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

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 9 |

Chapter Name |
Quadrilaterals |

Exercise |
Additional Questions |

Number of Questions Solved |
27 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 9 Quadrilaterals Additional Questions

**Multiple Choice Questions**

Question 1.

In the given Q figure, ABCD is a parallelogram (RBSESolutions.com) in which ∠ADC = 130°, then ∠CBE is equal to:

(A) 60°

(B) 130°

(C) 50°

(D) 70°

Solution

(C) 50°

Question 2.

In the given figure, PQR is a triangle in which X and Y are (RBSESolutions.com) the mid-point of the sides PR and QR respectively. If PQ = 6 cm, QR = 7 cm and PR = 8 cm, then XY is equal to:

(A) 12 cm

(B) 3.5 cm

(C) 3 cm

(D) 4 cm

Solution

(C) 3 cm

Question 3.

In figure, ABCD is a parallelogram then (RBSESolutions.com) the value of x is:

(A) 25°

(B) 60°

(C) 75°

(D) 45°

Solution

(D) 45°

Question 4.

If an angle of a parallelogram is two-third of its adjacent angle, then (RBSESolutions.com) the smallest angle of the parallelogram is:

(A) 54°

(B) 72°

(C) 84°

(D) 108°

Solution

(B) 72°

Question 5.

The ratio of the side and diagonal of a square is:

(A) 1:√2

(B) 3:√2

(C) √2:1

(D) √2:3

Solution

(A) 1:√2

Question 6.

In the given figure, ABCD is a rhombus. If ∠OAB = 35°, then (RBSESolutions.com) the value of x is:

(A) 25°

(B) 35°

(C) 55°

(D) 70°

Solution

(C) 55°

Question 7.

If the two adjacent angles of (RBSESolutions.com) a parallelogram are (3x – 20°) and (50° – x), then the value of x is :

(A) 55°

(B) 75°

(C) 20°

(D) 80°

Solution

(B) 75°

Question 8.

In the given figure, ABCD is a quadrilateral and P, Q, R and S are the mid-points (RBSESolutions.com) of the sides AB, BC, CD, DA respectively. If BD =12 cm, then length of QR is:

(A) 6 cm

(B) 8 cm

(C) 3 cm

(D) 4 cm

Solution

(A) 6 cm

Question 9.

In figure, D and E are mid-points (RBSESolutions.com) of the sides AB and AC respectively of ∆ABC, then measure of ∠B is:

(A) 50°

(B) 30°

(C) 80°

(D) 40°

Solution

(A) 50°

Question 10.

In a quadrilateral PQRS, ∠P, ∠Q, ∠R and ∠S are interior angles. If ∠P: ∠Q: ∠R: ∠S = 1 : 2 : 3 : 4, then (RBSESolutions.com) which angle is equal to 144°.

(A) ∠P

(B) ∠Q

(C) ∠R

(D) ∠S

Solution

(D) ∠S

**Very Short Answer Type Questions**

Question 1.

Two consecutive angles of a parallelogram are (RBSESolutions.com) in the ratio 1 : 3. Find the smallest angle.

Solution.

Let the two consecutive angles be x and 3x.

Since the sum of two consecutive angles of a parallelogram is 180°.

Therefore, x + 3x = 180°

⇒ 4x = 180°

⇒ x = 45°

Smallest angle = 45°.

Question 2.

In the given figure, ABCD is a rectangle (RBSESolutions.com) in which ∠APB = 100°. Find the value of x.

Solution.

In ΔPBC,

∠BPA + ∠BPC = 180° (linear pair of angles)

⇒ ∠BPC = 180° – 100° = 80°

⇒ ∠PCB = ∠PBC = x

⇒ x + x + 80° = 180°

⇒ 2x = 100°

⇒ x = 50°

Question 3.

The three angles of a quadrilateral measure 56°, 100° and 88°. Find the (RBSESolutions.com) measure of the fourth angle.

Solution.

Let the measure of the fourth angle be x.

As we know that sum of the angles of a quadrilateral is 360°.

56° + 100° + 88° + x = 360°

⇒ 244° + x = 360°

⇒ 360° – 244° = 116°

Hence, the measure of the fourth angle is 116°.

