{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["8"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai12/lv3/img\/H822_K1_1.jpg' \/><\/center> <br\/> H\u00ecnh ch\u1eef nh\u1eadt $ABCD$ c\u00f3 $AB= 48\\, cm$, $E$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $CD,$ \u0111i\u1ec3m $F$ thu\u1ed9c c\u1ea1nh $AB.$ T\u00ednh \u0111\u1ed9 d\u00e0i $BF$ bi\u1ebft r\u1eb1ng di\u1ec7n t\u00edch h\u00ecnh thang $BFEC$ b\u1eb1ng $\\dfrac{1}{3}$ di\u1ec7n t\u00edch h\u00ecnh ch\u1eef nh\u1eadt $ABCD$. <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $BF=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\,(cm)$ <\/span> ","hint":"T\u00ednh di\u1ec7n t\u00edch c\u1ee7a $ABCD$ v\u00e0 $BFEC$.","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai12/lv3/img\/H822_K1_1.jpg' \/><\/center> \u0110\u1eb7t $BF = x, BC= h.$ <br\/> Ta c\u00f3 $ABCD$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt n\u00ean $AB=CD=48\\,cm.$<br\/> M\u00e0 $EC=\\dfrac{1}{2}CD=24\\,(cm)$ <br\/>Khi \u0111\u00f3: <br\/> $\\begin{align} & {{S}_{ABCD}}=AB.BC=48h \\\\ & {{S}_{B\\text{FE}C}}=\\dfrac{\\left( BF+EC \\right)BC}{2} \\\\ & =\\dfrac{\\left( x+24 \\right)h}{2} \\\\ \\end{align}$ <br\/> M\u1eb7t kh\u00e1c, theo gi\u1ea3 thi\u1ebft: <br\/>$\\begin{align} & {{S}_{BFEC}}=\\dfrac{1}{3}{{S}_{ABCD}} \\\\ & \\Leftrightarrow \\dfrac{\\left( x+24 \\right)h}{2}=16h \\\\ & \\Leftrightarrow x+24=32 \\\\ & \\Leftrightarrow x=8\\,\\,\\left( cm \\right) \\\\ \\end{align}$<br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $8$. <\/span> <br\/> <b> L\u01b0u \u00fd: <\/b> <i> Di\u1ec7n t\u00edch h\u00ecnh thang b\u1eb1ng n\u1eeda t\u00edch \u0111\u01b0\u1eddng cao v\u1edbi t\u1ed5ng hai \u0111\u00e1y. <\/i> <\/span> "}]}],"id_ques":1611},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["20"],["40"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<span class='basic_left'> Cho h\u00ecnh b\u00ecnh h\u00e0nh $ABCD$ c\u00f3 di\u1ec7n t\u00edch $720\\,\\,c{{m}^{2}}$, $O$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u01b0\u1eddng ch\u00e9o. Kho\u1ea3ng c\u00e1ch t\u1eeb $O$ \u0111\u1ebfn $CD$ b\u1eb1ng $9 \\,cm$, kho\u1ea3ng c\u00e1ch t\u1eeb $O$ \u0111\u1ebfn $AD$ b\u1eb1ng $18 \\,cm.$ T\u00ednh c\u00e1c \u0111\u1ed9 d\u00e0i $AD$ v\u00e0 $CD$. <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> <br\/> $AD=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\,(cm)$ <br\/> $CD=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\,(cm)$ <\/span> ","hint":"H\u1ea1 $CK'\\bot AD$ t\u1ea1i $K';\\,\\,\\,\\,BH'\\bot CD $ t\u1ea1i $H'$. <br\/> T\u00ednh di\u1ec7n t\u00edch $ABCD$ theo hai \u0111\u01b0\u1eddng cao $CK'$ v\u00e0 $BH'$.","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai12/lv3/img\/H822_K1_2.jpg' \/><\/center> K\u1ebb $OH\\bot CD$ t\u1ea1i $H$; $OK\\bot AD$ t\u1ea1i $K$. <br\/> $\\Rightarrow OH=9\\,\\,\\left( cm \\right);\\,\\,OK=18\\,\\,\\left( cm \\right)$ <br\/> H\u1ea1 $CK'\\bot AD$ t\u1ea1i $K';\\,\\,\\,\\,BH'\\bot CD$ t\u1ea1i $H'$. <br\/> D\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c $CK'=2OK=36\\,(cm), BH'=2OH=18\\, (cm)$. <br\/> M\u1eb7t kh\u00e1c ${{S}_{ABCD}}=CK'.AD$. <br\/> $\\Leftrightarrow 720=AD.36$ <br\/> $\\Leftrightarrow AD=20\\,\\,\\left( cm \\right)$ <br\/> M\u00e0 ${{S}_{ABCD}}=BH'.DC$ <br\/> $\\Leftrightarrow 720=DC.18$ <br\/> $\\Leftrightarrow DC=40\\,\\,\\left( cm \\right)$ <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $20$ v\u00e0 $40$. <\/span> <br\/> <b> L\u01b0u \u00fd: <\/b> <i> Di\u1ec7n t\u00edch h\u00ecnh b\u00ecnh h\u00e0nh b\u1eb1ng t\u00edch \u0111\u01b0\u1eddng cao v\u1edbi \u0111\u00e1y t\u01b0\u01a1ng \u1ee9ng. <\/i> <\/span> "}]}],"id_ques":1612},{"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 h\u00ecnh thang c\u00f3 di\u1ec7n t\u00edch l\u00e0 $18\\,cm^2$, \u0111\u01b0\u1eddng cao l\u00e0 $3$. \u0110\u1ed9 d\u00e0i \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a h\u00ecnh thang \u0111\u00f3 l\u00e0: <\/span>","select":["A. $6\\,cm$ ","B. $8\\,cm$","C. $12\\,cm$","D. $15\\,cm$"],"explain":" <span class='basic_left'> T\u1ed5ng \u0111\u1ed9 d\u00e0i hai \u0111\u00e1y c\u1ee7a h\u00ecnh thang \u0111\u00e3 cho l\u00e0: <br\/> $(18.2):3=12\\,cm$ <br\/> \u0110\u1ed9 d\u00e0i \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a h\u00ecnh thang l\u00e0: <br\/> $12:2=6\\,cm$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0: A <\/span> <br\/> <b> L\u01b0u \u00fd: <\/b> <i> \u0110\u01b0\u1eddng trung b\u00ecnh c\u1ee7a h\u00ecnh thang c\u00f3 \u0111\u1ed9 d\u00e0i b\u1eb1ng n\u1eeda t\u1ed5ng \u0111\u1ed9 d\u00e0i hai \u0111\u00e1y. <\/i> <br\/> <b> K\u1ebft lu\u1eadn: <\/b> Trong b\u00e0i to\u00e1n n\u00e0y ta kh\u00f4ng c\u1ea7n t\u00ednh \u0111\u1ed9 d\u00e0i c\u1ee7a t\u1eebng \u0111\u00e1y m\u00e0 ch\u1ec9 c\u1ea7n t\u00ednh t\u1ed5ng \u0111\u1ed9 d\u00e0i hai \u0111\u00e1y r\u1ed3i suy ra \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a h\u00ecnh thang. <\/span> ","column":2}]}],"id_ques":1613},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["10"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<span class='basic_left'> Cho h\u00ecnh b\u00ecnh h\u00e0nh $ABCD$ c\u00f3 di\u1ec7n t\u00edch l\u00e0 $80\\,m^2$. G\u1ecdi $E$ v\u00e0 $F$ theo th\u1ee9 t\u1ef1 l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AD, BC.$ C\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng $BE, AF$ c\u1eaft nhau \u1edf $O$ v\u00e0 c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng $DC$ theo th\u1ee9 t\u1ef1 \u1edf $M$ v\u00e0 $N.$ <br\/> <b> C\u00e2u 1: <\/b> T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $OEF$. <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $S_{\\Delta OEF}=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\,m^2$ <\/span> ","hint":"T\u00ecm li\u00ean h\u1ec7 gi\u1eefa di\u1ec7n t\u00edch tam gi\u00e1c $OEF$ v\u00e0 di\u1ec7n t\u00edch $ABFE$.","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai12/lv3/img\/H822_K1_3.jpg' \/><\/center> Ta c\u00f3: $AD\/\/BC$ n\u00ean $AE\/\/ BF.