Bài tập nâng cao bài Một số hệ thức về cạnh và góc trong tam giác vuông

Home Lớp 9 Toán lớp 9 Một số hệ thức về cạnh và góc trong tam giác vuông Bài tập nâng cao

{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["7"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Cho h\u00ecnh thang c\u00e2n $ABCD \\,(AB\/\/CD)$. Bi\u1ebft $AB= 2 cm$, $CD= 5 cm$ v\u00e0 $\\widehat{A}=127^{o}$<br\/>Di\u1ec7n t\u00edch h\u00ecnh thang $ABCD$ l\u00e0_input_ $(cm^2)$<br\/>(K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb )","explain":"H\u01b0\u1edbng d\u1eabn<\/span><br\/> B\u01b0\u1edbc 1: <\/b> V\u1ebd hai \u0111\u01b0\u1eddng cao $AH$ v\u00e0 $BK$. Ch\u1ee9ng minh $AB=HK$<br\/>B\u01b0\u1edbc 2: <\/b> T\u00ednh \u0111\u1ed9 d\u00e0i c\u1ea1nh $HD$<br\/> B\u01b0\u1edbc 3: <\/b> T\u00ednh \u0111\u1ed9 d\u00e0i \u0111\u01b0\u1eddng cao $AH$<br\/> B\u01b0\u1edbc 4: <\/b> T\u00ednh di\u1ec7n t\u00edch h\u00ecnh thang<br\/>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai3/lv3/img\/H913_K14.png' \/><\/center><br\/>V\u1ebd $AH\\bot CD$, $BK\\bot CD$ ($H,K $ thu\u1ed9c $CD$). <br\/> V\u00ec $AH\\bot CD;\\, AB\/\/CD\\,\\, \\Rightarrow AH\\bot AB$.<br\/> T\u1ee9 gi\u00e1c $ABKH$ c\u00f3 $\\widehat A =\\widehat H = \\widehat K = 90^o$. <br\/>Suy ra $ABKH$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) $\\Rightarrow AB = HK = 2 cm $(t\u00ednh ch\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt)<br\/> Ta c\u00f3: $AD=BC$, $\\widehat D =\\widehat C$ (do $ABCD$ l\u00e0 h\u00ecnh thang c\u00e2n).<br\/>X\u00e9t $\\Delta ADH$ vu\u00f4ng t\u1ea1i $H$ v\u00e0 $\\Delta BCK$ vu\u00f4ng t\u1ea1i $K$ c\u00f3: <br\/>$\\left. \\begin{align} &AD =BC\\,\\, \\\\ &\\widehat {D} =\\widehat {C} \\,\\,\\\\ \\end {align}\\right\\}$$\\,\\,\\,$$\\Rightarrow\\Delta ADH =\\Delta BCK $ (c\u1ea1nh huy\u1ec1n $-$ g\u00f3c nh\u1ecdn) <br\/>Suy ra: $HD= KC= \\dfrac{DC-HK}{2}=1,5 \\,(cm)$<br\/>Ta c\u00f3 $AB \/\/ CD$ (gi\u1ea3 thi\u1ebft) $\\Rightarrow \\widehat{D}=180^{o}-\\widehat{A}=180^{o}-127^{o}=53^{o}$<br\/>X\u00e9t $\\Delta AHD$ c\u00f3: $ \\widehat{H}=90^{o}$ <br\/>$\\Rightarrow AH =HD.tg\\, D=1,5.tg\\, 53^{o} \\approx 2\\,cm$<br\/>$S_{ABCD}=\\dfrac{1}{2}(AB+CD).AH\\approx \\dfrac{1}{2}(2+5).2=7\\,(cm^2)$<br\/>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $7$<\/span>"}]}],"id_ques":1361},{"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":"T\u00ecm $x$ trong h\u00ecnh v\u1ebd sau<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai3/lv3/img\/H913_K13.png' \/><\/center>","select":["A. $53^o7'$","B. $50^o12'$","C. $30^o57'$","D. $36^o52'$"],"hint":"D\u00f9ng \u0111\u1ecbnh l\u00ed Ta-let t\u00ednh t\u1ec9 s\u1ed1 $\\dfrac{BM}{BN}$","explain":"H\u01b0\u1edbng