Question 4.

In a quadrilateral ABCD, ∠A = 3x, ∠B = 5x, ∠C = 20x and ∠D = 8x. Find the value of x.

Sol. The sum of the angles of a quadrilateral is 360°.

⇒ ∠A + ∠B + ∠C + ∠D = 360°

⇒ 3x + 5x + 20x + 8x = 360°

⇒ 36x = 360°

⇒ x = 10°

Question 5.

A diagonal of a rectangle is inclined to one side of the rectangle at 25°. Then find (RBSESolutions.com) the acute angle between the diagonals.

Solution.

According to figure,

In ΔOCD,

∠ODC + ∠OCD + ∠COD = 180°

⇒ 25° + 25° + ∠COD = 180°

⇒ ∠COD = 130°

Acute angle i.e. ∠DOA between the diagonals = 180° – ∠DOC = 180° – 130° = 50°

Question 6.

In the given figure, BCPQ and BCDA are two parallelograms on (RBSESolutions.com) the same base BC. Find the value of x + y.

Solution.

BCDA is a parallelogram

∠A = y = 100° (opposite angles of a parallelogram are equal)

Also x + 100° = 120° (exterior angle property)

⇒ x = 20°

x + y = 20 + 100 = 120°

Question 7.

“The opposite sides of (RBSESolutions.com) a parallelogram are equal”. Prove it.

Solution.

Given: A parallelogram ABCD

To prove: AB = CD and BC = DA

Construction: Join diagonal BD.

Proof: In parallelogram ABCD, AB || DC and AD || BC

Now in Δs ABD and CBD

∠ABD = ∠CDB (alt. ∠S)

∠ADB = ∠CBD (alt. ∠S)

BD = BD

ΔABD = ΔCDB (by AAS congruencey rule)

AB = CD (by c.p.c.t)

and AD = BC

Question 8.

In figure, D and E are the mid¬points (RBSESolutions.com) of the sides AB and AC respectively of ΔABC. If BC = 6.4 cm, find DE.

Solution.

Since, D and E are mid-points of sides AB and AC respectively. Therefore, from mid-point theorem,

DE = [latex]\frac { 1 }{ 2 }[/latex] BC

⇒ DE = [latex]\frac { 1 }{ 2 }[/latex] x 6.4 = 3.2

⇒ DE = 3.2 cm.

Question 9.

In the given figure, PQRS is a rectangle. If ∠RPQ = 30° then find the value of (x + y).

Solution.

Since, diagonal of a rectangle are equal and bisect (RBSESolutions.com) each other and vice versa.

OR = OS

⇒ ∠ORS = ∠OSR …(i)

Also PQ || SR

⇒ ∠OPQ = ∠ORS = 30° …(ii) (alt. angles)

In ΔORS,

∠ORS + ∠OSR + y = 180° (angle sum property)

⇒ 30° + 30° + y = 180° [using (i) and (ii)]

⇒ y = 120°

Also OR = OQ

⇒ ∠ORQ = ∠OQR = x

In ΔORQ,

x + x + ∠ROQ = 180°

⇒ 2x + 60° = 180° (∵ y + ∠ROQ = 180° i.e. linear pair of angles)

⇒ x = [latex]\frac { 120 }{ 2 }[/latex] = 60°

x + y = 60° + 120° = 180°

Question 10.

In the given figure, AD and BE are (RBSESolutions.com) medians of ΔABC and BE || DF. Prove that CF = [latex]\frac { 1 }{ 4 }[/latex] AC.

Solution.

In ΔBEC,

D is mid-point of BC and from converse (RBSESolutions.com) of mid-point theorem.

i.e. A line i.e. DF drawn through the mid-point D of BC and parallel to BE bisects the third side i.e. EC at F.

Now, F becomes the mid-point of CE.

⇒ CF = [latex]\frac { 1 }{ 2 }[/latex] CE

CF = [latex]\frac { 1 }{ 2 }[/latex] ([latex]\frac { 1 }{ 2 }[/latex] AC)

[E is mid-point of AC ⇒ AE = EC = [latex]\frac { 1 }{ 2 }[/latex] AC]

⇒ CF = [latex]\frac { 1 }{ 4 }[/latex] AC

Hence proved.

