Question

Construct ΔABC, in which BC = 6.2 cm, ∠ACB= 50°, AB + AC = 9.8 cm.

Answer 1

Rough figure:

Explanation:

As shown in the rough figure draw seg CB = 6.2 cm

Draw a ray CT making an angle of 50° with CB

Take a point D on ray CT, such that

CD = 9.8 cm

Now, CA + AD = CD    ...[C-A-D]

∴ CA + AD = 9.8 cm    …(i)

Also, AB + AC = 9.8 cm    ...(ii) [Given]

∴ CA + AD = AB + AC      ...[From (i) and (ii)]

∴ AD = AB

∴ Point A is on the perpendicular bisector of seg DB.

∴ The point of intersection of ray CT and perpendicular bisector of seg DB is point A.

Steps of constructions:

1. Draw seg BC of length 6.2 cm.
2. Draw ray CT, such that ∠BCT = 50°.
3. Mark point D on ray CT such that l(CD) = 9.8 cm.
4. Join points D and B.
5. Draw perpendicular bisector of seg DB intersecting ray CT. Name the point as A.
6. Join the points A and B.

Therefore, △ABC is the required triangle.