d\u1eabn<\/span><br\/> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $MN \/\/ AC$<br\/> B\u01b0\u1edbc 2: <\/b>D\u1ef1a v\u00e0o \u0111\u1ecbnh l\u00ed Talet t\u00ednh t\u1ec9 s\u1ed1 $\\dfrac{BM}{BN}$<br\/> B\u01b0\u1edbc 3: <\/b> T\u00ednh $cos \\,B$<br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/>$\\begin{align}&MN\\bot AB \\,\\, \\text{(gi\u1ea3 thi\u1ebft)}\\\\&AC\\bot AB \\,\\, \\text{(gi\u1ea3 thi\u1ebft)}\\\\ \\end{align}$$\\,\\Rightarrow MN \/\/AC $ (T\u1eeb vu\u00f4ng g\u00f3c \u0111\u1ebfn song song)<br\/>X\u00e9t $\\Delta ABC $ c\u00f3 $ MN\/\/AC$<br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Talet, ta c\u00f3:<br\/>$\\dfrac{BN}{NC}=\\dfrac{BM}{MA}\\Rightarrow \\dfrac{BN}{BM}=\\dfrac{NC}{MA}=\\dfrac{5}{3}$<br\/>X\u00e9t $ \\Delta BMN$ c\u00f3: $ \\widehat{M}=90^{o}$ <br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c gi\u1eefa c\u1ea1nh v\u00e0 g\u00f3c trong tam gi\u00e1c vu\u00f4ng $BMN$, ta c\u00f3:<br\/>$cos \\,B=\\dfrac{BM}{BN}=\\dfrac{3}{5}\\\\ \\Rightarrow x\\approx 53^o7'$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":4}]}],"id_ques":1362},{"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":"Cho h\u00ecnh thoi $ABCD$ c\u00f3 $AC=60;\\, BD=32$. S\u1ed1 \u0111o g\u00f3c $\\widehat{DAB}$ l\u00e0: ","select":["A. $48^o52'$","B. $50^o16'$","C. $51^o4'$","D. $56^o8'$"],"hint":"$AC \\cap BD$ t\u1ea1i $I$. T\u00ednh s\u1ed1 \u0111o g\u00f3c $\\widehat {ABI}$","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai3/lv3/img\/H913_K11.png' \/><\/center><br\/>G\u1ecdi $AC \\cap BD$ t\u1ea1i $I$. V\u00ec $ABCD$ l\u00e0 h\u00ecnh thoi n\u00ean ta c\u00f3:<br\/>$AI=\\dfrac{AC}{2}=30\\,(cm)\\\\IB=\\dfrac{BD}{2}=16\\,(cm)\\\\ \\widehat{BAI}= \\dfrac{\\widehat{DAB}}{2}$ (t\u00ednh ch\u1ea5t h\u00ecnh thoi)<br\/>X\u00e9t $\\Delta ABI$ c\u00f3: $\\widehat{I}=90^{o}$. Ta c\u00f3:<br\/>$ tg\\,\\widehat{BAI}=\\dfrac{BI}{AI}=\\dfrac{16}{30}=\\dfrac{8}{15}\\\\ \\Rightarrow \\widehat{BAI} \\approx 28^o4'$<br\/>Suy ra: $\\widehat {BAD}=2\\widehat{BAI}=56^o8'$<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":4}]}],"id_ques":1363},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["19,4"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Trong h\u00ecnh v\u1ebd sau, v\u1edbi c\u00e1c s\u1ed1 li\u1ec7u th\u1ef1c t\u1ebf \u0111\u01b0\u1ee3c ghi tr\u00ean \u0111\u00f3. T\u00ednh $AB$. (L\u00e0m tr\u00f2n \u0111\u1ebfn s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t) <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai3/lv3/img\/H913_k10.png' \/><\/center>\u0110\u00e1p s\u1ed1: $AB\\approx \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} (m)$<\/span>","hint":"K\u1ebb $AH \\bot DE$. T\u00ednh c\u1ea1nh $AC$","explain":"H\u01b0\u1edbng