**Short Answer Type Questions**

Question 1.

If the diagonals of a parallelogram are equal, then (RBSESolutions.com) show that it is a rectangle.

Sol.

Given: A parallelogram ABCD in which AC = BD.

To prove: Parallelogram ABCD is a rectangle

Proof: In ΔBAD and ΔABC,

AB = BA (common sides)

BD = AC (diagonals of a parallelogram are equal)

AD = BC (opposite sides of a parallelogram are equal)

ΔBAD = ΔABC (by SSS congruency rule)

⇒ ∠BAD = ∠ABC (by c.p.c.t)

But ∠BAD + ∠ABC = 180° (sum of interior angles (RBSESolutions.com) on the same side of a transversal is 180°)

⇒ 2 ∠BAD = 180° [∵ ∠BAD = ∠ABC]

⇒ ∠BAD = 90°

Hence, parallelogram ABCD is a rectangle.

Question 2.

In a trapezium ABCD, E is the mid-point of AD and EF || AB. Prove that EF = [latex]\frac { 1 }{ 2 }[/latex] (AB + DC).

Solution.

Given: ABCD is a trapezium in which E is the mid-point of AD and EF || AB.

To prove: EF = [latex]\frac { 1 }{ 2 }[/latex] (AB + DC)

Construction: Join BD which meets EF at G.

Proof: EF || AB (given)

EG || AB

Now in ΔADB,

E is the mid-point of AD and EG || AB.

EG bisects BD (by converse of mid-point theorem)

EG = [latex]\frac { 1 }{ 2 }[/latex] AB …(i) (from mid-point theorem)

Also GF || AB and AB || DC

⇒ GF || DC

Now in ΔBCD, G is mid-point of BD and GF || DC

⇒ GF bisects BC ⇒ F is mid-point of BC

⇒ GF = [latex]\frac { 1 }{ 2 }[/latex] DC …(ii)

Adding (i) and (ii), we get

EG + GF= [latex]\frac { 1 }{ 2 }[/latex] AB + [latex]\frac { 1 }{ 2 }[/latex] DC

⇒ EF = [latex]\frac { 1 }{ 2 }[/latex] (AB + DC)

Hence proved.

Question 3.

In the given figure, PQRS is (RBSESolutions.com) a parallelogram in which PQ is produced to T such that QT = PQ. Prove that ST bisects RQ.

Solution.

Sol. Given: PQRS is a parallelogram in which PQ is produced to T such that PQ = QT.

To prove: QO = OR i.e. ST bisects RQ.

Proof: PQRS is a parallelogram.

PQ || RS

Now, PQ || RS and (RBSESolutions.com) transversal QR intersects them

∠1 = ∠2 …(i) (alternate interior angles)

Now PQRS is a parallelogram

⇒ PQ = RS

⇒ QT = RS …(ii)

QT = PQ (given)

Thus in ΔQOT and ΔSOR, we have

∠1 = ∠2 [using (i)]

∠3 = ∠4 (vertically opposite angles)

and QT = RS [using (ii)]

ΔQOT = ΔROS (by AAS congruency rule)

⇒ QO = OR

⇒ O is the mid-point of QR

⇒ ST bisects RQ.

**Long Answer Type Questions**

Question 1.

Show that if the diagonals of a quadrilateral are (RBSESolutions.com) equal and bisect each other at right angles, then it is a square.

Solution.

Given: ABCD is a quadrilateral in which diagonals AC and BD are equal and bisect each other at right angles.

To prove: ABCD is a square.

Proof: ABCD is a quadrilateral whose diagonals bisect (RBSESolutions.com) each other. So it is a parallelogram. Also its diagonals bisect each other at right angles. So ABCD is a rhombus.

⇒ AB = BC = CD = DA

In ΔABC and ΔABD, we have

AB = AB (common sides)

BC = AD (side of a rhombus are equal)

AC = BD (given)

ΔABC = ΔBAD (by SSS congruency rule)

∠ABC = ∠BAD (by c.p.c.t)

But ∠ABC + ∠BAD = 180° (sum of the consecutive interior angles)

⇒ ∠ABC = ∠BAD = 90°

⇒ ∠A = ∠B = ∠C = ∠D = 90° (opposite angles of a parallelogram)

⇒ ABCD is a rhombus whose angles are of 90° each.

