Question

Construct PQR, such that QR = 6.5 cm, ∠ PQR= 40o and PQ – PR = 2.5 cm ​

Answer 1

Step-by-step explanation:

Here, PQ – PR = 2.5 cm 

∴ PQ > PR  As shown in the rough figure draw seg  QR = 6.5 cm .

Draw a ray QT making on angle of 40° with QR . Take a point S on ray QT, such that QS = 2.5 cm.  Now, PQ – PS = QS [Q - S - T] 

∴ PQ – PS = 2.5 cm ……(i) [Given] 

Also, PQ – PR = 2.5 cm …..(ii) [From (i) and (ii)] 

∴ PQ – PS = PQ – PR 

∴ 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 6.5 cm. 

ii. Draw ray QT, such that ∠RQT = 40°

iii. Mark point S on ray QT such that l(QS) = 2.5 cm.

iv. Join points S and R. 

v. Draw perpendicular bisector of seg SR intersecting ray QT. Name the point as P. 

vi. Join the points P and R. Hence, ∆PQR is the required triangle.