I have in mind a right-angled triangle ABC where C is the right angle (see picture below). Opposite angle C is side c. By Pythagoras, c^2=a^2+b^2=225+400=625 and c=25. In the triangle, a=BC=15, b=AC=20 and c=AB=25. Draw the perpendicular from C to AB and call the intersection point N. The line CN is the height of the triangle if AB is the base. Call CN side length d. Area of ABC is (1/2)AB*CN. But the area is also (1/2)AC*BC=(1/2)20*15=150. So (1/2)25*CN=150 and d=CN=300/25=12. If e=AN and f=NB, then in right-angled triangle ANC, N is the right angle. By Pythagoras, AC^2=CN^2+AN^2; 400=144+e^2, so e^2=(20-12)(20+12)=8*32=256 and e=16, so f=25-16=9.
The solution, according to this construction, is c=25, d=12, e=16 and f=9. The area of triangle ANC is (1/2)de=96, so if a perpendicular is dropped from N onto AC at M (not shown in the picture), MN is the height of ANC if AC is the base, so (1/2)b*MN=96, therefore MN=192/b=192/20=9.6. The question doesn't actually say what the letters represent, so if one of the letters c, d, e or f = MN then its length is 9.6. Other line segment lengths can be calculated using the principles already stated (Pythagoras and triangular areas). Line segments can also be calculated using the properties of similar triangles; for example, ANC is similar to ABC.
