Question

Construct ΔPQR, in which PQ - PR = 2.4 cm, QR = 6.4 cm and ∠PQR = 55°. 

Answer 1

Rough figure:

Explanation:

Here, PQ – PR = 2.4 cm

∴ PQ > PR

As shown in the rough figure, draw seg QR = 6.4 cm

Draw a ray QT making on angle of 55° with QR

Take a point S on ray QT, such that QS = 2.4 cm.

Now, PQ – PS = QS       ...[Q-S-P]

∴ PQ – PS = 2.4 cm       …(i)

Also, PQ – PR = 2.4 cm       ...(ii) [Given]

∴ PQ – PS = PQ – PR        ...[From (i) and (ii)]

∴ PS = PR

∴ Point P is on the perpendicular bisector of seg RS.

∴ Point P is the intersection of ray QT and the perpendicular bisector of seg RS.

​Steps of construction: 

1. Draw seg QR of length 6.4 cm.
2. Draw ray QT, such that ∠RQT = 55°.
3. Take point S on ray QT such that l(QS) = 2.4 cm.
4. Join the points S and R.
5. Draw perpendicular bisector of seg SR intersecting ray QT. Name that point as P.
6. Join the points P and R.

Therefore, △PQR is required triangle.