Question

If the sides of a right triangle are consecutive natural numbers, then the length of the shortest side is ___.

• A. 3

Answer 1

The correct option is A 3
Let the lengths of the sides of the right-triangle be $x-1,x$ and $x+1.$


Using Pythagoras' theorem, we have
$(x+1{)}^2=(x-1{)}^2+x^2.$
$\Rightarrow x^2+2x+1=x^2-2x+1+x^2$
$\Rightarrow x^2-4x=0$

The roots of a quadratic equation $a{x}^2+bx+c=0,$ where $a,b$ and $c$ are constants are given by

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$

Here, $a=1,b=-4$ and $c=0.$

$\Rightarrow x=\frac{-(-4)\pm \sqrt{(-4{)}^2-4(1\left)\right(0)}}{2\times 1}$

$\Rightarrow x=\frac{4\pm \sqrt{16}}{2}$

$\Rightarrow x=4,0$

Thus, the roots of the equation are 4 and 0.
But, since length cannot be zero, we have $x=4\text{units.}$

$\therefore x-1=3\text{units},x+1=5\text{units.}$

Thus, the length of the shortest side is $x-1=3\text{units.}$