Răspuns :
Hey !
Răspuns:
a)
G - centru de greutate => GT = 1/3 * AT
T - mijloc BC => BT=TC =20/2 = 10 cm
Δ ABT - Δ dreptunghic ⇒ TP => AB²=AT² + BT²
AB²=24²+10²=576+100=676 => AB=√676=26 cm
P ΔABC = AB+BC+AC = 26+26+20 = 52 + 20 = 72 cm
b)
fie E - mijloc AC => AE=EC
S=sim E fata de G => GE=ES
=> AGCS - paralelogram : AG ║ SC , AG=SC
AS ║ GC , AS=GC
AT ⊥ BC , G∈ AT => AG⊥BC } => SC ⊥ BC
G-centru de greutate => GE=1/3 * BE , BG=2/3 * BE
GS=2*1/3 BE => GS=2/3*BE
BG=GS =>G-mijloc BS , GC-mediana in Δ SCB
Aria ΔSGC = 1/2 * Aria ΔSBC
Aria ΔSBC = (16*20) / 2 = 320/2 = 160 cm²
Aria ΔSGC = 80 cm²
ΔSCB , ∡C = 90° , GC - mediana => Teorema medianei : GC=SB/2 => GC=BG
ΔGTB , ∡T=90° => TP : GB²=GT²+BT² => 64+100=164 => GB=√164=2√41 => GC=2√41 cm
Aria ΔSCB = GC * d(S,CG) / 2 = 80= (2√41 * d(S,CG)) / 2 = > √41 * d(S,CG) = 80 => d(S,CG) = 80/√41 = 80√41/ 41 => d(S,CG) = 80√41 / 41