Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 Textbook Exercise Questions and Answers.

## RBSE Class 9 Maths Solutions Chapter 8 Quadrilaterals Ex 8.1

Question 1.

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

Answer:

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

Then, 3x + 5x + 9x + 13x = 360°

or 3x = 360°

or x = \(\frac{360^{\circ}}{30}\) = 12

∴ The angles are (3 × 12)°, (5 × 12)°, (9 × 12)° and (13 × 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.

Answer:

A parallelogram ABCD in which AC = BD.

To show : ABCD is a rectangle.

In ∆ABC and ∆DCB, we have :

AB = DC (Opp. sides of a ∥gm)

BC = CB (Common)

and, AC = DB (Given)

∴ By SSS criterion of congruence, we have :

∆ABC = ∆DCB

So, ∠ABC = ∠DCB …………(1)

(Corresponding parts of congruent triangles are equal.) But AB ∥ DC and BC cuts them.

∴ ∠ABC + ∠DCB = 180° …………(2) (Sum of interior angles on the same side of the transversal is 180°.)

or 2∠ABC = 180° [From (1) and (2)]

⇒ ∠ABC = 90°

Thus, ∠ABC = ∠DCB = 90°

So, ABCD is a parallelogram one of whose angles is 90°.

Hence, ABCD is a rectangle. Hence proved.

Question 3.

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

Answer:

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

To show : ABCD is a rhombus.

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

∴ AC is the perpendicular bisector of the line segment BD.

A and C both are equidistant from B and D.

So, AB = AD and CB = CD

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

B and D both are equidistant from A and C.

So, 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.

Hence proved.

Alternate Method :

First we shall show that ABCD is a ∥gm.

In As AOD and COB, we have :

AO = OC (Given)

OD = OB (Given)

and ∠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 show that ∥gm ABCD is a rhombus.

In As AOD and COD, we have :

OA = OC (Given)

∠AOD = ∠COD (Both are right angles)

OD = OD , (Common)

By SAS criterion of congruence, we have :

∆AOD = ∆COD

AD = CD …..(2)

(Corresponding parts of congruent triangles are equal.) (Shown 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.

Hence proved.

Question 4.

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

Answer:

Given : A square ABCD.

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

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.

∴ ∠ABD – ∠CDB (∵ Alternate interior angles are equal.)

⇒ ∠ABO = ∠CDO ………..(2)

Now, in As AOB and COD, we have :

∠BAO = ∠DCO

AB = CD

and, ∠ABO = ∠CDO

By ASA congruence criterion, we have:

∆AOB = ∆COD

So, OA = OC and OB = OD

(Corresponding parts of congruent As are equal.)

Hence, the diagonals bisect each other.

In As ADB and BCA, we have :

AD = BC (Sides of a square are equal.)

∠BAD = ∠ABC (Each angle equal to 90°)

and, AB = BA (Common)

∴ By SAS criterion of congruence, we have :

∆ADB = ∆BCA

So, 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 a ∥gm bisect each other)

AB = AD (∵ Sides of a square are equal.)

and, AO = AO (Common)

∴ By SSS criterion of congruence, we have:

∆AOB = ∆AOD

So, ∠AOB = ∠AOD (Corresponding parts of congruent As are equal.)

But ∠AOB + ∠AOD = 180°

∴∠AOB = ∠AOD =90°

AO ⊥ BD AC ⊥ BD

Hence, diagonals intersect at right angles. Hence proved.

Question 5.

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

Answer:

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

To show : Quadrilateral ABCD is a square.

First, we shall show that ABCD is a parallelogram.

In As AOD and COB, we have :

OA = OC (Given)

OD = OB (Given)

and ∠AOD = ∠COB (Vertically opp. angles)

By SAS criterion of congruence,

∆AOD = ∆COB

So, ∠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 show that it is a square.

In As AOB and AOD, we have :

AO = BO (Common)

∠AOB = ∠AOD (Each = 90°, given)

and, OB =OD (v Diagonals of a ∥gm bisect each other)

∴ By SAS criterion of congruence, we have :

∆AOB = ∆AOD

So, AB =AD

(Corresponding parts of congruent triangles are equal.)

