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Construct ∆XYZ, in which YZ = 6 cm, XY + XZ = 9 cm, ∠XYZ = 50°.

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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.

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