RBSE Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 9 Quadrilaterals Exercise 9.1.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 9 |

Chapter Name |
Quadrilaterals |

Exercise |
Ex 9.1 |

Number of Questions Solved |
14 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 9 Quadrilaterals Ex 9.1

Question 1.

The angles of a quadrilateral are in (RBSESolutions.com) the ratio 3:5:9:13. Find all the angles of the quadrilateral.

Solution.

Suppose the four angles of the quadrilateral are 3x, 5x, 9x and 13x.

∴ 3x + 5x + 9x + 13x = 360°

(by angle sum property of a quadrilateral)

⇒ 30x = 360°

⇒ x = 12°

Therefore, angles of the quadrilateral are 3 x 12°, 5 x 12°, 9 x 12° and 13 x 12° i.e., 36°, 60°, 108° and 156°.

Question 2.

In a parallelogram ABCD, diagonals AC and BD intersect at O. If OA = 3 cm and OD = 2 cm, then find (RBSESolutions.com) the length of AC and BD.

Solution.

Since, we know that diagonals of a parallelogram bisect each other

∴ AC = 2OA = 2 x 3 cm = 6 cm

and BD = 2OD = 2 x 2 cm = 4 cm.

Question 3.

Diagonals of a parallelogram are (RBSESolutions.com) mutually perpendicular. Is this statement true? Give reason for your answer.

Solution.

No, diagonals of a parallelogram bisect each other.

Question 4.

Can the angles 110°, 80°, 70° and 95° be the angles of a quadrilateral? Why or why not?

Solution.

We know that, sum of the angles of a quadrilateral is 360°.

Here 110° + 80° +70° + 95° = 355°.

∵ Sum of all the angles is not equal to 360°. So it would not be a quadrilateral.

Question 5.

Can all the four angles a quadrilateral (RBSESolutions.com) are obtuse? Give reason for your answer.

Solution.

No, if all the angles will be obtuse then sum of angles of a quadrilateral will exceed to 360°.

Question 6.

One angle of a quadrilateral is 108° and the other three angles are equal. Find the value of each of the equal angle.

Solution.

Let each of the three equal angle be x.

∴ 108° + x + x + x = 360°

(by angle sum property of a quadrilateral)

⇒ 3x + 108° = 360°

⇒ 3x = 252°

\(x=\frac { 252 }{ 3 }84\)

∴ Each equal angle is 84°.

Question 7.

ABCD is a trapezium in which AB || DC and ∠A = ∠B = 45°, then find (RBSESolutions.com) the values of the angles C and D.

Solution.

∵ AB || DC

∴ ∠A + ∠D = 180°

(sum of the interior angle on the same side of a transversal is always 180°)

45 + ∠D = 180°

⇒ ∠D = 135°

Similarly ∠C = 135°.

Question 8.

The angle between two altitudes of a parallelogram through (RBSESolutions.com) the vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.

Solution.

Suppose PQRS is a parallelogram in which ∠S is obtuse.

Through the vertex S altitudes, ST and SU are drawn.

Also we have ∠TSU = 60°

Now in quadrilateral QTSU,

∠TSU + ∠STQ + ∠TQU + ∠SUQ = 360°

(sum of the angles of a quadrilateral)

or 60° + 90° + ∠TQU + 90° = 360°

⇒∠TQU = 360° – 240°

⇒∠TQU = 120°

⇒∠PQR = 120°

Now ∠PSR = ∠PQR= 120°

(opposite angles of a parallelogram)

Again ∠QPS + ∠PQR = 180°

(adjacent angles of a parallelogram are supplementary)

⇒∠QPS + 120° = 180°

⇒∠QPS = 60°

⇒∠QPS = ∠QRS = 60°

(opposite angles of a parallelogram)

Hence, angles of the parallelogram are 120°, 60°, 120° and 60°.

Question 9.

In a parallelogram ABCD, E and F are the points (RBSESolutions.com) on the diagonals AC in such a way that AE = CF then show that BFDE is a parallelogram.

Solution.

∵ ABCD is a parallelogram and E and F are the points on AC, such that AE = CF

Proof: In ∆ABE and ∆CDF

AB = DC

(opposite sides of a parallelogram)

AE = CF (given)

and ∠1 = ∠2 (alt. angles)

∴ ∆ABE ≅ ∆CDF

(by SAS property)

⇒ BE = BF (by c.p.c.t)

Similarly, ∆AED ≅ ∆BFC

(by SAS property)

DE = BF (by c.p.c.t)

⇒ BFDE is a parallelogram

Question 10.

