Răspuns :
[tex]\it AC=\ell\sqrt2=10\sqrt2\Rightarrow \ell=10\ cm\\ \\ Raza\ discului\ este\ R=\dfrac{\ell}{2}=\dfrac{10}{2}=5\ cm\\ \\ \mathcal{A}_{disc}=\pi R^2=\pi\cdot5^2=25\pi\ cm^2\\ \\ AM=\dfrac{AC-2R}{2}=\dfrac{10\sqrt2-10}{2}=\dfrac{10(\sqrt2-1)}{2}=5(\sqrty2-1)\ cm[/tex]
Răspuns: d=l√2=10√2 ; l=10 ; raza=10/2=5 ; A=πR²=25π ; AM=(10√2-10)/2
Explicație pas cu pas: