Quan hệ giữa đường vuông góc và đường xiên, đường xiên và hình chiếu level3

Home Lớp 7 Toán lớp 7 - Sách kết nối tri thức Quan hệ giữa đường vuông góc và đường xiên, đường xiên và hình chiếuBài tập nâng cao

{"common":{"save":0,"post_id":"1339","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"1851","post_id":"1339","mon_id":"0","chapter_id":"0","question":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","options":{"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":[[["5"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Cho $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 $AB = 3cm$, $AC = 4cm$. G\u1ecdi $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AC$, $d$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a \u0111o\u1ea1n $AC$ v\u00e0 $M$ l\u00e0 \u0111i\u1ec3m t\u00f9y \u00fd tr\u00ean $d$. H\u00e3y t\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $MA + MB$ <br\/> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $Min_{(MA + MB)}$ = _input_ $cm$ ","hint":"Ch\u1ec9 ra $MA + MB$ l\u1edbn h\u01a1n ho\u1eb7c b\u1eb1ng m\u1ed9t gi\u00e1 tr\u1ecb kh\u00f4ng \u0111\u1ed5i, t\u1eeb \u0111\u00f3 t\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1:<\/b> Ch\u1ee9ng minh $MA = MC$, khi \u0111\u00f3 $MA + MB = MB + MC$ <br\/>B\u01b0\u1edbc 2:<\/b> So s\u00e1nh $MB + MC$ v\u1edbi $BC$ r\u1ed3i t\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $MA + MB$ <br\/>B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='img\/H7C3B18_K01.png' \/><\/center><br\/> G\u1ecdi $J$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a \u0111\u01b0\u1eddng th\u1eb3ng $d$ v\u00e0 $BC$ <br\/> V\u00ec $AI = IC$ (gt) <br\/> $MI \\perp AC$ (gt) <br\/> $\\Rightarrow$ $MA = MC$ (H\u00ecnh chi\u1ebfu b\u1eb1ng nhau th\u00ec \u0111\u01b0\u1eddng xi\u00ean b\u1eb1ng nhau) <br\/> $\\Rightarrow$ $MA + MB = MB + MC$ (1) <br\/> M\u00e0 $MB + MC \\geq BC$ (b\u1ea5t \u0111\u1eb3ng th\u1ee9c h\u1ec7 \u0111i\u1ec3m) (2) <br\/> X\u00e9t $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$ c\u00f3: <br\/> $BC^{2} = AB^{2} + AC^{2} = 3^{2} + 4^{2} = 5^{2}$ (\u0111\u1ecbnh l\u00fd Pitago) <br\/> $\\Rightarrow$ $BC = 5(cm)$ (3) <br\/> T\u1eeb (1), (2) v\u00e0 (3) $\\Rightarrow$ $MA + MB \\geq 5$ <br\/> $\\Rightarrow$ $MA + MB$ nh\u1ecf nh\u1ea5t l\u00e0 b\u1eb1ng $5$ khi v\u00e0 ch\u1ec9 khi $MB + MC = BC$ <br\/> $\\Leftrightarrow$ $M$ n\u1eb1m tr\u00ean \u0111o\u1ea1n th\u1eb3ng $BC$ <br\/> $\\Leftrightarrow$ $M \\equiv J$ <br\/> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0: 5 <\/span> <br\/> Ch\u00fa \u00fd: \u0110\u1ec3 gi\u1ea3i b\u00e0i to\u00e1n c\u1ef1c tr\u1ecb h\u00ecnh h\u1ecdc \"T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a t\u1ed5ng $MA + MB$ \", ta ph\u1ea3i ch\u1ee9ng t\u1ecf $MA + MB$ l\u1edbn h\u01a1n ho\u1eb7c