C\u00d4NG TH\u1ee8C T\u00cdNH DI\u1ec6N T\u00cdCH TAM GI\u00c1C<\/h2>\n

C\u00e1c em h\u1ecdc sinh, sinh vi\u00ean ho\u1eb7c nh\u1eefng ng\u01b0\u1eddi th\u00edch h\u1ecdc To\u00e1n ch\u1eafc ch\u1eafn kh\u00f4ng th\u1ec3 qu\u00ean nh\u1eefng c\u00f4ng th\u1ee9c to\u00e1n h\u1ecdc quan tr\u1ecdng khi \u00e1p d\u1ee5ng v\u00e0o c\u00e1c b\u00e0i t\u1eadp \u1ee9ng d\u1ee5ng, v\u00ed d\u1ee5 nh\u01b0 c\u00f4ng th\u1ee9c t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c, h\u00ecnh vu\u00f4ng, h\u00ecnh b\u00ecnh h\u00e0nh,\u2026M\u1eb7c d\u00f9 v\u1eady trong m\u1ed7i h\u00ecnh, \u0111\u1eb7c bi\u1ec7t h\u00ecnh tam gi\u00e1c l\u1ea1i c\u00f3 r\u1ea5t nhi\u1ec1u c\u00e1c t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c kh\u00e1c nhau, \u0111\u01a1n c\u1eed nh\u01b0 c\u00e1ch t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c th\u01b0\u1edbng s\u1ebd kh\u00e1c so v\u1edbi khi t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c vu\u00f4ng, c\u00e2n ho\u1eb7c \u0111\u1ec1u.
\ncong thuc tinh dien tich tam giac
\nNh\u01b0 v\u1eady vi\u1ec7c ph\u00e2n t\u00e1ch r\u00f5 r\u00e0ng gi\u1eefa c\u00e1c c\u00f4ng th\u1ee9c t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c ch\u1eafc ch\u1eafn s\u1ebd h\u1ed7 tr\u1ee3 r\u1ea5t nhi\u1ec1u cho c\u00e1c b\u1ea1n h\u1ecdc sinh, sinh vi\u00ean khi ti\u1ebfp c\u1eadn v\u1edbi nh\u1eefng b\u00e0i to\u00e1n li\u00ean quan \u0111\u1ebfn h\u00ecnh tam gi\u00e1c.<\/p>\n

 <\/p>\n

<\/h2>\n

C\u00d4NG TH\u1ee8C T\u00cdNH DI\u1ec6N T\u00cdCH TAM GI\u00c1C<\/em><\/h3>\n

* C\u00f4ng Th\u1ee9c T\u00ednh Di\u1ec7n T\u00edch Tam Gi\u00e1c Th\u01b0\u1eddng Trong To\u00e1n H\u1ecdc
\nDi\u1ec7n t\u00edch tam gi\u00e1c th\u01b0\u1eddng \u0111\u01b0\u1ee3c t\u00ednh b\u1eb1ng c\u00e1ch nh\u00e2n chi\u1ec1u cao v\u1edbi \u0111\u1ed9 d\u00e0i \u0111\u00e1y, sau \u0111\u00f3 t\u1ea5t c\u1ea3 chia cho 2. N\u00f3i c\u00e1ch kh\u00e1c, di\u1ec7n t\u00edch tam gi\u00e1c th\u01b0\u1eddng s\u1ebd b\u1eb1ng 1\/2 t\u00edch c\u1ee7a chi\u1ec1u cao v\u00e0 chi\u1ec1u d\u00e0i c\u1ea1nh \u0111\u00e1y c\u1ee7a tam gi\u00e1c.
\n– C\u00f4ng th\u1ee9c t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c th\u01b0\u1eddng: S = (A X H)\/ 2<\/p>\n

<\/p>\n

C\u00f4ng th\u1ee9c t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c th\u01b0\u1eddng<\/p>\n

