[tex]\it a)\ \ \Delta ABD-dr \stackrel{T.Pitagora}{\Longrightarrow}\ BD^2=AB^2+AD^2 \Rightarrow\ BND^2=20^2+15^2=\\ \\ =400+225=625=25^2 \Rightarrow BD=25\ cm[/tex]
Fie AF - înălțimea corespunzătoare ipotenuzei în ΔDAB. Vom avea :
[tex]\it AF=\dfrac{AB\cdot AD}{BD}=\dfrac{20\cdot15}{25}=\dfrac{300}{25}=12\ cm[/tex]
[tex]\it \left.\begin{aligned}\it A'A \perp (ABD)\\ \\ \it AF \perp BD\\ \\ AF,\ BD\ \subset\ (ABD)\end{aligned}\right\}\ \stackrel{T3 \perp}{\Longrightarrow}\ A'F\perp BD\\ \\ \\ A'AF-dr.\ (\hat{A}=90^o)\ \stackrel{T.Pitagora}{\Longrightarrow}\ A'F^2=A'A^2+AF^2\ \Rightarrow\\ \\ \Rightarrow A'F^2=5^2+12^2=25+144=169=13^2 \Rightarrow A'F=13\ cm[/tex]
A'F este înălțime corespunzătoare bazei BD în ΔA'BD.
[tex]\it \mathcal{A}_{A'BD}=\dfrac{BD\cdot A'F}{2}=\dfrac{25\cdot13}{2}=\dfrac{325}{2}=162,5\ cm^2[/tex]
b) Ducem AO - înălțimea corespunzătoare ipotenuzei în ΔA'BD.
Acum, avem:
[tex]\it \left.\begin{aligned}\it AO\perp A'F\\ \\ A'F\subset (A'BD)\\ \\ OF\perp BD\\ \\ AF\perp BD\end{aligned}\right\}\ \stackrel{RT3\perp}{\Longrightarrow}\ AO\perp (A'BD) \Rightarrow d[A,\ (A'BD)]=AO[/tex]
[tex]\it d[A,\ (A'BD)]= AO=\dfrac{A'A\cdot AF}{A'F}=\dfrac{5\cdot12}{13}=\dfrac{60}{13}\approx\ 4,6\ cm[/tex]
