a) Xét ΔBDC và ΔHBC có
\(\widehat{BCD}\) chung
\(\widehat{DBC}=\widehat{BHC}\left(=90^0\right)\)
Do đó: ΔBDC∼ΔHBC(g-g)
b) Áp dụng định lí pytago vào ΔDBC vuông tại B, ta được:
\(DC^2=BD^2+BC^2\)
\(\Leftrightarrow BD^2=DC^2-BC^2=10^2-6^2=64\)
hay \(BD=\sqrt{64}=8cm\)
Ta có: ΔBDC∼ΔHBC(cmt)
⇒\(\frac{BC}{HC}=\frac{DC}{BC}\)
⇒\(HC=\frac{BC\cdot BC}{DC}=\frac{6\cdot6}{10}=\frac{36}{10}=3,6cm\)
Ta có: DC=HC+HD(H nằm giữa D và C)
hay HD=DC-HC=10-3,6=6,4(cm)
Vậy: HC=3,6cm; HD=6,4cm
c) Ta có: ΔBDC∼ΔHBC(cmt)
⇒\(\frac{BD}{HB}=\frac{BC}{HC}\)
hay \(BH=\frac{BD\cdot HC}{BC}=\frac{8\cdot3,6}{6}=4,8cm\)
Vậy: BH=4,8cm
a, Xét \(\Delta BDC\) và \(\Delta HBC\) có :
\(\widehat{C}:chung\)
\(\widehat{DBC}=\widehat{BHC}=90^o\)
\(\Rightarrow\) \(\Delta BDC\sim\Delta HBC\left(g.g\right)\)
b, Ta có : \(\Delta BDC\sim\Delta HBC\) ( câu a )
\(\Rightarrow\) \(\frac{DC}{BC}=\frac{BC}{HC}\) \(\Rightarrow\) \(HC=\frac{BC^2}{DC}=\frac{6^2}{10}=\frac{36}{10}=3,6\left(cm\right)\)
\(HD=DC-HC=10-3,6=6,4\left(cm\right)\)
c, \(\Delta BHC:\) \(\widehat{BHC}=90^o\)
\(\Rightarrow\) \(BC^2=BH^2+HC^2\) ( Định lý Py - ta - go )
\(\Rightarrow\) \(BH^2=BC^2-HC^2=6^2-\left(3,6\right)^2=23,04\)
\(\Rightarrow\) \(BH=4,8\left(cm\right)\)
d, Kẻ \(AK\perp CD\) ( \(K\in CD\) )
Xét \(\Delta ADK\) và \(\Delta BCH\) có :
AD = BC ( gt )
\(\widehat{ADK}=\widehat{BCH}\left(gt\right)\)
\(\widehat{AKD}=\widehat{BHC}=90^o\)
\(\Rightarrow\) \(\Delta ADK=\Delta BCH\left(g.c.g\right)\)
\(\Rightarrow\) DK = CH
AB = CD - 2CH = 10 - 2.3,6 = 2,8 ( cm )
\(\Rightarrow\) \(S_{ABCD}=\frac{\left(AB+CD\right)BH}{2}=\frac{\left(2,8+10\right)4,8}{2}=30,72\left(cm^2\right)\)
