{"segment":[{"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'> \u0110a gi\u00e1c \u0111\u1ec1u c\u00f3 t\u1ed5ng s\u1ed1 \u0111o c\u00e1c g\u00f3c trong l\u00e0 $2340^o$. S\u1ed1 c\u1ea1nh c\u1ee7a \u0111a gi\u00e1c n\u00e0y l\u00e0: <\/span>","select":["A. $12$","B. $13$","C. $14$","D. $15$ "],"explain":" <span class='basic_left'> T\u1ed5ng s\u1ed1 \u0111o c\u00e1c g\u00f3c c\u1ee7a \u0111a gi\u00e1c \u0111\u1ec1u $n$ c\u1ea1nh l\u00e0: <br\/> $(n-2).180^o=2340^o$ <br\/> $\\Rightarrow n-2=13$ <br\/> $\\Rightarrow n=15$ <br\/> V\u1eady \u0111a gi\u00e1c \u0111\u1ec1u \u0111\u00f3 c\u00f3 $15$ c\u1ea1nh. <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D. <\/span> <br\/> <b> L\u01b0u \u00fd: <\/b> <i> T\u1ed5ng s\u1ed1 \u0111o c\u00e1c g\u00f3c c\u1ee7a \u0111a gi\u00e1c \u0111\u1ec1u $n$ c\u1ea1nh l\u00e0: $(n-2).180^o$. <\/i> <\/span> ","column":4}]}],"id_ques":1641},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t ","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":" <span class='basic_left'> \u0110a gi\u00e1c l\u1ed3i c\u00f3 s\u1ed1 \u0111\u01b0\u1eddng ch\u00e9o l\u00e0 $65$. S\u1ed1 c\u1ea1nh c\u1ee7a \u0111a gi\u00e1c n\u00e0y l\u00e0: <\/span>","select":["A. $11$","B. $12$","C. $13$","D. $14$ "],"explain":" <span class='basic_left'> C\u00f4ng th\u1ee9c t\u00ednh s\u1ed1 \u0111\u01b0\u1eddng ch\u00e9o c\u1ee7a \u0111a gi\u00e1c l\u1ed3i $n$ c\u1ea1nh l\u00e0: $\\dfrac{n(n-3)}{2}\\,\\,(n \\in \\mathbb{N^*})$ <br\/> Theo b\u00e0i, ta c\u00f3: <br\/> $\\dfrac{n(n-3)}{2}=65$ <br\/> $\\Rightarrow n(n-3)=130$ <br\/> $\\Rightarrow n^2-3n-130=0$ <br\/> $\\Rightarrow n^2-13n+10n-130=0$ <br\/> $\\Rightarrow n(n-13)+10(n-13)=0$ <br\/> $\\Rightarrow (n-13)(n+10)=0$ <br\/> $\\begin{aligned} & \\Rightarrow \\left[ \\begin{aligned} & n-13=0 \\\\ & n+10=0 \\\\ \\end{aligned} \\right. \\\\ & \\Rightarrow \\left[ \\begin{aligned} & n=13\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ & n=-10 \\,(\\text{lo\u1ea1i}) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111a gi\u00e1c \u0111\u1ec1u \u0111\u00f3 c\u00f3 $13$ c\u1ea1nh. <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C. <\/span> <br\/> <b> L\u01b0u \u00fd: <\/b> <i> C\u00f4ng th\u1ee9c t\u00ednh s\u1ed1 \u0111\u01b0\u1eddng ch\u00e9o c\u1ee7a \u0111a gi\u00e1c l\u1ed3i $n$ c\u1ea1nh l\u00e0: $\\dfrac{n(n-3)}{2}$ <\/i> <\/span> ","column":4}]}],"id_ques":1642},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["10"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ c\u00f3 di\u1ec7n t\u00edch b\u1eb1ng $48 \\,cm^2$, $E$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AC$. L\u1ea5y \u0111i\u1ec3m $D$ tr\u00ean c\u1ea1nh $BC$ sao cho $BD=\\dfrac {1}{3}BC$, l\u1ea5y \u0111i\u1ec3m $G$ tr\u00ean c\u1ea1nh $AE$ sao cho $AG=\\dfrac{1}{3}AE$. \u0110o\u1ea1n th\u1eb3ng $AD$ c\u1eaft $BG,\\, BE$ theo th\u1ee9 t\u1ef1 t\u1ea1i $M,\\,N$. T\u00ednh di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c $MNEG$. <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $S_{MNEG}=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\,(cm^2)$ <\/span> ","hint":"T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $AMG$ theo di\u1ec7n t\u00edch tam gi\u00e1c $ANE$. <br\/>T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $ANE$ theo di\u1ec7n t\u00edch tam gi\u00e1c $ABC$<br\/>Sau \u0111\u00f3, t\u00ecm di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c $MNEG$ theo di\u1ec7n t\u00edch tam gi\u00e1c $ABC$. ","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai13/lv3/img\/H823_K1_2a.png' \/><\/center> G\u1ecdi $F$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $DC$. <br\/>Suy ra, $EF\/\/AD$ ($EF$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c $ADC$).<br\/>D\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c $ND$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c $BEF$ hay $N$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BE$.<br\/>G\u1ecdi $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $GE$. <br\/>Suy ra, $NI\/\/BG\\, \\Rightarrow NI\/\/MG\\, \\Rightarrow AM=MN.$ <br\/>Ta c\u00f3: <br\/> $S_{AMG}=\\dfrac{1}{2}S_{ANG}=\\dfrac{1}{2}S_{AGN}=\\dfrac{1}{6}S_{ANE}$(v\u00ec $AM=\\dfrac{1}{2}AN;\\,AG=\\dfrac{1}{3}AE$)<br\/>$S_{ANE}=\\dfrac{1}{2}S_{ABE}$(v\u00ec $NE=\\dfrac{1}{2}BE$)<br\/>$S_{ABE}=\\dfrac{1}{2}S_{ABC}$(v\u00ec $AE=\\dfrac{1}{2}AC$)<br\/>N\u00ean $S_{MNEG}=S_{ANE}-S_{AMG}=\\dfrac{5}{6}S_{ANE}=\\dfrac{5}{12}S_{ABE}=\\dfrac{5}{24}S_{ABC}$<br\/>$\\Rightarrow S_{MNEG}=\\dfrac{5}{24}.48=10\\,(cm^2)$<br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $10$. <\/span> <\/span> "}]}],"id_ques":1643},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["9"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<span class='basic_left'> $\\Delta ABC$ c\u00f3 $BC=15\\,cm$, \u0111\u01b0\u1eddng cao $AH=10\\,cm$. M\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng $d$ song song v\u1edbi $BC$ c\u1eaft c\u00e1c c\u1ea1nh $AB, AC$ theo th\u1ee9 t\u1ef1 t\u1ea1i $D$ v\u00e0 $E$. <br\/> <b> C\u00e2u 1: <\/b> N\u1ebfu kho\u1ea3ng c\u00e1ch t\u1eeb $D$ \u0111\u1ebfn c\u1ea1nh $BC$ b\u1eb1ng $4 \\,cm$ th\u00ec \u0111\u1ed9 d\u00e0i c\u1ea1nh $DE$ l\u00e0 bao nhi\u00eau? <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $DE=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\,(cm)$ <\/span> ","hint":"\u0110\u1eb7t $DE=x$. K\u1ebb $EK\\bot BC$ t\u1ea1i $K$ <br\/> $S_{ADE}+S_{BDEC}=S_{ABC}$.","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai13/lv3/img\/H823_K1_1.png' \/><\/center> G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AH$ v\u00e0 $DE$ <br\/> \u0110\u1eb7t $DE=x$ <br\/> K\u1ebb $EK\\bot BC$ t\u1ea1i $K,$ ta c\u00f3 kho\u1ea3ng c\u00e1ch t\u1eeb $D$ t\u1edbi $BC$ ch\u00ednh l\u00e0 \u0111\u1ed9 d\u00e0i $EK$ <br\/> Suy ra, $EK=4\\,cm$ <br\/> $S_{ADE}+S_{BDEC}=S_{ABC}$ <br\/>$\\dfrac{1}{2}AI.DE+\\dfrac{1}{2}EK.(DE+BC)=\\dfrac{1}{2}AH.BC$<br\/> $\\Rightarrow \\dfrac{1}{2}.x.6+\\dfrac{1}{2} (x+15).4=\\dfrac{1}{2}.15.10$ <br\/> $\\Rightarrow 3x+2x+30=75$ <br\/> $\\Rightarrow x=9$ <br\/> V\u1eady $DE=9\\,cm$ <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $9$. <\/span> <\/span> "}]}],"id_ques":1644},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["64"],["161"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<span class='basic_left'> $\\Delta ABC$ c\u00f3 $BC=15\\,cm$, \u0111\u01b0\u1eddng cao $AH=10\\,cm$. M\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng $d$ song song v\u1edbi $BC$ c\u1eaft c\u00e1c c\u1ea1nh $AB, AC$ theo th\u1ee9 t\u1ef1 t\u1ea1i $D$ v\u00e0 $E$. <br\/> <b> C\u00e2u 2: <\/b> N\u1ebfu \u0111\u1ed9 d\u00e0i $DE=8\\,cm$. T\u00ednh t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch tam gi\u00e1c $ADE$ v\u00e0 di\u1ec7n t\u00edch h\u00ecnh thang $BDEC$. <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $\\dfrac{S_{ADE}}{S_{BDEC}}=\\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ <\/span> ","hint":"T\u00ecnh $S_{ADE}$ v\u00e0 $S_{BDEC}$ r\u1ed3i suy ra t\u1ec9 s\u1ed1.","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai13/lv3/img\/H823_K1_1b.png' \/><\/center> G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AH$ v\u00e0 $DE$. <br\/>V\u00ec $AH\\bot BC;\\, DE\/\/BC$(gi\u1ea3 thi\u1ebft)<br\/>$\\Rightarrow AI\\bot BD$ <br\/> \u0110\u1eb7t $AI=x\\,(cm)\\,(x<10) \\Rightarrow IH=10-x\\,(cm)\\, (AH=AI+IH)$ <br\/>$S_{ADE}+S_{BDEC}=S_{ABC}$ <br\/>$\\dfrac{1}{2}AI.DE+\\dfrac{1}{2}IH.(DE+BC)=\\dfrac{1}{2}AH.BC$<br\/> $\\dfrac{1}{2}.x.8+\\dfrac{1}{2}.(8+15).(10-x)=75$($S_{ABC}=75\\,cm^2$) <br\/> $15x=80$ <br\/> $x=\\dfrac{16}{3}$ <br\/> Khi \u0111\u00f3, ta c\u00f3 $AI=\\dfrac{16}{3}\\,(cm); IH=\\dfrac{14}{3}\\,(cm)$ <br\/>V\u1eady $S_{ADE}=\\dfrac{1}{2}.\\dfrac{16}{3}.8=\\dfrac{64}{3}\\,(cm^2);\\,\\,S_{BDEC}=75-S_{ADE}=\\dfrac{161}{3}\\,(cm^2)$<br\/>Suy ra $\\dfrac{S_{ADE}}{S_{BDEC}}=\\dfrac{64}{161}$<br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $\\dfrac{64}{161}$. <\/span> <\/span> "}]}],"id_ques":1645},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["6"]]],"list":[{"point":10,"width":40,"type_input":"","ques":"<span class='basic_left'> $\\Delta ABC$ c\u00f3 $BC=15\\,cm$, \u0111\u01b0\u1eddng cao $AH=10\\,cm$. M\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng $d$ song song v\u1edbi $BC$ c\u1eaft c\u00e1c c\u1ea1nh $AB, AC$ theo th\u1ee9 t\u1ef1 t\u1ea1i $D$ v\u00e0 $E$. <br\/> <b> C\u00e2u 3: <\/b> N\u1ebfu $DE$ b\u1eb1ng kho\u1ea3ng c\u00e1ch t\u1eeb $d$ \u0111\u1ebfn $BC$ th\u00ec \u0111\u1ed9 d\u00e0i $DE$ l\u00e0 bao nhi\u00eau? <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $DE=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\,(cm)$ <\/span> ","hint":"\u0110\u1eb7t $DE=x$. K\u1ebb $EK\\bot BC$ t\u1ea1i $K$ <br\/> $S_{ADE}+S_{BDEC}=S_{ABC}$","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai13/lv3/img\/H823_K1_1c.jpg' \/><\/center> Ta k\u1ebb $EK\\bot BC$ t\u1ea1i $K$ khi \u0111\u00f3 kho\u1ea3ng c\u00e1ch t\u1eeb $d$ t\u1edbi $BC$ ch\u00ednh l\u00e0 \u0111\u1ed9 d\u00e0i $EK$ <br\/> \u0110\u1eb7t $DE=EK=x$ $\\Rightarrow AI=10-x$ <br\/> Ta c\u00f3: <br\/> $S_{ADE}+S_{BDEC}=S_{ABC}$ <br\/> $\\Rightarrow \\dfrac{1}{2}AI.DE+\\dfrac{1}{2}EK.(DE+BC)=\\dfrac{1}{2}AH.BC$ <br\/> $\\Rightarrow \\dfrac{1}{2}.x.(10-x)+\\dfrac{1}{2}(x+15).x=75$ ($S_{ABC}=75\\,cm^2$) <br\/> $\\Rightarrow 10x-x^2+x^2+15x=150$ <br\/> $\\Rightarrow 25x=150$ <br\/> $\\Rightarrow x=6$ <br\/> V\u1eady $DE=6\\,cm$ <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $6$. <\/span> <\/span> "}]}],"id_ques":1646},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0110\u00fang ho\u1eb7c Sai ","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho tam gi\u00e1c $ABC$. Tr\u00ean c\u1ea1nh $BC, AC$ v\u00e0 $AB$ l\u1ea5y c\u00e1c \u0111i\u1ec3m t\u01b0\u01a1ng \u1ee9ng ${{A}_{1}};\\,{{B}_{1}};\\,\\,{{C}_{1}}$ sao cho $B{{A}_{1}}=\\dfrac{1}{6}BC;\\,\\,C{{B}_{1}}=\\dfrac{1}{3}CA;\\,\\,A{{C}_{1}}=\\dfrac{1}{2}AB.$ Hai c\u1ea1nh $AA_1$ v\u00e0 $C{{C}_{1}}$ c\u1eaft nhau t\u1ea1i $P$, \u0111\u1ed3ng th\u1eddi hai \u0111\u01b0\u1eddng th\u1eb3ng \u0111\u00f3 c\u1eaft $B{{B}_{1}}$ l\u1ea7n l\u01b0\u1ee3t t\u1ea1i $M$ v\u00e0 $N$. <br\/> Ta ch\u1ee9ng minh \u0111\u01b0\u1ee3c: ${{S}_{MNP}}={{S}_{{{A}_{1}}MB}}+{{S}_{{{B}_{1}}CN}}+{{S}_{{{C}_{1}}AP}}.$ <\/span>","select":["\u0110\u00fang","Sai "],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai13/lv3/img\/H823_K1_2.jpg' \/><\/center> \u0110\u1eb7t $S_{A_1MB}=S_1;\\, S_{B_1NC}=S_2;\\, S_{C_1PA}=S_3$<br\/>Hai tam gi\u00e1c $ABA_1$ v\u00e0 $ABC$ c\u00f3 chung \u0111\u01b0\u1eddng cao xu\u1ea5t ph\u00e1t t\u1eeb $A$ v\u00e0 $B{{A}_{1}}=\\dfrac{1}{6}BC$ <br\/> $\\Rightarrow {{S}_{AB{{A}_{1}}}}=\\dfrac{1}{6}{{S}_{ABC}}$ <br\/> Hai tam gi\u00e1c $BCB_1$ v\u00e0 $ABC$ c\u00f3 chung \u0111\u01b0\u1eddng cao xu\u1ea5t ph\u00e1t t\u1eeb $B$ v\u00e0 $C{{B}_{1}}=\\dfrac{1}{3}CA$ <br\/> $\\Rightarrow {{S}_{BC{{B}_{1}}}}=\\dfrac{1}{3}{{S}_{ABC}}$ <br\/> Hai tam gi\u00e1c $ACC_1$ v\u00e0 $ABC$ c\u00f3 chung \u0111\u01b0\u1eddng cao xu\u1ea5t ph\u00e1t t\u1eeb $C$ v\u00e0 $C{{A}_{1}}=\\dfrac{1}{2}AB$ <br\/> $\\Rightarrow {{S}_{AC{{C}_{1}}}}=\\dfrac{1}{2}{{S}_{ABC}}$ <br\/> Do \u0111\u00f3 ${{S}_{AB{{A}_{1}}}}+{{S}_{BC{{B}_{1}}}}+{{S}_{AC{{C}_{1}}}}$<br\/> $=\\dfrac{1}{6}{{S}_{ABC}}+\\dfrac{1}{3}{{S}_{ABC}}+\\dfrac{1}{2}{{S}_{ABC}}$ <br\/> $={{S}_{ABC}}$<br\/> T\u1ee9c l\u00e0: ${{S}_{AB{{A}_{1}}}}+{{S}_{BC{{B}_{1}}}}+{{S}_{AC{{C}_{1}}}}={{S}_{ABC}}$ <br\/> $\\left( S_3+{{S}_{P{{C}_{1}}MB}}+S_1 \\right)$$+\\left( S_1+{{S}_{M{{A}_{1}}CN}}+S_2 \\right)$$+\\left( S_2+{{S}_{N{{B}_{1}}AP}}+S_3 \\right)$$={{S}_{ABC}}$ <br\/> $\\Rightarrow {{S}_{ABC}}-{{S}_{NPM}}$$+\\left( S_1+S_2+S_3 \\right)$$={{S}_{ABC}}$ <br\/> Suy ra ${{S}_{MNP}}={{S}_{{{A}_{1}}MB}}+{{S}_{{{B}_{1}}CN}}+{{S}_{{{C}_{1}}AP}}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang <\/span> <\/span> ","column":2}]}],"id_ques":1647},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t ","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho b\u00e1t gi\u00e1c \u0111\u1ec1u $ABCDEFGH$ c\u00f3 c\u1ea1nh $1\\,cm$. G\u1ecdi $I, K, M, N$ theo th\u1ee9 t\u1ef1 l\u00e0 giao \u0111i\u1ec3m c\u1ee7a c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng $AB$ v\u00e0 $CD, CD$ v\u00e0 $EF, EF$ v\u00e0 $GH, GH$ v\u00e0 $AB$. <br\/> <b> C\u00e2u 1: <\/b> T\u1ee9 gi\u00e1c $IKMN$ l\u00e0 h\u00ecnh g\u00ec? <\/span> ","select":["A. H\u00ecnh b\u00ecnh h\u00e0nh","B. H\u00ecnh ch\u1eef nh\u1eadt","C. H\u00ecnh thoi","D. H\u00ecnh vu\u00f4ng "],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai13/lv3/img\/H823_K1_3.jpg' \/><\/center> Ta c\u00f3 m\u1ed7i g\u00f3c c\u1ee7a b\u00e1t gi\u00e1c \u0111\u1ec1u c\u00f3 s\u1ed1 \u0111o l\u00e0: $\\dfrac{\\left( 8-2 \\right){{.180}^{o}}}{8}={{135}^{o}}$ <br\/>Khi \u0111\u00f3, Ta c\u00f3 $\\widehat{{{A}_{1}}}+\\widehat{{{A}_{2}}}={{180}^{o}}$ <br\/> $\\Rightarrow \\widehat{{{A}_{2}}}={{180}^{o}}-{{135}^{o}}={{45}^{o}}$ <br\/> $\\widehat{{{H}_{1}}}+\\widehat{{{H}_{2}}}={{180}^{o}}$ <br\/> $\\Rightarrow \\widehat{{{H}_{2}}}$$={{180}^{o}}-{{135}^{o}}={{45}^{o}}$ <br\/> $\\Rightarrow \\Delta ANH$ vu\u00f4ng c\u00e2n t\u1ea1i $N$ <br\/> $\\Rightarrow \\widehat{ANH}={{90}^{o}};\\,\\,NA=HN$ <br\/> T\u01b0\u01a1ng t\u1ef1 $\\Delta BIC;\\,\\,\\Delta DEK;\\,\\,\\Delta G\\text{EF}$ vu\u00f4ng c\u00e2n <br\/> $\\Rightarrow \\widehat{N}=\\widehat{I}=\\widehat{K}=\\widehat{M}={{90}^{o}}$ <br\/> D\u1ec5 d\u00e0ng ch\u1ee9ng minh \u0111\u01b0\u1ee3c: $NI = IK$ <br\/> Do \u0111\u00f3 t\u1ee9 gi\u00e1c $IKMN$ l\u00e0 h\u00ecnh vu\u00f4ng.