d\u1eabn<\/span><br\/> B\u01b0\u1edbc 1: <\/b> K\u1ebb $AH\\bot DE$. Ch\u1ee9ng minh t\u1ee9 gi\u00e1c $AHDC$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt<br\/>B\u01b0\u1edbc 2: <\/b> T\u00ednh $EH, HD$ v\u00e0 $AC$<br\/> B\u01b0\u1edbc 3: <\/b> T\u00ednh c\u1ea1nh $AB$ d\u1ef1a v\u00e0o h\u1ec7 th\u1ee9c c\u1ea1nh v\u00e0 g\u00f3c trong tam gi\u00e1c vu\u00f4ng $ABC$<br\/>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai3/lv3/img\/H913_K10a.png' \/><\/center><br\/>H\u1ea1 $AH \\bot ED$ t\u1ea1i $ H$<br\/>X\u00e9t t\u1ee9 gi\u00e1c $AHDC$ c\u00f3 $\\widehat H = \\widehat C =\\widehat D =90^o$. Suy ra $AHDC$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt. (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft) <br\/>$\\Rightarrow AC\/\/HD, AC=HD, \\, AH=CD$ (t\u00ednh ch\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt)<br\/>$ AC\/\/ED \\Rightarrow \\widehat{E}=\\widehat{BAC}={{60}^{o}}$ (hai g\u00f3c \u0111\u1ed3ng v\u1ecb)<br\/>X\u00e9t $\\Delta AEH$ c\u00f3: $\\widehat{H}={{90}^{o}}$ <br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c li\u00ean h\u1ec7 gi\u1eefa c\u1ea1nh v\u00e0 g\u00f3c trong tam gi\u00e1c vu\u00f4ng $AEH$, ta c\u00f3:<br\/>$\\Rightarrow EH=AH.cotg\\,{{60}^{o}}=4.cotg \\,{{60}^{o}}\\approx 2,3\\left( m \\right)$<br\/>Suy ra, $AC=HD=ED-EH\\approx 12-2,3=9,7 \\,(m)$<br\/>X\u00e9t $\\Delta ABC$ c\u00f3: $\\widehat{C}={{90}^{o}}$ <br\/>$\\Rightarrow AC=AB.cos\\,\\widehat{BAC}$$\\Rightarrow AB\\approx \\dfrac{AC}{cos\\,{{60}^{o}}}\\approx 19,4\\,\\left( m \\right)$ <br\/>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $19,4$<\/span>"}]}],"id_ques":1364},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["43,5"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"\u0110\u1ed9 d\u00e0i hai \u0111\u01b0\u1eddng ch\u00e9o c\u1ee7a m\u1ed9t t\u1ee9 gi\u00e1c l\u00e0 $9$ v\u00e0 $13$. G\u00f3c nh\u1ecdn gi\u1eefa hai \u0111\u01b0\u1eddng ch\u00e9o l\u00e0 $48^{o}$.<br\/>Di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c l\u00e0_input_(\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>(L\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n th\u1ee9 nh\u1ea5t)","hint":"","explain":"H\u01b0\u1edbng d\u1eabn<\/span><br\/> B\u01b0\u1edbc 1: <\/b> T\u00ednh $AH$ v\u00e0 $CK$ theo $OA$ v\u00e0 $OC$<br\/>B\u01b0\u1edbc 2: <\/b> T\u00ednh $S_{ABCD}$ theo $S_{ADB}$ v\u00e0 $S_{CDB}$<br\/>B\u00e0i gi\u1ea3i<\/span><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai3/lv3/img\/H913_K8.png' \/><\/center><br\/>V\u1ebd $AH \\bot BD$; $CK \\bot BD$<br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c gi\u1eefa c\u1ea1nh v\u00e0 g\u00f3c trong tam gi\u00e1c vu\u00f4ng $AHO$ v\u00e0 $CKO$, ta c\u00f3: <br\/> $AH=OA.sin O_{1}=OA.sin 48^{o}\\\\CK=OC.sin O_{2}=OC.sin 48^{o}$<br\/>$\\begin{align} S_{ABCD}&=S_{ABD}+S_{CBD}\\\\&=\\dfrac{1}{2}.BD.