Hence, ABCD is a square.

Question 2.

In a parallelogram ABCD, E and F are the mid-points (RBSESolutions.com) of sides AB and CD respectively (see figure). Show that the line segment, AF and EC trisect the diagonal BD.

Solution.

Given: ABCD is a parallelogram in which E and F are (RBSESolutions.com) the mid-points of sides AB and CD respectively.

To prove: BQ = QP = PD

Proof: ABCD is a parallelogram

⇒ AB || DC

⇒ AE || AF and AE = CF (∵ [latex]\frac { 1 }{ 2 }[/latex] AB = [latex]\frac { 1 }{ 2 }[/latex] CD)

⇒ AECF is a parallelogram.

⇒ EC || AF …(i)

If ΔAPB, E is mid-point of AB (given)

and EQ ||AP [using (ii)]

⇒ Q is the mid-point PB

⇒ BQ = PQ …(ii)

Similarly, in ΔDQC, we can show that

DP = PQ

From (ii) and (iii), we have BQ = PQ = PD

Hence, line segment AF and EC trisect the diagonal BD.

Question 3.

Show that the line segment joining the mid-points (RBSESolutions.com) of the opposite sides of a quadrilateral bisect each other.

Solution.

Given: ABCD is a quadrilateral in which PR and SQ are the line segments joining the mid-points of the opposite sides.

To prove: PR and SQ bisect each other.

Construction: Join PQ, QR, RS, SP and AC.

Proof: In ΔABC, P and Q are mid-points of sides AB and BC respectively.

From mid-point theorem

PQ || AC and PQ = [latex]\frac { 1 }{ 2 }[/latex] AC …(i)

In ΔADC, we have

S and R are mid-points of sides AD and CD respectively.

Therefore, from mid-point theorem

SR || AC and SR = [latex]\frac { 1 }{ 2 }[/latex] AC …(ii)

From (i) and (ii), we get

PQ || SR and PQ = SR

⇒ PQRS is a parallelogram.

Reason: A quadrilateral is a parallelogram if its (RBSESolutions.com) one pair of opposite sides is equal and parallel.

Now PR and SQ are diagonals of the parallelogram PQRS.

PR and SQ bisect each other. (∵ Diagonals of a parallelogram bisect each other)

Question 4.

P, Q and R are respectively, the mid-points of sides BC, CA and AB of a triangle ABC. PR and BQ meet at X. CR and PQ meet at Y.

Prove that XY = [latex]\frac { 1 }{ 4 }[/latex] BC.

Solution.

Given: ABC is a triangle in which P, Q and R are respectively the mid-points of sides BC, CA and AB. PR and BQ meet at X and CR and PQ meets at Y.

Construction:

Join X to Y.

To prove: XY = [latex]\frac { 1 }{ 4 }[/latex] BC

Proof: R and Q are the mid-points of AB and AC respectively.

Therefore, RQ || BC and PQ = [latex]\frac { 1 }{ 2 }[/latex] BC …(i) (by mid-point theorem)

⇒ RQ || BP and RQ = BP (∵ P is mid-point of BC)

⇒ BPQR is a parallelogram.

∵ Diagonals is (RBSESolutions.com) a parallelogram bisect each other.

∵ X is the mid-point of PR.

Similarly, PCQR is a parallelogram.

⇒ Y is the mid-point of PQ

Now in ΔPRQ, X and Y are mid-points of sides PR and PQ respectively.

⇒ XY = [latex]\frac { 1 }{ 2 }[/latex] RQ …… (ii)

From (i) and (ii), we get

XY = [latex]\frac { 1 }{ 2 }[/latex] RQ

= [latex]\frac { 1 }{ 2 }[/latex] ([latex]\frac { 1 }{ 2 }[/latex] BC)

= [latex]\frac { 1 }{ 4 }[/latex] BC

Hence, XY = [latex]\frac { 1 }{ 4 }[/latex] BC.

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

## Leave a Reply