Ta có: \(\dfrac{HB}{HC}=\dfrac{1}{4}=>HB=\dfrac{1}{4}HC\)
Áp dụng hệ thức lượng cho tam giác ABC ta có:
\(AH^2=HB\cdot HC\\
=>\dfrac{1}{4}HC\cdot HC=14^2=196\\
=>HC^2=196:\dfrac{1}{4}=784\\
=>HC=\sqrt{784}=28\left(cm\right)=>HB=\dfrac{1}{4}\cdot28=7\left(cm\right)\)
\(BC=HB+HC=28+7=35\left(cm\right)\)
\(AB^2=BC\cdot HB=>AB=\sqrt{BC\cdot HB}=7\sqrt{5}\left(cm\right)\)
\(AC^2=BC\cdot HC=>AC=\sqrt{BC\cdot HC}=\sqrt{35\cdot28}=14\sqrt{5}\left(cm\right)\)
\(=>C_{\Delta ABC}=AB+AC+BC=7\sqrt{5}+14\sqrt{5}+35=21\sqrt{5}+35\) (cm)
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