Bài tập nâng cao bài Định lí Py-ta-go

Home Lớp 7 Toán lớp 7 - Sách cũ Định lí Py-ta-go Bài tập nâng cao

{"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":"T\u00ecm c\u00e2u tr\u1ea3 l\u1eddi sai: <\/b><br\/> Cho h\u00ecnh v\u1ebd, bi\u1ebft: <br\/> $ AB = 1cm, BC = \\sqrt{2} cm , \\widehat{ADC} = 45^{o} $. Khi \u0111\u00f3: <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai14/lv3/img\/H7B14-03.png' \/><\/center> ","select":["A. $\\widehat{ABC} = \\widehat{ACB} = \\widehat{CAD} $","B. $\\widehat{BAC} = 90^{o}, \\widehat{BAD} = 135^{o} $","C. $AB \/\/ CD, AD \/\/ BC $ ","D. $AD = CD = 1cm $ "],"explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai14/lv3/img\/H7B14-03.png' \/><\/center> Ta c\u00f3 $\\triangle ABC $ vu\u00f4ng t\u1ea1i A n\u00ean theo \u0111\u1ecbnh l\u00ed Py-ta-go c\u00f3: <br\/> $BC^{2} = AB^{2} + AC^{2} \\\\ \\Rightarrow AC^{2} = BC^{2} - AB^{2} \\\\ = (\\sqrt{2})^{2} - (1)^{2} \\\\ = 1^{2} \\\\ \\Rightarrow AC = 1 cm $ <br\/> C\u00f3 $AB = AC = 1cm$ n\u00ean $\\triangle ABC $ vu\u00f4ng c\u00e2n t\u1ea1i A (1) <br\/> M\u00e0 trong $\\Delta ADC$ vu\u00f4ng t\u1ea1i C v\u00e0 c\u00f3 $\\widehat{ADC}=45^o$ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow \\widehat{DAC}=45^o$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) suy ra $\\widehat{ABC} = \\widehat{ACB} = \\widehat{CAD} (= 45^{o}) \\Rightarrow $ A \u0111\u00fang. <br\/> Tam gi\u00e1c $ACD$ vu\u00f4ng t\u1ea1i C, c\u00f3 $\\widehat{CDA} = 45^{o} $ <br\/> $\\Rightarrow \\widehat{CAD} = 45^{o} $ <br\/> $\\Rightarrow \\widehat{BAD} = 90^{o} + 45^{o} = 135^{o} \\Rightarrow $ B \u0111\u00fang. <br\/> C\u00f3 $\\widehat{BAC}= \\widehat{ACD} (= 90^{o}) \\Rightarrow AB \/\/ CD$ (so le trong) <br\/> C\u00f3 $\\widehat{CAD}= \\widehat{ACB} (= 45^{o}) \\Rightarrow AD \/\/ BC$ (so le trong) <br\/> $ \\Rightarrow $ C \u0111\u00fang <br\/> Tam gi\u00e1c ACD vu\u00f4ng c\u00e2n t\u1ea1i C n\u00ean: <br\/> $CA = CD = 1cm $ (v\u00ec AC = AB = 1cm) <br\/> $AD^{2} = AC^{2} + CD^{2} = 1^{2} + 1^{1} = 2 \\\\ \\Rightarrow AD = \\sqrt{2} $ <br\/> $\\Rightarrow $ D sai. <br\/> <br\/> C\u00e2u tr\u1ea3 l\u1eddi sai l\u00e0 D. <\/span> ","column":2}]}],"id_ques":1721},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["20"],["21"]]],"list":[{"point":10,"width":20,"content":"","type_input":"","ques":" Cho tam gi\u00e1c nh\u1ecdn $ABC$.<br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai14/lv3/img\/H7B14-10.png' \/><\/center> <br\/> K\u1ebb $AH$ vu\u00f4ng g\u00f3c v\u1edbi $BC$ $(H \\in BC)$. <br\/> Cho bi\u1ebft $AB = 13cm, AH = 12cm, HC = 16cm$. <br\/> T\u00ednh \u0111\u1ed9 d\u00e0i c\u00e1c c\u1ea1nh $AC; BC$. <br\/> \u0110\u00e1p \u00e1n: <\/b> $AC = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} (cm) \\\\ BC = \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} (cm)$ ","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> - D\u1ef1a v\u00e0o tam gi\u00e1c vu\u00f4ng $AHC$ t\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh $AC$. <br\/> - D\u1ef1a v\u00e0o tam gi\u00e1c vu\u00f4ng $AHB$ t\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh $BH$. <br\/> - T\u00ednh $BC = BH + HC$ <br\/> B\u00e0i