Question

Construct ∆, such that XY + XZ = 10.3 cm, YZ = 4.9 cm, ∠ = 450


Urgent☹️​

Answer 1

Answer:

Step-by-step explanation:

Draw segYZ = 4.9cm

Draw a ray YT making an angle of 45° with YZ

Take a point W on ray YT, such that YW = 10.3 cm

Now, YX + XW = YW [Y-X-W]

∴ YX + XW = 10.3 cm …..

Also, XY + X ∠10.3cm ……(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 point X

Steps of construction: i. Draw seg YZ of length 4.9 cm.  

ii. Draw ray YT, such that ∠ZYT = 75°.

iii. Mark point W on ray YT such that l(YW) = 10.3 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.

HOPE IT HELPS!!

Answer 2

I hope you will understand it