Question

Construct a $△ABC$ in which: $AB=5.4cm$, $\angle CAB={45}^0$ and $AC+BC=9cm$.

Then the length of $AC$ (in $cm.$) is:

Answer 1

1. Draw $AB$ of $5.4cm.$

2. Construct angle ${45}^{\circ }$ at $A$.

3. With radius as $9cm$ at $A$, cut an arc along the ${45}^{\circ }$ line and mark the point as $D$. Join $BD$.

4. Draw a perpendicular bisector of $BD$ and mark its intersection with $AD$ as $C$.

5. Join $AC$.

6. $ABC$ is the required triangle.