6
punti
Livello 0
AC(20/3-1) = 17AC/3 = 68
AC = 68/17*3 = 12 cm
BD = 68+12 = 80 cm
lato AB = √(BD/2)^2+(AC/2)^2 = √40^2+6^2 = √1636 = 40,4475 cm
perimetro 2p = AB*4 = 161,790 cm
area A = AC*BD/2 = 40*12 = 480 cm^2
142442
punti
Genius III
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Differenza diagonali $\small = BD-AC= 68\,cm;$
rapporto tra le diagonali $\small = \dfrac{20}{3};$
diagonale maggiore $\small D (BD)= \dfrac{68}{20-3}×20 = \dfrac{68}{17}×20 = 4×20 = 80\,cm;$
diagonale minore $\small d (AC)= \dfrac{68}{20-3}×3 = \dfrac{68}{17}×3 = 4×3 = 12\,cm;$
lato $\small l= \sqrt{\left(\dfrac{D}{2}\right)^2+\left(\dfrac{d}{2}\right)^2}= \sqrt{\left(\dfrac{80}{2}\right)^2+\left(\dfrac{12}{2}\right)^2}= \sqrt{40^2+6^2} = 40,4475\,cm$ (teorema di Pitagora);
area $\small A= \dfrac{D×d}{2} = \dfrac{80×\cancel{12}^6}{\cancel2_1} = 80×6 = 480\,cm^2;$
perimetro $\small 2p= 4×l = 4×40,4475 \approx{161,8}\,cm.$
68277
punti
Genius I