In figure, ABCD is a parallelogram and lines AQ and CP are (RBSESolutions.com) the bisector of ∠A – ∠C respectively. Prove the APCQ is a parallelogram.

Solution.

Given: ABCD is a parallelogram in which AQ and CP are the bisectors of ∠A and ∠C.

To prove: APCQ is a parallelogram.

Proof: In a parallelogram ABCD

∠A = ∠C

(opposite angles of a parallelogram)

\(\frac { 1 }{ 2 }\)∠A = \(\frac { 1 }{ 2 }\)∠C

∠PAQ = ∠PCQ

(as AQ bisect ∠A and PC bisect ∠C)

AB || CD

⇒ AP || CQ

APCQ is a parallelogram.

Hence proved.

Question 11.

In figure, ABCD and ABEF are parallelograms. Prove that CDFE is (RBSESolutions.com) also a parallelogram.

Solution.

∵ ABCD is a parallelogram

∴ AB = DC

and AB || DC …(i)

Reason: Opposite sides of a parallelogram (RBSESolutions.com) are equal and parallel. Again ABEF is a parallelogram

⇒ AB = FE

and AB || FE …(ii)

From (i) and (ii)

DC = FE and DC || FE

Therefore, CDFE is a parallelogram.

Hence proved.

Question 12.

ABCD is a parallelogram and AP and CQ are the perpendiculars from A and C on its diagonal BD, respectively. Prove that AP = CQ.

Solution.

Given: ABCD is a parallelogram in which BD is (RBSESolutions.com) diagonal and AP and CQ are drawn perpendiculars to BD from A and C respectively.

To prove: AP = CQ

Proof: Since diagonal of a parallelogram y divides its into two triangles equal in area.

∴ ar (∆ABD) = ar (∆BDC)

⇒ \(\frac { 1 }{ 2 }\) x BD x AP = \(\frac { 1 }{ 2 }\) x BD x QC

⇒ AP = QC

Hence proved.

Question 13.

In figure, ABCD is a quadrilateral in (RBSESolutions.com) which AB || CD and AD = BC, then prove that ∠A = ∠B.

Solution.

Given: ABCD is a quadrilateral in which AB || CD and AD = BC.

Construction: Produce AB and draw a line CE parallel to AD.

To prove: ∠A = ∠B

Proof: Since AD || CE and AE cuts (RBSESolutions.com) them at A and E respectively.

∴ ∠A + ∠E=180° …(i)

Since AB || CD and AD || CE

Therefore, AECD is a parallelogram

⇒ AD = CE

⇒ BC = CE (∵ AD = BC)

Thus, in ∆BCE, we have

BC = CE (proved above)

⇒ ∠CBE = ∠CEB

(angles opposite to equal sides)

⇒ 180°-∠B = ∠E

⇒ ∠B + ∠E=180° …(ii)

From (i) and (ii), we get

∠A = ∠B

Hence proved.

Question 14.

In figure. ABCD is a parallelogram. P and Q are the mid-points (RBSESolutions.com) of opposite sides AB and DC of a parallelogram ABCD. Prove that PRQS is a parallelogram.

Solution.

Given: ABCD is a parallelogram. P and Q are respectively (RBSESolutions.com) the mid-points of opposite sides AB and DC of a parallelogram ABCD. AQ and DP are joined intersecting in S and CP and BQ are joined intersecting in R.

To prove: Quadrilateral PQRS is a parallelogram.

Proof: ∵ ABCD is a parallelogram.

∴ AB = DC and AB || DC

⇒ \(\frac { 1 }{ 2 }\)AB = \(\frac { 1 }{ 2 }\)DC

⇒ AP = QC and AP || QC

⇒ APCQ is a parallelogram.

Similarly. PBQD is also a parallelogram. Since in parallelogram APCQ.

AQ || PC

(opposite sides of a parallelogram)

∴ SQ || PR

Now in parallelogram PBQD

PD || BQ

∴ SP || QR

Now in quadrilateral PRQS, SQ || PR and SP || QR.

Therefore, PRQS is (RBSESolutions.com) a parallelogram.

Hence proved.

We hope the given RBSE Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 9 Maths Solutions Chapter 9 Quadrilaterals Exercise 9.1, drop a comment below and we will get back to you at the earliest.