tham khảo của đỗ chí dũng câu hỏi của chi khánh

Lời giải:
 Vì $HB:HC=1:4$ nên đặt $HB=a; HC=4a$ với $a>0$

Áp dụng HTL trong tam giác vuông:
$AH^2=BH.CH$

$14^2=a.4a$

$4a^2=196$

$a^2=49\Rightarrow a=7$ (do $a>0$)

Khi đó:

$BH=a=7$ (cm); $CH=4a=28$ (cm)

$BC=BH+CH=7+28=35$ (cm)

$AB=\sqrt{AH^2+BH^2}=\sqrt{14^2+7^2}=7\sqrt{5}$ (cm)

$AC=\sqrt{AH^2+CH^2}=\sqrt{14^2+28^2}=14\sqrt{5}$ (cm)

Chu vi tam giác $ABC$:

$P=AB+BC+AC=7\sqrt{5}+14\sqrt{5}+35=21\sqrt{5}+35$ (cm)

Hình vẽ:

1: AB/AC=5/7

=>HB/HC=(AB/AC)^2=25/49

=>HB/25=HC/49=k

=>HB=25k; HC=49k

ΔABC vuông tại A có AH là đường cao

nên AH^2=HB*HC

=>1225k^2=15^2=225

=>k^2=9/49

=>k=3/7

=>HB=75/7cm; HC=21(cm)

\(HB:HC=2:3\Rightarrow\dfrac{HB}{2}=\dfrac{HC}{3}\Rightarrow HB=\dfrac{2}{3}HC\)

Áp dụng HTL:

\(AH^2=BH\cdot HC\Rightarrow24^2=\dfrac{2}{3}HC^2\Rightarrow HC^2=576\cdot\dfrac{3}{2}=864\\ \Rightarrow HC=12\sqrt{6}\left(cm\right)\\ \Rightarrow HB=\dfrac{2}{3}\cdot12\sqrt{6}=8\sqrt{6}\left(cm\right)\\ \Rightarrow BC=HB+HC=20\sqrt{6}\left(cm\right)\\ \Rightarrow S_{ABC}=\dfrac{1}{2}AH\cdot BC=\dfrac{1}{2}\cdot24\cdot20\sqrt{6}=240\sqrt{6}\left(cm^2\right)\)

\(\dfrac{HB}{HC}=\dfrac{2}{5}\\ \Rightarrow HB=\dfrac{2}{5}HC\)

Xét tam giác ABC vuông tại A
\(AH^2=BH.CH\\ \Rightarrow16^2=\dfrac{2}{5}HC.HC\\ \Rightarrow HC^2=640\\ \Rightarrow HC=8\sqrt{10}\)

\(\Rightarrow HB=\dfrac{2}{5}.8\sqrt{10}=\dfrac{16\sqrt{10}}{5}\)

\(BC=HC+HB=8\sqrt{10}+\dfrac{16\sqrt{10}}{5}=\dfrac{56\sqrt{10}}{5}\)

\(AB^2=BH.BC\\ \Rightarrow AB=\sqrt{\dfrac{16\sqrt{10}}{5}.\dfrac{56\sqrt{10}}{5}}=\dfrac{16\sqrt{35}}{5}\)

\(AC^2=CH.BC\\ \Rightarrow AC=\sqrt{8\sqrt{10}.\dfrac{56\sqrt{10}}{5}}=8\sqrt{14}\)

Chu vi : \(AB+AC+BC==8\sqrt{14}+\dfrac{56\sqrt{10}}{5}+\dfrac{16\sqrt{35}}{5}=84,28\)

\(\dfrac{HB}{HC}=\dfrac{2}{5}\Rightarrow\dfrac{HB}{2}=\dfrac{HC}{5}=\dfrac{HB.HC}{2.5}=\dfrac{AH^2}{10}=\dfrac{256}{10}=\dfrac{128}{5}\)

\(\Rightarrow HB=\dfrac{128}{5}.2=\dfrac{256}{5}\left(cm\right);HC=\dfrac{128}{5}.5=128\left(cm\right)\)

\(\Rightarrow BC=HB+HC=\dfrac{256}{5}+128=\dfrac{896}{5}\left(cm\right)\)

\(AC^2=AH^2+HC^2=256+\left(\dfrac{256}{2}\right)^2=256\left(1+\dfrac{256}{4}\right)\Rightarrow AC=16\sqrt[]{1+\dfrac{256}{4}}=16\sqrt[]{\dfrac{260}{4}}=16.\dfrac{1}{2}.2\sqrt[]{65}=16\sqrt[]{65}\left(cm\right)\)

\(AB^2=AH^2+BH^2=256+\left(\dfrac{256}{5}\right)^2=256\left(1+\dfrac{256}{25}\right)\Rightarrow AB=16\sqrt[]{1+\dfrac{256}{25}}=\dfrac{16}{5}\sqrt[]{281}\left(cm\right)\)

Chu vi tam giác ABC là : \(AB+AC+BC\)

\(=\dfrac{16}{5}\sqrt[]{281}+16\sqrt[]{65}+\dfrac{896}{5}\)

\(=16\left(\dfrac{1}{5}\sqrt[]{281}+\sqrt[]{65}+\dfrac{56}{5}\right)\)

\(=16\left(\sqrt[]{65}+\dfrac{56+\sqrt[]{281}}{5}\right)\left(cm\right)\)