Trong \u0111\u00f3:
\n+ a: Chi\u1ec1u d\u00e0i \u0111\u00e1y tam gi\u00e1c (\u0111\u00e1y l\u00e0 m\u1ed9t trong 3 c\u1ea1nh c\u1ee7a tam gi\u00e1c t\u00f9y theo quy \u0111\u1eb7t c\u1ee7a ng\u01b0\u1eddi t\u00ednh)
\n+ h: Chi\u1ec1u cao c\u1ee7a tam gi\u00e1c, \u1ee9ng v\u1edbi ph\u1ea7n \u0111\u00e1y chi\u1ebfu l\u00ean (chi\u1ec1u cao tam gi\u00e1c b\u1eb1ng \u0111o\u1ea1n th\u1eb3ng h\u1ea1 t\u1eeb \u0111\u1ec9nh xu\u1ed1ng \u0111\u00e1y, \u0111\u1ed3ng th\u1eddi vu\u00f4ng g\u00f3c v\u1edbi \u0111\u00e1y c\u1ee7a m\u1ed9t tam gi\u00e1c).
\n– C\u00f4ng th\u1ee9c suy ra: H= (Sx2)\/ A ho\u1eb7c a= (Sx2)\/ H
\n– V\u00ed d\u1ee5: Cho m\u1ed9t h\u00ecnh tam gi\u00e1c ABC, trong \u0111\u00f3 c\u00f3 chi\u1ec1u cao n\u1ed1i t\u1eeb \u0111\u1ec9nh \u1ea2nh xu\u1ed1ng \u0111\u00e1y BC b\u1eb1ng 3, chi\u1ec1u d\u00e0i \u0111\u00e1y BC b\u1eb1ng 6. T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c th\u01b0\u1eddng ABC? (\u0110\u01a1n v\u1ecb t\u00ednh: cm)
\ncach tinh dien tich tam giac
\n\u0110\u00e1p \u00e1n: G\u1ecdi a =6 v\u00e0 h=3.
\nSuy ra S = (a x h)\/ 2 = (6×3)\/2 ho\u1eb7c 1\/2 x (6×3) = 9 cm
\n* Ch\u00fa \u00dd: Tr\u01b0\u1eddng h\u1ee3p kh\u00f4ng cho c\u1ea1nh \u0111\u00e1y ho\u1eb7c chi\u1ec1u cao, m\u00e0 cho tr\u01b0\u1edbc di\u1ec7n t\u00edch v\u00e0 c\u1ea1nh c\u00f2n l\u1ea1i, c\u00e1c b\u1ea1n h\u00e3y \u00e1p d\u1ee5ng c\u00f4ng th\u1ee9c suy ra \u1edf tr\u00ean \u0111\u1ec3 t\u00ednh to\u00e1n.
\n* C\u00f4ng Th\u1ee9c T\u00ednh Di\u1ec7n T\u00edch Tam Gi\u00e1c Vu\u00f4ng Trong To\u00e1n H\u1ecdc<\/p>\n

\n– Di\u1ec5n gi\u1ea3i: C\u00f4ng th\u1ee9c t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c vu\u00f4ng t\u01b0\u01a1ng t\u1ef1 v\u1edbi c\u00e1ch t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c th\u01b0\u1eddng, \u0111\u00f3 l\u00e0 b\u1eb1ng1\/2 t\u00edch c\u1ee7a chi\u1ec1u cao v\u1edbi chi\u1ec1u d\u00e0i \u0111\u00e1y. M\u1eb7c d\u00f9 v\u1eady h\u00ecnh tam gi\u00e1c vu\u00f4ng s\u1ebd kh\u00e1c bi\u1ec7t h\u01a1n so v\u1edbi tam gi\u00e1c th\u01b0\u1eddng do th\u1ec3 hi\u1ec7n r\u00f5 chi\u1ec1u cao v\u00e0 chi\u1ec1u d\u00e0i c\u1ea1nh \u0111\u00e1y, v\u00e0 b\u1ea1n kh\u00f4ng c\u1ea7n v\u1ebd th\u00eam \u0111\u1ec3 t\u00ednh chi\u1ec1u cao tam gi\u00e1c.
\n– C\u00f4ng th\u1ee9c t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c vu\u00f4ng: S = (A X H)\/ 2<\/p>\n

