cho tam giác ABC có góc vuông ở A ; cạnh AB =30cm ; AC = 36cm
M là điểm bất kì trên AB sao cho AM = 20cm
qua M kẻ đường thẳng song song với cạnh BC cắt cạnh AC tại N. tính :
a) diện tích hình tam giác BCM
b) diện tích hình thang BCNM.
TH
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Cho tam giác ABC vuông ở A và tam giác DEF vuông ở D có AB = DE và góc ABC = góc DEF. Chứng minh tam giác ABC = tam giác DEF.
xét 2 tam giác vuông ABC và tam giác EDF, ta có:
cạnh góc vuông : AB = DE
góc nhọn : ABC = DEF
=> tam giác ABC = tam giác DEF ( cgv - gn )
Lý thuyết : Cạnh góc vuông - góc nhọn: Nếu một cạnh góc vuông và một góc nhọn kề cạnh ấy của tam giác vuông này bằng một cạnh góc vuông và một góc nhọn kề cạnh ấy của tam giác vuông kia thì hai tam giác đó bằng nhau (cgv-gn)
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xét 2 tam giác vuông ABC và tam giác EDF, ta có:
cạnh góc vuông : AB = DE
góc nhọn : ABC = DEF
=> tam giác ABC = tam giác DEF ( cgv - gn )
Lý thuyết : Cạnh góc vuông - góc nhọn: Nếu một cạnh góc vuông và một góc nhọn kề cạnh ấy của tam giác vuông này bằng một cạnh góc vuông
và một góc nhọn kề cạnh ấy của tam giác vuông kia thì hai tam giác đó bằng nhau (cgv-gn)
1.cho tam giác ABC vuông ở A. kẻ AH vuông góc với BC (H thuộc BC). kẻ HM vuông góc với AB. CM:GÓC B=GÓC MAH=GÓC HAC?
2.cho tam giác ABC có góc B=góc C=40 độ. vì AC là tia phân giác của góc ngoài ở đỉnh A. CM:Ax //BC.
2.tự vẽ hình nhe
xét tam giác abc có
Góc CAx= góc B+góc C =40 + 10=80<đlí góc ngoài tam giác>
Vì Ac là phân giác của A
Góc A1=A2=1/2A=40
Ta có A2=C=40
Mà hai góc này ở vị trí so le trong
suy ra ax song song BC
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cho tam giác abc có góc a tù.vẽ tam giác vuông cân abd có góc b =90 độ(a và d ở 2 phía của bc)vẽ tam giác vuông cân cbg có b =90 độ(a và g ở cùng phía với bc)chứng minh: ga vuông góc bc
cho hình tam giác ABC vuông ở A có chu vi 30 cm , biết cạnh góc vuông AB = 5/12 cạnh góc vuông AC ,cạnh BC = 13 cm .Tính diện tích tam giác vuông ABC?
Tổng độ dài hai cạnh AB và AC:
30 - 13 = 17 (cm)
Tổng số phần bằng nhau:
5 + 12 = 17 (phần)
Cạnh AB dài:
17 . 5 : 17 = 5 (cm)
Cạnh AC dài:
17 . 12 : 17 = 12 (cm)
Diện tích tam giác ABC:
5 . 12 : 2 = 30 (cm²)
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Tổng độ dài 2 đáy AB và AC là :
30 - 13 = 17 ( cm )
Tổng số phần bằng nhau là
5 + 12 = 17 ( phần )
Cạnh AB dài là
17 : 17 x 5 = 5 ( cm )
Cạnh AC dài là :
17 - 5 = 12 ( cm )
Diện tích hình tam giác vuông ABC là
12 x 5 : 2 = 30 ( m2)
Đáp số : 30 m2
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cho tam giác ABC vuông ở A,có góc B=50 độ.So sánh các cạnh của tam giác ABC
góc C=90-50=40độ
Vì góc A>góc B>góc C
nên BC>AC>AB
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Tam giác ABC vuông tại A \(\Rightarrow\widehat{A}=90^o\)
Xét tam giác ABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)
\(\Rightarrow\widehat{C}=180^o-50^o-90^o=40^o\)
Vậy \(\widehat{A}>\widehat{B}>\widehat{C}\)
\(\Rightarrow\text{BC>AC>AB}\)
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Vì \(\widehat{A}\) là góc vuông, \(\widehat{B}=50^0\)
`->` \(\widehat{C}=40^0\)
`->` \(\widehat{A}>\widehat{B}>\widehat{C}\)
`->`\(BC>AC>AB\)
