Construction 2

Q. The construction of $\Delta$ABC, given that BC = 12 cm, is possible when the difference of AB and AC equals

1. 8 cm
2. 14 cm
3. 12 cm
4. 15 cm

Q. Arrange the steps to construct a △ABC where BC and difference of sides AB-AC (AB>AC) and ∠B is given:

1. Construct ∠B
2. Draw line BC
3. Join point D & C
4. Mark D as AB-AC
5. Bisect a ⊥ to DC

Q. Construct a triangle ABC in which BC = 6 cm, $\angle B={45}^{\circ }$ and AB - AC = 2.5 cm

Steps of construction:
Step 1: Draw BC = 6 cm
Step 2: Construct $\angle YBC={45}^{\circ }$
Step 3 will be -

1. Join AC, $\Delta ABC$ is the required triangle.
2. Draw perpendicular bisector of CD intersecting BY at A.
3. From ray BY, cut-off line segement BD = 2.5 cm
4. Join CD

Q.

What is the number of arcs that need to be drawn during the method of triangle construction 2?

1. 1

2. 2

3. 3

4. 4

Q.

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

1. Instead of having B as the base angle, have the same construction with C as the base angle

2. Instead of drawing the perpendicular bisector of line segment BC, we’ll draw the angle bisector of angle B in this case

3. 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.

4. None of the above

Q. To construct a $\Delta PQR$, in which QR= 5.5 cm, $\angle Q={60}^{\circ }$ and PR - PQ = 2.5 cm, the steps of construction are given below. Complete the third step

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 $\angle XQR={60}^{\circ }$
Step 3 will be

1. Cut line segment QS = PR - PQ = 2 cm from the line QX extended on opposite side of linesegment QR
2. Cut line segment QS = PR - PQ = 2.5 cm from the line QX extended on opposite side of line segment QR
3. Cut line segment QS = PR - PQ = 2.5 cm from the ray QX
4. Draw the perpendicular bisector LM of SR.

Q. 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

1. Let PQ intersect BX at A.Then, join AC
2. Join DC and draw $PQ\perp BC$
3. Join DC and draw $PQ\perp BY$
4. Join DC and draw $PQ\perp DC$

Q.

The steps to draw a triangle with the base BC, a base angle $\angle$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:

1. Let PQ intersect BX at a point A. Join AC.

2. Draw angle bisector of $\angle$C

3. Draw perpendicular bisector of AD

4. Draw angle bisector of $\angle$ A