Question

"Question 3 Construct a triangle PQR in which QR = 6 cm, ∠Q = 60° and PR − PQ = 2 cm

Class 9 - Math - Constructions Page 195"

Answer 1

[ fig. is in the attachment]

Given: 

In ∆PQR, QR= 6 cm, PR-PQ=2 cm & ∠Q= 60°

Steps of construction:

1. Draw the base QR=6 cm.

At Point Q draw a ray QX making an  ∠XQR=60°

Here , PR -PQ= 2cm

PR>PQ

The side containing the base angle Q is less than third side.

 

2. Cut the line segment QS equal to PR-PQ=2 cm , from the ray QX extended on  opposite side of base QR.

3. Join SR and draw its perpendicular bisector ray AB which intersect SR at M.

4. Let P be the intersection point of SX and perpendicular bisector AB. Then join PR.

Thus ∆PQR is the required Triangle.

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Answer 2

Steps of construction:-
1)Draw a ray QX and cut off a line segment QR =6cm from it.

2))Construct a ray QY making an angle of 60 degree with QR and produce YQ To form a line YQY.

3)Cut off a line segment QS=2cm from QY.

4)Join RS

5(Draw perpendicular bisectors of Rs intersecting QY at a point P.

6)Join PR ..

Then, PQR is the required ∆.

Hope helped!!!