b\u1eb1ng m\u1ed9t s\u1ed1 kh\u00f4ng \u0111\u1ed5i v\u00e0 ph\u1ea3i ch\u1ec9 r\u00f5 d\u1ea5u b\u1eb1ng c\u00f3 th\u1ec3 x\u1ea3y ra. \u0110i\u1ec1u ki\u1ec7n x\u1ea3y ra d\u1ea5u b\u1eb1ng l\u00e0 r\u1ea5t quan tr\u1ecdng. <br\/> Ch\u1eb3ng h\u1ea1n v\u1edbi b\u00e0i to\u00e1n tr\u00ean ta lu\u00f4n c\u00f3: $MA + MB \\geq AB = 3$. Tuy nhi\u00ean d\u1ea5u \"=\" kh\u00f4ng th\u1ec3 x\u1ea3y ra v\u00ec $M$ kh\u00f4ng th\u1ec3 n\u1eb1m tr\u00ean \u0111o\u1ea1n th\u1eb3ng $AB$. V\u00ec v\u1eady kh\u00f4ng c\u00f3 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $MA + MB$ l\u00e0 3 <\/i> <\/span> "}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:04"},{"id":"1852","post_id":"1339","mon_id":"0","chapter_id":"0","question":"\u0110i\u1ec1n d\u1ea5u (\">\", \"<\" ho\u1eb7c \"=\") th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","options":{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u (\">\", \"<\" ho\u1eb7c \"=\") th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[[">"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Cho $\\triangle{ABC}$ vu\u00f4ng t\u1ea1i $A$. G\u1ecdi $H$ l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $A$ tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng $BC$. H\u00e3y so s\u00e1nh $AH + BC$ v\u1edbi $AB + AC$ <br\/> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $AH + BC$ _input_ $AB + AC$ ","hint":"Tr\u00ean $AC$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $CA = CE$, tr\u00ean $AB$ l\u1ea5y \u0111i\u1ec3m $F$ sao cho $AH = AF$","explain":"Ph\u00e2n t\u00edch b\u00e0i to\u00e1n:<\/span><br\/> \u0110\u1ec3 so s\u00e1nh \u0111\u01b0\u1ee3c $AH + BC$ v\u1edbi $AB + AC$ ta c\u1ea7n v\u1ebd th\u00eam \u0111\u01b0\u1eddng ph\u1ee5 l\u00e0m xu\u1ea5t hi\u1ec7n \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c, c\u00e1c \u0111\u01b0\u1eddng xi\u00ean v\u00e0 h\u00ecnh chi\u1ebfu \u0111\u1ec3 \u00e1p d\u1ee5ng \u0111\u01b0\u1ee3c \u0111\u1ecbnh l\u00fd quan h\u1ec7 gi\u1eefa \u0111\u01b0\u1eddng xi\u00ean v\u00e0 \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c, \u0111\u01b0\u1eddng xi\u00ean v\u00e0 h\u00ecnh chi\u1ebfu. C\u1ee5 th\u1ec3 ta c\u1ea7n l\u1ea5y \u0111i\u1ec3m $E$ tr\u00ean $BC$ sao cho $CA = CE$, tr\u00ean $AB$ l\u1ea5y \u0111i\u1ec3m $F$ sao cho $AH = AF$. Khi \u0111\u00f3 b\u00e0i to\u00e1n chuy\u1ec3n v\u1ec1 so s\u00e1nh $BE$ v\u00e0 $BF$ <br\/>B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='img\/H7C3B18_K02.png' \/><\/center><br\/> $\\blacktriangleright$ K\u1ebb $AH \\perp BC$, ($H \\in BC$) <br\/> Tr\u00ean tia $CB$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $CE = CA$ (1) <br\/> Tr\u00ean tia $AB$ l\u1ea5y \u0111i\u1ec3m l\u1ea5y \u0111i\u1ec3m $F$ sao cho $AF = AH$ <br\/> $\\blacktriangleright$ V\u00ec $AH \\perp BC$ n\u00ean \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c $CH$ nh\u1ecf h\u01a1n \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng xi\u00ean $CA$ <br\/> $\\Rightarrow$ $CH < CE$ (2) <br\/> M\u1eb7t kh\u00e1c do $CA \\perp AB$ n\u00ean $CA < CB$ (\u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c nh\u1ecf h\u01a1n \u0111\u01b0\u1eddng xi\u00ean) (3) <br\/> T\u1eeb (1), (2) v\u00e0 (3) $\\Rightarrow$ $CH < CE < CB$, suy ra $E$ n\u1eb1m gi\u1eefa $H$ v\u00e0 $B$ <br\/> V\u00ec $AH < AB$ (\u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c nh\u1ecf h\u01a1n \u0111\u01b0\u1eddng xi\u00ean) <br\/> $\\Rightarrow$ $AF < AB$ (v\u00ec $AH = AF$) <br\/> $\\Rightarrow$ \u0110i\u1ec3m $F$ n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m $A$ v\u00e0 $B$ <br\/> $\\blacktriangleright$ L\u1ea1i c\u00f3 $CE = CA$ $\\Rightarrow$ $\\triangle{ACE}$ c\u00e2n t\u1ea1i $C$ $\\Rightarrow$ $\\widehat{CAE} = \\widehat{E_{1}}$ <br\/> M\u00e0 $\\begin{cases} \\widehat{CAE} + \\widehat{A_{1}} = 90^{o} \\\\ \\widehat{E_{1}} + \\widehat{A_{2}} = 90^{o} \\end{cases}$ <br\/> $\\Rightarrow$ $\\widehat{A_{1}} = \\widehat{A_{2}}$ <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{AHE}$ v\u00e0 $\\triangle{AFE}$ c\u00f3: <br\/> $\\begin{cases} AE \\hspace{0,2cm} \\text{chung} \\\\ \\widehat{A_{1}} = \\widehat{A_{2}} (cmt) \\\\ AH = AF (\\text{c\u00e1ch} \\hspace{0,2cm} \\text{l\u1ea5y} \\hspace{0,2cm} \\text{\u0111i\u1ec3m} F) \\end{cases}$ <br\/> $\\Rightarrow \\triangle{AHE} = \\triangle{AFE}$ (c.g.c) <br\/> $\\Rightarrow$ $ \\widehat{AFE} = \\widehat{AHE} = 90^{o}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> Hay $EF \\perp AB$ <br\/> $\\blacktriangleright$ Ta c\u00f3: $AH + BC = AF + CE + BE$ (4) <br\/> $AB + AC = AF + BF + CE$ (5) <br\/> M\u00e0 $BE > BF$ (\u0111\u01b0\u1eddng xi\u00ean l\u1edbn h\u01a1n \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c) (6) <br\/> T\u1eeb (4), (5), (6) $\\Rightarrow$ $AH + BC > AB + AC$ <br\/> V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 d\u1ea5u: > <\/span> <br\/> Nh\u1eadn x\u00e9t: \u1ede b\u00e0i to\u00e1n n\u00e0y ta \u0111\u00e3 s\u1eed d\u1ee5ng m\u1ed9t ph\u01b0\u01a1ng ph\u00e1p quen thu\u1ed9c c\u1ee7a l\u1edbp $6$: \u0110\u1ec3 ch\u1ee9ng minh \u0111i\u1ec3m $E$ n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m $H$ v\u00e0 $B$ ta ch\u1ee9ng minh $CH < CE < CB$, t\u01b0\u01a1ng t\u1ef1 v\u1edbi \u0111i\u1ec3m $F$ <\/i> <\/span> "}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:04"},{"id":"1853","post_id":"1339","mon_id":"0","chapter_id":"0","question":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","options":{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Cho $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$. T\u1eeb m\u1ed9t \u0111i\u1ec3m $E$ b\u1ea5t