Question

Exercitiul 1 va rog!!

Answer 1

a)
ducem FM inaltime in triunghiul echilateral ABE
inaltimea in triunghiul echilateral= L√3/2
FM=6√3/2 =3√3
A(ABE)=6× 3√3/2=9√3 cm²

b)
AD=FM=3√3
AC=√[9²+(3√3²]=√(81+27)=√108=6√3 cm

c)
DF=AM=3 cm
FC=9-3=6 cm
EB=9-6=3 cm
triunghiurile ADF si EBC sunt congruente    caz  CC
adica AD=BC=3√3 cm
DF=EB=3 cm
rezulta ca si EC=AF
FCIIAE                            rezulta ca AFCE este romb si in romb diagonalele sunt perpendiculare deci AC_I_EF

Answer 2

[tex]a.\\\·AEF-echilateral=\ \textgreater \ A _{\·AEF}=\frac{l^2 \sqrt{3} }{4} = \frac{AE^2 \sqrt{3} }{4} = \frac{36 \sqrt{3} }{4}= \9 \sqrt{3} ~cm\\\\ b.\\ \·ABC-dreptunghic\\ =\ \textgreater \ T.P=\ \textgreater \ BC^2+AB^2=AC^2\\ Dar, nu ~il ~stim~pe ~AD.=\ \textgreater \ \\ Ducem~ h ~din ~F=\ \textgreater \ FN= \frac{l \sqrt{3} }{2}= \frac{6 \sqrt{3} }{2}=3 \sqrt{3} cm\\ =\ \textgreater \ FN=AD\\ AC^2= 9^2+(3 \sqrt{3})^2\\ AC^2=81+27\\ AC^2=108\\ AC= \sqrt{108}\\ AC= 6\sqrt{3} \\ [/tex]