a) AD⊥BC⇒AD-h (inaltime)
Aria ΔABC=[tex]\frac{b*h}{2} =\frac{BC*AD}{2} =\frac{8*5}{2} =\frac{40}{2} =20cm^2[/tex]
b) Fie d(B,AC)=BP, BP⊥AC, P∈AC⇒BP-h
O sa aflam BP cu aria in doua moduri, cu bazele si inaltimile corespunzatoare:
Aria ΔABC=[tex]\frac{b*h}{2} =\frac{BC*AD}{2} =\frac{8*5}{2} =\frac{40}{2} =20cm^2[/tex] din punctul a)
*O vom inlocui*
AriaΔABC=[tex]\frac{b*h}{2}[/tex]
20=[tex]\frac{AC*BP}{2}[/tex]
20=[tex]\frac{10*BP}{2}[/tex]
20=5BP
BP=20:5
BP=4 cm
