Question

To construct a $\Delta ABC$ in which BC = 5.7 cm, $\angle B={60}^{\circ }$ and AB - AC = 3cm, the steps of construction are given below.

Here, AB > AC, i.e. the side containing the base angle B is greater than third side
Steps of construction:
Step 1: Draw the base BC = 5.7 cm and draw a ray BY making an $\angle YBC={60}^{\circ }$
Step 2: Cut the line segment BD = 3 cm from the ray BY.
Step 3 will be

• A. Let PQ intersect BX at A.Then, join AC
• B. Join DC and draw $PQ\perp DC$
• C. Join DC and draw $PQ\perp BY$
• D. Join DC and draw $PQ\perp BC$

Answer 1

The correct option is B Join DC and draw $PQ\perp DC$

Steps of construction to construct a triangle given its base, a base angle and the difference of the other two sides:

Step 1: Draw the base BC = 5.7 cm and draw a ray BY making an $\angle YBC={60}^{\circ }$
Step 2: Cut the line segment BD = 3 cm from the ray BY.
Step 3 : Join DC and draw $PQ\perp DC$
Step 4: Draw the perpendicular bisector of CD. Let it meet ray BDY at a point A.
Step 5: Join AC to get the required triangle ABC.