Two circles of radii 20 cm and 5 cm, respectively, touch each other externally at the point P, AB is the direct common tangent of those two circles of centres R and S, respectively. The length of AB is equal to:

AB is tangent to both the circles.

$$\Rightarrow$$  AB$$\bot\ $$AR

Let SC be the line parallel to AB

$$\Rightarrow$$  SC$$\bot\ $$AR and AC = BS = 5 cm and AB = SC

Radius of bigger circle AR = 20 cm

$$\Rightarrow$$  AC + CR = 20

$$\Rightarrow$$  5 + CR = 20

$$\Rightarrow$$  CR = 15 cm

Since circles touch externally as shown in figure

RS = RP + PS = 20 + 5 = 25 cm

In $$\triangle$$RCS,

CR$$^2$$ + SC$$^2$$ = RS$$^2$$

$$\Rightarrow$$  15$$^2$$ + SC$$^2$$ = 25$$^2$$

$$\Rightarrow$$  225 + SC$$^2$$ = 625

$$\Rightarrow$$  SC$$^2$$ = 400

$$\Rightarrow$$  SC = 20 cm

$$\therefore\ $$Length of AB = SC = 20 cm

Hence, the correct answer is Option D