gi\u1ea3i:<\/span> <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai14/lv3/img\/H7B14-10.png' \/><\/center>$\\triangle AHC $ vu\u00f4ng t\u1ea1i H n\u00ean theo \u0111\u1ecbnh l\u00ed Py-ta-go ta c\u00f3: <br\/> $AC^{2}$$ = AH^{2} + HC^{2} \\\\ = 12^{2} + 16^{2} \\\\ = 144 + 256 \\\\ = 400 = 20^{2} $ <br\/> $\\Rightarrow AC = 20 (cm) $ <br\/> $\\triangle AHB $ vu\u00f4ng t\u1ea1i B n\u00ean: <br\/> $BH^{2} $$= AB^{2} - AH^{2} \\\\ = 13^{2} - 12^{2} \\\\ = 169 - 144 \\\\ = 25 = 5^{2} $ <br\/> $\\Rightarrow BH = 5(cm) $ <br\/> Suy ra $BC = BH + HC = 5 + 16 = 21$ (cm) <\/span> <br\/> <br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 20 v\u00e0 21 <\/span>"}]}],"id_ques":1722},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["100"]]],"list":[{"point":10,"width":20,"content":"","type_input":"","ques":"Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i A, bi\u1ebft $AB = 8cm, BC = 17cm.$ <br\/> Tr\u00ean n\u1eeda m\u1eb7t ph\u1eb3ng b\u1edd $AC$ kh\u00f4ng ch\u1ee9a \u0111i\u1ec3m $B$, v\u1ebd tia $CD \\perp AC $ v\u00e0 $CD = 36cm.$ <br\/> T\u00ednh t\u1ed5ng \u0111\u1ed9 d\u00e0i c\u00e1c \u0111o\u1ea1n th\u1eb3ng $AB + BC + CD + DA$ <br\/> \u0110\u00e1p \u00e1n: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}(cm)$ <\/span> ","hint":"T\u00ednh $AC, AD$","explain":" H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> T\u00ednh c\u1ea1nh AC. <br\/> B\u01b0\u1edbc 2: <\/b> T\u00ednh AD. <br\/> B\u01b0\u1edbc 3: <\/b> T\u00ednh t\u1ed5ng \u0111\u1ed9 d\u00e0i: $ AB + BC + CD + DA$ <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai14/lv3/img\/H7B14-40.png' \/><\/center> Tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i A n\u00ean theo \u0111\u1ecbnh l\u00ed Py-ta-go ta c\u00f3: <br\/> $BC^2=AB^2+AC^2$ <br\/> $AC^{2}$$ = BC^{2} - AB^{2} \\\\ = 17^{2} - 8^{2} \\\\ = 289 - 64 \\\\ = 225 = 15^{2} $ <br\/> $\\Rightarrow AC = 15(cm) $ <br\/> Tam gi\u00e1c ACD vu\u00f4ng t\u1ea1i C n\u00ean theo \u0111\u1ecbnh l\u00ed Py-ta-go ta c\u00f3: <br\/> $AD^{2}$$ = AC^{2} + CD^{2} \\\\ = 15^{2} + 36^{2} \\\\ = 1521 $<br\/> $\\Rightarrow BC = \\sqrt{1521} = 39 (cm) $ <br\/> Suy ra: $AB + BC + CD + DA$$ = 8 + 17 + 36 + 39 = 100$ (cm) <br\/><br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0: 100 <\/span>"}]}],"id_ques":1723},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["9"],["12"],["15"]]],"list":[{"point":10,"width":20,"content":"","type_input":"","ques":"T\u00ednh c\u00e1c c\u1ea1nh c\u1ee7a m\u1ed9t tam gi\u00e1c vu\u00f4ng bi\u1ebft t\u1ec9 s\u1ed1 c\u00e1c c\u1ea1nh g\u00f3c vu\u00f4ng l\u00e0 $3 : 4$, chu vi c\u1ee7a tam gi\u00e1c vu\u00f4ng b\u1eb1ng $36cm$. <br\/> \u0110\u00e1p \u00e1n: $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}(cm); \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}(cm); \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}(cm)$ <\/span> ","explain":" G\u1ecdi a v\u00e0 b l\u00e0 \u0111\u1ed9 d\u00e0i c\u00e1c c\u1ea1nh g\u00f3c vu\u00f4ng, c l\u00e0 \u0111\u1ed9 d\u00e0i c\u1ea1nh huy\u1ec1n (cm) $(a, b, c > 0)$ <br\/> Ta c\u00f3: $\\dfrac{a}{3} = \\dfrac{b}{4} $ <br\/> $\\Rightarrow {{\\left( \\dfrac{a}{3} \\right)}^{2}}={{\\left( \\dfrac{b}{4} \\right)}^{2}} = \\dfrac{{{a}^{2}}+{{b}^{2}}}{9+16} = \\dfrac{{{c}^{2}}}{25} = {{\\left( \\dfrac{c}{5} \\right)}^{2}}$ <br\/> Theo t\u00ednh ch\u1ea5t d\u00e3y t\u1ec9 s\u1ed1 b\u1eb1ng nhau: <br\/> $\\dfrac{a}{3} = \\dfrac{b}{4} = \\dfrac{c}{5} = \\dfrac{a+b+c}{3+4+5} = \\dfrac{36}{12} = 3$ <br\/> Do \u0111\u00f3: $a = 3.3 = 9 (cm) \\\\ b = 4.3 = 12 (cm) \\\\ c = 5.3 = 15 (cm) $ <br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng c\u00e1c s\u1ed1 l\u00e0: 9; 12 v\u00e0 15 <\/span> <\/span> "}]}],"id_ques":1724},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["7"]]],"list":[{"point":10,"width":20,"content":"","type_input":"","ques":"T\u00ednh $x$ trong h\u00ecnh v\u1ebd sau: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai14/lv3/img\/H7B14-20A.png' \/><\/center> \u0110\u00e1p \u00e1n: <\/b> $x= \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} $ ","hint":"K\u1ebb $AH \\perp BC. $","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> - K\u1ebb $AH \\perp BC. $ <br\/> - Ch\u1ee9ng minh $AB = 2BH$ <br\/> - T\u00ednh $AH, HC$ $\\Rightarrow AC.$ <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/><center> <img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai14/lv3/img\/H7B14-20B.png' \/><\/center> K\u1ebb $AH \\perp BC $ <br\/> Tam gi\u00e1c $AHB$ c\u00f3 $\\widehat{H} = 90^{o}, \\widehat{ABH} = 60^{o} \\Rightarrow \\widehat{BAH} = 30^{o} $ <br\/> $\\Rightarrow AB = 2BH $ <b\/> ( Ta c\u00f4ng nh\u1eadn: Trong tam gi\u00e1c vu\u00f4ng c\u00f3 m\u1ed9t g\u00f3c b\u1eb1ng $30^o$ th\u00ec c\u1ea1nh g\u00f3c vu\u00f4ng \u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c $30^o$ \u0111\u00f3 c\u00f3 \u0111\u1ed9 d\u00e0i b\u1eb1ng m\u1ed9t n\u1eeda c\u1ea1nh huy\u1ec1n). <br\/> m\u00e0 $ AB = 3 \\Rightarrow BH = 1,5 \\\\ HC = BC + BH = 5 + 1,5 = 6,5 $ <br\/> Tam gi\u00e1c AHB vu\u00f4ng t\u1ea1i H c\u00f3: <br\/> $AH^{2}$ $ = AB^{2} - BH^{2} \\\\ = 3^{2} - 1,5^{2} \\\\ = 6,75 $ <br\/> Tam gi\u00e1c AHC vu\u00f4ng t\u1ea1i H c\u00f3: <br\/> $AC^{2} $ $= AH^{2} + HC^{2} \\\\ = 6,75 + 6,5^{2} \\\\ = 49 = 7^{2} \\\\ \\Rightarrow AC = 7 $ <br\/> <br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0: 7 <\/span> "}]}],"id_ques":1725},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["12"]]],"list":[{"point":10,"width":20,"content":"","type_input":"","ques":" Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i A. K\u1ebb $AH$ vu\u00f4ng g\u00f3c v\u1edbi $BC$ $(H \\in BC).$ <br\/> Bi\u1ebft $HB = 9cm, HC = 16cm$. T\u00ednh \u0111\u1ed9 d\u00e0i $AH$. <br\/> \u0110\u00e1p \u00e1n:<\/b> $ \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}(cm)$ ","explain":"<center> <img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai14/lv3/img\/H7B14-21.png' \/><\/center> \u0110\u1eb7t $AH = x$, ta \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Py - ta - go l\u1ea7n l\u01b0\u1ee3t v\u00e0o hai tam gi\u00e1c vu\u00f4ng $AHB$ v\u00e0 $AHC$: <br\/> $AB^{2} = 9^{2} + x^{2} = 81 + x^{2} \\\\ AC^{2} = 16^{2} + x^{2} = 256 + x^{2} $ <br\/> Suy ra: $AB^{2} + AC^{2} = 337 + 2x^{2}\\,\\, (1) $ <br\/> Ta l\u1ea1i \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Py - ta - go v\u00e0o tam gi\u00e1c vu\u00f4ng $ABC$: <br\/> $AB^{2} + AC^{2} = BC^{2} = 25^{2} = 625\\,\\, (2) $ <br\/> T\u1eeb (1) v\u00e0 (2) suy ra: <br\/> $2x^{2} + 337 = 625 \\\\ \\Rightarrow 2x^{2} = 288 \\\\ \\Rightarrow x^{2} = 144 \\\\ \\Rightarrow x = 12 $ <br\/> V\u1eady $AH = 12cm$ <br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0: 12<\/span> "}]}],"id_ques":1726},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["5"]]],"list":[{"point":10,"width":20,"content":"","type_input":"","ques":"T\u00ecm $x$ trong h\u00ecnh v\u1ebd sau: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai14/lv3/img\/H7B14-18A.jpg' \/><\/center> \u0110\u00e1p \u00e1n: <\/b> $x= \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} $ ","hint":"K\u1ebb $AH \\perp BC$","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> - K\u1ebb $AH \\perp BC. $ <br\/> - Ch\u1ee9ng minh $\\triangle ABH $ vu\u00f4ng c\u00e2n t\u1ea1i H. <br\/> - T\u00ednh $AH, BH$ $\\Rightarrow AC.$ <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> <img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai14/lv3/img\/H7B14-18B.jpg' \/><\/center> K\u1ebb $AH \\perp BC $ <br\/> Khi \u0111\u00f3 $\\triangle AHB $ vu\u00f4ng c\u00e2n t\u1ea1i H ($\\widehat{ABH} = \\widehat{BAH} = 45^{o} $) <br\/> $\\Rightarrow HB = HA $ <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Py-ta-go cho tam gi\u00e1c vu\u00f4ng AHB, ta \u0111\u01b0\u1ee3c: <br\/> $HB^{2} + HA^{2} = AB^{2} \\\\ \\Leftrightarrow 2HB^{2} = (\\sqrt{18})^{2} \\\\ \\Rightarrow HB^{2} = 9 \\\\ \\Rightarrow HB = 3 = HA $ <br\/> $\\Rightarrow HC = BC - HB = 7 - 3 = 4 $ <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Py-ta-go cho tam gi\u00e1c vu\u00f4ng AHC, ta \u0111\u01b0\u1ee3c: <br\/> $AC^{2} $ $= AH^{2} + HC^{2} \\\\= 3^{2} + 4^{2} \\\\ = 25 = 5^{2} \\\\ \\Rightarrow AC = 5 $ <br\/> Do \u0111\u00f3: $x=5$ <br\/> <br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0: 5 <\/span> "}]}],"id_ques":1727},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["12"]]],"list":[{"point":10,"width":20,"content":"","type_input":"","ques":"T\u00ecm $x$ trong h\u00ecnh v\u1ebd sau: <br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai14/lv3/img\/H7B14-17A.png' \/><\/center> \u0110\u00e1p \u00e1n: <\/b> $x= \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} $ ","hint":"K\u1ebb $CH \\perp AD. $","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> - K\u1ebb $CH \\perp AD. $ <br\/> - Ch\u1ee9ng minh $AB = CH, AH = BC$ <br\/> - T\u00ednh $CH \\Rightarrow AB.$ <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai14/lv3/img\/H7B14-17B.png' \/><\/center> K\u1ebb $CH \\perp AD \\Rightarrow CH \/\/ AB. $ <br\/> X\u00e9t $\\Delta ABH$ v\u00e0 $\\Delta CHB$ c\u00f3: <br\/> + $\\widehat{A}=\\widehat{C}=90^o$ <br\/> + $BH$ l\u00e0 c\u1ea1nh chung <br\/> + $\\widehat{B_1}=\\widehat{C_1}$ (hai g\u00f3c so le trong) <br\/> $\\Rightarrow \\triangle ABH = \\triangle CHB $ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow AB = CH = x \\\\ AH = BC = 10 $ <br\/> Do \u0111\u00f3 $HD = AD - AH = 15 - 10 = 5$ <br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Py-ta-go cho tam gi\u00e1c vu\u00f4ng CHD, ta \u0111\u01b0\u1ee3c: <br\/> $CH^{2} $ $ = CD^{2} - HD^{2} \\\\ = 13^{2} - 5^{2} \\\\ = 144 = 12^{2} $ <br\/> $\\Rightarrow CH = 12 = AB $ <br\/> V\u1eady $AB = x = 12$ <br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0: 12 <\/span> "}]}],"id_ques":1728},{"time":14,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["13"]]],"list":[{"point":10,"width":50,"type_input":"","input_hint":["sqrt"],"ques":"Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i A c\u00f3 $AB + AC = 17cm, AB - AC = 7cm.