Question

Urgentt aflati restul de diagonale a cubului si fete
Bc' =l radical 2= 12 lat.

Answer 1

bc'=l√2=12

l√2=12

l=12/√2

l=12√2/2

l=6√2

d = l√3

d = 6√2√3

d = 6√6

Acub=6l² =6* 6√2*6√2 =6*36*2=6*72=432

Vcub=l³=6√2 * 6√2 *6√2 = 72*6√2= 432√2

Answer 2

BC' $= \it l\sqrt{2} = 12~cm$

l $= ~ ^{\sqrt{2} )} \it \frac{12}{\sqrt{2} } = \frac{12\sqrt{2} }{2} = 6\sqrt{2}  ~cm$

d $\it = l\sqrt{3} = \it 6\sqrt{3} \cdot \sqrt{2} = 6\sqrt{6}~cm$

$A_{t} = \it 6l^{2} = 6 \cdot (6\sqrt{2} )^{2} = 6 \cdot 72 = 432~cm^{2}$

$A_{l} = \it 4l^{2} = 4 \cdot (6\sqrt{2})^{2} = 4 \cdot 72 = 288~cm^{2}$

$V_{cub} = \it l^{3} = (6\sqrt{2} )^{3} = 216 \cdot  2\sqrt{2} = 432\sqrt{2} ~cm^{3}$