Ôn tập chương III - hình 7

Home Lớp 7 Toán lớp 7 - Sách kết nối tri thức Ôn tập chương IIIBài tập nâng cao

{"common":{"save":0,"post_id":"1511","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"2031","post_id":"1511","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 tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$. Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $BA$ l\u1ea5y \u0111i\u1ec3m $D$ v\u00e0 tr\u00ean c\u1ea1nh $AC$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $BD = CE$. H\u00e3y so s\u00e1nh $BC$ v\u00e0 $DE$. <br\/> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $BC$ _input_ $DE$ ","hint":"S\u1eed d\u1ee5ng quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong tam gi\u00e1c","explain":" <center><img src='img\/H7C3B24_K1.png' \/><\/center><br\/> $\\blacktriangleright$ $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ $\\Rightarrow$ $\\widehat{ABC} = \\widehat{ACB}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> X\u00e9t $\\triangle{BCD}$, c\u00f3: $\\widehat{ABC} > \\widehat{BDC}$ (t\u00ednh ch\u1ea5t g\u00f3c ngo\u00e0i c\u1ee7a tam gi\u00e1c) <br\/> M\u1eb7t kh\u00e1c: $\\widehat{DCE} > \\widehat{ACB} = \\widehat{ABC}$ <br\/> T\u1eeb \u0111\u00f3 suy ra $\\widehat{DCE} > \\widehat{BDC}$ (t\u00ednh ch\u1ea5t b\u1eafc c\u1ea7u) <br\/> $\\blacktriangleright$ X\u00e9t $\\triangle{BCD}$ v\u00e0 $\\triangle{CDE}$, c\u00f3: <br\/> $\\begin{cases} BD = CE \\hspace{0,2cm} (gt) \\\\ \\widehat{BDC} < \\widehat{DCE} \\hspace{0,2cm} (cmt) \\\\ CD \\hspace{0,2cm} \\text{chung} \\end{cases}$ <br\/> $\\Rightarrow$ $BC < DE$ (quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong hai tam gi\u00e1c) <br\/> V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0: $<$ <\/span> <br\/> Nh\u1eadn x\u00e9t: T\u1eeb b\u00e0i to\u00e1n n\u00e0y ta suy ra k\u1ebft qu\u1ea3 sau: <br\/> Trong tam gi\u00e1c $ABC$ c\u00f3 g\u00f3c $A$ v\u00e0 $AB + AC$ kh\u00f4ng \u0111\u1ed5i, tam gi\u00e1c c\u00e2n (t\u1ea1i $A$) l\u00e0 tam gi\u00e1c c\u00f3 chu vi nh\u1ecf nh\u1ea5t <br\/> Quan h\u1ec7 gi\u1eefa g\u00f3c v\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n trong hai tam gi\u00e1c: N\u1ebfu hai tam gi\u00e1c c\u00f3 hai c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng b\u1eb1ng nhau nh\u01b0ng c\u00e1c g\u00f3c xen gi\u1eefa kh\u00f4ng b\u1eb1ng nhau th\u00ec \u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c l\u1edbn h\u01a1n l\u00e0 c\u1ea1nh l\u1edbn h\u01a1n. \u0110\u1ea3o l\u1ea1i \u0111\u1ed1i di\u1ec7n v\u1edbi c\u1ea1nh l\u1edbn h\u01a1n l\u00e0 g\u00f3c l\u1edbn h\u01a1n <\/i><\/span> "}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:07"},{"id":"2032","post_id":"1511","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_random","correct":[[["<"],["="]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Cho $\\widehat{xOy} = 60^{o}$, $A$ l\u00e0 \u0111i\u1ec3m tr\u00ean tia $Ox$, $B$ l\u00e0 \u0111i\u1ec3m tr\u00ean tia $Oy$ ($A, B$ kh\u00f4ng tr\u00f9ng v\u1edbi $O$). H\u00e3y so s\u00e1nh $OA + OB$ v\u1edbi $2AB$ <br\/> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $OA + OB$ _input_ $2AB$ ho\u1eb7c $OA + OB$ _input_ $2AB$ ","hint":"S\u1eed d\u1ee5ng quan h\u1ec7 gi\u1eefa \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c - \u0111\u01b0\u1eddng xi\u00ean - h\u00ecnh chi\u1ebfu <br\/> V\u1ebd tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $xOy$","explain":" <center><img src='img\/H7C3B24_K2.png' \/><\/center><br\/> $\\blacktriangleright$ K\u1ebb tia ph\u00e2n gi\u00e1c $Ot$ c\u1ee7a $\\widehat{xOy}$. <br\/> G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 $Ot$; $H, K$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a $A, B$ tr\u00ean $Ot$ <br\/> $\\blacktriangleright$ V\u00ec $Ot$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{xOy}$ n\u00ean $\\widehat{xOt} = \\widehat{yOt} = \\dfrac{\\widehat{xOy}}{2} = \\dfrac{60^{o}}{2} = 30^{o}$ <br\/> X\u00e9t $\\triangle{OAH}$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: <br\/> $\\widehat{AOH} = 30^{o}$ n\u00ean $OA = 2AH$ (trong tam gi\u00e1c vu\u00f4ng c\u1ea1nh \u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c $30^{o}$ b\u1eb1ng n\u1eeda c\u1ea1nh huy\u1ec1n) <br\/> V\u00ec $AH, AI$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c, \u0111\u01b0\u1eddng xi\u00ean k\u1ebb t\u1eeb $A$ \u0111\u1ebfn \u0111\u01b0\u1eddng th\u1eb3ng $Ot$ n\u00ean $AH \\leq AI$ <br\/> Do v\u1eady: $OA \\leq 2AI$ (1) <br\/> $\\blacktriangleright$ Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u00f3 <br\/> $OB = 2BK \\leq 2BI$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) ta c\u00f3: $OA + OB \\leq 2AI + 2BI = 2(AI + BI) = 2AB$ <br\/> D\u1ea5u \"=\" x\u1ea3y ra khi $H \\equiv I \\equiv K$ hay $AB \\perp Ot$ <br\/> V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0: $<$ v\u00e0 $=$ <\/span> <br\/> Nh\u1eadn x\u00e9t: Ch\u00eca kh\u00f3a \u0111\u1ec3 gi\u1ea3i b\u00e0i to\u00e1n tr\u00ean l\u00e0 vi\u1ec7c v\u1ebd tia ph\u00e2n gi\u00e1c $Ot$. Nh\u1eefng b\u01b0\u1edbc ti\u1ebfp theo tr\u1edf n\u00ean \u0111\u01a1n gi\u1ea3n h\u01a1n v\u1edbi vi\u1ec7c s\u1eed d\u1ee5ng t\u00ednh ch\u1ea5t c\u1ee7a tam gi\u00e1c vu\u00f4ng c\u00f3 g\u00f3c $30^o$ v\u00e0 quan h\u1ec7 gi\u1eefa \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c v\u00e0 \u0111\u01b0\u1eddng xi\u00ean <\/i><\/span> "}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:07"},{"id":"2033","post_id":"1511","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 tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$. G\u1ecdi $H$ l\u00e0 h\u00ecnh chi\u1ebfu c\u1ee7a \u0111i\u1ec3m $A$ tr\u00ean \u0111\u01b0\u1eddng th\u1eb3ng $BC$. H\u00e3y so s\u00e1nh $AH + BC$ v\u00e0 $AB + AC$ <br\/> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $AH + BC$ _input_ $AB + AC$ ","hint":"S\u1eed d\u1ee5ng quan h\u1ec7 gi\u1eefa \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c - \u0111\u01b0\u1eddng xi\u00ean","explain":" <center><img src='img\/H7C3B24_K3.png' \/><\/center><br\/> $\\blacktriangleright$ K\u1ebb \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c $AH$ xu\u1ed1ng $BC$ $\\Rightarrow$ $H \\in BC$ <br\/> Tr\u00ean tia $CB$ l\u1ea5y \u0111i\u1ec3m $E$ sao cho $CE = CA$, tr\u00ean tia $AB$ l\u1ea5y \u0111i\u1ec3m $F$ sao cho $AF = AH$ <br\/> $\\blacktriangleright$ Ta c\u00f3: $AH < CA$ (quan h\u1ec7 \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c - \u0111\u01b0\u1eddng xi\u00ean) <br\/> $\\Rightarrow$ $CH < CE$ (v\u00ec $CE = CA$) <br\/> M\u1eb7t kh\u00e1c: $CA < CB$ (quan h\u1ec7 \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c - \u0111\u01b0\u1eddng xi\u00ean) <br\/> $\\Rightarrow$ $CE < CB$ (v\u00ec $CE = CA$) <br\/> T\u1eeb \u0111\u00f3 ta c\u00f3: $CH < CE < CB$ hay $E$ n\u1eb1m gi\u1eefa $H$ v\u00e0 $B$ <br\/> V\u00ec $AH < AB$ (quan h\u1ec7 \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c - \u0111\u01b0\u1eddng xi\u00ean) <br\/> $\\Rightarrow$ $AF < AB$ $\\Rightarrow$ $F$ n\u1eb1m gi\u1eefa $A$ v\u00e0 $B$ <br\/> $\\blacktriangleright$ V\u00ec $CE = CA$ $\\Rightarrow$ $\\triangle{ACE}$ c\u00e2n t\u1ea1i $C$ (\u0111\u1ecbnh ngh\u0129a) <br\/> $\\Rightarrow$ $\\widehat{CAE} = \\widehat{E_{1}}$ (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n) <br\/> M\u00e0 $\\widehat{CAE} + \\widehat{A_{1}} = 90^{o}$ <br\/> $\\widehat{E_{1}} + \\widehat{A_{2}} = 90^{o}$ ($\\triangle{AHE}$ vu\u00f4ng t\u1ea1i $E$) <br\/> $\\Rightarrow$ $\\widehat{A_{1}} = \\widehat{A_{2}}$ <br\/> Do \u0111\u00f3 d\u1ec5 d\u00e0ng ch\u1ec9 ra \u0111\u01b0\u1ee3c $\\triangle{AHE} = \\triangle{AFE}$ (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: $AH + BC = AF + CE + BE$ <br\/> $AB + AC = AF + BF + CE$ <br\/> M\u00e0 $BE > BF$ (quan h\u1ec7 \u0111\u01b0\u1eddng vu\u00f4ng g\u00f3c - \u0111\u01b0\u1eddng xi\u00ean) <br\/> $\\Rightarrow$ $AH + BC > AB + AC$ <br\/> V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0: $>$ <\/span> "}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:07"}]}