i. As shown in the figure, take point D and E on line BC, such that
BD = AB and CE = AC ……(i)
BD + BC + CE = DE [D-B-C, B-C-E]
∴ AB + BC + AC = DE …..(ii)
Also, AB + BC + AC= 11.2 cm ….(iii) [Given]
∴ DE = 11.2 cm [From (ii) and (iii)]
ii. In ∆ADB
AB = BD [From (i)]
∴ ∠BAD = ∠BDA = x° ….(iv) [Isosceles triangle theorem]
In ∆ABD, ∠ABC is the exterior angle.
∴ ∠BAD + ∠BDA = ∠ABC [Remote interior angles theorem]
x + x = 70° [From (iv)]
∴ 2x = 70° x = 35°
∴ ∠ADB = 35°
∴ ∠D = 35°
Similarly, ∠E = 30°
iii. Now, in ∆ADE
∠D = 35°, ∠E = 30° and DE = 11.2 cm
Hence, ∆ADE can be drawn.
iv. Since, AB = BD
∴ Point B lies on perpendicular bisector of seg AD.
Also AC = CE
∴ Point C lies on perpendicular bisector of seg AE.
∴ Points B and C can be located by drawing the perpendicular bisector of AD and AE respectively.
∴ ∆ABC can be drawn.
Steps of construction:
i. Draw seg DE of length 11.2 cm.
ii. From point D draw ray making angle of 35°.
iii. From point E draw ray making angle of 30°.
iv. Name the point of intersection of two rays as A.
v. Draw the perpendicular bisector of seg DA and seg EA intersecting seg DE in B and C respectively.
vi. Join AB and AC.
Hence, ∆ABC is the required triangle.