k\u00ec tr\u00ean c\u1ea1nh $AB$, k\u1ebb m\u1ed9t \u0111\u01b0\u1eddng th\u1eb3ng song song v\u1edbi \u0111\u00e1y $BC$, \u0111\u01b0\u1eddng th\u1eb3ng n\u00e0y c\u1eaft $AC$ t\u1ea1i $F$. Khi \u0111\u00f3 $BF = \\dfrac{EF + BC}{2}$. \u0110\u00fang <\/b> hay sai <\/b>? ","select":[" A. \u0110\u00daNG"," B. SAI"],"hint":"K\u1ebb $AH \\perp BC$ l\u00e0m xu\u1ea5t hi\u1ec7n \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c, \u0111\u01b0\u1eddng xi\u00ean v\u00e0 h\u00ecnh chi\u1ebfu \u0111\u1ec3 so s\u00e1nh","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> K\u1ebb $AH \\perp BC$, g\u1ecdi $J$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AH$ v\u00e0 $EF$, g\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BF$ v\u00e0 $AH$ <br\/> So s\u00e1nh $HB$ v\u00e0 $HC$ <br\/> B\u01b0\u1edbc 2: <\/b> So s\u00e1nh $JE$ v\u00e0 $JF$ <br\/> B\u01b0\u1edbc 3: <\/b> So s\u00e1nh $IB$ v\u1edbi $HB$, $IF$ v\u1edbi $JF$ r\u1ed3i so s\u00e1nh $BC$ v\u1edbi $\\dfrac{EF + BC}{2}$ <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='img\/H7C3B18_K03.png' \/><\/center> <br\/> $\\blacktriangleright$ K\u1ebb $AH \\perp BC$, do $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ n\u00ean $\\widehat{B}$ v\u00e0 $\\widehat{C}$ nh\u1ecdn, suy ra \u0111i\u1ec3m $H$ n\u1eb1m gi\u1eefa hai \u0111i\u1ec3m $B$ v\u00e0 $C$ <br\/> V\u00ec $EF \/\/ BC$ (gt) $\\Rightarrow$ $AH \\perp EF$, gi\u1ea3 s\u1eed t\u1ea1i \u0111i\u1ec3m $J$ <br\/> G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $BF$ v\u00e0 $AH$ <br\/> $\\blacktriangleright$ Ta c\u00f3: $AB = AC$ ($\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$) <br\/> $\\Rightarrow$ $HB =HC$ (\u0111\u01b0\u1eddng xi\u00ean b\u1eb1ng nhau th\u00ec h\u00ecnh chi\u1ebfu b\u1eb1ng nhau) (1) <br\/> $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ n\u00ean $\\widehat{B} = \\widehat{C}$ <br\/> $\\Rightarrow$ $\\widehat{AEF} = \\widehat{B} = \\widehat{C} = \\widehat{AFE}$ (\u0111\u1ed3ng v\u1ecb) <br\/> $\\Rightarrow$ $\\triangle{AEF}$ c\u00e2n t\u1ea1i $A$ $\\Rightarrow$ $AE = AF$ <br\/> $\\Rightarrow$ $JE = JF$ (\u0111\u01b0\u1eddng xi\u00ean b\u1eb1ng nhau th\u00ec h\u00ecnh chi\u1ebfu b\u1eb1ng nhau) (2) <br\/> $\\blacktriangleright$ M\u1eb7t kh\u00e1c ta c\u00f3: $BI > BH$ (\u0111\u01b0\u1eddng xi\u00ean l\u1edbn h\u01a1n \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c) (3) <br\/> $FI > FJ$ (\u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c l\u1edbn h\u01a1n h\u00ecnh chi\u1ebfu) (4) <br\/> T\u1eeb (1), (2), (3), (4) $\\Rightarrow$ $BI + IF > BH + JF = \\dfrac{1}{2}BC + \\dfrac{1}{2}EF$ <br\/> Hay $BF > \\dfrac{BC + EF}{2}$ <br\/> V\u1eady kh\u1eb3ng \u0111\u1ecbnh $BF = \\dfrac{EF + BC}{2}$ l\u00e0 SAI <\/span> ","column":2}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:04"}]}