Question

Va rog, cine poate sa mi rezolve acest exercitiu!

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\it Dac\breve a\ not\breve am\ Ab=x,\ atunci\ BC=x\sqrt2\\ \\ Th.\ Pitagora\ \Rightarrow\ 3x^2=72|_{:3} \Rightarrow x^2=24=4\cdot6=2^2\cdot6 \Rightarrow x=2\sqrt6\\ \\ AB=x=2\sqrt6\ dam,\ \ BC=x\sqrt2=2\sqrt6\cdot\sqrt2=2\sqrt{12}=2\sqrt{4\cdot3}=4\sqrt3dam$

$\it \mathcal{A}=\dfrac{c_1\cdot c_2}{2}=\dfrac{AB\cdot BC}{2}=\dfrac{2\sqrt6\cdot4\sqrt3}{2}=4\sqrt{18}=4\sqrt{9\cdot2}=12\sqrt2dam^2(ari)\\ \\ \\ \mathcal{A}^2=(12\sqrt2)^2=288<289=17^2 \Rightarrow \mathcal{A}<17<34ari$