+ a: Chi\u1ec1u d\u00e0i \u0111\u00e1y tam gi\u00e1c vu\u00f4ng (\u0111\u00e1y l\u00e0 m\u1ed9t trong 3 c\u1ea1nh c\u1ee7a tam gi\u00e1c v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi m\u1ed9t c\u1ea1nh c\u00f2n l\u1ea1i)
\n+ h: Chi\u1ec1u cao c\u1ee7a tam gi\u00e1c, \u1ee9ng v\u1edbi ph\u1ea7n \u0111\u00e1y chi\u1ebfu l\u00ean (chi\u1ec1u cao tam gi\u00e1c b\u1eb1ng \u0111o\u1ea1n th\u1eb3ng h\u1ea1 t\u1eeb \u0111\u1ec9nh xu\u1ed1ng \u0111\u00e1y, \u0111\u1ed3ng th\u1eddi vu\u00f4ng g\u00f3c v\u1edbi \u0111\u00e1y c\u1ee7a m\u1ed9t tam gi\u00e1c).
\n– C\u00f4ng th\u1ee9c suy ra: H=(Sx2)\/ A ho\u1eb7c A= (Sx2)\/ H<\/p>\n

– V\u00ed d\u1ee5: C\u00f3 m\u1ed9t h\u00ecnh tam gi\u00e1c vu\u00f4ng ABC, vu\u00f4ng g\u00f3c nhau t\u1ea1i \u0111i\u1ec3m B, chi\u1ec1u d\u00e0i c\u1ea1nh \u0111\u00e1y BC l\u00e0 5 cm, chi\u1ec1u cao l\u00e0 2 cm. H\u1ecfi di\u1ec7n t\u00edch c\u1ee7a h\u00ecnh tam gi\u00e1c vu\u00f4ng ABC b\u1eb1ng bao nhi\u00eau? \u0110\u01a1n v\u1ecb t\u00ednh: cm.
\ntinh dien tich tam giac thuong
\n\u0110\u00e1p \u00e1n: G\u1ecdi a =5 v\u00e0 h=2.
\nSuy ra S = (a x h)\/ 2 = (5×2)\/2 ho\u1eb7c 1\/2 x (5×2) = 5 cm
\nT\u01b0\u01a1ng t\u1ef1 n\u1ebfu d\u1eef li\u1ec7u h\u1ecfi ng\u01b0\u1ee3c v\u1ec1 c\u00e1ch t\u00ednh chi\u1ec1u d\u00e0i c\u1ea1nh \u0111\u00e1y ho\u1eb7c chi\u1ec1u cao, c\u00e1c b\u1ea1n c\u00f3 th\u1ec3 s\u1eed d\u1ee5ng c\u00f4ng th\u1ee9c suy ra \u1edf tr\u00ean.<\/p>\n

* C\u00f4ng Th\u1ee9c T\u00ednh Di\u1ec7n T\u00edch Tam Gi\u00e1c C\u00e2n Trong To\u00e1n H\u1ecdc<\/p>\n