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Cho tam giác ABC. Vẽ ở phía ngoài tam giác ABC các tam giác vuông tại A và ABD, ACE có AB = AD, AC = AE. Kẻ AH vuông góc với BC, DM vuông góc với AH, EN vuông góc với AH. Chứng minh rằng: DM = AH
Ta có: ∠(BAH) +∠(BAD) +∠(DAM) =180o(kề bù)
Mà ∠(BAD) =90o⇒∠(BAH) +∠(DAM) =90o(1)
Trong tam giác vuông AMD, ta có:
∠(AMD) =90o⇒∠(DAM) +∠(ADM) =90o(2)
Từ (1) và (2) suy ra: ∠(BAH) =∠(ADM)
Xét hai tam giác vuông AMD và BHA, ta có:
∠(BAH) =∠(ADM)
AB = AD (gt)
Suy ra: ΔAMD= ΔBHA(cạnh huyền, góc nhọn)
Vậy: AH = DM (hai cạnh tương ứng) (3)
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Cho tam giác ABC vuông ở A có góc B bằng 60 độ . Tia phân giác của góc ABC cắt AC ở E. Kẻ EH vuông góc với BC
a) Chứng minh tam giác ABE bằng tam giác HBE
b) Chứng minh HB = HC
c) Từ H kẻ đường thẳng song song với BE cắt AC ở K. Chứng minh tam giác EHK đều
d) Gọi I là giao điểm của BA và HE. Chứng minh IE > EH
a) Xét \(\Delta ABE\) và \(\Delta HBE\):
BE chung
\(\widehat{ABE}=\widehat{EBH}\)
\(\widehat{EAB}=\widehat{EHB}=90^o\)
\(\Rightarrow\Delta ABE=\Delta HBE\left(ch-gn\right)\)
b) \(\widehat{EBH}=\dfrac{1}{2}\widehat{B}=30^o\)
\(\widehat{ACB}=90^o-\widehat{B}=30^o\)
\(\Rightarrow\Delta EBC\) cân tại E
Mà EH vuông góc BC
\(\Rightarrow HB=HC\)
c) \(\widehat{HEB}=90^o-\widehat{EBH}=60^o\)
\(KH//BE\Rightarrow\widehat{KHE}=\widehat{HEB}=60^o\)
\(\widehat{HEB}+\widehat{AEB}=60^o+60^o=120^o\)
\(\Rightarrow\widehat{KEH}=180^o-120^o=60^o\)
\(\Rightarrow\Delta EHK\) đều
d) Theo phần a. \(\Delta ABE=\Delta HBE\Rightarrow AE=EH\)
\(\Delta IAE\) vuông ở A \(\Rightarrow IE>AE\)
\(\Rightarrow IE>EH\)
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a) Xét ΔABEΔABE và ΔHBEΔHBE:
BE chung
ˆABE=ˆEBHABE^=EBH^
ˆEAB=ˆEHB=90oEAB^=EHB^=90o
⇒ΔABE=ΔHBE(ch−gn)⇒ΔABE=ΔHBE(ch−gn)
b) ˆEBH=12ˆB=30oEBH^=12B^=30o
ˆACB=90o−ˆB=30oACB^=90o−B^=30o
⇒ΔEBC⇒ΔEBC cân tại E
Mà EH vuông góc BC
⇒HB=HC⇒HB=HC
c) ˆHEB=90o−ˆEBH=60oHEB^=90o−EBH^=60o
KH//BE⇒ˆKHE=ˆHEB=60oKH//BE⇒KHE^=HEB^=60o
ˆHEB+ˆAEB=60o+60o=120oHEB^+AEB^=60o+60o=120o
⇒ˆKEH=180o−120o=60o⇒KEH^=180o−120o=60o
⇒ΔEHK⇒ΔEHK đều
d) Theo phần a. ΔABE=ΔHBE⇒AE=EHΔABE=ΔHBE⇒AE=EH
ΔIAEΔIAE vuông ở A ⇒IE>AE
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Cho tam giác ABC vuông ở A có A B = 10 c m , A C = 24 c m . So sánh các góc của tam giác ABC
A. A < B < C
B. A > B > C
C. B < A < C
D. C < A < B
Do tam giác ABC vuông tại A nên góc A là góc lớn nhất
Có AB < AC ⇒ C < B . Từ đó suy ra ∠C < ∠B < ∠A hay ∠A > ∠B > ∠C . Chọn B
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1. Cho tam giác ABC vuông ở A có AB<AC. AH vuông góc với BC tại H, D là điểm trên cạnh BC sao cho AD=AB. Vẽ DE vuông góc với BC tại E. Chứng mih rằng AH=HE.
2. Cho tam giác ABC vuông cân tại A.. Qua A vẽ đường thẳng d ở ngoài tam giác ABC . Vẽ BD vuông góc với d taị D. CE vuông góc với d tại E. M là trung điểm CB. Chứng minh rằng:
a) BD + CE = DE
b) Tam giác MDE là tam giác vuông cân
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Xem thêm câu trả lời
Bài 3 : Cho tam giác ABC vuông ở A có chu vi 24 cm, cạnh góc vuông thứ nhất bằng ¾ cạnh góc vuông thứ hai. Tìm diện tích tam giác ABC, biết cạnh BC= 10cm.
BC=10cm nên AB+AC=14cm
mà AB=3/4AC
nên 7/4AC=14cm
=>AC=8(cm)
=>AB=6(cm)
\(S_{ABC}=\dfrac{8\cdot6}{2}=24\left(cm^2\right)\)
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