Question

The hypotenuse of a right-angled triangle exceeds one side by $1cm$ and the other side by $18cm$ find the lengths of the hypotenuse of the triangle.

Answer 1

Let the hypotenuse of a

right angled $△$ be x cm

Base = (x-1) cm

Perpendicular = (x-18) cm

By Pythagoras Theorem,

$(Hypotenuse{)}^2=(Base{)}^2+(Perp.{)}^2$

$(x{)}^2=(x-1{)}^2+(x-18{)}^2$

$x^2=x^2+1-2x+x^2+324-36x$

$x^2=2x^2-38x+325$

$x^2-38x+325$

using Quadratic formula

$x=\frac{38\pm \sqrt{1444-1300}}{2}$

$=\frac{38\pm 12}{2}$

x = 25 cm or x = 13 cm

As side can't be negative

$x\ne 13cm$

So, x= 25 cm