$ <br\/> T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh $BC$. <br\/> \u0110\u00e1p \u00e1n:<\/b> _input_ (cm)","explain":" <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai14/lv3/img\/H7B14-33.png' \/><\/center> Ta c\u00f3 $AB + AC = 17 \\\\ AB - AC = 7 $ <br\/> $\\Rightarrow 2AB = 17 + 7 = 24 \\Rightarrow AB = 12 \\, (cm) \\\\ 2AC = 17 - 7 = 10 \\Rightarrow AC = 5 \\,(cm) $ <br\/> Tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i A n\u00ean theo \u0111\u1ecbnh l\u00ed Py-ta-go, ta c\u00f3: <br\/> $BC^{2} $$ = AB^{2} + AC^{2} \\\\ = 12^{2} + 5^{2} \\\\ = 169 = 13^{2} $ <br\/> $\\Rightarrow BC = 13\\, (cm) $"}]}],"id_ques":1729},{"time":24,"part":[{"title":"\u0110i\u1ec1n t\u1eeb th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["150"]]],"list":[{"point":10,"width":20,"content":"","type_input":"","ques":" Cho tam gi\u00e1c $ABC$ \u0111\u1ec1u c\u00f3 \u0111i\u1ec3m M n\u1eb1m b\u00ean trong tam gi\u00e1c \u0111\u00f3 th\u1ecfa m\u00e3n: $MA^{2} = MB^{2} + MC^{2} $ <br\/> T\u00ednh s\u1ed1 \u0111o g\u00f3c $\\widehat{BMC} $ <br\/> \u0110\u00e1p \u00e1n:<\/b> $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}^{o}$","hint":"D\u1ef1ng $\\triangle BMD $ \u0111\u1ec1u. <br\/> Ch\u1ee9ng minh tam gi\u00e1c $CMD$ vu\u00f4ng t\u1ea1i M. ","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> - D\u1ef1ng $\\triangle BMD $ \u0111\u1ec1u. <br\/> - Ch\u1ee9ng minh $\\triangle CDM $ vu\u00f4ng t\u1ea1i M <br\/> - T\u00ednh $\\widehat{BMC} = \\widehat{BMD} + \\widehat{DMC} $ <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/hinhhoc/bai14/lv3/img\/H7B14-34.png' \/><\/center> D\u1ef1ng $\\triangle BMD $ \u0111\u1ec1u. <br\/> X\u00e9t $\\triangle BAM $ v\u00e0 $\\triangle BCD $ c\u00f3: <br\/> + $BA = BC $ ($\\triangle ABC $ \u0111\u1ec1u ) <br\/> + $\\widehat{ABM} = \\widehat{CBD} ( + \\widehat{MBC} = 60^{o}) $ <br\/> + $BM = BD $ ($\\triangle BMD $ \u0111\u1ec1u) <br\/> $\\Rightarrow \\triangle BAM = \\triangle BCD (c.g.c) $ <br\/> $\\Rightarrow AM = CD $ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> M\u00e0 $MA^{2} = MB^{2} + MC^{2} $ (gi\u1ea3 thi\u1ebft) <br\/> $\\Rightarrow CD^{2} = MB^{2} + MC^{2} \\\\ \\Rightarrow CD^{2} = MD^{2} + MC^{2} $ <br\/> $\\Rightarrow \\triangle CDM $ vu\u00f4ng t\u1ea1i M. <br\/> Suy ra $\\widehat{BMC}$$ = \\widehat{BMD} + \\widehat{DMC} \\\\ = 60^{o} + 90^{o} \\\\ = 150^{o} $ <\/span> <br\/> <br\/> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng: 150 <\/span> <\/span> "}]}],"id_ques":1730}],"lesson":{"save":0,"level":3}}

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Trả lời đúng trong khoảng thời gian quy định bạn sẽ được + số điểm như sau:
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Trong khoảng 2 phút -> 3 phút	+ 3 điểm
Trong khoảng 3 phút -> 4 phút	+ 2 điểm
Trong khoảng 4 phút -> 5 phút	+ 1 điểm

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