BSEB Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 8 Quadrilaterals Ex 8.1 Book Answers |

## Bihar Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.1 Textbooks Solutions PDF

Bihar Board STD 9th Maths Chapter 8 Quadrilaterals Ex 8.1 Books Solutions with Answers are prepared and published by the Bihar Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of BSEB Class 9th Maths Chapter 8 Quadrilaterals Ex 8.1 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.1 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Bihar Board STD 9th Maths Chapter 8 Quadrilaterals Ex 8.1 Textbooks. These Bihar Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.1 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Bihar Board.## Bihar Board Class 9th Maths Chapter 8 Quadrilaterals Ex 8.1 Books Solutions

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## BSEB Class 9th Maths Chapter 8 Quadrilaterals Ex 8.1 Textbooks Solutions with Answer PDF Download

Find below the list of all BSEB Class 9th Maths Chapter 8 Quadrilaterals Ex 8.1 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:## BSEB Bihar Board Class 9th Maths Solutions Chapter 8 Quadrilaterals Ex 8.1

Question 1.

The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Solution:

Let the angles be (3x)°, (5x)°, (9x)° and (13x)°

Then, 3x + 5x + 9x + 13x = 360

⇒ 30x = 360

⇒ x = 36030 = 12

∴ The angles are (3 x 12)°, (5 x 12)°, (9 x 12)° and (13 x 12)° i.e., 36°, 60°, 108° and 156°.

Question 2.

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

Solution:

A parallelogram ABCD in which AC = BD.

To prove : ABCD is a rectangle.

Proof :

In ∆s ABC and DCB, we have

AB = DC [Opp.sides of a ||gm]

BC = BC [Common]

and, AC = DB [Given]

∴ By SSS criterion of congruence.

∆ ABC ≅ ∆ DCB

⇒ ∠ABC = ∠DCB … (1)

[Corresponding parts of congruent triangles are equal]

But AB || DC and BC cuts them.

∴ ∠ACB + ∠DCB = 180° … (2)

⇒ 2∠ABC = 180° [Sum of consecutive interior angles is 180°]

⇒ ∠ABC = 90°

Thus, ∠ABC = ∠DCB= 90°

⇒ ABCD is a parallelogram one of whose angle is 90°.

Hence, ABCD is a rectangle.

Question 3.

Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Solution:

A quadrilateral ABCD in which the diagonals AC and BD intersect at O such that AO = OC,

BO = OD and AC ⊥ BD.

To prove : ABCD is a rhombus.

Proof:

Since the diagonals AC and BD of quadrilateral ABCD bisect each other at right angles.

∴ AC is the perpendicular bisector of the segment BD.

⇒ A and C both are equidistant from B and D.

⇒ AB = AD and CB = CD … (1)

Also, BD is the perpendicular bisector of line segment AC.

⇒ B and D both are equidistant from A and C.

⇒ AB = BC and AD = DC … (2)

From (1) and (2), we get

AB = BC = CD = AD

Thus, ABCD is a quadrilateral whose diagonals bisect each other at right angles and all four sides are equal.

Hence, ABCD is a rhombus. .

Second Proof: First we shall prove that ABCD is a ||gm

In ∆s AOD and COB, we have

AO = OC

OD = OB [Given]

∠AOD = ∠COB [∵ertically opp. angles]

By SAS criterion of congruence,

∆ AOD ≅ ∆ COB

⇒ ∠OAD = ∠OCB … (1)

[Corresponding parts of congruent triangles are equal] Now, line AC intersects AD and BC at A and C respectively such that

∠OAD =∠OCB [From (1)]

i.e., alternate interior angles are equal.

∴ AD || BC Similarly,

AB || CD Hence, ABCD is a parallelogram.

Now, we shall prove that ||gm ABCD is a rhombus.

In ∆s AOD and COD, we have

OA = OC [Given]

∠AOD = ∠COD [Both are right angles]

OD = OD [Common]

∴By SAS criterion of congruence

∆ AOD ≅ ∆ COD

⇒ AD = CD … (2)

[Corresponding parts of congruent triangles are equal] Now, ABCD is a ||gm [Proved above]

⇒ AB = CD and AD = BC

[Opp. sides of a ||gm are equal] ⇒ AB = CD = AD = BC [Using (2)]

Hence, quadrilateral ABCD is a rhombus.

Question 4.

Show that the diagonals of a square are equal and bisect each other at right angles.

