Question

va rog mult sa ma ajutati la cele 2 probleme,multumesc frumos

Answer 1

1) a+b+c=1032
b=3a
c= a/2 + 60

a+3a +a/2 +60 =1032
4a +a/2 = 972
(8a+a)/2 =972
9a= 1944
a=1944/9
⇒a = 216
   b= 3×216 = 648
   c= 216/2 +60 ⇒c=168

2)
a) Trapezul ABCD , mas (D) = mas(A) = 90 grade
   m(C) = 360 - (90+90+45) = 135 grade

b) Ducem CM perpendicular pe AB,  (CM = AD=2 cm)
   In Δ BCM, m(M) = 90 grade, m(B) = 45 grade ⇒m(C) = 45 grade
   rezulta ca Δ BCM este isoscel (CM=MB= 2 cm)

   BC= ipotenuza
   BC² = MB² +CM²
   BC²= 2²+2²
   BC² = 8
   BC = 2√2 cm

c) A = (B+b)/2 × h
   h=AD=2 cm
   B= AB = AM +MB = 4 + 2 = 6 cm
   b = DC = 4 cm
  
A= (6+4)/2 ×2 = 10 cm²
P = AB +AD+DC+BC = 6+2+4+2√2 = 12 +2√2  cm