cho hình chữ nhật ABCD. Về phía ngoài hình chữ nhật dựng tam giác BCE vuông tại C có góc CBE= 45 độ và dựng tam giác ABF vuông tại F có À=6cm, BF=8cm. Biết BE = căn 18. TÍnh chu vi ngũ giác ADEBF
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NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 8 2021
Áp dụng định lý Pitago cho tam giác vuông ABC
\(AC=\sqrt{AB^2+BC^2}=10\left(cm\right)\)
Áp dụng hệ thức lượng cho tam giác vuông ABC với đường cao BE:
\(AB^2=AE.AC\Rightarrow AE=\dfrac{AB^2}{AC}=6,4\left(cm\right)\)
\(AB.AC=BE.AC\Rightarrow AE=\dfrac{AB.AC}{BC}=4,8\left(cm\right)\)
b.
Ta có: \(EC=AC-AE=3,6\left(cm\right)\)
Do AB song song CF, theo định lý Talet:
\(\dfrac{CF}{AB}=\dfrac{CE}{AE}\Rightarrow CF=\dfrac{AB.CE}{AE}=4,5\left(cm\right)\)
\(\Rightarrow DF=DC-CF=8-4,5=3,5\left(cm\right)\)
Áp dụng định lý Pitago cho tam giác vuông ADF:
\(AF=\sqrt{AD^2+DF^2}=\dfrac{\sqrt{193}}{2}\left(cm\right)\)
Pitago tam giác vuông BCF:
\(BF=\sqrt{BC^2+CF^2}=7,5\left(cm\right)\)
Kẻ FH vuông góc AB \(\Rightarrow ADFH\) là hình chữ nhật (tứ giác 3 góc vuông)
\(\Rightarrow FH=AD=6\left(cm\right)\)
\(S_{ABF}=\dfrac{1}{2}FH.AB=\dfrac{1}{2}.6.8=24\left(cm^2\right)\)
4 tháng 7 2023
a: Xét ΔBCE vuông tại C và ΔDBE vuông tại B có
góc E chung
=>ΔBCE đồng dạng với ΔDBE
b: Xét ΔCBD vuông tại C và ΔHCB vuông tại H có
góc CBD=góc HCB
=>ΔCBD đồng dạng với ΔHCB
=>CB/HC=BD/CB
=>BC^2=HC*BD
c: CE=6^2/8=4,5cm
CH//DB
=>ΔEHC đồng dạng với ΔEBD
=>S EHC/S EBD=(EC/ED)^2=(4,5/12,5)^2=81/625
8 tháng 4 2022
a: Xét ΔBCE vuông tại C và ΔDBE vuông tại B có
góc E chung
Do đó: ΔBCE\(\sim\)ΔDBE
b: Đề sai rồi bạn
15 tháng 6 2022
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KT
25 tháng 3 2017
a)xét tam giác BCE và tam giác DCE có:
\(\widehat{DBE}=\widehat{BCE}=90^o\)
\(\widehat{BEC}:chung\)
nên tam giác BCE ~ tam giác DBE(g-g)
SO
28 tháng 3 2021
a,Xét tam giác BDE và tam giác DCE có:
+)chung góc E
+)góc BDE=DCE=90độ
suy ra tam giác BDE đồng dạng tam giác DCE(g-g)
b,Xét tam giác CHD và tam giác DCB có:
+)góc DCH=góc BDC
+)góc DHC=góc BCD
suy ra tam giác CHD đồng dạng tam giác DCB
c,Do BD vuông DE và HC vuông DE
=>BD//HC
=>CK/OB=EK/EO=HK/OD(bn suy ra từ ta-lét)
Mà OB=OD =>CK=HK=>K là trung điểm của CH.
Tỉ số bn dựa vào phần a,b
d,Gọi F là giao điểm của KF và DC(Bây h mình k vt hẳn chữ góc ra nx)
Vì HC//BD nên:
=>HCBD là hình thang
=>BH và DC là 2 đường chéo cắt nhau tại F(*)
Xét tam giác OFD và tam giác KFC,có:
+) ECK= ODF(do BD//CH)
+)DÒF=CKE(Do OD//KC và 2 góc ở vị trí sole trong)
Suy ra tam giác OFD đồng dạng tam giác KFC(g-g)
=>OFD=KFC mà 2 góc ở vị trí đối đỉnh nên
=> DC cắt OK tại F
=>BOK+OKC=180độ(2 góc trong cùng phía)
mà BOK=OKC(do KC//BO) mà 2 góc ở vị trí đồng vị nên
=>CKE+OKC=180 độ
=>O;K;E thẳng hàng mà DC cắt OK tại F nên
=>DC cắt OF tại F(**)
từ (*) và (**) suy ra:
OE;CD;BH thẳng hàng.