Răspuns :
exista o formula in carte care ne spune ca diagonala la patrat = L²+l²+h² si aici inlocuim ceea ce ne da problema
100=36+16+h²⇒h²=100-36-16=48⇒h=√48=4√3
aria laterala =perimetrul bazei ×inaltimea
deci aria laterala =2(L+l)×h=2×10×4√3=80√3
aria laterala are forula ...2Ll+2Lh+2lh si inlociesc
2×6×4+2×6×4√3+2×4×4√3=48+48√3+32√3=48+80√3
V=L×l×h=6×4×4√3=96√3 ..sper ca ai inteles
100=36+16+h²⇒h²=100-36-16=48⇒h=√48=4√3
aria laterala =perimetrul bazei ×inaltimea
deci aria laterala =2(L+l)×h=2×10×4√3=80√3
aria laterala are forula ...2Ll+2Lh+2lh si inlociesc
2×6×4+2×6×4√3+2×4×4√3=48+48√3+32√3=48+80√3
V=L×l×h=6×4×4√3=96√3 ..sper ca ai inteles
Demonstratie ///
a)
Folosim formula:
L la pătrat + l la pătrat + h la pătrat rezultă:
100 = 36 + 16 + h × h
h × h = 100 - 36 - 18 = 48 => :
[tex]h = \sqrt{48 } = > h = 4 \sqrt{3} [/tex]
b) 2 (L + l) × h =>
[tex]2 \times 10 \times 4 \sqrt{3} = 80 \sqrt{3} [/tex]
c) V = L × l × h =>
[tex]6 \times 4 \times 4 \sqrt{3} = 96 \sqrt{3} [/tex]
a)
Folosim formula:
L la pătrat + l la pătrat + h la pătrat rezultă:
100 = 36 + 16 + h × h
h × h = 100 - 36 - 18 = 48 => :
[tex]h = \sqrt{48 } = > h = 4 \sqrt{3} [/tex]
b) 2 (L + l) × h =>
[tex]2 \times 10 \times 4 \sqrt{3} = 80 \sqrt{3} [/tex]
c) V = L × l × h =>
[tex]6 \times 4 \times 4 \sqrt{3} = 96 \sqrt{3} [/tex]