1. Avem unghiurile congruente:
<C≡<R
<O≡<A
<N≡<S
si laturile respectiv proportionale:
[tex] \frac{CO}{RA} = \frac{ON}{AS} = \frac{NC}{SR} [/tex]
2. [tex] \frac{20}{25} = \frac{x}{18} [/tex], de unde:
x=[tex] \frac{18*20}{25} [/tex]=[tex] \frac{18*4}{5} [/tex]=[tex] \frac{72}{5} [/tex]=14,4
3. Avem triunghiurle asemenea:
ΔFEH≈ΔJGH (cazul U.U.) deoarece:
<(FEH)≡<(JGH) (din desen)
<(FHE)≡<(JHG) (unghi comun)
4. MN=15 cm se imparte in segmentele de lungimi: a, b, c (deci a+b+c=MN) astfel incat:
[tex] \frac{a}{3} = \frac{b}{3} = \frac{c}{4} = \frac{a+b+c}{3+3+4} = \frac{15}{10} = \frac{3}{2} [/tex], deci:
a=[tex] \frac{3*3}{2} = \frac{9}{2} [/tex]=4,5 cm
b=[tex] \frac{3*3}{2} = \frac{9}{2} [/tex]=4,5 cm
c=[tex] \frac{3*4}{2} [/tex]=3*2=6 cm
5. [tex] \frac{CE}{ED} = \frac{5}{10} = \frac{1}{2} [/tex], deci:
ED=2CE, iar CE+ED=CD=10 cm, deci:
CE+2CE=10
3CE=10
CE=[tex] \frac{10}{3} [/tex] cm
ED=2*[tex] \frac{10}{3} [/tex] = [tex] \frac{20}{3} [/tex] cm
Construim CF perpendicular pe AE (F se afla pe preungirea lui AE), deci CF este distanta de la C la dreapta AE si observam ca avem triunghiurile dreptunghice asemenea:
ΔADE≈ΔCFE (U.U.), deoarece:
m(<ADE)=m(<CFE)=90 grade
<(AED)≡<(CEF) (ca unghiuri opuse la varf)
Deci avem relatiile de asemanare:
[tex] \frac{CF}{AD} = \frac{CE}{AE} [/tex]
[tex] \frac{CF}{10} = \frac{CE}{AE} [/tex]
Aflam AE cu Teorema lui Pitagora in triunghiul dreptunghic ΔADE, in care AE este ipotenuza:
[tex] AE^{2} = AD^{2} + DE^{2} [/tex]
[tex] AE^{2} = 10^{2} + ( \frac{20}{3} )^{2} [/tex]
[tex] AE^{2} = 100 + \frac{400}{9} [/tex]
[tex] AE^{2} = \frac{13*100}{9} [/tex]
AE=[tex] \frac{10 \sqrt{13} }{3} [/tex] deci:
[tex] \frac{CF}{10} = \frac{ \frac{10}{3} }{ \frac{10 \sqrt{13} }{3} } [/tex]
CF=[tex] \frac{10}{ \sqrt{13} } [/tex]
CF=[tex] \frac{10 \sqrt{13} }{13} [/tex] cm
