{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["45"],["80"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Cho h\u00ecnh v\u1ebd sau. T\u00ecm $x;y$. <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai1/lv3/img\/H911_K9.png' \/><center> <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $x=$ _input_; $y=$ _input_","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b> B\u01b0\u1edbc 1:<\/b> \u0110\u1eb7t $AB= 3t; AC=4t$ <br\/><b> B\u01b0\u1edbc 2: <\/b> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago t\u00ednh $t$ <br\/><b> B\u01b0\u1edbc 3: <\/b> V\u1eadn d\u1ee5ng h\u1ec7 th\u1ee9c $b^2=a.b'$ \u0111\u1ec3 t\u00ednh $x$ v\u00e0 $y$.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> Ta c\u00f3: $\\dfrac{AB}{AC}=\\dfrac{3}{4}\\Rightarrow \\dfrac{AB}{3}=\\dfrac{AC}{4}=t\\,\\,\\left( t>0 \\right)$ <br\/> Suy ra: $AB= 3t, AC= 4t$ <br\/> X\u00e9t $\\Delta ABC$ vu\u00f4ng t\u1ea1i $A$ (gi\u1ea3 thi\u1ebft) <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago ta c\u00f3: <br\/> $\\begin{align} & \\,\\,\\,\\,\\,\\,\\,\\,A{{B}^{2}}+A{{C}^{2}}=B{{C}^{2}} \\\\ & \\Leftrightarrow {{\\left( 3t \\right)}^{2}}+{{\\left( 4t \\right)}^{2}}={{125}^{2}} \\\\ & \\Leftrightarrow 25{{t}^{2}}=15625 \\\\ & \\Leftrightarrow t^2=625\\\\ & \\Leftrightarrow t=25 \\\\ \\end{align}$ <br\/> $\\Rightarrow AB=75;\\,\\,AC=100$ <br\/> \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c gi\u1eefa c\u1ea1nh g\u00f3c vu\u00f4ng v\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a n\u00f3 tr\u00ean c\u1ea1nh huy\u1ec1n ${{b}^{2}}=b'.a$ , ta \u0111\u01b0\u1ee3c: <br\/> $A{{B}^{2}}=x.BC\\,$$\\Rightarrow x=\\dfrac{A{{B}^{2}}}{BC}$$=\\dfrac{{{75}^{2}}}{125}=45$ <br\/>$\\Rightarrow y=125-x\\,$$= 125 - 45 = 80$ <br\/><span class='basic_pink'>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $45$ v\u00e0 $80$. <\/span><\/span>"}]}],"id_ques":1301},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{3}{2}$","B. $\\dfrac{1}{3}$","C. $\\dfrac{1}{2}$"],"ques":"Cho tam gi\u00e1c $ABC$ c\u00f3 $\\dfrac{AB}{AC}=\\dfrac{2}{3}$ v\u00e0 \u0111\u01b0\u1eddng cao $AH$. G\u1ecdi $M$ v\u00e0 $N$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $H$ tr\u00ean c\u1ea1nh $AB$ v\u00e0 $AC.$ Khi \u0111\u00f3 t\u1ec9 s\u1ed1 $\\dfrac{AM}{AN}$ l\u00e0 ?","hint":"\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c v\u1ec1 c\u1ea1nh v\u00e0 \u0111\u01b0\u1eddng cao trong tam gi\u00e1c vu\u00f4ng $AHB$ v\u00e0 $AHC$","explain":"<span class='basic_left'><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai1/lv3/img\/H911_K8.png' \/><\/center> \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c v\u1ec1 c\u1ea1nh v\u00e0 \u0111\u01b0\u1eddng cao trong tam gi\u00e1c vu\u00f4ng $AHB$ v\u00e0 $AHC$ ta c\u00f3:<br\/> $AM.AB=AH^2\\\\AN.AC=AH^2$ <br\/> $\\Rightarrow AM.AB=AN.AC\\,$ <br\/> $\\Leftrightarrow\\dfrac{AM}{AN}=\\dfrac{AC}{AB}=\\dfrac{3}{2}$ <\/span>"}]}],"id_ques":1302},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"select":["A. $3\\sqrt{13}$","B. $2\\sqrt{13}$","C. $\\sqrt{13}$"],"ques":"Cho tam gi\u00e1c $DEF$ vu\u00f4ng t\u1ea1i $D$ c\u00f3 hai trung tuy\u1ebfn $DM=2,5cm$; $EN= 4cm$. T\u00ednh \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng $DE$. <br\/> <b> \u0110\u00e1p s\u1ed1: <\/b> $DE = $ ?$(cm)$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b> B\u01b0\u1edbc 1:<\/b> T\u00ednh c\u1ea1nh $EF$. <br\/><b> B\u01b0\u1edbc 2: <\/b> V\u1eadn d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago v\u00e0o hai tam gi\u00e1c $DEF$ v\u00e0 $DEN$ \u0111\u1ec3 t\u00ednh $DN$. <br\/><b> B\u01b0\u1edbc 3: <\/b> V\u1eadn d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago v\u00e0o tam gi\u00e1c $ABC$ \u0111\u1ec3 t\u00ednh $DE$. <br\/><span class='basic_green'> B\u00e0i gi\u1ea3i <\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai1/lv3/img\/H911_K7.png' \/><\/center> X\u00e9t tam gi\u00e1c $EDF$ vu\u00f4ng t\u1ea1i $D$ c\u00f3 $DM$ l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh huy\u1ec1n $EF$<br\/> Suy ra $EM=MF\\Rightarrow EF=2DM=5\\, (cm)$ <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago v\u00e0o hai tam gi\u00e1c vu\u00f4ng $DEN$ v\u00e0 $DEF$ c\u00f3: <br\/> $E{{D}^{2}}+D{{N}^{2}}=E{{N}^{2}}\\,$$\\Rightarrow E{{D}^{2}}+D{{N}^{2}}=16\\,\\,\\,\\left( 1 \\right)$ <br\/> $D{{E}^{2}}+D{{F}^{2}}=E{{F}^{2}}\\,$$\\Rightarrow D{{E}^{2}}+4D{{N}^{2}}=25\\,\\,\\,\\left( 2 \\right)$ (do $DF=2DN$) <br\/> L\u1ea5y (2) \u2013(1) ta \u0111\u01b0\u1ee3c : <br\/> $3D{{N}^{2}}=9$ <br\/> $\\Rightarrow DN=\\sqrt{3}\\,$ <br\/> $\\Rightarrow DF=2\\sqrt{3}\\, (cm)$ <br\/> X\u00e9t $\\Delta DEF$ c\u00f3: $\\widehat{D}={{90}^{o}}$ <br\/> $\\,\\,\\,\\,\\,D{{E}^{2}}+D{{F}^{2}}=EF^2$ (\u0111\u1ecbnh l\u00ed Pytago) <br\/> $\\begin{align} & \\Rightarrow D{{E}^{2}}=25-{{\\left( 2\\sqrt{3} \\right)}^{2}} \\\\ & \\Rightarrow DE=\\sqrt{13}\\,(cm) \\\\\\end{align}$ <\/span>"}]}],"id_ques":1303},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0110\u00fang hay Sai ","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$, \u0111\u01b0\u1eddng cao $AH$, G\u1ecdi $D$ v\u00e0 $E$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $H$ tr\u00ean $AB, AC.