<br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D. <\/span> <\/span> ","column":4}]}],"id_ques":1648},{"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 b\u00e1t gi\u00e1c \u0111\u1ec1u $ABCDEFGH$ c\u00f3 c\u1ea1nh $1\\,cm$. G\u1ecdi $I, K, M, N$ theo th\u1ee9 t\u1ef1 l\u00e0 giao \u0111i\u1ec3m c\u1ee7a c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng $AB$ v\u00e0 $CD, CD$ v\u00e0 $EF, EF$ v\u00e0 $GH, GH$ v\u00e0 $AB$. <br\/> <b> C\u00e2u 2: <\/b> \u0110\u1ed9 d\u00e0i c\u1ea1nh c\u1ee7a t\u1ee9 gi\u00e1c $IKMN$ l\u00e0: <br\/> <i> (l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t)<\/i> <\/span> ","select":["A. $2,4\\,cm$","B. $3,2\\,cm$","C. $1,5\\,cm$","D. $4,2\\,cm$ "],"explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai13/lv3/img\/H823_K1_3.jpg' \/><\/center> Theo c\u00e2u 1, ta \u0111\u00e3 ch\u1ee9ng minh \u0111\u01b0\u1ee3c $IKMN$ l\u00e0 h\u00ecnh vu\u00f4ng. <br\/> \u0110\u1eb7t $AN= x$ <br\/> X\u00e9t $\\Delta NAH$ vu\u00f4ng t\u1ea1i $N$, c\u00f3: $N{{A}^{2}}+N{{H}^{2}}=A{{H}^{2}}$ (\u0111\u1ecbnh l\u00ed Pi - ta - go) <br\/> $\\Rightarrow 2{{x}^{2}}=1$ <br\/> $\\Rightarrow x=\\dfrac{1}{\\sqrt{2}}\\approx 0,7$ <br\/>C\u1ea1nh c\u1ee7a h\u00ecnh vu\u00f4ng $IKMN$ l\u00e0: <br\/> $0,7+1+0,7\\approx 2,4\\,\\,\\left( cm \\right)$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A. <\/span> <\/span> ","column":4}]}],"id_ques":1649},{"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 b\u00e1t gi\u00e1c \u0111\u1ec1u $ABCDEFGH$ c\u00f3 c\u1ea1nh $1\\,cm$. G\u1ecdi $I, K, M, N$ theo th\u1ee9 t\u1ef1 l\u00e0 giao \u0111i\u1ec3m c\u1ee7a c\u00e1c \u0111\u01b0\u1eddng th\u1eb3ng $AB$ v\u00e0 $CD, CD$ v\u00e0 $EF, EF$ v\u00e0 $GH, GH$ v\u00e0 $AB$ <br\/> <b> C\u00e2u 3: <\/b> Di\u1ec7n t\u00edch b\u00e1t gi\u00e1c \u0111\u1ec1u $ABCDEFGH$ l\u00e0: <br\/> <i> (l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t)<\/i> <\/span> ","select":["A. $5,8\\,cm^2$","B. $7,2\\,cm^2$","C. $4,8\\,cm^2$","D. $2,4\\,cm^2$ "],"hint":"${{S}_{ABCDEFGH}}={{S}_{MKIN}}-4{{S}_{NAH}}$","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/hinhhoc/bai13/lv3/img\/H823_K1_3.jpg' \/><\/center> Theo c\u00e2u 3, ta \u0111\u00e3 t\u00ednh \u0111\u01b0\u1ee3c c\u1ea1nh c\u1ee7a h\u00ecnh vu\u00f4ng $IKMN$ c\u00f3 \u0111\u1ed9 d\u00e0i l\u00e0: $2,4\\,cm$ <br\/> Ta c\u00f3: <br\/> ${{S}_{MKIN}}=N{{I}^{2}}={{2,4}^{2}}=5,76\\,\\,\\left( c{{m}^{2}} \\right)$ <br\/> ${{S}_{NAH}}=\\dfrac{1}{2}NA.NH$ <br\/> $=\\dfrac{1}{2}.\\dfrac{1}{\\sqrt{2}}.\\dfrac{1}{\\sqrt{2}}$$=\\dfrac{1}{4}\\,\\left( c{{m}^{2}} \\right)$ <br\/> $\\Rightarrow {{S}_{ABCDEFGH}}={{S}_{MKIN}}-4{{S}_{NAH}}$$=5,76-1\\approx 4,8\\,\\left( c{{m}^{2}} \\right)$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C. <\/span> <\/span> ","column":4}]}],"id_ques":1650}],"lesson":{"save":0,"level":3}}