Constructions of Triangles — Maths STD 9 — Question

Perimeter of ∆ABC = AB + BC + AC
∴ 10 = AB + 3.2 + AC
∴ AB + AC = 10 – 3.2
∴ AB + AC = 6.8 cm
Now, In ∆ABC
BC = 3.2 cm, ∠ACB = 45° and AB + AC = 6.8 cm ….(i)
As shown in the rough figure draw j seg BC = 3.2 cm
Draw a ray CT making an angle of 45° with CB
Take a point D on ray CT, such that
CD = 6.8 cm
Now, CA + AD = CD [C-A-D]
∴ CA + AD = 6.8 cm …(ii)
Also, AB + AC = 6.8 cm ….(iii) [From (i)]
∴ CA + AD = AB + AC [From (ii) and (iii)]
∴ 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 construction:
i. Draw seg BC of length 3.2 cm.
ii. Draw ray CT, such that ∠BCT = 45°.
iii. Mark point D on ray CT such l(CD) = 6.8 cm. that
iv. Join points D and B.
V. Draw perpendicular bisector of seg DB intersecting ray CT. Name the point as A.
vi. Join the points A and B.
Hence, ∆ABC is the required triangle.