Question

Fie N mijlocul muchiei [D'C'] in cubul ABCDA'B'C'D'. Daca aria triunghiului NAB este 8√2 cm², calculati:
a) aria totala si volumul cubului;

b) aria triunghiului ANC.

Answer 1

a)

ΔNAB isoscel, NA=NB

A=8√2 cm²

Fie NN'⊥AB

Notam AB=a

$A_{NAB}=\frac{NN'\times AB}{2}\\\\ 8\sqrt{2}=\frac{NN'\times a}{2} \\\\NN'=\frac{16\sqrt{2} }{a}$

Fie NM⊥DC

NM=N'M=a

In ΔNN'M, avem Pitagora

NN'²=MN'²+NM²

NN'²=a²+a²

$\frac{512}{a^2} =2a^2\\\\512=2a^4\\\\256=a^4\\\\a=4\ cm$

a= 4 cm

$A_t=6a^2=6\times 16=96\ cm^2\\\\V=a^3=64\ cm^3$

b)

AC=a√2=4√2 cm

NC²=NC'²+CC'²

NC²=4+16=20

NC=2√5 cm

$NN'=\frac{16\sqrt{2} }{a} =4\sqrt{2} \ cm$

In ΔANN' dr in N'

AN²=NN'²+AN'²

AN²=32+4=36

AN=6 cm

NM⊥DC

DC⊂(ABCD)

MO⊥AC

AC⊂(ABCD)⇒ NO⊥AC

MO=2 cm

• In ΔNOM aplicam Pitagora

NO²=NM²+OM²

NO²=16+4=20

NO=2√5 cm

$A_{ANC}=\frac{NO\times AC}{2} =\frac{2\sqrt{5} \times 4\sqrt{2} }{2} =4\sqrt{10} \ cm^2$