Question

Perimeter of $∆$ABC is 14 cm, AB = 4.5 cm and $\angle$A = 80$°$ construct $∆$ABC.

Answer 1

$$\begin{gathered} \mathrm{Perimeter}\left(△\mathrm{ABC}\right)=14\mathrm{cm} \\ \Rightarrow \mathrm{AB}+\mathrm{BC}+\mathrm{AC}=14 \\ \Rightarrow \mathrm{BC}+\mathrm{AC}=14-4.5=9.5\mathrm{cm} \end{gathered}$$

Steps of construction:
1. Draw AB = 4.5 cm.
2. Draw ray AX, such that $\angle \mathrm{BAX}=80°$.
3. With A as centre and radius 9.5 cm, draw an arc intersecting ray AX at point D.
4. Join D and B.
5. Draw the perpendicular bisector of seg BD intersecting seg AD at point C.
6. Join C and B.
7. Hence, $△$ABC is the required triangle.