Question

Construct a right angle triangle $△$PQR in which PQ=6 cm, $\angle$QPR = 40$°$ and $\angle$PRQ = 90$°$. Draw circumcircle of $△$PQR.

Answer 1

$$\begin{gathered} \angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}=180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow \angle \mathrm{Q}=180°-\angle \mathrm{P}-\angle \mathrm{R} \\ \Rightarrow \angle \mathrm{Q}=180°-40°-90°=50° \end{gathered}$$
Steps of construction:
1. Draw a line segment PQ = 6 cm.
2. Draw $\angle \mathrm{PQX}=50°$.
3. Draw another angle $\angle \mathrm{QPY}=40°$. The ray PY and QX intersects at point R.
4. Then $∆$PQR is the required triangle.
5. Draw perpendicular bisectors of sides PQ and PR (any two sides) which intersect at O.
6. Taking OP or OQ or OR as the radius, draw the required circumcircle.