question_answer

Construct \[\Delta \]PQR, if PQ = 5 cm, m \[\angle \]PQR = \[105{}^\circ \] and m \[\angle \]QRP = \[40{}^\circ \]. (Hint: Recall angle-sum property of a triangle.)

Answer:

\[PQ=5\text{ }cm,\,\,\angle PQR=105{}^\circ ,\angle QRP=40{}^\circ \]

\[\because \]Sum of angles of a triangle is 180°

\[~\therefore \angle PQR+\angle QRP+\angle QPR=180{}^\circ \]

\[\therefore 105{}^\circ +40{}^\circ +\angle QPR=180{}^\circ \]

\[\therefore 145{}^\circ +\angle QPR=180{}^\circ \]

\[\angle QPR\text{ }=\text{ }180{}^\circ 145{}^\circ \]

\[\angle QPR\text{ }=\text{ }35{}^\circ \]

Steps of Construction:

(a) Draw a line segment PQ of 5 cm.

(b) From point P draw an angle of 35°, such that

\[\angle QPX=35{}^\circ \]

(c) From point Q draw an angle of 105°, such that

\[\angle PQY\text{ }=\text{ }150{}^\circ \]

(d) These lines PX and QY cut at R.

PQR is required triangle.