The area of a right-angled triangle is 20 sq.cm and one of the sides containing the right angle is 4 cm. The altitude on the hypotenuse is :

Let the length of the other side containing the right angle be x cm
Then,
$$\eqalign{ & \frac{1}{2} \times 4 \times x = 20 \cr & \Rightarrow x = 10{\text{ cm}} \cr} $$
Hypotenuse :
$$\eqalign{ & = \sqrt {{{10}^2} + {4^2}} {\text{ cm}} \cr & = \sqrt {116} {\text{ cm}} \cr & = 2\sqrt {29} \,{\text{cm}} \cr} $$
Let the altitude on the hypotenuse be h cm
Then,
$$\eqalign{ & \frac{1}{2} \times 2\sqrt {29} \times h = 20 \cr & \Rightarrow h = \frac{{20}}{{\sqrt {29} }}{\text{ cm}} \cr} $$