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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)
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)
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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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 )
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:
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)
\(S_{ABD}=\dfrac{1}{2}.AB.KD=\dfrac{1}{2}.4.1,5=3cm^2\)
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...............
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:
CD .
a) Ta có : AB//DM (gt) (1)
Xét tam giác ABH và tam giácDMH có
BHA^=DHA^(đối đỉnh)
AH=HD(A đx D qua H)
BAH^=HDM^(so le trong)
=> tam giác ABH=tam giácDMH (g-c-g)
=>AB=DM ( 2 cạnh tương ứng) (2)
Tử (1)(2) => ABDM là hbh
Vì M thuộc BC
mà AH vuông BC => AH vuông BM
Xét hbh ABDM có
AH vuông BM
=> hbh ABDM là hình thoi