In $ \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} $. If $ \mathrm{AC}-\mathrm{BC}=4 $ and $ \mathrm{BC}-\mathrm{AB}=4 $, find all the three sides of $ \triangle \mathrm{ABC} $.

Given:

In \( \triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ} \). If \( \mathrm{AC}-\mathrm{BC}=4 \) and \( \mathrm{BC}-\mathrm{AB}=4 \)

To do:

We have to find all the three sides of \( \triangle \mathrm{ABC} \).

Solution:

$BC=AC-4$

$AB=BC-4=AC-4-4=AC-8$

By Pythagoras theorem,

$AC^2=AB^2+BC^2$

$AC^2=(AC-8)^2+(AC-4)^2$

$AC^2=AC^2+64-16AC+AC^2+16-8AC$

$AC^2-24AC+80=0$

$AC^2-20AC-4AC+80=0$

$AC(AC-20)-4(AC-20)=0$

$AC-4=0$ or $AC-20=0$
$AC=4$ or $AC=20$
AC cannot be 4

This implies,

$AC=20$

$AB=20-8=12$

$BC=20-4=16$