Solution:

Given : A square ABCD.

To prove : AC = BD, AC ⊥ BD and OA = OC, OB = OD.

Proof:

Since ABCD is a square. Therefore, AB || DC and AD || BC.

Now, AB || DC and transversal AC intersects them at A and C respectively.

∴ ∠BAC = ∠DCA

[Alternate interior angles are equal]

⇒ ∠BAO =∠DCO … (1)

Again, AB || DC and BD intersects them at B and D respectively.

[∵ Alternate interior angles are equal] … (2)

Now, in ∆s AOB and ∠COD, we have

∠BAO = ∠DCO [From (1)]

AB = CD [Opposite sides of a ||gm are equal]

and, ∠ABO = ∠CDO [From (2)]

By ASA congruence criterion

∆ AOB ≅ ∆ COD

⇒ OA = OC and OB = OD

[Corresponding parts of congruent As are equal] Hence, the diagonals bisect each other.

In ∆s ADB and BCA, we have

AD = BC [Sides of a square are equal]

∠BAD = ∠ABC [Each equal to 90°}

and, AB = BA [Common]

∴ By SAS criterion of congruence

∆ ADB ≅ ∆ BCA

⇒ AC = BD

[∵ Corresponding parts of congruent As are equal]

Hence, the diagonals are equal.

Now in As AOB and AOD, we have OB = OD

[∵ Diagonals of ||gm bisect each other]

AB = AD

[∵ Sides of a square are equal]

and, AO = AO [Common]

∴ By SSS criterion of congruence

∆ AOB ≅ ∆ AOD

⇒ ∠AOB – ∠AOD

[Corresponding parts of congruent ∆s are equal]

But ∠AOB + ∠AOD = 180°

∴∠AOB = ∠AOD = 90°

⇒ AO ⊥ BD

⇒ AC ⊥ BD .

Hence, diagonals intersect at right angles.

Question 5.

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

Solution:

Given : A quadrilateral ABCD in which the diagonals AC = BD, AO = OC, BO = OD and AC ⊥ BD.

To prove : Quadrilateral ABCD is a square.

Proof :

First, we shall prove that ABCD is a parallelogram.

In ∆s AOD and COB, we have AO = OC OD = OB

∠AOD = ∠COB [∵Vertically opp. angles]

By SAS criterion of congruence,

∆ AOD ≅ ∆ COB

⇒ ∠OAD = ∠OCB …(1)

[Corresponding parts of congruent triangles are equal] Now, line AC intersects AD and BC at A and C respectively such that

∠OAD = ∠OCB [From (1)]

i.e., alternate interior angles are equal.

∴ AD || BC

Similarly, AB || CD

Hence, ABCD is a parallelogram.

Now, we shall prove that it is a square.

In ∆s AOB and ∠AOD, we have

AO = BO [Common]

∠AOB = ∠AOD [Each = 90°, given]

and, OB = OD [∵ Diagonals of a ||gm bisect each other]

∴By SAS criterion of congruence

∆ ABD ≅ ∆ BAC

⇒ AB = AD

[Corresponding parts of congruent triangles are equal]

But AB = CD and AD = BC

[Opp. sides of a ||gm are equal]

∴ AB = BC = CD = AD … (2)

Now, in ∆s ABD and BAC, we have

AB = BA [Common]

AD = BC [Opp. sides of a ||gm are equal] and, BD = AC [Given]

∴ By SSS criterion of congruence

∆ ABD ≅ ∆ BAC

⇒ ∠DAB = ∠CBA [Corresponding parts of congruent As are equal]

Question 6.

Diagonal AC of a nparallelogram ABCD bisects ∠A see figure). Show that

(i) it bisects ∠C also,

(ii) ABCD is a rhombus.

Solution:

(i) Given : A parallelogram ABCD in which diagonal AC bisects ∠A.

To prove : That AC bisects ∠C.

Proof:

Since ABCD is a ||gm. Therefore, AB || DC.

Now, AB || DC and AC intersects them

∴ ∠1 = ∠3 … (1) [Alternate interior angles]

Again, AD || BC and AC intersects them

∴ ∠2 = ∠4 … (2) [Alternate interior angles]

But it is given that AC is the bisector of ∠A.

∴ ∠1 = ∠2 … (3)

From (1), (2) and (3), we have ∠3 = ∠4

Hence, AC bisects ∠C.

