Notam cu AD inaltimea
intr-un triunghi dreptunghic aria se poate calcula in 2 moduri
(produsul catetelor) / 2
sau (ipotenuza ×inaltimea )/2
[tex]A= \frac{AB*AC}{2} \\ \\ A= \frac{BC*AD}{2} \\ \\39= \frac{13*AD}{2}\\ \\AD= \frac{2*39}{13} \\ \\AD=6cm [/tex]
perimetrul=AB+BC+AC
Teorema lui Pitagora
BC²=AB²+AC²
13²=AB²+AC²
169=AB²+AC²
(AB+AC)²=AB²+AC²+2AB×AC ⇒(AB+AC)²=169+2AB×AC
dar A=(AB×AC)/2 deci AB×AC=2×A=2×39=78
(AB+AC)²=169+2AB×AC devine
(AB+AC)²=169+2×78
(AB+AC)²=169+156
[tex] (AB+AC)= \sqrt{325} =5\sqrt{13} [/tex]
perimetrul= AB+AC+BC= 5√13+13 cm
