Area of triangle=1/2base*height. Let b=base, h=height, a=area, so a=1/2bh. Pythagoras says: b^2+h^2=25, so b=sqrt(25-h^2) and a=1/2h*sqrt(25-h^2). differentiating by parts, da/dh=1/2sqrt(25-h^2)+h/4(25-h^2)^-1/2.(-2h)=1/2sqrt(25-h^2)-h^2/2sqrt(25-h^2). At max or min da/dh=0, so 1/2(25-h^2)=h^2/2 and 2h^2=25, so h=5/sqrt(2). Therefore b=sqrt(25-25/2)=5/sqrt(2)=h. And the maximum area=1/2.25/2=25/4=6.25sq cm.
A simpler approach is to realise that the maximum area is when height and base are of equal length, so if x is that length, by Pythagoras, 2x^2=25 and area=1/2x^2=25/4=6.25sq cm.