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a) Vì tam giác BAC vuông tại A
=> AB^2 + AC^2 = BC^2 ( đl pytago )
=> BC^2 = 5^2 + 7^2 = 74
=> BC = căn bậc 2 của 74
b)
Xét tam giác ABE; tam giác DBE có :
AB = DB ( gt)
góc ABE = góc DBE ( gt)
BE chung
=> tam giác ABE = tam giác DBE (c.g.c) - đpcm
c)
Vì tam giác ABE = tam giác DBE (câu b)
=> AE = DE
Xét tg AEF ⊥ tại A; tg DEC ⊥ tại D:
AE = DE (c/m trên)
g AEF = g DEC (đối đỉnh)
=> tg AEF = tg DEC (cgv - gn) - đpcm
=> EF = EC
d)
Do tam giác AEF = tam giác DEC (câu c)
=> AE = DE
=> E ∈ đường trung trực của AD (1)
Lại do AB = BD (gt)
=> B ∈ đường trung trực của AD (2)
Từ (1) và (2) => BE là đường trung trực của AD. - đpcm
a)tg BAC vuông tại A suy ra AB^2+AC^2=BC^2(định lý pi-ta-go)
suy ra BC^2=5^2+7^2=74
suy ra BC=\(\sqrt{74}\)
b)tg ABE=tgDBE(ch cgv)suy ra AE=ED
c)tg AEF=DEC(g c g) suy ra EF=EC(2 cạnh tương ứng )
d)gọi I là giao điểm của AD và BE
ta có AB=BD suy ra tgABD cân tại B
tg ABE=DBE(cmt) suy ra góc ABE=DBE mà BE nằm giữa 2 tia AB và BD suy ra BE là tia phân giác của góc ABD
tg cân ABD có BI là tia phân giác của góc ABD suy ra BI còn là đường trung trực của AD suy ra BE là đường trung trực của AD
a) Áp dụng pytago .
b) Xét t/g ABE; tg DBE:
AB = DB ( gt)
g ABE = DBE (suy từ gt)
BE chung
=> tg ABE = tg DBE (c.g.c)
c) Vì tg ABE = tg DBE (câu b)
=> AE = DE
Xét tg AEF ⊥⊥ tại A; tg DEC ⊥⊥ tại D:
AE = DE (c/m trên)
g AEF = g DEC (đối đỉnh)
=> tg AEF = tg DEC (cgv - gn)
=> EF = EC
d) Do tg AEF = tg DEC (câu c)
=> AE = DE
=> E ∈∈ đg trung trực của AD (1)
Lại do AB = BD (gt)
=> B ∈ đg trung trực của AD (2)
Từ (1) và (2) => BE là đg trung trực của AD.
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a/ Chiều dài thực của sân vận động đó là:
15 x 1000 = 15000 ﴾cm﴿
Chiều rộng thực của sân vận động đó là:
12 x 1000 = 12000 ﴾cm﴿
Đổi: 15000 cm = 150 m; 12000 cm = 120 m
Chu vi thực của sân vận động đó là:
﴾150 + 120﴿ x 2 = 540 ﴾m﴿
b/ Diện tích thực của sân vận động đó là:
150 x 120 = 18000 ﴾m2﴿
Đáp số: a/ 540 m b/ 18000 m2
hình tự vẽ:
xét hai tam giác vuông ABE và DBE:
ab=ad(gt); be là cạnh huyền chung
=>\(\Delta\) ABE = \(\Delta\)DBE
mình sẽ giải tiếp
a) theo đinh j lý pitago : tam giác abc vuông tại A
=> \(AB^2+AC^2=BC^2\)THAY SỐ TA ĐƯỢC \(5^2+7^2=BC^2\) TA ĐƯỢC \(74=BC^2\) =>BC =
8.6023
a) tam giác ABC vuông tại A
=> AB2 + AC2 = BC2
=> 52 + 72 = BC2
=> BC2 = 25 + 49 = 74
=> BC = \(\sqrt{74}cm\)
hình như bn ghi sai đề rùi làm sao làm bài b) !!!!!!!1
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