1/ 

a, 

$\Delta$ AHB và $\Delta$ BCD có:

$\widehat{AHB}= \widehat{BCD}= 90^o$

$\widehat{ABH}= \widehat{BDC}$ (SLT) 

=> $\Delta$ AHB $\backsim$ $\Delta$ BCD (g.g)    (*) 

b, 

$\Delta$ BCD vuông tại C có BD= $\sqrt{9^2+12^2}= 15cm$ 

(*) => $\frac{AH}{BC}= \frac{AB}{BD}= \frac{BH}{CD}$

=> $\frac{AH}{9}= \frac{12}{15}= \frac{BH}{12}$

=> $AH= 7,2; BH= 9,6$ 

=> $S_{AHB}= 0,5AH.BH= 34,56 (cm^2)$  

2/  

$\widehat{HAB} = \widehat{BAC}-\widehat{HAC}= 90^o - \widehat{HAC}$(*) 

$\widehat{HCA}= 90^o - \widehat{HAC}$(**)

(*)(**) => $\widehat{HAB}= \widehat{HCA}$ (1) 

Mà $\widehat{BHA}= \widehat{AHC}= 90^o$  (2) 

(1)(2)=> $\Delta$ BHA $\backsim$ $\Delta$ AHC (g.g)

=> $\frac{BH}{AH}= \frac{AH}{HC}$

=> $AH^2= BH.HC= 9.16= 144$

=> $AH= 12$ 

$BC= 9+16= 25$  

Theo Pytago:

$AB= \sqrt{AH^2+ BH^2}= 15$

$AC= \sqrt{AH^2+ HC^2}= 20$

Đáp án:

1/ 

a, 

ΔΔ AHB và ΔΔ BCD có:

ˆAHB=ˆBCD=90oAHB^=BCD^=90o

ˆABH=ˆBDCABH^=BDC^ (SLT) 

=> ΔΔ AHB ∽∽ ΔΔ BCD (g.g)    (*) 

b, 

ΔΔ BCD vuông tại C có BD= √92+122=15cm92+122=15cm 

(*) => AHBC=ABBD=BHCDAHBC=ABBD=BHCD

=> AH9=1215=BH12AH9=1215=BH12

=> AH=7,2;BH=9,6AH=7,2;BH=9,6 

=> SAHB=0,5AH.BH=34,56(cm2)SAHB=0,5AH.BH=34,56(cm2)  

2/  

ˆHAB=ˆBAC−ˆHAC=90o−ˆHACHAB^=BAC^−HAC^=90o−HAC^(*) 

ˆHCA=90o−ˆHACHCA^=90o−HAC^(**)

(*)(**) => ˆHAB=ˆHCAHAB^=HCA^ (1) 

Mà ˆBHA=ˆAHC=90oBHA^=AHC^=90o  (2) 

(1)(2)=> ΔΔ BHA ∽∽ ΔΔ AHC (g.g)

=> BHAH=AHHCBHAH=AHHC

=> AH2=BH.HC=9.16=144AH2=BH.HC=9.16=144

=> AH=12AH=12 

BC=9+16=25BC=9+16=25  

Theo Pytago:

AB=√AH2+BH2=15AB=AH2+BH2=15

AC=√AH2+HC2=20

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