Gọi \(AB=x\) có \(\dfrac{AB}{AC}=\dfrac{20}{21}=\dfrac{x}{AC}=>AC=\dfrac{21.x}{20}\)
ΔABC vuông có AH là đường cáo \(=>\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}=>\dfrac{1}{420^2}=\dfrac{1}{x^2}+\dfrac{1}{\left(\dfrac{21.x}{20}\right)^2}\)
\(=>x=580=AB=>AC=609=>BC=\sqrt{AB^2+AC^2}=841\)
=> Chu vi ΔABC là \(\dfrac{AB+AC+BC}{2}=1015\)
5) \(\dfrac{AB}{AC}=\dfrac{20}{21}\Rightarrow AC=\dfrac{21}{20}AB\)
ΔABC vuông tại A, AH là đường cao.
\(\Rightarrow\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\)
\(\Rightarrow\dfrac{1}{AB^2}+\dfrac{1}{\left(\dfrac{21}{20}AB\right)^2}=\dfrac{1}{420^2}\)
\(\Rightarrow\dfrac{1}{AB^2}+\dfrac{1}{AB^2}.\dfrac{20^2}{21^2}=\dfrac{1}{420^2}\)
\(\Rightarrow\dfrac{1}{AB^2}.\dfrac{20^2+21^2}{21^2}=\dfrac{1}{420^2}\)
\(\Rightarrow\dfrac{1}{AB^2}=\dfrac{1}{580^2}\)
\(\Rightarrow AB=580\left(cm\right)\)
\(\Rightarrow AC=\dfrac{21}{20}AB=\dfrac{21}{20}.580=609\left(cm\right)\)
\(BC^2=AB^2+AC^2\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{580^2+609^2}=841\left(cm\right)\)
\(P_{ABC}=AB+AC+BC=580+609+841=2030\left(cm\right)\)
