a)A triunghi = [tex] \frac{bc \times h}{2} = \frac{(2 \sqrt{10} - 4) \times (\sqrt{10} + 2)}{2} = \frac{2 \sqrt{10} \times \sqrt{10} + 2 \times 2 \sqrt{10} - 4 \sqrt{10} - 8 }{2} = \frac{2 \times 10 + 4 \sqrt{10} - 4 \sqrt{10} - 8}{2} = \frac{20 - 8}{2} = \frac{12}{2} = 6cm^{2} [/tex]
b)Fie AD inaltime in ∆ABC, AD€[BC]=>AD mediana si AD_|_BC
AD mediana=>BD=DC=BC/2=(2√10-4)/2=√10-2
AD_|_BC=>∆ADC dreptunghic=>AD^2+DC^2=AC^2
(√10+2)^2+(√10-2)^2=AC^2
10+4√10+4+10-4√10+4=AC^2
10+4+10+4=AC^2
28=AC^2
AC=√28=2√7=AB
P=AB+AC+BC=4√7+2√10-4
