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)
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)
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
Cho tam giác ABC vuông ở A ( AB
Bài 1: Cho tam giac ABC, M là trung điểm cua AB. Đường thẳng qua M và song song với BC cắt AC ở I và song song với AB cắt BC ở k. Chứng minh rằng: a) AM=IK b) Tam giác AMI bằng tam giác IKC c) AI=IC Bài 2: Cho tam giác ABC vuông tại A. Gọi I là trung điểm BC. Trên tia đối của tia IA lấy điểm D sao cho ID=IA a) CMR tam giác BID bằng tam giác CIA b) CMR : BD vuông góc với AB c) Qua A kẻ đường thẳng song song với BC cắt đường thẳng BD tại M. C/M tam giác BAM bằng tam giác ABC d) CMR: AB là tia phân giác cuả góc DAM Bài 3: Cho tam giác ABC vuông ở A và AB=AC.Gọi K là trung điểm của BC a) C/M: tam giác AKB bằng tam giác AKC b) C/M: AK vuông góc với BC c) từ C vẽ đường vuông góc với BC cắt đường thẳng AB tại E.C/M EK song song với AK Bài 4: Cho tam giác ABC có AB=AC, kẻ BD vuông góc với AC, CE vuông góc với AB(D thuộc AC, E thuộc AB). Gọi O là giao điểm của BD và CE. CMR a) BD= CE b) tam giác OEB bằng tam giác ODC c) AO là tia phân giác cua góc BAC
1. Câu hỏi của 1234567890 - Toán lớp 7 - Học toán với OnlineMath
Cho tam giác vuông ABC vuông ở A có chu vi 237,6cm.Cạnh AB dài hơn cạnh AC 19,8dm.Cạnh BC dài 99dm.Tính diện tích hình tam giác vuông ABC.
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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Cho tam giác ABC vuông tại A và có BC = 2AB, AB = a. Ở phía ngoài tam giác, ta vẽ hình vuông BCDE, tam giác đều ABF và tam giác đều AGC. Tính các góc B, C, cạnh AC và diện tích tam giác ABC.
Gọi M là trung điểm của BC, ta có:
AM = MB = 1/2 BC = a (tính chất tam giác vuông)
Suy ra MA = MB = AB = a
Suy ra ∆ AMB đều ⇒ ∠ (ABC) = 60 0
Mặt khác: ∠ (ABC) + ∠ (ACB) = 90 0 (tính chất tam giác vuông)
Suy ra: ∠ (ACB) = 90 0 - ∠ (ABC) = 90 0 – 60 0 = 30 0
Trong tam giác vuông ABC, theo Pi-ta-go, ta có: B C 2 = A B 2 + A C 2
⇒ A C 2 = B C 2 - A B 2 = 4 a 2 - a 2 = 3 a 2 ⇒ AC = a 3
Vậy S A B C = 1/2 .AB.AC
= 1 2 a . a 3 = a 2 3 2 ( đ v d t )
Tam giác `ABC` có đường AH thỏa mãn `AH^2 = CH.BH` thì khẳng định nào đúng?
`\triangle ABC` vuông ở `A`
`AB^2 = BH.BC`
`\triangle AHB` đồng dạng `\triangle CHA`
`AB^2 +AC^2 = BC^2`
Cho tam giác ABC vuông ở `A,AB=3;AC=4`. Đường cao `AH`. Tính `AH`?
Câu 1: Cả 4 câu đều đúng
Câu 2:
ΔABC vuông tại A
=>\(AB^2+AC^2=BC^2\)
=>\(BC^2=3^2+4^2=25\)
=>BC=5
Xét ΔABC vuông tại A có AH là đường cao
nên \(AH\cdot BC=AB\cdot AC\)
=>\(AH\cdot5=3\cdot4=12\)
=>AH=2,4
Bài 6: Cho tam giác ABC vuông tại A, AB = 4cm, AC = 3 cm, trung tuyến AD, kẻ DK vuông góc với với AB, kẻ DH vuông góc với AC
a. Tứ giác AKDH là hình gì? Vì sao?
b. Tính độ dài AD
c. Tính diện tích tam giác ABD
Bài 7: Cho 
ABC vuông ở A (AB < AC ), đường cao AH. Gọi D là điểm đối xứng của A qua H. Đường thẳng kẻ qua D song song với AB cắt BC và AC lần lượt ở M và N. Chứng minh:
a. Tứ giác ABDM là hình thoi.
b. AM 
CD .
c. Gọi I là trung điểm của MC; chứng minh IN HN.
Bài 6:
a: Xét tứ giác AKDH có
\(\widehat{AKD}=\widehat{AHD}=\widehat{KAH}=90^0\)
Do đó: AKDH là hình chữ nhật
b: Ta có: ΔABC vuông tại A
mà AD là đường trung tuyến
nên AD=BC/2=2,5(cm)
a. Tứ giác AKDH là hình chữ nhật , vì có góc \(DKA=KAH=DHA=90^o\)
b, áp dụng đl pytago vào tam giác vuông ABC có :
\(BC^2=AB^2+AC^2\Leftrightarrow BC=\sqrt{4^2+3^2}=5cm\)
vì AD là trung tuyến tam giác vuông ABC nên :
\(AD=\dfrac{1}{2}BC=\dfrac{1}{2}.5=2,5cm\)
c,vì AKDH là hình chữ nhật nên : DH//KA
mà D là trung điểm BC
=>H là trung điểm AC
<=>AH=\(\dfrac{1}{2}AC=\dfrac{1}{2}.3=1,5cm\)
vì AH = 1,5 cm nên => KD cũng = 1,5cm (AKDH là hình chữ nhật)
\(S_{ABD}=\dfrac{1}{2}.AB.KD=\dfrac{1}{2}.4.1,5=3cm^2\)
cho tam giác abc vuông ở A có chu vi = 24m có cạch ab = 3/4 ac ,ab= 10m .Tính diện tích tam giác abc.
Tổng độ dài hai cạnh AB và AC là :
24 - 10 = 14 ( cm )
Độ dài cạnh AB là :
14 : ( 3 + 4 ) x 3 = 6 ( cm )
Độ dài cạnh AC là :
14 - 6 = 9 ( cm )
Diện tích hình tam giác ABC là :
6 x 9 : 2 = 27 ( cm2)
Đáp số : 27 cm2
tổng độ dài hai cạnh là
24-10=14 cm
độ dại cạnh AB là
14:(3+4).3=6 cm
độ dài cạnh AC là
14-6=8 cm
diện tích là
6.7:2=27cm2
đáp số...............
Thảo Mai bạn tham khảo đây nhé:
Câu hỏi của Tran Quynh Anh - Toán lớp 5 - Học toán với OnlineMath
......
Thảo Mai
