Question

"Question 1 Construct a triangle ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.

Class 9 - Math - Constructions Page 195"

Answer 1

[Fig. is in the attachment]

Given: In ∆ABC,

 BC= 7cm, AB+AC=13cm & ∠B= 60°

Steps of construction:

1. First draw the base BC=7 cm

Now draw a ray BX such that ∠XBC= ∠B= 75°

2. Here sum of two sides = AB+AC=13cm

So cut the line segment BD=13cm  from ray  BX.

 

3. Join DC.

4. Draw perpendicular bisector of DC , say PQ, which intersects BD at A.

5. Now join AC thus we get a ∆ABC which is the required Triangle.

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Hope this will help you.....

Answer 2

Steps of Construction: 
Step 1: A line segment BC of 7 cm is drawn.
Step 2: At point B, an angle ∠XBC is constructed such that it is equal to 75°.
Step 3: A line segment BD = 13 cm is cut on BX (which is equal to AB+AC).
Step 3: DC is joined and ∠DCY = ∠BDC is made.
Step 4: Let CY intersect BX at A.
Thus, ΔABC is the required triangle.