a)
[tex]\it Ducem\ AD\perp BC.\ \ Not\breve am\ BD=x\ \Rightarrow CD=28-x.\\ \\ \Delta DAB,\ Th.\ Pitagora \Rightarrow AD^2=AB^2-BD^2=26^2-x^2\ \ \ \ (1)\\ \\ \Delta DBC,\ Th.\ Pitagora \Rightarrow AD^2=AC^2-CD^2=30^2-(28-x)^2\ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow 26^2-x^2=30^2-28^2+56x-x^2\Rightarrow 56x=676+784-900\Rightarrow \\ \\ \Rightarrow 56x=560\Rightarrow x=10\\ \\ AD^2=26^2-x^2=26-10^2=(26-10)(26+10)=16\cdot36=4^2\cdot6^2\Rightarrow \\ \\ \Rightarrow AD=4\cdot6=24\ cm[/tex]
b)
[tex]\it \mathcal{A}_{ABC}=\dfrac{BC\cdot AD}{2}=\dfrac{28\cdot24}{2}=14\cdot24=336\ cm^2[/tex]
c)
[tex]\it \mathcal{A}_{ABC} =\dfrac{AC\cdot d(B,\ AC)}{2} \Rightarrow 336=\dfrac{30\cdot d(B,\ AC)}{2} \Rightarrow 336=15\cdot d(B,\ AC) \Rightarrow \\ \\ \\ \Rightarrow d(B,\ AC)=\dfrac{\ 336^{(3}}{15}=\dfrac{^{2)}112}{\ 5}=\dfrac{224}{10}=22,4\ cm[/tex]