<\/p>\n

Tam gi\u00e1c c\u00e2n l\u00e0 tam gi\u00e1c trong \u0111\u00f3 c\u00f3 hai c\u1ea1nh b\u00ean v\u00e0 hai g\u00f3c b\u1eb1ng nhau. Trong \u0111\u00f3 c\u00e1ch t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c c\u00e2n c\u0169ng t\u01b0\u01a1ng t\u1ef1 c\u00e1ch t\u00ednh tam gi\u00e1c th\u01b0\u1eddng, ch\u1ec9 c\u1ea7n b\u1ea1n bi\u1ebft chi\u1ec1u cao tam gi\u00e1c v\u00e0 c\u1ea1nh \u0111\u00e1y.
\n– Di\u1ec5n gi\u1ea3i: Di\u1ec7n t\u00edch tam gi\u00e1c c\u00e2n b\u1eb1ng T\u00edch c\u1ee7a chi\u1ec1u cao n\u1ed1i t\u1eeb \u0111\u1ec9nh tam gi\u00e1c \u0111\u00f3 t\u1edbi c\u1ea1nh \u0111\u00e1y tam gi\u00e1c, sau \u0111\u00f3 chia cho 2.
\n– C\u00f4ng th\u1ee9c t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c c\u00e2n: S = (A X H)\/ 2
\n+ a: Chi\u1ec1u d\u00e0i \u0111\u00e1y tam gi\u00e1c c\u00e2n (\u0111\u00e1y l\u00e0 m\u1ed9t trong 3 c\u1ea1nh c\u1ee7a tam gi\u00e1c)
\n+ h: Chi\u1ec1u cao c\u1ee7a tam gi\u00e1c (chi\u1ec1u cao tam gi\u00e1c b\u1eb1ng \u0111o\u1ea1n th\u1eb3ng h\u1ea1 t\u1eeb \u0111\u1ec9nh xu\u1ed1ng \u0111\u00e1y).
\n– V\u00ed d\u1ee5: Cho m\u1ed9t tam gi\u00e1c c\u00e2n ABC c\u00f3 chi\u1ec1u cao n\u1ed1i t\u1eeb \u0111\u1ec9nh A xu\u1ed1ng \u0111\u00e1y BC b\u1eb1ng 7 cm, chi\u1ec1u d\u00e0i \u0111\u00e1y cho l\u00e0 6 cm. H\u1ecfi di\u1ec7n t\u00edch c\u1ee7a tam gi\u00e1c c\u00e2n ABC b\u1eb1ng bao nhi\u00eau.
\ntinh dien tich tam giac vuong
\n\u0110\u00e1p \u00e1n: G\u1ecdi a =6 v\u00e0 h=7.
\nSuy ra S = (a x h)\/ 2 = (6×7)\/2 ho\u1eb7c 1\/2 x (6×7) = 21 cm
\n* C\u00f4ng Th\u1ee9c T\u00ednh Di\u1ec7n T\u00edch Tam Gi\u00e1c \u0110\u1ec1u Trong To\u00e1n H\u1ecdc<\/p>\n

