According to the question,
We have to draw a triangle PQR such that QR = 3cm, ∠ PQR = 45° and QP – PR = 2 cm
Steps of construction:
1. Draw a ray OX and cut off a line segment QR = 3 cm.
2. AT Q, construction ∠YQR = 45o.
3. From QY, cut off QS = 2 cm.
4. Join RS.
5. Draw perpendicular bisector of RS to Meet QY at P.
6. Join PR. Then PQR is the required triangle.