Q5.The altitude of right triangle is 7 cm less than its base. If, hypotenuse is 13 cm. Find the other two sides.

Let the base of the right triangle be $$x$$.
The altitude can be represented by $$x - 7$$.
In a right triangle,
$$Hypotenuse^2 = Base^2 + Altitude^2$$ (Pythagoras Theorem)
$$13^2 = x^2 + (x - 7)^2$$
$$169 = x^2 + x^2 - 14x + 49$$
$$2x^2 - 14x - 120 = 0$$
$$x^2 - 7x - 60 = 0$$
(Splitting -7x as 5x - 12x )
$$x^2 + 5x - 12x - 60 = 0$$
$$x(x + 5) - 12(x + 5) = 0$$
$$(x - 12)(x + 5) = 0$$
The roots of this equation are the values of x for which $$(x - 12)(x + 5) = 0$$
which are,
$$x + 5 = 0$$ or $$x - 12 = 0$$
Thus, $$x = -5$$ or $$x = 12$$

Since the length of any side cannot be negative, we reject $$x = -5$$.
Thus, the length of the base is 12 cm and the length of the altitude is 12 - 7 = 5 cm.