Question
Construct ∆PQR, in which PQ – PR = 2.4 cm, QR = 6.4 cm and ∠PQR = 55°.
✓
Answer
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:
i. Draw seg QR of length 6.4 cm.
ii. Draw ray QT, such that ∠RQT = 55°.
iii. Take point S on ray QT such that l(QS) = 2.4 cm.
iv. Join the points S and R.
v. Draw perpendicular bisector of seg SR intersecting ray QT.
Name that point as P.
vi. Join the points P and R.
Hence, ∆PQR is the required triangle.
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