a. Diện tích của Δ ABC là:
\(\dfrac{1}{2}\) . 6 . 8 = 24 cm2
b. Ta có: Δ ABC vuông tại A
Theo đ/lí Py - ta - go
BC2 = AB2 + AC2
BC2 = 62 + 82
BC2 = 100
\(\Rightarrow\) BC = \(\sqrt{100}\) = 10 cm
Vì AD là tia phân giác của \(\widehat{A}\)
\(\dfrac{AB}{AC}=\dfrac{DB}{DC}\)
\(\Rightarrow\) \(\dfrac{6}{8}\) = \(\dfrac{DB}{10-DB}\)
\(\Rightarrow\) \(\dfrac{3}{4}=\dfrac{DB}{10-DB}\)
\(\Rightarrow\) 3 . (10 - DB) = 4DB
\(\Rightarrow\) 30 - 3DB - 4DB = 0
\(\Rightarrow\) 30 - 7DB = 0
\(\Rightarrow\) DB = \(\dfrac{30}{7}\) \(\approx\) 4,3 cm
Ta có: DC = 10 - DB
\(\Rightarrow\) DC = 10 - 4,3
\(\Rightarrow\) DC = 5,7 cm
c. Xét ΔABC và ΔHBA:
\(\widehat{A}=\widehat{H}\) = 900 (gt)
\(\widehat{B}\) chung
\(\Rightarrow\) ΔABC \(\sim\) ΔHBA (g.g)
Ta có: ΔABC \(\sim\) ΔHBA
\(\dfrac{AB}{HB}=\dfrac{BC}{BA}\)
\(\Rightarrow\) AB2 = BH . BC
Vì ΔABC vuông tại A
SΔABC = \(\dfrac{AH.BC}{2}\) = \(\dfrac{AB.AC}{2}\) \(\Rightarrow\) AB . AC
\(\Leftrightarrow\) AH = \(\dfrac{AB.AC}{BC}\) = \(\Leftrightarrow\) \(\dfrac{1}{AH}\) = \(\dfrac{AH}{AB.AC}\)
\(\Leftrightarrow\) \(\dfrac{1}{AB^2}\) = \(\dfrac{BC^2}{AB^2.AC^2}\)
Mặt khác theo đ/lí Py - ta - go:
BC2 = AB2 + AC2
\(\Rightarrow\) \(\dfrac{1}{AH^2}\) = \(\dfrac{AB^2+AC^2}{AB^2.ÂC^2}\) = \(\dfrac{1}{AB^2}\) + \(\dfrac{1}{AC^2}\)
\(\Rightarrow\) \(\dfrac{1}{AH^2}\) = \(\dfrac{1}{AB^2}\) + \(\dfrac{1}{AC^2}\) (dpcm)
nhớ tick cho cj nha
