As shown in the rough figure draw seg YZ = 6 cm
Draw a ray YT making an angle of 50° with YZ
Take a point W on ray YT, such that YW = 9 cm
Now, YX + XW = YW [Y-X-W]
∴ YX + XW = 9 cm ….(i)
Also, XY + XZ = 9 cm ….(ii) [Given]
∴ YX + XW = XY + XZ [From (i) and (ii) ]
∴ XW = XZ
∴ Point X is on the perpendicular bisector of seg WZ
∴ The point of intersection of ray YT and perpendicular bisector of seg WZ is j point X.
Steps of construction:
i. Draw seg YZ of length 6 cm.
ii. Draw ray YT, such that ∠ZYT = 50°.
iii. Mark point W on ray YT such that l(YW) = 9 cm.
iv. Join points W and Z.
v. Draw perpendicular bisector of seg WZ intersecting ray YT. Name the point as X.
vi. Join the points X and Z.
Hence, ∆XYZ is the required triangle.