\u00f3c $\\widehat{AIB}$ trong h\u00ecnh v\u1ebd l\u00e0 bao nhi\u00eau \u0111\u1ed9 n\u1ebfu $\\text{s\u0111}\\overset\\frown{AmB}={{70}^{o}},\\,\\text{s\u0111}\\overset\\frown{BnC}={{170}^{o}}$s\u0111 <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $\\widehat{AIB} = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} ^o$ ","explain":" <span class='basic_left'> <br\/> Ta c\u00f3: $\\text{s\u0111}\\overset\\frown{AC}+\\text{s\u0111}\\overset\\frown{AmB}+\\text{s\u0111}\\overset\\frown{BnC}={{360}^{o}}$ (\u0111\u1ecbnh ngh\u0129a s\u1ed1 \u0111o cung) <br\/> $\\Rightarrow \\text{s\u0111}\\overset\\frown{AC}={{360}^{o}}-\\text{s\u0111}\\overset\\frown{AmB}-\\text{s\u0111}\\overset\\frown{BnC}\\\\ ={{360}^{o}}-{{70}^{o}}-{{170}^{o}}={{120}^{o}}$ <br\/> $\\widehat{AIB}=\\dfrac{\\text{s\u0111}\\overset\\frown{AC}-\\text{s\u0111}\\overset\\frown{AmB}}{2}=\\dfrac{{{120}^{o}}-{{70}^{o}}}{2}={{25}^{o}}$ (\u0111\u1ecbnh l\u00ed g\u00f3c c\u00f3 \u0111\u1ec9nh n\u1eb1m ngo\u00e0i \u0111\u01b0\u1eddng tr\u00f2n) <br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $25$ <\/span><\/span><\/span> "}],"id_ques":1660}],"id_ques":0}],"lesson":{"save":1,"level":3,"time":44}}