16. Avem AC=4√2
AB=BC (sunt laturile patratului)
Facem Pitagora in ΔABC-dr:
AC²=AB²+BC²
(cum AB=BC, putem inlocui)
AC²=AB²+AB²
(4√2)²=2AB²
32=2AB²
AB²=32:2
AB²=16
AB=√16
AB=4
P=4AB
P=4*4
P=16 cm
17. Linia mijlocie in trapez este egala cu jumatate din suma lungimilor celor doua baze. MN-baza mare, PQ-baza mica. Deci:
DE-linie mijlocie (l.m.)
DE=[tex]\frac{MN+PQ}{2}[/tex]
[tex]\frac{6}{1}[/tex]=[tex]\frac{MN+PQ}{2}[/tex]
Acum, inmultim in diagonala
(MN+PQ)*1=6*2
MN+PQ=12 cm
Deci, suma lungimilor bazelor trapezului MNPQ este de 12 cm.
18. [tex]40+\frac{25}{100} *40[/tex] (simplificam 25 si 100 prin 25)
[tex]=40+\frac{1}{4} *40\\\\=40+\frac{40}{4} \\\\=40+10\\\\=50 lei[/tex]
19.
[tex]80-\frac{10}{100} *80\\\\=80-\frac{10*80}{100} \\\\=80-\frac{800}{100} \\\\=80-8\\\\=72lei[/tex]
