Question

construct ∆PQR, in which QR=4.2cm,m angle Q=40°and PQ+PR=8.5 cm.​

Answer 1

Answer:

As shown in the rough figure draw seg QR = 4.2

Draw a ray QT making an angle of 40° with QR

Take a point S on ray QT, such that QS = 8.5

Now, QP + PS = QS [Q - P - S]

∴ QP + PS = 8.5 cm …….(i)

Also, PQ + PR = 8.5 cm ……(ii) [Given]

∴ QP + PS = PQ + PR [From (i) and (ii)]

∴ PS = PR

∴ Point P is on the perpendicular bisector of seg SR

∴ The point of intersection of ray QT and perpendicular bisector of seg SR is point P.

Steps of construction:

i. Draw seg QR of length 4.2 cm.

ii. Djraw ray QT, such that ∠RQT = 40°.

iii. Mark point S on ray QT such that l(QS) = 8.5 cm.

iv. Join points R and S.

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

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

I hope it will help u