$ Bi\u1ec3u th\u1ee9c $AB.AD=AC.AE$$=HB.HC$,<b> \u0111\u00fang<\/b> hay <b>sai<\/b>? ","select":["A. \u0110\u00fang","B. Sai"],"explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai1/lv3/img\/H911_K6.png' \/><\/center><br\/> \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c gi\u1eefa c\u1ea1nh g\u00f3c vu\u00f4ng v\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a n\u00f3 tr\u00ean c\u1ea1nh huy\u1ec1n v\u00e0o c\u00e1c tam gi\u00e1c vu\u00f4ng $HAB$ v\u00e0 $HAC$ ta \u0111\u01b0\u1ee3c: <br\/>$AH^2=AB.AD\\,\\,\\,\\,(1)\\\\AH^2=AC.AE\\,\\,\\,\\,(2)$ <br\/> \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c $h^2=b'.c'\\,$ v\u00e0o tam gi\u00e1c vu\u00f4ng $ABC$ ta \u0111\u01b0\u1ee3c: <br\/>$AH^2=HB.HC\\,\\,\\,\\,(3)$<br\/>T\u1eeb (1), (2), (3) suy ra: $AB.AD=AC.AE$$=HB.HC$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00fang. <br\/><span class='basic_pink'>Do \u0111\u00f3 \u0111\u00e1p \u00e1n l\u00e0 A. <\/span><br\/><span class='basic_green'>L\u01b0u \u00fd:<\/span>$AH$ \u0111\u00f3ng vai tr\u00f2 kh\u00e1c nhau trong t\u1eebng tam gi\u00e1c n\u00ean ta ph\u1ea3i lu\u00f4n v\u1eadn d\u1ee5ng linh ho\u1ea1t c\u00e1c h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c vu\u00f4ng, th\u00edch h\u1ee3p v\u1edbi t\u1eebng \u0111i\u1ec1u ki\u1ec7n c\u1ee5 th\u1ec3 <\/span>","column":2}]}],"id_ques":1304},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{60}{7}$","B. $\\dfrac{62}{7}$","C. $\\dfrac{61}{7}$"],"ques":"<span class='basic_left'>Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$, \u0111\u01b0\u1eddng cao $AH$. Bi\u1ebft $AB =12\\, cm$; $AC= 16\\, cm$. K\u1ebb ph\u00e2n gi\u00e1c $AD$ ($D \\,\\in BC$).<br\/>\u0110\u1ed9 d\u00e0i c\u1ea1nh $BD$ l\u00e0 ? $(cm)$<\/span>","hint":"T\u00ednh ch\u1ea5t tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c <br\/> $\\dfrac{BD}{DC}=\\dfrac{AB}{AC}$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b> B\u01b0\u1edbc 1:<\/b> S\u1eed d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago t\u00ednh c\u1ea1nh $BC$ <br\/><b> B\u01b0\u1edbc 2: <\/b> V\u1eadn d\u1ee5ng t\u00ednh ch\u1ea5t tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $BAC$ \u0111\u1ec3 t\u00ednh c\u1ea1nh $BD$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai1/lv3/img\/H911_K5.png' \/><\/center>X\u00e9t $\\Delta ABC$ c\u00f3: $\\widehat{A}={{90}^{o}}.$ <br\/> Theo \u0111\u1ecbnh l\u00ed Pytago ta c\u00f3: <br\/> $B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}\\,$ <br\/> $={{12}^{2}}+{{16}^{2}}$ <br\/> $=400\\,$ <br\/> $\\Rightarrow BC=20\\left( cm \\right)$ <br\/> Do $AD$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $A$ <br\/> $\\dfrac{BD}{DC}=\\dfrac{AB}{AC}=\\dfrac{12}{16}\\,\\,$ (t\u00ednh ch\u1ea5t tia ph\u00e2n gi\u00e1c) <br\/> $\\Rightarrow \\dfrac{BD}{12}=\\dfrac{DC}{16}\\,$$=\\dfrac{BD+DC}{12+16}=\\dfrac{20}{28}=\\dfrac{5}{7}$ (t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau) <br\/> Suy ra $BD=\\dfrac{5}{7}.12=\\dfrac{60}{7}\\,\\,\\left( cm \\right)$<\/span>"}]}],"id_ques":1305},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["6,58"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'><span class='basic_left'>Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$, \u0111\u01b0\u1eddng cao $AH$. Bi\u1ebft $AB =12\\, cm$; $AC= 16\\, cm$. K\u1ebb ph\u00e2n gi\u00e1c $AD$ ($D \\,\\in BC$).