Cho Tam giac ABC vuong tai A, duong cao AH va biet HC-HB=AB
a) CM rang \(\frac{BC^2-2.AC^2}{HB^2-HC^2}=1\)
b) Tính số đo góc C
cho tam giac abc vuong tai a, co duong cao ah ( h thuoc bc ), biet ah=6cm,hc-hb=9cm.Tinh hb,hc
Ta có: \(HC-HB=9\Rightarrow HC=9+HB\)
tam giác ABC vuông tại A có đường cao AH nên áp dụng hệ thức lượng
\(\Rightarrow AH^2=HB.HC=HB\left(HB+9\right)\Rightarrow HB^2+9HB=36\)
\(\Rightarrow HB^2+9HB-36=0\Rightarrow\left(HB-3\right)\left(HB+12\right)=0\)
mà \(HB>0\Rightarrow HB=3\left(cm\right)\Rightarrow HC=3+9=12\left(cm\right)\)
Ta có: HC-HB=9(gt)
nên HB=HC-9
Áp dụng hệ thức lượng trong tam giác vuông vào ΔABC vuông tại A có AH là đường cao ứng với cạnh huyền BC, ta được:
\(HB\cdot HC=AH^2\)
\(\Leftrightarrow HC\left(HC-9\right)-36=0\)
\(\Leftrightarrow HC^2-9HC-36=0\)
\(\Leftrightarrow HC^2-12HC+3HC-36=0\)
\(\Leftrightarrow\left(HC+3\right)\left(HC-12\right)=0\)
\(\Leftrightarrow HC=12\left(cm\right)\)
\(\Leftrightarrow HB=HC-9=12-9=3\left(cm\right)\)
cho tam giac abc vuong tai a, duong cao ah
a. chung minh tam giac hba dong dang voi tam giac abc
b. chung minh ah^2 =hb* hc
c. tia phan giac cua goc ahc cat ac tai d . chung minh \(\frac{hb}{hc}\)=\(\frac{ad^2}{dc^2}\)
cho tam giac ABC vuong tai A, AB< AC , AH la duong cao
a)chung minh Tam giac HAC va Tam giac ABC đồng dạng
b) cm HA2= HB . HC
c goi D, E lan luot la trung diem cua AB, BC chung minh CH . CB = 4DE2
cho tam giac ABC vuong tai A duong cao AH. Biet AB=4cm, AC=7,5cm. Tinh HB, HC
1. cho tam giac ABC vuong tai A , duong cao AH . I,K lan luot la trung diem cua AB va AC. Tinh HB, HC,AH va dien tich tu giac AIHK biet HI 9cm, HK= 12cm
1.cho tam giac ABC vuong tai A ,goi AH la duong cao .biet rang \(\frac{AC}{AB}=\frac{5}{6'},BC=122cm\)
a)tinh BH,CH
b)tinh AH
2.cho tam giac ABC vuong o A,phan giac AD,duong cao AH.bietCD=68cm,BD=51cm.tinh BH,HC.
cho tam giac ABC vuong tai A , duong cao AH ,HB = 3,6 cm ,HC = 6,4 cm . Tinh AB , AC ,AH
Ta có BC=HB+HC=3,6+6,4=10(cm)
Xét △ABC vuông tại A đường cao AH:
AB2=BC.HB=10.3,6=36⇒AB=6(cm)
AC2=BC.HC=10.6,4=64⇒AC=8(cm)
\(AC.AB=BC.AH\Rightarrow AH=\dfrac{AC.AB}{BC}=\dfrac{6.8}{10}=4,8\left(cm\right)\)
cho tam giac abc vuong tai a, duong cao ah
a. chung minh tam giac hba dong dang voi tam giac abc
b. chung minh ah^2 =hb* hc
c. tia phan giac cua goc ahc cat ac tai d . chung minh \(\frac{hb}{hc}\)=\(\frac{ad^2}{dc^2}\)
a) Xét \(\Delta HBA\)và \(\Delta ABC\)có :
\(\widehat{AHB}=\widehat{BAC}=90^o;\widehat{B}\left(chung\right)\)
\(\Rightarrow\)\(\Delta HBA\)\(\approx\)\(\Delta ABC\)( g.g )
b) Xét \(\Delta HBA\)và \(\Delta HAC\)có :
\(\widehat{AHB}=\widehat{AHC}=90^o\)
\(\widehat{BAH}=\widehat{ACH}\left(cung-phu-\widehat{B}\right)\)
\(\Rightarrow\Delta HBA\approx\Delta HAC\left(g.g\right)\)
\(\Rightarrow\frac{BH}{AH}=\frac{AH}{HC}\Rightarrow AH^2=BH.HC\)
cho tam giac ABC vuong tai,duong cao AH,biet HB=25cm,HC=36cm,AH=30cm.
a/ chung minh tam giac HBA dong dang voi tam giac HAC.
b/tinh do dai cac doan thang AB,BC,AC
a) Ta có: \(\widehat{HAB}+\widehat{HBA}=90^0\)
\(\widehat{HAB}+\widehat{HAC}=90^0\)
suy ra: \(\widehat{HBA}=\widehat{HAC}\)
Xét 2 tam giác vuông: \(\Delta HBA\) và \(\Delta HAC\) có:
\(\widehat{BHA}=\widehat{AHC}=90^0\)
\(\widehat{HBA}=\widehat{HAC}\) (CMT)
suy ra: \(\Delta HBA~\Delta HAC\)
b) \(BC=BH+HC=25+36=61\)cm
\(\Delta HBA~\Delta HAC\) \(\Rightarrow\)\(\frac{HB}{HA}=\frac{AB}{AC}\)
\(\Rightarrow\)\(\frac{AB}{AC}=\frac{5}{6}\)\(\Leftrightarrow\)\(\frac{AB}{5}=\frac{AC}{6}\)\(\Leftrightarrow\)\(\frac{AB^2}{25}=\frac{AC^2}{36}=\frac{AB^2+AC^2}{25+36}=\frac{BC^2}{61}=\frac{61^2}{61}=61\)
suy ra: \(\frac{AB^2}{25}=61\) \(\Leftrightarrow\) \(AB=\sqrt{1525}\) cm
\(\frac{AC^2}{36}=61\)\(\Leftrightarrow\) \(AC=\sqrt{2196}\)cm
p/s: tham khảo