(ii) To prove : That ABCD is a rhombus.

From part (i) : (1), (2) and (3) give ∠1 = ∠2 = ∠3 = ∠4 Now in ∆ ABC,

∠1 = ∠4

AB = BC [Sides opp. to equal angles in a ∆ are equal]

∴AD = DC

Similarly, in ∆ ADC, we have

AD = DC

Also, ABCD is a ||gm

∴ AB = CD, AD = BC

Combining these, we get

AB = BC = CD = DA

Hence, ABCD is a rhombus.

Question 7.

ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Solution:

Given : A rhombus ABCD.

To prove that (i) Diagonal t AC bisects ∠A as well ∠C.

(ii) Diagonal BD bisects ∠B as well as ∠D.

Proof:

In ∆ ADC, AD = DC

[Sides of a rhombus are equal]

⇒ ∠DAC = ∠DCA …(1)

[Angles opp. to equal sides of a triangle are equal]

Now AB || DC and AC intersects them

∠BCA = ∠DAC …(2) [Alternate angles]

From (1) and (2), we have

∠DCA = ∠BCA

⇒ AC bisects ∠C

In ∆ ABC, AB = BC [Sides of a rhombus are equal]

⇒ ∠BCA = ∠BAC …(3)

[Angles opp. to equal sides of a triangle are equal]

From (2) and (3), we have

∠BAC = ∠DAC

⇒ AC bisects ∠A

Hence, diagonal AC bisects ∠A as well as ∠C.

Similarly, diagonal BD bisects ∠B as well as ∠D.

Question 8.

ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that : (i) ABCD is a square (ii) diagonal BD bisects ∠B as well as ∠D.

Solution:

Given : ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.

To prove that (i) ABCD is a square

(ii) diagonal BD bisects ∠B as well as ∠D.

Proof:

(i) Since AC bisects ∠A as well as ∠C in the rectangle ABCD.

∴ ∠1 = ∠2 = ∠3 = ∠4 [∵ Each = 90°2 = 45°]

∴ In ∆ ADC, ∠2 = ∠4

⇒ AD = CD [Sides opposite to equal angles]

Thus, the rectangle ABCD is a square.

(ii) In a square, diagonals bisect the angles.

So, BD bisects ∠B as well as ∠D.

Question 9.

In parallelogram ABCD, two points P and Q are taken dn diagonal BD such that DP = BQ (see figure). Show that:

(i) ∆ APD ≅ ∆ CQB

(ii) AP = CQ

(iii) ∆ AQB ≅ ∆ CPD

(iv) AQ = CP

(v) APCQ is a parallelogram

Solution:

ABCD is a parallelogram. P and Q are points on the diagonal BD such that DP = BQ.

To prove :

(i) ∆ APD ≅ ∆ CQB

(ii) AP = CQ

(ii) ∆ AQB ≅ ∆ CPD

(iv) AQ = CP

(v) APCQ is a parallelogram

Construction : Join AC to meet BD in O.

Proof:

We know that the diagonals of a parallelogram bisect each other. Therefore, AC and BD bisect each other at O.

OB = OD

But BQ = DP [Given]

⇒ OB – BQ = OD – DP ⇒ OQ = OP

Thus, in quadrilateral APCQ diagonals AC and PQ are such that OQ = OP and OA = OC. i.e., the diagonals AC and PQ bisects each other.

Hence, APCQ is a parallelogram, which prove the (v) part.

(i) In ∆s APD and CQB, we have

AD = CB [Opp. sides of a ||gm ABCD]

AP = CQ [Opp. sides of a ||gm APCQ]

DP = BQ [Given]

∴ By SSS criterion of congruence

∆ APD ≅ ∆ CQB

(ii) AP = CQ [Opp. sides of a ||gm APCQ]

(iii) In As AQB and CPD, we have

AB = CD [Opp. sides of a ||gm ABCD]

AQ = CP [Opp. sides of a ||gm APCQ]

BQ = DP [Given]

∴ By SSS criterion of congruence

∆ AQB ≅ ∆ CPD

(iv) AQ = CP [Opp. sides of a ||gm APCQ]

Question 10.

ABCD is a parallelogram and AP and CQ are perpendiculars fi-om vertices A and C on diagonal BD respectively (see figure). Show that

(i) ∆ APB ≅ ∆ CQD

(ii) AP = CQ

Solution:

(i) Since ABCD is a parallelogram. Therefore,

DC || AB.