<br\/><b> C\u00e2u 2: <\/b> T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $AHD$. (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 hai) <br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $S_{AHD} = $ _input_ $(cm^2)$<\/span>","hint":"\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c v\u1ec1 c\u1ea1nh v\u00e0 \u0111\u01b0\u1eddng cao: $b^2=a.b'$ v\u00e0 $a.h=b.c$ \u0111\u1ec3 t\u00ednh c\u1ea1nh $BH$ v\u00e0 $AH$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b> B\u01b0\u1edbc 1:<\/b> T\u00ednh c\u1ea1nh $BH$ v\u00e0 $AH$ <br\/><b> B\u01b0\u1edbc 2: <\/b> T\u00ednh c\u1ea1nh $HD$ <br\/> <b> B\u01b0\u1edbc 3: <\/b> T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $AHD$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai1/lv3/img\/H911_K5.png' \/><\/center>X\u00e9t tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 \u0111\u01b0\u1eddng cao $AH$.<br\/> \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c v\u1ec1 c\u1ea1nh v\u00e0 \u0111\u01b0\u1eddng cao v\u00e0o tam gi\u00e1c vu\u00f4ng $ABC$ ta c\u00f3: <br\/> $\\bullet AB^2=BH.BC\\,$ <br\/> $\\Rightarrow BH=\\dfrac{AB^2}{BC}\\,$ $=\\dfrac{12^2}{20}=\\dfrac{36}{5} (cm)$ <br\/> $\\bullet AH.BC=AB.AC \\,$ <br\/> $\\Rightarrow AH=\\dfrac{AB.AC}{BC}\\,$$=\\dfrac{12.16}{20}=\\dfrac{48}{5}\\, (cm)$ <br\/> Theo c\u00e2u tr\u00ean, ta c\u00f3 $BD=\\dfrac{60}{7}\\,$ <br\/> $\\Rightarrow HD=BD-BH\\,$$=\\dfrac{60}{7}-\\dfrac{36}{5}\\,$$=\\dfrac{48}{35}\\,(cm)$ <br\/> Di\u1ec7n t\u00edch tam gi\u00e1c $AHD$ l\u00e0: <br\/> $ \\dfrac{AH.HD}{2}\\,$$=\\dfrac{\\dfrac{48}{5}.\\dfrac{48}{35}}{2}$$\\approx 6,58 \\,(cm^2)$ <br\/><span class='basic_pink'>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $6,58$<\/span><\/span>"}]}],"id_ques":1306},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["25"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Trong m\u1ed9t tam gi\u00e1c vu\u00f4ng, \u0111\u01b0\u1eddng cao \u1ee9ng v\u1edbi c\u1ea1nh huy\u1ec1n chia tam gi\u00e1c th\u00e0nh hai ph\u1ea7n c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng $54 \\,cm^2$ v\u00e0 $96\\,cm^2$. T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh huy\u1ec1n. <br\/><b> \u0110\u00e1p s\u1ed1: <\/b> \u0110\u1ed9 d\u00e0i c\u1ea1nh huy\u1ec1n l\u00e0: _input_ $(cm)$ <\/span>","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/> Ta gi\u1ea3i theo s\u01a1 \u0111\u1ed3 t\u1eeb d\u01b0\u1edbi l\u00ean tr\u00ean: <br\/> $\\begin{aligned} & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,BC \\\\ & \\,\\,\\,\\,\\,\\,\\,\\nearrow \\,\\,\\,\\,\\,\\,\\nwarrow \\\\ & {{S}_{\\Delta ABC}}\\,\\,\\,\\,\\,\\,\\,\\,\\,AH^2=HB.HC \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\ & \\,\\,\\,\\,\\,\\, \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\nearrow \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\nwarrow \\\\ & \\,\\,\\,\\,\\,HB=\\frac{2.{{S}_{ABH}}}{AH}\\,\\,\\,\\,\\,\\,\\,\\,HC=\\frac{2.{{S}_{ACH}}}{AH} \\\\ \\end{aligned}$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai1/lv3/img\/H911_K4.png' \/><\/center> $\\Delta ABC$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 \u0111\u01b0\u1eddng cao $AH$ chia tam gi\u00e1c th\u00e0nh $\\Delta ABH$ c\u00f3 di\u1ec7n t\u00edch l\u00e0 $54 cm^2$ v\u00e0 $\\Delta AHC$ c\u00f3 di\u1ec7n t\u00edch l\u00e0 $90 cm^2.$<br\/>Ta c\u00f3: <br\/> ${{S}_{ABH}}=\\dfrac{1}{2}AH.BH$$\\Rightarrow BH=\\dfrac{2{{S}_{ABH}}}{AH}$ <br\/> ${{S}_{ACH}}=\\dfrac{1}{2}AH.CH$$\\Rightarrow CH=\\dfrac{2{{S}_{ACH}}}{AH}$ <br\/> X\u00e9t $\\Delta ABC$ c\u00f3: $\\widehat{A}=90^o$, \u0111\u01b0\u1eddng cao $AH$ <br\/> ${A{H}^{2}}=BH.CH$ (h\u1ec7 th\u1ee9c v\u1ec1 c\u1ea1nh v\u00e0 \u0111\u01b0\u1eddng cao trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\Rightarrow AH^2 =\\dfrac{2{{S}_{ABH}}.2{{S}_{ACH}}}{A{{H}^{2}}}$ <br\/> $\\Rightarrow A{{H}^{4}}$$=4S_{ABH}.S_{ACH}=4.54.96=20736 $ <br\/> $\\Rightarrow AH= 12\\,(cm)$ <br\/> Ta c\u00f3: ${{S}_{ABC}}={{S}_{ABH}}+{{S}_{ACH}}$$=54 + 96$$= 150(cm^2)$ <br\/> M\u1eb7t kh\u00e1c: ${{S}_{ABC}}=\\dfrac{1}{2}AH.BC$ <br\/> $\\Rightarrow BC=\\dfrac{2{{S}_{ABC}}}{AH}$$=\\dfrac{2.150}{12}$$=25 \\,(cm)$ <br\/><span class='basic_pink'>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $25$.<\/span><\/span>"}]}],"id_ques":1307},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2"],["5"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"H\u00ecnh thang c\u00e2n $ABCD$ c\u00f3 \u0111\u00e1y l\u1edbn $CD= 10 cm$, \u0111\u00e1y nh\u1ecf b\u1eb1ng \u0111\u01b0\u1eddng cao, \u0111\u01b0\u1eddng ch\u00e9o vu\u00f4ng g\u00f3c v\u1edbi c\u1ea1nh b\u00ean. T\u00ednh chi\u1ec1u cao c\u1ee7a h\u00ecnh thang. <br\/><b> \u0110\u00e1p s\u1ed1: <\/b> Chi\u1ec1u cao c\u1ee7a h\u00ecnh thang l\u00e0 $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}.\\sqrt {\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$","hint":"K\u1ebb $AH, BK$ vu\u00f4ng g\u00f3c v\u1edbi $CD$. \u0110\u1eb7t $AB=x$. T\u00ednh $DH, HC$ theo $x$.