\nTam gi\u00e1c \u0111\u1ec1u l\u00e0 tam gi\u00e1c c\u00f3 3 c\u1ea1nh b\u1eb1ng nhau v\u00e0 m\u1ed7i g\u00f3c trong tam gi\u00e1c \u0111\u1ec1u c\u00f3 g\u00f3c b\u1eb1ng 60 \u0111\u1ed9, v\u00e0 b\u1ea5t c\u1ee9 tam gi\u00e1c n\u00e0o c\u00f3 ba g\u00f3c b\u1eb1ng nhau c\u0169ng \u0111\u01b0\u1ee3c coi l\u00e0 m\u1ed9t tam gi\u00e1c \u0111\u1ec1u.
\n– C\u00f4ng th\u1ee9c t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c \u0111\u1ec1u: S = A2 X (\u221a3)\/4
\nTrong \u0111\u00f3:
\n+ a: chi\u1ec1u d\u00e0i m\u1ed9t c\u1ea1nh b\u1ea5t k\u1ef3 trong tam gi\u00e1c \u0111\u1ec1u.
\n– V\u00ed d\u1ee5: C\u00f3 m\u1ed9t tam gi\u00e1c \u0111\u1ec1u ABC v\u1edbi chi\u1ec1u d\u00e0i c\u00e1c c\u1ea1nh b\u1eb1ng nhau l\u00e0 9 cm, bi\u1ebft c\u00e1c g\u00f3c c\u1ee7a tam gi\u00e1c n\u00e0y \u0111\u1ec1u b\u1eb1ng 60 \u0111\u1ed9. H\u1ecfi di\u1ec7n t\u00edch tam gi\u00e1c \u0111\u1ec1u ABC b\u1eb1ng bao nhi\u00eau?
\ntinh dien tich tam giac can
\n\u0110\u00e1p \u00e1n: Do m\u1ed7i c\u1ea1nh AB = AC = BC = 9 n\u00ean ta c\u00f3 chi\u1ec1u d\u00e0i c\u1ea1nh a = 9.
\nThay v\u00e0o c\u00f4ng th\u1ee9c t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c \u0111\u1ec1u ta c\u00f3: S = a2 x (\u221a3)\/4 = S = 92 x (\u221a3)\/4 = 81 x (\u221a3)\/4 = 81 x (1,732\/4) = 35,07 cm
\nNgo\u00e0i ra c\u00f2n r\u1ea5t nhi\u1ec1u c\u00e1ch t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c kh\u00e1c khi ng\u01b0\u1eddi d\u00f9ng bi\u1ebft \u0111\u01b0\u1ee3c t\u1ea5t c\u1ea3 c\u00e1c c\u1ea1nh, v\u00ed d\u1ee5 nh\u01b0 s\u1eed d\u1ee5ng c\u00f4ng th\u1ee9c Heron, t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c b\u1eb1ng g\u00f3c v\u00e0 h\u00e0m l\u01b0\u1ee3ng gi\u00e1c.
\nD\u00f9 s\u1eed d\u1ee5ng c\u00f4ng th\u1ee9c t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c n\u00e0o \u0111i ch\u0103ng n\u1eefa th\u00ec c\u00e1c b\u1ea1n, c\u00e1c em h\u1ecdc sinh, sinh vi\u00ean c\u1ea7n hi\u1ec3u r\u1eb1ng, kh\u00f4ng ph\u1ea3i l\u00fac chi\u1ec1u cao c\u0169ng n\u1eb1m trong tam gi\u00e1c, l\u00fac n\u00e0y c\u1ea7n v\u1ebd th\u00eam m\u1ed9t chi\u1ec1u cao v\u00e0 c\u1ea1nh \u0111\u00e1y b\u1ed5 sung. V\u00e0 quan tr\u1ecdng khi t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c, c\u1ea7n ch\u00fa \u00fd chi\u1ec1u cao ph\u1ea3i \u1ee9ng v\u1edbi c\u1ea1nh \u0111\u00e1y n\u01a1i n\u00f3 chi\u1ebfu xu\u1ed1ng.
\nTr\u00ean \u0111\u00e2y l\u00e0 c\u00e1c c\u00f4ng th\u1ee9c t\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c th\u01b0\u1eddng, vu\u00f4ng, c\u00e2n \u0111\u1ec1u, c\u00e1c em tham kh\u1ea3o nh\u00e9.<\/p>\n","protected":false},"excerpt":{"rendered":"

C\u00d4NG TH\u1ee8C T\u00cdNH DI\u1ec6N T\u00cdCH TAM GI\u00c1C C\u00e1c em h\u1ecdc sinh, sinh vi\u00ean ho\u1eb7c nh\u1eefng ng\u01b0\u1eddi th\u00edch h\u1ecdc To\u00e1n ch\u1eafc ch\u1eafn kh\u00f4ng th\u1ec3 qu\u00ean nh\u1eefng c\u00f4ng th\u1ee9c to\u00e1n h\u1ecdc quan… <\/p>\n","protected":false},"author":1,"featured_media":4865,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[405],"tags":[],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/posts\/4860"}],"collection":[{"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/comments?post=4860"}],"version-history":[{"count":0,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/posts\/4860\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/media\/4865"}],"wp:attachment":[{"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/media?parent=4860"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/categories?post=4860"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/tags?post=4860"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}