PSEB 9th Class Maths Solutions Chapter 11 Constructions Ex 11.2

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 11 Constructions Ex 11.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2

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

Steps of construction:

1. Draw any ray BX. With centre B and radius 7 cm draw an arc to intersect BX at C.
2. At B, construct ∠YBC with measure 75°.
3. With centre B and radius 13 cm, draw an arc to intersect BY at M.
4. Draw line segment MC. Draw the perpendicular bisector of MC to intersect BM at A.
5. Draw line segment AC.
   Then, ∆ ABC is the required triangle.

Question 2.
Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 3.5 cm.
Answer:

Steps of construction:

1. Draw any ray BX and from that obtain the line segment BC of length 8 cm.
2. At B, draw ray BY such that ∠YBC = 45°.
3. With centre B and radius 3.5 cm, draw an arc to intersect ray BY at D.
4. Draw line segment DC. Draw the perpendicular bisector of DC to intersect ray BY at A.
5. Draw line segment AC.
   Then, ∆ ABC is the required triangle.

Question 3.
Construct a triangle PQR in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm.
Answer:

Steps of construction:

1. Draw any ray QX and from that obtain the line segment QR of length 6 cm.
2. At Q, construct ray QY such that Z YQR = 60°.
3. Produce ray QY on the side of Q to obtain ray QZ. Obtain point S on ray QZ such that QS = 2 cm.
4. Draw line segment RS. Draw the perpendicular bisector of RS to intersect QY at E
5. Draw line segment PR.
   Then, ∆ PQR is the required triangle.

Question 4.
Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
Answer:

Steps of construction:

1. Draw any ray AP and from that obtain the line segment AB of length 11 cm.
2. Construct ray AL such that ∠LAB = 30°.
3. Construct ray BM such that ∠MBA = 90°.
4. Draw the bisectors of ∠LAB and ∠MBA to intersect each other at X.
5. Draw line segment XB. Draw the perpendicular bisector of XB to intersect AB at Z.
6. Draw line segment XA. Draw the perpendicular bisector of XA to intersect AB at Y.
7. Draw line segments XY and XZ.
   Then, ∆ XYZ is the required triangle.

Question 5.
Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.
Answer:

Steps of construction:

1. Draw any ray BX and from that obtain the line segment BC of length 12 cm.
2. Construct ray BY such that ∠YBC = 90°.
3. Taking B as centre and radius 18 cm, draw an arc to intersect BY at M.
4. Draw line segment CM. Draw the perpendicular bisector of CM to intersect BM at A.
5. Draw line segment AC.
   Then, ∆ ABC is the require triangle in which ∠B is a right angle, BC = 12 cm and AB + AC = 18 cm.