Hướng dẫn giải bài tập Toán 8

By Trang Ánh Nam -

Giải hộ e bài này ạ
Cho tam giác ABC vuông tại A có AH là đường cao. Vẽ HD vuông góc với AB và HEvuông góc với AC .AB = 12 cm AC = 16 cm. Chứng minh
a) ∆HAC~∆ABC
b) AH^2=AD.AB
c) AD.AB=AE.AC
d) Tính S ADE/ S ABC
Giải:


a.  Xét tam giác ABC và tam giác HAC, ta có:
ü Góc A = góc H = 90 độ.
ü Góc C chung
ð ∆HAC~∆ABC (góc-góc)
b.  Xét tam giác BAH và tam giác HAD, ta có:
ü Góc A chung
ü Góc D = góc H = 90 độ
=>∆AHD~∆ABH
=>  AH/AB=AD/AH
=> AH2=AD.AB  (*)
c. xét tam giác AHE và tam giác AHC, ta có:
ü góc A chung
ü góc H=góc E= 90 độ
=>∆AHE~∆ACH
=>AH/AC=AE/AH
=> AH2=AC.AE  (**)
Từ (*) với (**) suy ra: AD.AB=AE.AC
d. Áp dụng pitago cho tam giác ABC ta có:
BC2=AB2+AC2=122+162
=>BC=20
Lại có:
ü SABC=0,5.AH.BC
ü SABC=0,5.AB.AC
=>AH.BC=AB.AC
=>AH=12.16/20=9,6
Mà AH = DE (ADHE là hình chử nhật)
Nên DE=9,6
Xét tam giác ADE và tam giác ABC, ta có:
ü AD.AB=AE.AC
ü Góc A chung
=>∆ADE~∆ACB
=>tỉ số đồng dạng t=DE/BC=9,6/20=0,48
=> SADE/ SABC = t2=0,482=0,2304

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