$ <br\/> L\u1ea1i c\u00f3: $AE=BF=\\dfrac{AD}{2}$ (gi\u1ea3 thi\u1ebft) <br\/> Suy ra: $ABFE$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh. <br\/> $\\Rightarrow {{S}_{ABEF}}={{S}_{DCFE}}=\\dfrac{{{S}_{ABCD}}}{2}$$=\\dfrac{80}{2}=40\\,\\,\\left( {{m}^{2}} \\right)$ <br\/>M\u1eb7t kh\u00e1c: <br\/> ${{S}_{\\text{OEF}}}=\\dfrac{1}{2}{{S}_{AEF}}$$=\\dfrac{1}{4}{{S}_{ABFE}}$$=40:4=\\,\\,10\\,\\,\\left( {{m}^{2}} \\right)$<br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $10$. <\/span> <\/span> "}]}],"id_ques":1614},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["90"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<span class='basic_left'> Cho h\u00ecnh b\u00ecnh h\u00e0nh $ABCD$ c\u00f3 di\u1ec7n t\u00edch l\u00e0 $80\\,m^2$. G\u1ecdi $E$ v\u00e0 $F$ theo th\u1ee9 t\u1ef1 l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AD, BC.$ C\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng $BE, AF$ c\u1eaft nhau \u1edf $O$ v\u00e0 c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng $DC$ theo th\u1ee9 t\u1ef1 \u1edf $M$ v\u00e0 $N.$ <br\/> <b> C\u00e2u 2: <\/b> T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $OMN$. <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $S_{\\Delta OMN}=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\,m^2$ <\/span> ","hint":"${{S}_{OMN}}={{S}_{\\text{OEF}}}+{{S}_{EMD}}+{{S}_{\\text{EF}CD}}+{{S}_{FCN}}$","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai12/lv3/img\/H822_K1_3.jpg' \/><\/center> Theo c\u00e2u 1, ta c\u00f3: $S_{ABEF}=40\\,(m^2)$ v\u00e0 $S_{\\Delta OEF}=10\\,(m^2)$ <br\/> Ta c\u0169ng d\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c: <br\/> $\\Delta ABE=\\Delta DME;\\,\\,\\Delta ABF=\\Delta NCF\\mbox{ (g.c.g)}$ <br\/> $\\Rightarrow {{S}_{DME}}={{S}_{ABE}}$$=\\dfrac{{{S}_{ABEF}}}{2}=40:2=20\\left( c{{m}^{2}} \\right)$ <br\/> ${{S}_{NCF}}={{S}_{ABF}}$$=\\dfrac{{{S}_{ABEF}}}{2}=\\,20\\,\\left( {{m}^{2}} \\right)$<br\/>$\\Rightarrow {{S}_{OMN}}={{S}_{\\text{OEF}}}+{{S}_{EMD}}+{{S}_{\\text{EF}CD}}+{{S}_{FCN}}$ <br\/>$=10+20+40+20=90\\,\\,\\left( {{m}^{2}} \\right)$<br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $90$. <\/span> <\/span> "}]}],"id_ques":1615},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t ","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho h\u00ecnh ch\u1eef nh\u1eadt $ABCD$ c\u00f3 $AB= 12\\, cm, AD = 6\\,cm$. H\u00ecnh thoi $EFGH$ c\u00f3 $E, F, G, H$ thu\u1ed9c c\u00e1c c\u1ea1nh $AB, BC, CD, DA$ sao cho $AE=AH=CF=CG$. \u0110\u1ed9 d\u00e0i $AE$ l\u00e0: <\/span>","select":["A. $8\\,cm$ ","B. $7\\,cm$","C. $6\\,cm$","D. $5\\,cm$"],"hint":" \u0110\u1eb7t $AE = AH = CF = CG= x$ <br\/> D\u1ef1a v\u00e0o t\u00ednh ch\u1ea5t h\u00ecnh thoi $EFGH$ \u0111\u1ec3 t\u00ecm $x$. ","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai12/lv3/img\/H822_K1_4.jpg' \/><\/center> Ta c\u00f3 $ABCD$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt. <br\/> $\\Rightarrow AB=CD=12\\,(cm);$$\\,\\,AD=BC=6\\,(cm)$ <br\/> \u0110\u1eb7t $AE = AH = CF = CG= x$ <br\/> $\\Rightarrow BE=DG=12-x,$$\\,\\,BF=DH=6-x$ <br\/> Do $EFGH$ l\u00e0 h\u00ecnh thoi n\u00ean: <br\/> $ EH=\\text{EF}\\Rightarrow E{{H}^{2}}=\\text{E}{{\\text{F}}^{2}}$ <br\/> $\\begin{align} & \\Rightarrow 2{{x}^{2}}={{\\left( 12-x \\right)}^{2}}+{{\\left( 6-x \\right)}^{2}} \\\\ & \\Rightarrow 2{{x}^{2}}=180-36x+2{{x}^{2}} \\\\ & \\Rightarrow 36x=180\\\\ & \\Rightarrow x=5 \\\\ \\end{align}$ <br\/> V\u1eady $AE = 5\\,cm$. <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D. <\/span> <\/span> ","column":4}]}],"id_ques":1616},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["21"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<span class='basic_left'> T\u1ee9 gi\u00e1c $ABCD$ c\u00f3 $O$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a hai \u0111\u01b0\u1eddng ch\u00e9o. Bi\u1ebft di\u1ec7n t\u00edch tam gi\u00e1c $AOB, BOC, COD$ theo th\u1ee9 t\u1ef1 b\u1eb1ng $2, 5, 10$ $c{{m}^{2}}$. T\u00ednh di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c $ABCD$. <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> <br\/> $S_{ABCD}=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\,(cm^2)$ <\/span> ","hint":"H\u1ea1 $AH$ $\\bot DB$ <br\/> Suy ra, ${{S}_{ABCD}}={{S}_{AOB}}+{{S}_{BOC}}+{{S}_{DOC}}+{{S}_{AOD}}$.","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai12/lv3/img\/H822_K1_5.jpg' \/><\/center> H\u1ea1 $AH \\bot DB (H \\in DB)$. <br\/> $ \\Rightarrow {{S}_{AOD}}=\\dfrac{1}{2}AH.OD$$;\\,\\,{{S}_{AOB}}=\\dfrac{1}{2}AH.OB$ <br\/> Do \u0111\u00f3 : $\\dfrac{{{S}_{AOD}}}{{{S}_{AOB}}}$$=\\dfrac{OD}{OB}=\\dfrac{{{S}_{COD}}}{{{S}_{COB}}}$$=\\dfrac{10}{5}=2$ <br\/> $\\Rightarrow {{S}_{AOD}}=2{{S}_{AOB}}$$=2.2=\\,\\,4\\,\\,\\left( c{{m}^{2}} \\right)$ <br\/> Do \u0111\u00f3 ${{S}_{ABCD}}={{S}_{AOB}}+{{S}_{BOC}}+{{S}_{DOC}}+{{S}_{AOD}}$ <br\/> $=2+5+10+4=21\\,\\,\\left( c{{m}^{2}} \\right)$ <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $21$. <\/span> <\/span> "}]}],"id_ques":1617},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t ","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho t\u1ee9 gi\u00e1c $ABCD$ c\u00f3 di\u1ec7n t\u00edch $60\\,\\,{{m}^{2}}$. Tr\u00ean c\u1ea1nh $AB$ l\u1ea5y c\u00e1c \u0111i\u1ec3m $E$ v\u00e0 $F$ sao cho $AE=EF=FB$. Tr\u00ean c\u1ea1nh $CD$ l\u1ea5y c\u00e1c \u0111i\u1ec3m $G$ v\u00e0 $H$ sao cho $CG=GH=HD$ <br\/> <b> C\u00e2u 1: <\/b> T\u1ed5ng di\u1ec7n t\u00edch c\u00e1c tam gi\u00e1c $ADH$ v\u00e0 $CBF$ l\u00e0: <\/span>","select":["A. $20\\,m^2$","B. $30\\,m^2$","C. $40\\,m^2$","D. $50\\,m^2$"],"hint":" T\u00ednh $S_{ADH}$ theo $S_{ADC}$ v\u00e0 t\u00ednh $S_{CBF}$ theo $ S_{ABC}$","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai12/lv3/img\/H822_K1_6.jpg' \/><\/center> Ta c\u00f3 : $DH = HG = GC$ <br\/> $\\Rightarrow DH=\\dfrac{1}{3}DC\\Rightarrow {{S}_{ADH}}=\\dfrac{1}{3}{{S}_{ADC}}$ (chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb $A$ \u0111\u1ebfn $DH$ v\u00e0 $DC$). <br\/> M\u1eb7t kh\u00e1c $AE=EF=FB$ <br\/> $\\Rightarrow BF=\\dfrac{1}{3}AB$ <br\/> $\\Rightarrow {{S}_{CBF}}=\\dfrac{1}{3}{{S}_{CAB}}$ (chung \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb $C$ \u0111\u1ebfn $FB$ v\u00e0 $AB$) <br\/> $\\Rightarrow {{S}_{ADH}}+{{S}_{CBF}}=\\dfrac{1}{3}{{S}_{ADC}}+\\dfrac{1}{3}{{S}_{ABC}}$ <br\/> $=\\dfrac{1}{3}{{S}_{ABCD}}=60:3=20\\,\\,\\left( {{m}^{2}} \\right)$<br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span> <\/span> ","column":4}]}],"id_ques":1618},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t ","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho t\u1ee9 gi\u00e1c $ABCD$ c\u00f3 di\u1ec7n t\u00edch $60\\,\\,{{m}^{2}}$. Tr\u00ean c\u1ea1nh $AB$ l\u1ea5y c\u00e1c \u0111i\u1ec3m $E$ v\u00e0 $F$ sao cho $AE=EF=FB$. Tr\u00ean c\u1ea1nh $CD$ l\u1ea5y c\u00e1c \u0111i\u1ec3m $G$ v\u00e0 $H$ sao cho $CG=GH=HD$ <br\/> <b> C\u00e2u 2: <\/b> Di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c $AFCH$ l\u00e0: <\/span>","select":["A. $20\\,m^2$","B. $30\\,m^2$","C. $40\\,m^2$","D. $50\\,m^2$"],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai12/lv3/img\/H822_K1_6.jpg' \/><\/center> Theo c\u00e2u 1, ta c\u00f3: ${{S}_{ADH}}+{{S}_{CBF}}=S_1+S_6=20\\,cm^2$ <br\/> Ta c\u00f3 ${{S}_{\\text{AF}CH}}={{S}_{ABCD}}-\\left( {{S}_{1}}+{{S}_{6}} \\right)$ $=60-20=40\\,\\left( {{m}^{2}} \\right)$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C. <\/span> <\/span> ","column":4}]}],"id_ques":1619},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t ","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho t\u1ee9 gi\u00e1c $ABCD$ c\u00f3 di\u1ec7n t\u00edch $60\\,\\,{{m}^{2}}$. Tr\u00ean c\u1ea1nh $AB$ l\u1ea5y c\u00e1c \u0111i\u1ec3m $E$ v\u00e0 $F$ sao cho $AE=EF=FB$. Tr\u00ean c\u1ea1nh $CD$ l\u1ea5y c\u00e1c \u0111i\u1ec3m $G$ v\u00e0 $H$ sao cho $CG=GH=HD$ <br\/> <b> C\u00e2u 3: <\/b> Di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c $EFGH$ l\u00e0: <\/span>","select":["A. $40 \\,m^2$","B. $30\\,m^2$","C. $20\\,m^2$","D. $10\\,m^2$"],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai12/lv3/img\/H822_K1_6.jpg' \/><\/center> Theo c\u00e2u 2, ta c\u00f3: $S_{AFCH}=40\\,\\,\\left( c{{m}^{2}} \\right)$ <br\/> M\u00e0 $AE=EF\\Rightarrow S_2=S_3$ (c\u00f9ng \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb $H$ \u0111\u1ebfn $AE$ v\u00e0 $EF$) (1) <br\/> Ta c\u00f3: $HG=GC\\Rightarrow S_4=S_5$ (c\u00f9ng \u0111\u01b0\u1eddng cao h\u1ea1 t\u1eeb $F$ \u0111\u1ebfn $HG$ v\u00e0 $GC$) (2) <br\/> T\u1eeb (1) v\u00e0 (2), ta c\u00f3: $S_3+S_4=S_2+S_5$. <br\/> Do \u0111\u00f3 ${{S}_{\\text{EF}GH}}=\\dfrac{1}{2}{{S}_{\\text{AF}CH}}$$=40:2=\\,20\\,\\left( {{m}^{2}} \\right)$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C. <\/span> <\/span> ","column":4}]}],"id_ques":1620}],"lesson":{"save":0,"level":3}}