Chapter 3.7, Problem 11PSA

To determine

We need to find whether $\Delta ABC$ is isosceles or not

Explanation of Solution

Given information: $\overline{AC}\perp \overline{BC}$ , $\angle C={\left(3x\right)}^{\circ }$ , $BC=x+20$ and $AC=2x-20$ is provided in the question which will help to find the solution.

Proof: Here, $\angle C={90}^{\circ }$

  →${\left(3x\right)}^{\circ }={90}^{\circ }$

  →$x={30}^{\circ }$

And, $BC=30+20=50$

  $AC=60-20=40$

So that, $AC\ne BC$

Hence, $\Delta ABC$ is not isosceles.