Question

de la 5 la 10 repede va rog frumos​

Answer 1

1)Detali:Pb=18√3 Pf=28√3

a)l=18√3/3=6√3cm

b)m=(28√3:2)-6√3=14√3-6√3=8√3cm (muchia laterala)

c)Ab=l²√3/4=(6√3)²√3/4=27√3cm²

d)Af=L·l=8√3·6√3=144cm²

2)Detali:Ab=72cm² Af=72√3cm²

a)Pb=patrat

l=√72=6√2cm

b)m=72√3/6√2=72√6/12=6√6cm

c)Pb=4·6√2=24√2

d)Pf=(6√2+6√6)·2=(12√2+12√6)cm

e)In ΔABD, m(∡A)=90°⇒BD²=AB²+AD²

BD²=(6√2)²+(6√2)²=72+72=144

BD=12

f)InΔC1CB,m(∡C)=90°⇒BC1²=CC1²+BC²

BC1²=(6√6)²+(6√2)²=216+72=288

BC1²=12√2 (la demonstratii si la exercitii de genu ,se foloseste teorema celor trei perpendiculare ca introducere)

3)Detalii:corpul este o prisma triunghiulara regulata ,Ab=12√3cm si dm(diagonala unei fete laterale)=8cm

a)Ab=l²√3/4

l=√[(12√3·4):√3]=√(48√3:√3)=√48=4√3cm

b)Pb=4√3·3=12√3cm

c)In ΔBB1C ,m(∡B)=90°⇒BB1²=B1C²-BC²

BB1²=8²-(4√3)²=64-48=16

BB1=4cm

c)Pf=(4+4√3)·2=(8+8√3)cm

d)Af=4·4√3=16√3 cm²