Construction 2
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- 8 cm
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- Construct ∠B
- Draw line BC
- Join point D & C
- Mark D as AB-AC
- Bisect a ⊥ to DC

Steps of construction:
Step 1: Draw BC = 6 cm
Step 2: Construct ∠YBC=45∘
Step 3 will be -
- Join AC, ΔABC is the required triangle.
- Draw perpendicular bisector of CD intersecting BY at A.
- From ray BY, cut-off line segement BD = 2.5 cm
- Join CD
What is the number of arcs that need to be drawn during the method of triangle construction 2?
1
2
3
4
Given a base angle B and the difference of the sides AC and AB (not necessarily in this order), what is the difference in the steps of constructions when
i) AC > AB
ii) AC < AB
Instead of having B as the base angle, have the same construction with C as the base angle
Instead of drawing the perpendicular bisector of line segment BC, we’ll draw the angle bisector of angle B in this case
Instead of plotting the point D on the ray BX as (AC – AB) = say 2units, we rewrite it as (AB – AC) = -2 units and plot the point D 2 units down south of point B on the same ray BX.
None of the above

Here, PR > PQ,
i.e., the side containing base angle is less than the third side.
Steps of construction:
Step 1: Draw the base QR = 5.5 cm
Step 2: At the point Q, make an ∠XQR=60∘
Step 3 will be
- Cut line segment QS = PR - PQ = 2 cm from the line QX extended on opposite side of linesegment QR
- Cut line segment QS = PR - PQ = 2.5 cm from the line QX extended on opposite side of line segment QR
- Cut line segment QS = PR - PQ = 2.5 cm from the ray QX
- Draw the perpendicular bisector LM of SR.

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 ∠YBC=60∘
Step 2: Cut the line segment BD = 3 cm from the ray BY.
Step 3 will be
- Let PQ intersect BX at A.Then, join AC
- Join DC and draw PQ⊥BC
- Join DC and draw PQ⊥BY
- Join DC and draw PQ⊥DC
The steps to draw a triangle with the base BC, a base angle ∠B and the difference of other two sides is given below.
1. Draw the base BC and at point B make an angle say XBC equal to the given angle.
2. Cut the line segment BD equal to AB - AC on the reflection of ray BX (i.e. BX').
3. Join DC and draw the perpendicular bisector, say PQ of DC.
The next step will be:
Let PQ intersect BX at a point A. Join AC.
Draw angle bisector of ∠C
Draw perpendicular bisector of AD
Draw angle bisector of ∠ A