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> \u0110\u1eb7t c\u1ea1nh $AB= x$ <br\/><b> B\u01b0\u1edbc 2: <\/b>T\u00ednh c\u1ea1nh $HD$ v\u00e0 $HC$ theo $x$ <br\/><b> B\u01b0\u1edbc 3: <\/b> V\u1eadn d\u1ee5ng h\u1ec7 th\u1ee9c $h^2=a'.b'$ v\u00e0o tam gi\u00e1c $ADC$ \u0111\u1ec3 t\u00ednh c\u1ea1nh $AH$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai1/lv3/img\/H911_K3.jpg' \/><\/center><br\/>G\u1ecdi $AH, BK$ l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a h\u00ecnh thang. <br\/> \u0110\u1eb7t $AB=x$. Suy ra $AH=BK =x$.<br\/> D\u1ec5 th\u1ea5y $ABKH$ l\u00e0 h\u00ecnh vu\u00f4ng n\u00ean $AB=HK=x$ <br\/>X\u00e9t $\\Delta AHD$ vu\u00f4ng t\u1ea1i $H$ v\u00e0 $\\Delta BKC$ vu\u00f4ng t\u1ea1i $K$ c\u00f3: <br\/>$AD=BC$ (gi\u1ea3 thi\u1ebft).<br\/> $\\widehat{D}=\\widehat{C}$ (gi\u1ea3 thi\u1ebft).<br\/> $\\Rightarrow \\Delta AHD=\\Delta BKC$ (c\u1ea1nh huy\u1ec1n_g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow DH=CK=\\dfrac{DC-HK}{2}=\\dfrac{10-x}{2}\\,(cm)$ <br\/> $\\Rightarrow HC=HK+KC=\\dfrac{10-x}{2}+x\\,$$=\\dfrac{10+x}{2}\\,(cm)$ <br\/> X\u00e9t tam gi\u00e1c $ADC$ vu\u00f4ng t\u1ea1i $A$ ta c\u00f3: <br\/> $A{{H}^{2}}=HD.HC$ (h\u1ec7 th\u1ee9c v\u1ec1 c\u1ea1nh v\u00e0 \u0111\u01b0\u1eddng cao trong tam gi\u00e1c vu\u00f4ng) <br\/> $\\begin{aligned} & \\Rightarrow {{x}^{2}}=\\dfrac{10-x}{2}.\\dfrac{10+x}{2} \\\\ & \\Leftrightarrow {{x}^{2}}=\\dfrac{100-{{x}^{2}}}{4} \\\\ & \\Leftrightarrow 5x^2=100 \\\\ & \\Leftrightarrow x^2=\\dfrac{100}{5}\\\\ & \\Leftrightarrow x^2=20 \\\\ & \\Leftrightarrow x=2\\sqrt {5} (cm) \\\\ \\end{aligned}$ <br\/><span class='basic_pink'> V\u1eady chi\u1ec1u cao c\u1ee7a h\u00ecnh thang b\u1eb1ng $2\\sqrt {5} \\,(cm)$<\/span><\/span>"}]}],"id_ques":1308},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["40"],["30"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 \u0111\u01b0\u1eddng cao $AH$, \u0111\u01b0\u1eddng trung tuy\u1ebfn $AM$. T\u00ednh \u0111\u1ed9 d\u00e0i hai c\u1ea1nh g\u00f3c vu\u00f4ng bi\u1ebft $HB:HC=16:9$ v\u00e0 $AM = 25 \\,cm$ <br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $AB$ _input_ $(cm)$$\\,\\,\\,\\,\\,$$AC=$_input_ $(cm)$","hint":"T\u00ednh c\u1ea1nh $HB$ v\u00e0 $HC$.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $BC$. <br\/> <b> B\u01b0\u1edbc 2: <\/b>D\u1ef1a v\u00e0o d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau t\u00ednh c\u1ea1nh $HB$ v\u00e0 $HC$. <br\/><b> B\u01b0\u1edbc 3: <\/b> V\u1eadn d\u1ee5ng h\u1ec7 th\u1ee9c $b^2=a.b'$ \u0111\u1ec3 t\u00ednh c\u1ea1nh $AB$ v\u00e0 $AC$ <br\/><span class='basic_green'> B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai1/lv3/img\/H911_K2.jpg' \/><\/center> X\u00e9t tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 $AM$ l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn. <br\/> Ta c\u00f3: $AM = 25\\, cm$ $\\Rightarrow BC=$$ 2AM= 50\\, (cm)$ <br\/>Ta l\u1ea1i c\u00f3: $\\dfrac{HB}{HC}=\\dfrac{16}{9}\\,$$\\Rightarrow \\dfrac{HB}{16}=\\dfrac{HC}{9}\\,$$=\\dfrac{HB+HC}{25}=\\dfrac{50}{25}=2$ (t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau). <br\/> Suy ra $HB =2.16$$= 32\\,(cm)$; $HC=2.9$$=18\\, (cm)$<br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c ${{b}^{2}}=a.b'$ v\u00e0o tam gi\u00e1c vu\u00f4ng $ABC$ ta \u0111\u01b0\u1ee3c: <br\/> $A{{B}^{2}}=BC.BH$$\\Rightarrow AB=\\sqrt{50.32}=40\\,\\left( cm \\right) $ <br\/> $ A{{C}^{2}}=BC.HC\\,$$\\Rightarrow AC=\\sqrt{50.18}=30\\,\\left( cm \\right)$<br\/><span class='basic_pink'>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $40$ v\u00e0 $30$.<\/span><\/span>"}]}],"id_ques":1309},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["13"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"T\u00ednh c\u1ea1nh \u0111\u00e1y $BC$ c\u1ee7a tam gi\u00e1c c\u00e2n $ABC$. Bi\u1ebft \u0111\u01b0\u1eddng cao \u1ee9ng v\u1edbi c\u1ea1nh \u0111\u00e1y b\u1eb1ng $15,6\\,cm$ v\u00e0 \u0111\u01b0\u1eddng cao \u1ee9ng v\u1edbi c\u1ea1nh b\u00ean b\u1eb1ng $12\\,cm$.<br\/><b> \u0110\u00e1p s\u1ed1 :<\/b> $BC=$ _input_ $(cm)$","hint":"G\u1ecdi $AH, BK$ l\u00e0 c\u00e1c \u0111\u01b0\u1eddng cao c\u1ee7a $\\Delta ABC$. <br\/> T\u00ednh $AC$ theo $BC$ (d\u1ef1a v\u00e0o c\u00f4ng th\u1ee9c t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c)","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $AC$ theo $BC$ <br\/> <b> B\u01b0\u1edbc 2: <\/b> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pytago v\u00e0o tam gi\u00e1c $AHC$ \u0111\u1ec3 t\u00ednh c\u1ea1nh $BC$.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai1/lv3/img\/H911_K1.jpg' \/><\/center> G\u1ecdi $AH, BK$ l\u00e0 c\u00e1c \u0111\u01b0\u1eddng cao c\u1ee7a $\\Delta ABC$ ta c\u00f3: <br\/> $2S_{ABC} = AC.BK $$ = BC.AH$ <br\/> $\\Rightarrow AC=\\dfrac{BC.AH}{BK}\\,$$=\\dfrac{BC.15,6}{12}=1,3BC$ <br\/> Tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$, $AH$ vu\u00f4ng g\u00f3c v\u1edbi $BC$ n\u00ean $AH$ l\u00e0 trung tuy\u1ebfn. (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n). <br\/>\u0110\u1eb7t $BC=x \\Rightarrow HC = \\dfrac{x}{2};$$ AC=1,3x.$ <br\/> Theo \u0111\u1ecbnh l\u00fd Pytago, ta c\u00f3: <br\/> $A{{H}^{2}}+H{{C}^{2}}=A{{C}^{2}}$ <br\/> $\\begin{align} & \\Rightarrow 15,{{6}^{2}}+{{\\left( \\dfrac{x}{2} \\right)}^{2}}={{\\left( 1,3x \\right)}^{2}} \\\\ & \\Leftrightarrow 243,36=1,44{{x}^{2}} \\\\ & \\Leftrightarrow {{x}^{2}}=\\dfrac{243,36}{1,44}=169 \\\\ & \\Leftrightarrow x=13\\,\\left( cm \\right) \\\\ \\end{align}$ <br\/>V\u1eady c\u1ea1nh \u0111\u00e1y $BC = 13\\,(cm)$.<br\/><span class='basic_pink'>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $13$.<\/span><\/span>"}]}],"id_ques":1310}],"lesson":{"save":0,"level":3}}