Let x = altitude (height) of the triangle
x+4 = base of the triangle
A = 16 dm²
[tex]A_\Delta = \frac{1}{2}bh \\ \\ 16 = \frac{1}{2}(x)(x+4) \\ \\ 32 = x^2+4x \\ \\ 0=x^2+4x-32 \\ \\ 0=(x+8)(x-4) \\ \\ x+8=0 ; x-4=0 \\ \\ x=-8 ; \boxed{x=4}[/tex]
Altitude = 4 dm
Base = 8 dm (from x+4)