(AH+CK)\\\\&=\\dfrac{1}{2}BD.sin 48^{o}(OA+OC)\\\\ &=\\dfrac{1}{2}.BD.AC.sin 48^{o}\\\\& \\approx 43,5 \\, \\text{(\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)}\\\\ \\end{align}$ <br\/>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $43,5$<br\/><\/span>L\u01b0u \u00fd:<\/span> M\u1ed9t c\u00e1ch t\u1ed5ng qu\u00e1t, ta ch\u1ee9ng minh \u0111\u01b0\u1ee3c <br\/>Di\u1ec7n t\u00edch t\u1ee9 gi\u00e1c b\u1eb1ng n\u1eeda t\u00edch c\u1ee7a hai \u0111\u01b0\u1eddng ch\u00e9o nh\u00e2n v\u1edbi sin c\u1ee7a g\u00f3c nh\u1ecdn xen gi\u1eefa hai \u0111\u01b0\u1eddng ch\u00e9o<\/span>"}]}],"id_ques":1365},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["10"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Tam gi\u00e1c $ABC$ c\u00f3 $AB= 16 cm$, $AC= 14 cm$ v\u00e0 $\\widehat{B}=60^{o}$<br\/>\u0110\u1ed9 d\u00e0i c\u1ea1nh $BC$ l\u00e0 _input_ $(cm)$","hint":"H\u1ea1 $AH \\bot BC$ t\u1ea1i $ H$. T\u00ednh $HC$ v\u00e0 $HB$","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai3/lv3/img\/H913_K6.png' \/><\/center><br\/>H\u1ea1 $AH \\bot BC$ t\u1ea1i $H$<br\/>X\u00e9t $\\Delta ABH $c\u00f3: $ \\widehat {H}=90^{o}$<br\/>$BH=AB.cos B=16.cos 60^{o}=8 \\,(cm)$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Pitago v\u00e0o hai tam gi\u00e1c vu\u00f4ng $AHB$ v\u00e0 $AHC$ ta c\u00f3:<br\/>$AB^2=AH^2+HB^2\\\\AC^2=AH^2+HC^2\\\\ \\Rightarrow AB^2-HB^2=AC^2-HC^2\\\\ \\Rightarrow 16^2-8^2=14^2-HC^2\\\\ \\Rightarrow HC^2=4 \\\\ \\Rightarrow HC=2\\,(cm)$ <br\/> Suy ra: $BC= HC+ HB= 2 + 8= 10\\, (cm)$<br\/>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 10<\/span><\/span>"}]}],"id_ques":1366},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["60"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"M\u1ed9t h\u00ecnh b\u00ecnh h\u00e0nh c\u00f3 hai c\u1ea1nh l\u00e0 $10 cm$ v\u00e0 $12 cm$, g\u00f3c t\u1ea1o b\u1edfi hai c\u1ea1nh \u0111\u00f3 b\u1eb1ng $150^o$.<br\/> Di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh b\u00ecnh h\u00e0nh l\u00e0_input_($cm^2$)","hint":"K\u1ebb \u0111\u01b0\u1eddng cao $CH$, t\u00ednh \u0111\u1ed9 d\u00e0i $CH$ r\u1ed3i t\u00ednh di\u1ec7n t\u00edch h\u00ecnh b\u00ecnh h\u00e0nh $ABCD$","explain":"<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai3/lv3/img\/H913_K5.png' \/><\/center><br\/>Gi\u1ea3 s\u1eed h\u00ecnh b\u00ecnh h\u00e0nh $ABCD$ c\u00f3 $AB= 10cm$ v\u00e0 $AD= 12 cm$; $\\widehat{A}=150^o$<br\/>Khi \u0111\u00f3, $CD = AB=10 cm$ v\u00e0 $\\widehat{D}=180^o-\\widehat{A}=180^o-150^o=30^o$ (t\u00ednh ch\u1ea5t h\u00ecnh b\u00ecnh h\u00e0nh)<br\/>K\u1ebb \u0111\u01b0\u1eddng cao $CH$, $(H\\in AD)$. <br\/>X\u00e9t tam gi\u00e1c $CHD$ vu\u00f4ng t\u1ea1i $H$ c\u00f3:<br\/> $CH=CD.sin \\,D=10.sin {30^o}=10.