Question

Va rog ajutatima la exercitiile 6, 7, 8, 9, 10. Cu figura. Imi trebuie acum, va rog mult! URGENT​

Answer 1

6) P = 90 = 2L + 2l

avem ca L = 2l

inlocuim

2·2l + 2l = 90

4l + 2l = 90

6l = 90

l = 90 : 6

l = 15 cm

L = 2l = 2·15

L = 30 cm

A = 15×30 = 450 cm²

7) A = L² = 36 cm²

L = √36 = 6 cm

P = 4L = 4·6 = 24 cm

8) A(trapez) = Lm×h = (B+b):2×h

A = (AB+CD):2×AD

A = (20cm+12cm):2×8cm

A = 32cm:2×8cm

A = 16cm×8cm

A = 128 cm²

9) A = l×h : 2

l = P/3 = 18cm/3 = 6 cm

A = 6cm×3√3cm/2 = 3cm×3√3cm = 9√3 cm²

10) intr-un triunghi isocel dreptunghic inaltimea corespunzatoare ipotenuzie are jumatate din valoarea ipotenuzei

h = Ip/2 = 10cm/2 = 5 cm

A = Ip×h/2

A = 10cm×5cm/2

A = 50cm²/2

A = 25cm²

Answer 2

6)

P = 2L + 2l = 90

L = 2l

2×2l + 2l = 90

6l = 90

l = 15 cm

L = 30 cm

A = 15×30 = 450 cm²

7)

A = 36 = l²

l² = 36 => l = √36 = 6 cm

P = 4l = 24 cm

8)

A = (AB+CD)/2×AD

= (20+12)/2×8

= 32×4

= 128 cm²

9)

P = 18 cm => L = 18/3 = 6 cm

A = L×h/2 = 3√3×6/2 = 18√3/2 = 9√3 cm²

10)

CnfT45° => h = ip/2 = 5 cm

A = Ip×h/2

= 10×5/2

= 50/2

= 25 cm²

=