Question

Construct an isosceles right-angled triangle $\mathrm{ABC}$, where $\mathrm{m}\angle \mathrm{ACB}=90°$ and $\mathrm{AC}=6\mathrm{cm}$.

Answer 1

Given, $∆\mathrm{ABC}$

where

$\mathrm{m}\angle \mathrm{ACB}=90°$and $\mathrm{AC}=6\mathrm{cm}$.

Also triangle $\mathrm{ABC}$is an isosceles.

Now in a right angled triangle, hypotenuse is the largest side.

So, it can not be equal to any other side.

$\therefore \mathrm{BC}=\mathrm{AC}=6\mathrm{cm}$.

Let's construct it by following the steps below:

1.Draw a line segment $\mathrm{AC}=6\mathrm{cm}$.

2. Draw $90°$ from point $\mathrm{C}$.

3. Considering $\mathrm{C}$ as a center and radius $6\mathrm{cm}$, draw an arc that cuts the ray at $\mathrm{B}$.

4. Join $\mathrm{AB}$.

Thus, $∆\mathrm{ABC}$ obtained is the required right angled triangle.