\\dfrac 1 2 = 5\\,(cm)$<br\/>Ta c\u00f3: <br\/>$S_{ABCD}=CH.AD=5.12=60\\,(cm^2)$<br\/>V\u1eady di\u1ec7n t\u00edch $ABCD$ b\u1eb1ng $60\\,cm^2$<br\/>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $60$<\/span>"}]}],"id_ques":1367},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["5"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Cho tam gi\u00e1c $ABC$ v\u1edbi \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c trong c\u1ee7a g\u00f3c $BAC$ l\u00e0 $AD$. Bi\u1ebft $AB= 5, AC= 8$ v\u00e0 $\\widehat{A}=72^{o}$<br\/> \u0110\u1ed9 d\u00e0i c\u1ea1nh $AD \\approx$_input_<br\/>(K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn ch\u1eef s\u1ed1 h\u00e0ng \u0111\u01a1n v\u1ecb)<\/span> ","hint":"T\u00ednh $AD$ t\u1eeb \u0111\u1eb3ng th\u1ee9c $S_{ABC}=S_{ABD}+S_{ADC}$. ","explain":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai3/lv3/img\/H913_K4.png' \/><\/center><br\/>V\u00ec $AD$ l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $A$ n\u00ean $\\widehat {A_1}=\\widehat{A_2}=\\dfrac{72^o}{2}=36^o$<br\/> G\u1ecdi di\u1ec7n t\u00edch c\u00e1c tam gi\u00e1c $ABD, ADC$ v\u00e0 $ABC$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 $S_{1},\\,S_{2}$ v\u00e0 $S$. Ta c\u00f3: <br\/>$S_{1}=\\dfrac{1}{2}AB.AD.sin A_{1}$$=\\dfrac{1}{2}.5.AD.sin 36^{o}$<br\/>$S_{2}=\\dfrac{1}{2}AC.AD.sin A_{2}=\\dfrac{1}{2}.8.AD.sin 36^{o}$<br\/>$S=\\dfrac{1}{2}AB.AC.sin A=\\dfrac{1}{2}.5.8.sin 72^{o}$<br\/>V\u00ec $S_{1}+S_{2}=S$ n\u00ean <br\/>$\\dfrac{1}{2}.5.AD.sin 36^{o}$$+\\dfrac{1}{2}.8.AD.sin 36^{o}$$=\\dfrac{1}{2}.5.8.sin72^{o}$<br\/>$\\Rightarrow 13.AD.sin 36^{o}=40.sin 72^{o}\\\\ \\Rightarrow AD \\approx 5$<br\/>V\u1eady k\u1ebft qu\u1ea3 c\u1ea7n \u0111i\u1ec1n l\u00e0 $5$<\/span><br\/>Ch\u00fa \u00fd:<\/span> Di\u1ec7n t\u00edch tam gi\u00e1c b\u1eb1ng n\u1eeda t\u00edch hai c\u1ea1nh v\u1edbi sin c\u1ee7a g\u00f3c xen gi\u1eefa.<\/span>"}]}],"id_ques":1368},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"Cho tam gi\u00e1c nh\u1ecdn $ABC$ c\u00f3 di\u1ec7n t\u00edch $S_{ABC}=1$. D\u1ef1ng ba \u0111\u01b0\u1eddng cao $AD, BE$ v\u00e0 $CF$. <br\/>Khi \u0111\u00f3 $S_{AEF}+S_{BDF}+S_{CDE}=\\dfrac{1}{2}(\\cos^2A+\\cos^2B+\\cos^2C)$","select":["A. \u0110\u00fang","B. Sai"],"hint":"T\u00ednh t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch c\u00e1c tam gi\u00e1c $AEF,\\,BDF,\\,CDE $ v\u1edbi di\u1ec7n t\u00edch tam gi\u00e1c $ABC$. <br\/>S\u1eed d\u1ee5ng: Di\u1ec7n t\u00edch tam gi\u00e1c b\u1eb1ng n\u1eeda t\u00edch hai c\u1ea1ch nh\u00e2n v\u1edbi sin c\u1ee7a g\u00f3c xen