RBSE Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.3 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 9 Quadrilaterals Ex 9.3.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 9 |

Chapter Name |
Quadrilaterals |

Exercise |
Ex 9.3 |

Number of Questions Solved |
9 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 9 Quadrilaterals Ex 9.3

Question 1.

Construct a quadrilateral ABCD, when AB = 3.5 cm, BC = 4.8 cm, CD = 5.1 cm, AD = 4.4 cm and diagonal AC = 5.9 cm.

Solution.

Steps of construction:

1. Draw AB = 3.5 cm as base.

2. With centre A, draw an arc AC = 5.9 cm.

3. With B as centre, draw an arc BC = 4.8 cm. These (RBSESolutions.com) arcs intersect at C.

4. Join A to C and B to C.

5. Now take C as centre and draw an arc of radius 5.1 cm and with A as centre and radius 4.4 cm mark an arc. These arcs intersect each other at D.

6. Join A to D and D to C.

Hence, ABCD is the required quadrilateral.

Question 2.

Construct a quadrilateral PQRS, when PQ = 4 cm, QR = 3 cm, QS = 4.8 cm, PS = 3.5 and PR = 5 cm.

Solution.

Steps of construction:

1. Draw PQ = 4 cm as base.

2. With P as centre and radius 5 cm draw an arc.

3. With Q as centre, draw another arc QR = 3 cm. These arcs intersect at R.

4. With centre P, draw an arc SP = 3.5 cm and (RBSESolutions.com) with centre Q draw an arc QS = 4.8 cm. These arcs intersect at point S. Join QR, RS, SP, PR and SQ.

Hence, PQRS is required quadrilateral.

Question 3.

Construct a quadrilateral ABCD, when AB = 4 cm, BC = 4.5 cm, CD = 3.5 cm, AD = 3 cm and ∠A = 60°.

Solution.

Steps of construction:

1. Draw AB = 4 cm as base.

2. At A draw an angle of 60°.

3. By taking P as centre, draw an arc of radius 3 cm which gives point D.

4. Again by taking D as centre, draw an arc (RBSESolutions.com) of radius 3.5 cm and from B drawn an arc of 4.5 cm which intersects at C.

5. Join CD and BC.

6. Hence, ABCD is the required quadrilateral.

Question 4.

Construct a quadrilateral ABCD, when AB = 3.5 cm, BC = 3 cm, AD = 2.5 cm, AC = 4.5 cm and BD = 4 cm.

Solution.

Steps of construction:

1. Draw AB = 3.5 cm as base.

2. By taking A and B as centres, draw an arc (RBSESolutions.com) of radius 4.5 cm and 4 cm respectively.

3. Also from B, draw an arc of radius 3 cm which intersects the previous arc of radius 4.5 cm at C.

4. Again from A, draw an arc of 2.5 cm which intersects the arc of 4 cm at D.

5. Join A to D, B to C, B to D and C to D.

6. Hence, ABCD is the required quadrilateral.

Question 5.

To construct (RBSESolutions.com) a quadrilateral PQRS when PQ = 3.0 cm, QR = 4.0 cm, PS = 4.5 cm, QS = 5.5 cm and PR = 6 cm.

Solution.

Steps of construction:

1. Draw PQ = 3.0 cm as base.

2. With P as centre and radius 6 cm draw an arc.

3. With Q as centre, draw another arcs of radius 4 cm, these arcs intersect at R.

4. With centre P, draw an arc PS = 4.5 cm and with centre Q, draw an arc QS = 5.5 cm. These arcs intersect at point S. Join QR, RS, SP, PR and SQ. Hence, PQRS is (RBSESolutions.com) the required quadrilateral.

Question 6.

Construct a quadrilateral ABCD, when AB = BC = 3 cm, AD = 5 cm, ∠A = 90° and ∠B = 120°.

Solution.

Steps of construction:

1. Draw AB = 3 cm as base.

2. At A make an angle of 90°.

3. At B make an angle of 120°.

4. With A and B as centres and radius 5 cm and 3 cm, draw arcs which (RBSESolutions.com) cut line segments at D and C.

5. Finally, join D to C.

Hence, ABCD is the required quadrilateral.

Question 7.

To construct a quadrilateral ABCD, when AB = 3.8 cm, BC = 2.5 cm, CD = 4.5 cm, ∠B = 30° and ∠C = 150°.

Solution.

Steps, of construction:

1. Draw BC = 2.5 cm as base.

2. At B and C, make ∠B = 30° and ∠C = 150°.

3. With B and C as centres, (RBSESolutions.com) draw an arc of 3.8 cm and 4.5 cm respectively.

4. Join A to D.

Hence, ABCD is a required quadrilateral.

Question 8.

Construct a quadrilateral PQRS, when PQ = 3 cm, QR = 3.5 cm, ∠Q = 90° and ∠P = 105°, ∠R = 120°.

Solution.

Steps of construction:

1. Draw a line segment PQ = 3 cm as base.

2. At P draw a ray PX such that ∠QPS = 105°.

3. At Q draw a ray QZ such that ∠PQZ = 90° and mark an arc (RBSESolutions.com) of radius 3.5 cm on it. Mark this point as R.

4. At R draw ∠QRS = 120° which cuts PX at S.

Hence, PQRS is the required quadrilateral.

Question 9.

Construct a quadrilateral PQRS, when PQ = 2.5 cm, QR = 3.7 cm, ∠Q = 120°, ∠S = 60° and ∠R = 90°.

Solution.

Steps of construction:

∵∠P + ∠Q + ∠R + ∠S = 360°

∴ ∠P = 360° – (120° + 90° + 60°)

=> ∠P = 90°

1. Draw a line segment PQ = 2.5 cm.

2. At P draw a ray PX such that ∠XPQ = 90° and at Q draw a ray QY such that ∠PQY = 120°.

3. With Q as centre, take an arc (RBSESolutions.com) of radius 3.7 cm and mark it as R.

4. Also at point R construct an angle of 90° which will meet ray PX at S.

Hence, PQRS is the required quadrilateral.

We hope the given RBSE Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.3 will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.3, drop a comment below and we will get back to you at the earliest.

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