Question

Construct $∆$ PQR such that $\angle$P = ${70}^{°}$ , $\angle$R = ${50}^{°}$ , QR = 7.3 cm and constructs its circumcircle.

Answer 1

By angle sum property of triangles,
In $∆$PQR,
$$\begin{gathered} \angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}=180° \\ \Rightarrow 70°+\angle \mathrm{Q}+50°=180° \\ \Rightarrow \angle \mathrm{Q}=60° \end{gathered}$$
Steps of construction:
1. Draw a line QR = 7.3 cm.
2. With Q as centre, draw an angle of 60$°$ using the protractor. Similarly, draw an angle of 50$°$ with R as centre.
Where the rays RY and XQ meet, name the point as P.
Thus, $∆$PQR is obtained.
3. Draw the perpendicular bisectors of the lines PR and QR. Let these perpendicular bisectors meet at point O.
4. With O as centre and OP as radius, construct a circle touching all the vertices of the $∆$PQR.
This circle is thus the required circumcircle.