gi\u1eefa","explain":"<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/hinhhoc/bai3/lv3/img\/H913_K18.png' \/><\/center><br\/>\u00c1p d\u1ee5ng t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c trong hai tam gi\u00e1c vu\u00f4ng $AEB$ v\u00e0 $AFB$, ta c\u00f3: <br\/> $\\cos\\,A= \\dfrac{AE}{AB}=\\dfrac{AF}{AC}$ <br\/> L\u1ea1i c\u00f3: $\\begin{align} \\dfrac{{{S}_{AEF}}}{{{S}_{ABC}}}& =\\dfrac{\\dfrac{1}{2}AE.AF.sin\\,A}{\\dfrac{1}{2}AB.AC.sin \\,A} \\\\ & =\\dfrac{AE}{AB}.\\dfrac{AF}{AC} \\\\ & =cos\\,A.cos\\,A \\\\ & =co{{s}^{2}}A \\\\ \\end{align}$ <br\/>Suy ra: $S_{AEF}=\\cos^2A$<br\/>T\u01b0\u01a1ng t\u1ef1, ta t\u00ednh \u0111\u01b0\u1ee3c: $\\dfrac{S_{BDF}}{S_{ABC}}=\\cos^2B$; $\\dfrac{S_{CDE}}{S_{ABC}}=\\cos^2C$<br\/>$\\Rightarrow S_{BDF}=\\cos^2B; S_{CDE}=\\cos^2C$<br\/>$\\Rightarrow S_{AEF}+S_{BDF}+S_{CDE}$$=\\cos^2A+\\cos^2B+\\cos^2C$<br\/>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 Sai<br\/>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":1369},{"time":24,"part":[{"time":3,"title":"S\u1eafp x\u1ebfp l\u1ea1i th\u1ee9 t\u1ef1 \u0111\u1ec3 ch\u1ee9ng minh b\u00e0i to\u00e1n","title_trans":"Cho tam gi\u00e1c $ABC$ c\u00f3 ba g\u00f3c nh\u1ecdn, $BC=a$, $AC=b$, $AB= c$. Ch\u1ee9ng minh r\u1eb1ng $\\dfrac{a}{sin A }=\\dfrac{b}{sin B}=\\dfrac{c}{sin C}$","temp":"sequence","correct":[[[3],[1],[5],[2],[4]]],"list":[{"point":10,"image":"img\/H913_K1.png","left":["Do \u0111\u00f3, $\\dfrac{sin B}{sin C}=\\dfrac{AC}{AB}=\\dfrac{b}{c}$$\\Rightarrow \\dfrac{b}{sin B}=\\dfrac{c}{sin C}$","V\u1ebd $AH \\bot BC$ t\u1ea1i $H$. $\\Delta HAB$ vu\u00f4ng t\u1ea1i $H$ n\u00ean $sin B=\\dfrac{AH}{AB}$","V\u1eady $\\dfrac{a}{sin A }=\\dfrac{b}{sin B}=\\dfrac{c}{sin C}$","$\\Delta HAC$ vu\u00f4ng t\u1ea1i $H$ n\u00ean $sin C=\\dfrac{AH}{AC}$","Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1, ta c\u00f3: $\\dfrac{a}{sin A}=\\dfrac{b}{sin B}$"],"top":85,"hint":"V\u1ebd $AH\\bot BC$ t\u1ea1i $H$. T\u00ednh $sin\\,B$, $sin\\,A$, $sin \\,C$.","explain":"V\u1ebd $AH \\bot BC$ t\u1ea1i $H$.<br\/>Ta c\u00f3:<br\/> $\\Delta HAB$ vu\u00f4ng t\u1ea1i $H$ n\u00ean $sin B=\\dfrac{AH}{AB}$ (t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c)<br\/>$\\Delta HAC$ vu\u00f4ng t\u1ea1i $H$ n\u00ean $sin C=\\dfrac{AH}{AC}$ (t\u1ec9 s\u1ed1 l\u01b0\u1ee3ng gi\u00e1c)<br\/>Do \u0111\u00f3, $\\dfrac{sin B}{sin C}=\\dfrac{AC}{AB}=\\dfrac{b}{c}$$\\Rightarrow \\dfrac{b}{sin B}=\\dfrac{c}{sin C}$<br\/>Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1, ta c\u00f3: $\\dfrac{a}{sin A}=\\dfrac{b}{sin B}$<br\/>V\u1eady $\\dfrac{a}{sin A }=\\dfrac{b}{sin B}=\\dfrac{c}{sin C}$<\/span>"}]}],"id_ques":1370}],"lesson":{"save":0,"level":3}}

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