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the base of a triangle is 4 dm longer than the altitude and its area is 16 dm^2. find the length of the base and the altitude.
thanks please show your solution ..


Sagot :

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)
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