Question

Ajutati-ma va rog . ​

Answer 1

$\it T.h \Rightarrow AD^2=BD\cdotDC=8\cdot18=144=12^2 \Rightarrow AD=12\ cm\\ \\ BC=BD+DC=8+18=26\ cm\\ \\ \Delta BDA-dr,\ \hat D=90^o,\ \stackrel{T.P.}{\Longrightarrow}\ AB^2=AD^2+BD^2=12^2+8^2=144+64=208\\ \\ AB=\sqrt{208}= \sqrt{16\cdot13}=4\sqrt{13}\ cm$

$\Delta ADC-dr,\ \hat D=90^o,\ \stackrel{T.P.}{\Longrightarrow }\ AC^2=AD^2+DC^2=12^2+18^2=144+324=468\\ \\ AC=\sqrt{468}=\sqrt{36\cdot13}=6\sqrt{13}\ cm$

$\it \mathcal{P}_{ABC}=AB+AC+BC=4\sqrt{13}+6\sqrt{13}+26=10\sqrt{13}+26\ cm$