câu a là HB/HC ạ , em quên viết hoa

nhờ các bạn giải giúp hộ mình vs ạ mình cần gấp

Xét \(\Delta AHB\) có : \(\widehat{AHB}=90^0\)

\(\Leftrightarrow AB^2=HB^2+AH^2\) (định lí Py ta go)

\(\Leftrightarrow AH^2=AB^2-HB^2\)

\(\Leftrightarrow AH^2=12^2-5^2\)

\(\Leftrightarrow AH^2=119\)

\(\Leftrightarrow AH=\sqrt{199}cm\)

Ta có :

\(BC=BH+HC\)

\(\Leftrightarrow HC=BC-BH\)

\(\Leftrightarrow HC=20-5\)

\(\Leftrightarrow HC=15cm\)

Xét \(\Delta AHC\) có : \(\widehat{AHC}=90^0\)

\(\Leftrightarrow AC^2=AH^2+HC^2\)

\(\Leftrightarrow AC^2=\left(\sqrt{199}\right)^2+15^2\)

\(\Leftrightarrow AC^2=424\)

\(\Leftrightarrow AC=\sqrt{424}cm\)

mơn nhìu nha...

>3 >3 >3

Xét \(\Delta ABC\) và \(\Delta HBA\) có:

\(\widehat{BAC}=\widehat{BHA}=90^o\)

\(\widehat{B}\) chung

\(\Rightarrow\Delta ABC\sim\Delta HBA\left(g.g\right)\) (1)

\(\Rightarrow\dfrac{AB}{BC}=\dfrac{HB}{AB}\) hay \(\dfrac{AB}{4+9}=\dfrac{4}{AB}\Rightarrow AB^2=52\Rightarrow AB=\sqrt{52}=2\sqrt{13}cm\)

Xét \(\Delta\text{A}BC\) và \(\Delta HAC\) có:

\(\widehat{BAC}=\widehat{AHC}=90^o\)

\(\widehat{C}\) chung

\(\Rightarrow\Delta ABC\sim\Delta HAC\left(g.g\right)\) (2)

Từ (1) và (2) \(\Rightarrow\Delta HAB\sim\Delta HCA\)

\(\Rightarrow\dfrac{AH}{HC}=\dfrac{HB}{AH}\) hay \(\dfrac{AH}{9}=\dfrac{4}{AH}\Rightarrow AH^2=36\Rightarrow AH=\sqrt{36}=6\left(cm\right)\)

Ta có \(\Delta ABC\) vuông tại A.

Áp dụng đinh lý Py-ta-go ta có:

\(AC=\sqrt{BC^2-AB^2}=\sqrt{\left(4+9\right)^2-\left(2\sqrt{13}\right)^2}=3\sqrt{13}cm\)

b) Diện tích tam giác ABC là:

\(S_{ABC}=\dfrac{1}{2}\cdot BC\cdot AH=\dfrac{1}{2}\cdot\left(4+9\right)\cdot6=39\left(cm^2\right)\)

a, \(BC=BH+HC=5\left(cm\right)\)

Áp dụng HTL: \(\left\{{}\begin{matrix}AB=\sqrt{BH\cdot BC}=2\sqrt{5}\left(cm\right)\\AC=\sqrt{CH\cdot BC}=\sqrt{5}\left(cm\right)\end{matrix}\right.\)

b, Áp dụng HTL: \(AH=\dfrac{AB\cdot AC}{BC}=2\left(cm\right)\)

a: BC=4+1=5(cm)

\(\Leftrightarrow\left\{{}\begin{matrix}AB=2\sqrt{5}\left(cm\right)\\AC=\sqrt{5}\left(cm\right)\end{matrix}\right.\)

b: \(AH=\sqrt{HB\cdot HC}=2\left(cm\right)\)

AB/AC=4/3

=>HB/HC=16/9

=>HB/16=HC/9=k

=>HB=16k; HC=9k

AH^2=HB*HC

=>144k^2=24^2=576

=>k=2

=>HB=32cm; HC=18cm

AB=căn 32*50=40cm

AC=căn 18*50=30cm

a: AB/AC=3/4

=>BH/CH=9/16

=>BH/9=CH/16=(BH+CH)/(9+16)=125/25=5

=>BH=45cm; CH=80cm

b: AB/AC=3/7

=>HB/HC=(3/7)^2=9/49

=>HB/9=HC/49=k

=>HB=9k; HC=49k

AH^2=HB*HC

=>9k*49k=42^2

=>k=2

=>HB=18cm; HC=98cm

c: Đặt HB/9=HC/16=k

=>HB=9k; HC=16k

AH^2=HB*HC

=>144k^2=48^2

=>k=4

=>HB=36cm; HC=64cm

BC=36+64=100cm

AB=căn 36*100=60cm

AC=căn 64*100=80cm