Question

DAU COROANA VA ROG!!!!​

Answer 1

$\it a)\ \ \Delta ABC-dr.\ (m(\hat B)=90^o), \stackrel{T.P.}{\Longrightarrow}\ BC^2=AC^2-AB^2 \Rightarrow \\ \\ \Rightarrow BC^2=(16\sqrt2)^2-16^2=256\cdot2-256=256=16^2 \Rightarrow BC=16\ m\\ \\ AB=BC=16m \Rightarrow ABCD-p\breve atrat \Rightarrow A'B'C'D'-\ p\breve atrat\\ \\ \mathcal{A}_b=16^2=256m^2,\ \ BB'=h=10m\\ \\ \mathcal{V}=\mathcal{A}_b\cdot h=256\cdot10=2560m^3=2560000dm^3=2560000\ell$

b)

$\it \left.\begin{aligned} \it AA'\perp(A'B'C')\ \ \ \ \ \ \ \ .\\ \\ \it A'O'\perp B'D'\ \ \ \ \ \ \ \ \ \ \ \ .\\ \\ \it A'O',\ B'D'\subset(A'B'C') \end{aligned} \right\} \stackrel{T3\perp}{\Longrightarrow} AO'\perp B'D' \Rightarrow d(A,\ B'D')=AO'$

$\it AA'=BB'=10m,\ \ A'O'=\dfrac{A'C'}{2}=\dfrac{AC}{2}=\dfrac{16\sqrt2}{2}=8\sqrt2m\\ \\ \\ \Delta AA'O\ -dr.\ (m(\hat A)=90^o) \stackrel{T.P.}{\Longrightarrow} (AO')^2=(AA')^2+(A'O')^2=\\ \\ \\ =10^2+(8\sqrt2)^2=100+128=228 \Rightarrow AO'=\sqrt{228}\ \approx\sqrt{225}=15m$

c)

Fie h' înălțimea la care se ridică apa în bazin.

$\it 2048000\ \ell = 2048000\ dm^3=2048m^3\\ \\ \mathcal{A}_b\cdot h'=2048 \Rightarrow 256\cdot h'=2048|_{:256} \Rightarrow h'=8m$