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The hypotenuse of a right angled triangle is \( 5 \mathrm{~cm} \) and the other two sides differ by \( 1 \mathrm{~cm} \). Find the other two sided of the triangle.
Given:
The hypotenuse of a right angled triangle is \( 5 \mathrm{~cm} \) and the other two sides differ by \( 1 \mathrm{~cm} \).
To do:
We have to find the other two sides of the triangle.
Solution:
Let the other two sides of the triangle be $x$ cm and $x-1$ cm.
Therefore, using Pythagoras theorem, we get,
$x^2+(x-1)^2=5^2$
$x^2+x^2+1-2x=25$
$2x^2-2x+1-25=0$
$2x^2-2x-24=0$
$x^2-x-12=0$
$x^2-4x+3x-12=0$
$x(x-4)+3(x-4)=0$
$(x+3)(x-4)=0$
$x=4$ or $x=-3$ which is not possible because the length cannot be negative.
$\Rightarrow x=4$
$\Rightarrow x-1=4-1=3$
The other two sides of the triangle are $3\ cm$ and $4\ cm$.