The area of a right-triangle is 600 cm2. If the base of the triangle exceeds the altitude by 10 cm,

Solution:

Let the altitude of the triangle be x cm.
Therefore, the base of the triangle will be (x + 10) cm.

Area of triangle $=\frac{1}{2} x(x+10)=600$

$\Rightarrow x(x+10)=1200$

$\Rightarrow x^{2}+10 x-1200=0$

$\Rightarrow x^{2}+(40-30) x-1200=0$

$\Rightarrow x^{2}+40 x-30 x-1200=0$

$\Rightarrow x(x+40)-30(x+40)=0$

$\Rightarrow(x+40)(x-30)=0$

$\Rightarrow x=-40$ or $x=30$

$\Rightarrow x=30 \quad[\because$ Altitude cannot be negative $]$

Thus, the altitude and base of the triangle are $30 \mathrm{~cm}$ and $(30+10=40) \mathrm{cm}$, respectively.

$\because$ (Hypotenuse) $^{2}=$ (Altitude) $^{2}+$ (Base) $^{2}$

$\Rightarrow(\text { Hypotenuse })^{2}=(30)^{2}+(40)^{2}$

$\Rightarrow(\text { Hypotenuse })^{2}=900+1600=2500$

$\Rightarrow(\text { Hypotenuse })^{2}=(50)^{2}$

$\Rightarrow($ Hypotenuse $)=50$

Thus, the dimensions of the triangle are:

Hypotenuse $=50 \mathrm{~cm}$

Altitude $=30 \mathrm{~cm}$

Base $=40 \mathrm{~cm}$