Question

Draw $△$ABC in which AB = 5.5cm, $\angle$C =40$°$ and BC $-$ AC = 2.5 cm.

Answer 1

Steps of construction:
1. Draw ray BY. Mark point D, such that BD = 2.5 cm.
2. Draw ray DX, such that $\angle \mathrm{BDX}=110°$.

$$\begin{gathered} \left[\angle \mathrm{BCD}=40° \\ \mathrm{And}\angle \mathrm{CDB}=\angle \mathrm{BDC}=\frac{180°-40°}{2}\left(\because \mathrm{BC}+\mathrm{CD}\right) \\ =70° \\ \therefore \angle \mathrm{BDX}=180°-\angle \mathrm{BDC}=180°-70°=110°\right] \end{gathered}$$

3. With B as centre and radius 5.5 cm, draw an arc, intersecting ray DX at point A.
4. Join B and A.
5. Draw the perpendicular bisector of seg AD, intersecting BY at point C.
6. Hence, $△$ABC is the required triangle.