Question

Construct a triangle ABC in which BC = 3cm, B = 30° and AB + AC = 5.2 cm.​

Answer 1

Answer:

Given, in ∆ABC, base BC = 3 cm, ∠B = 30° and AB+ AC = 5.2 cm

Steps of construction

(i) First, draw base BC = 3 cm.

(ii) Construct∠YBC = 30° at the point B.

(iii) From ray BY, cut-off line segment BD = AB+ AC = 5.2 cm.

(iv) Join CD.

(v) Draw the perpendicular bisector of CD, which meets BY at A

(vi) Join AC. Thus, we get the required ∆ABC.

Justification Since, A lies on the perpendicular bisector of CD.

∴ AC = AD

Now, BD = 5.2 cm =>BA+ AD = 52 cm

∴ BA+ AC = 5.2 cm

Answer 2

Given, in ∆ABC, base BC = 3 cm, ∠B = 30° and AB+ AC = 5.2 cm

Steps of construction

(i) First, draw base BC = 3 cm.

(ii) Construct∠YBC = 30° at the point B.

(iii) From ray BY, cut-off line segment BD = AB+ AC = 5.2 cm.

(iv) Join CD.

(v) Draw the perpendicular bisector of CD, which meets BY at A

(vi) Join AC. Thus, we get the required ∆ABC.

Justification Since, A lies on the perpendicular bisector of CD.

∴ AC = AD

Now, BD = 5.2 cm =>BA+ AD = 52 cm

∴ BA+ AC = 5.2 cm

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