Now, DC || AB and transversal BD intersects them at B and D.

∴ ∠ABD = ∠BDC [Alternate interior angles]

Now, in As APB and CQD, we have

∠ABP = ∠QDC [∵ ∠ABD = ∠BDC]

∠APB = ∠CQD [Each = 90°]

and, AB = CD [Opp. sides of a ||gm]

∴ By AAS criterion of congruence

∆ APB ≅ ∆ CQD

(ii) Since ∆ APB ≅ ∆ CQD

∴ AP = CQ

[∵ Corresponding parts of congruent triangles are equall

Question 11.

In ∆ ABC and ∆ DEF, AB = DE, AB || DE, BC = EF and. BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see figure). Show that

(i) quadrilateral ABED is a parallelogram

(ii) quadrilateral BEFC is a parallelogram

(iii) AD || CF and AD = CF

(iv) quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) ∆ ABC ≅ ∆ DEF.

Solution:

Given : Two ∆s ABC and DEF such that AB = DE and AB || DE. Also BC = EF and BC || EF.

To prove that: (i) quadrilateral ABED is a parallelogram.

(ii) quadrilateral BEFC is a parallelogram.

(iii) AD || CF and AD = CF.

(iv) quadrilateral ACFD is a parallelogram.

(v) AC || DF and AC = DF

(vi) ∆ ABC = ∆ DEF.

Proof: (i) Consider the quadrilateral ABED

We have, AB = DE and AB || DE

⇒ One pair of opposite sides are equal and parallel.

⇒ ABED is a parallelogram.

(ii) Now, consider quadrilateral BEFC, we have

BC = EF and BC || EF

One pair of opposite sides are equal and parallel.

⇒ BEFC is a parallelogram.

(iii) Now, AD = BE and AD || BE … (1) [∵ ABED is a ||gm]

and CF = BE and CF || BE … (2) [∵ BEFC is a ||gm]

From (1) and (2), we have

AD = CF and AD || CF.

(iv) Since AD = CF and AD || CF

⇒ One pair of opposite sides are equal and parallel

⇒ ACFD is a parallelogram.

(v) Since ACFD is a parallelogram.

∴ AC = DF [Opp. sides of a ||gm ACFD]

(vi) In ∆s ABC and DEF, we have

AB = DE [Opp. sides of a ||gm ABED]

BC = EF [Opp. sides of a ||gm BEFC]

and, CA = FD [Opp. sides of ||gm ACFD]

∴ By SSS criterion of congruence ∆ ABC ≅ ∆ DEF.

Question 12

ABCD is a trapezium in which AB || CD and AD = BC (see figure). Show that

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) ∆ ABC ≅ ∆ BAD

(iv) diagonal AC = diagonal BD

Solution:

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

To prove that:

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) ∆ ABC ≅ ∆ BAD

(iv) diagonal AC = diagonal BD.

Construction : Produce AB and draw a line CE || AD.

Proof:

(i) Since AD || CE and transversal AE cuts them at A and E respectively.

∴ ∠A + ∠E = 180° … (1)

Since AB || CD and AD || CE. Therefore, AECD is a parallelogram.

⇒ AD = CE

BC = CE [∵ AD = BC (given)]

Thus, in A BCE, we have BC = CE

⇒ ∠CBE = ∠CEB

⇒ 180 – ∠B = ∠E

⇒ 180 – ∠E = ∠B

From (1) and (2), we get ∠A = ∠B

(ii) Since ∠A = ∠B ⇒ ∠BAD = ∠ABD

⇒ 180° – ∠BAD = 180° – ∠ABD

⇒ ∠ADB = ∠BCD

⇒ ∠D = ∠C i.e., ∠C = ∠D.

(iii) In ∆s ABC and BAD, we have

BC = AD [Given]

AB = BA [Common]

∠A = ∠B [Proved]

∴ By SAS criterion of congruence ∆ ABC ≅ ∆ BAD

(iv) Since ∆ ABC = ∆ BAD

∴ AC = BD

[Corresponding parts of congruent triangles are equal]

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- BSEB Class 9 Maths Chapter 14 Statistics Ex 14.1 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 14 Statistics Ex 14.1 Book Answers
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- BSEB Class 9 Maths Chapter 15 Probability Ex 15.1 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 15 Probability Ex 15.1 Book Answers

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