The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm.

Solution:

Let the base and altitude of the right-angled triangle be $x$ and $y \mathrm{~cm}$, respectively.

Therefore, the hypotenuse will be $(x+2) \mathrm{cm}$.

$\therefore(x+2)^{2}=y^{2}+x^{2} \quad \ldots$ (i)

Again, the hypotenuse exceeds twice the length of the altitude by $1 \mathrm{~cm}$.

$\therefore h=(2 y+1)$

$\Rightarrow x+2=2 y+1$

$\Rightarrow x=2 y-1$

Putting the value of $x$ in (i), we get:

$(2 y-1+2)^{2}=y^{2}+(2 y-1)^{2}$

$\Rightarrow(2 y+1)^{2}=y^{2}+4 y^{2}-4 y+1$

$\Rightarrow 4 y^{2}+4 y+1=5 y^{2}-4 y+1$

$\Rightarrow-y^{2}+8 y=0$

$\Rightarrow y^{2}-8 y=0$

$\Rightarrow y(y-8)=0$

$\Rightarrow y=8 \mathrm{~cm}$

$\therefore x=16-1=15 \mathrm{~cm}$

$\therefore h=16+1=17 \mathrm{~cm}$

Thus, the base, altitude and hypotenuse of the triangle are 15 cm, 8 cm and 17 cm, respectively.