But AB = CD and AD = BC

AB = BC = CD = AD

Now, in As ABD and BAC, we have :

AB = BA AD =BC

and, BD = AC

By SSS criterion of congruence, we have :

∆ABD = ∆BAC

So, ∠DAB = ∠CBA

(Corresponding parts of congruent As are equal.)

But ∠DAB + ∠CBA = 180° (Adjacent angles of a parallelogram)

Similarly, other angles ∠ADC and ∠BCD are each equal to 90°.

Hence, ABCD is a square.

Hence Proved.

Question 6.

Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that:

(i) it bisects ∠C also,

(ii) ABCD is a rhombus.

Answer:

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

To show: That AC bisects ∠C.

Since ABCD is a ∥gm, therefore AB ∥ DC.

Now, AB ∥ DC and AC intersects them.

∠1 = ∠3 [alternate interior angles] ……………(1)

Again, AD ∥ BC and AC intersects them.

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

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 show: That ABCD is a rhombus.

From part (i): (1), (2) and (3) give ∠1 = ∠2 = ∠3 = ∠4

Now in ∆ABC, ∠1 = ∠4

So, AB = BC (Sides opp. to equal angles in a A are equal.)

Similarly, in AADC, we have :

AD = DC

Also, ABCD is a ∥gm.

∴ AB = CD and AD = BC.

Combining these, we get AB = BC = CD = DA

Hence, ABCD is a rhombus.

Hence proved.

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.

Answer:

Given : A rhombus ABCD.

To show that: (i) Diagonal AC bisects ∠A as well ∠C.

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

In ∆ADC, AD = DC

(Sides of a rhombus are equal) B

So, ∠DAC = ∠DCA …(1)

(Angles opp. to equal sides of a triangle are equal)

Now AB ∥ DC and AC intersects them.

∴ ∠BCA = ∠DAC (Alternate angles) ……………(2)

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, it can be shown that diagonal BD bisects ∠B as well as ∠D.

Hence proved.

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.

Answer:

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

To show that: (i) ABCD is a square.

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

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

∠1 = ∠2 = ∠3 = ∠4 [∵ Each = \(\frac{90^{\circ}}{2}\) = 45°]

In ∆ADC, ∠2 = ∠4

So, AD = CD (Sides opposite to equal angles of a triangle)

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 on 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

Answer:

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

To show :

(i) ∆APD ≅ ∆CQB

(ii) AP = CQ

(iii) ∆AQB ≅ ∆CPD

(iv) AQ = CP

(v) APCQ is a parallelogram

Construction : Join AC to meet BD in O,

(i) AD ∥ BC and BD is a transversal (ABCD is a ∥gm)

∠ADP = ∠CBQ (Alternate angles) …………. (1)

In As APD and CQB, we have :

AD = BC (Opposite sides of a ∥gm)

DP =BQ (Given)

and ∠ADP = ∠CBQ [From (1)]

∴ ∆APD = ∆CQB (SAS)

(ii) AP=CQ [CPCT in (i)]

(iii) AB ∥DC and BD is a transversal. (ABCD is a ∥gm)

∠ABQ = ∠CDP (Alternate angles)

In As AQB and CPD, we have :

AB = DC

BQ = DP

and ∠ABQ = ∠CDP

So, ∆AQB ≅ ∆CPD (SAS)

(iv) AQ = CP [CPCT in (iii)]

(v) OA = OC (AC andBD bisect each other) ………..(1)

and OB = OD

So, OB – BQ = OD – DP (BQ = DP, given) ………………. (2)

or OQ = OP [From (1) and (2)]

Thus, AC and PQ bisect each other So, APCQ is a ∥gm.

Aliter: 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 or 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 bisect each other.

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

(i) In As 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, we have :

∆APD = ∆CQB

(ii) AP = CQ (Corresponding parts of congruent triangles)

(iii) In As AQB and CPD, we have :

AB = CD AQ = CP BQ = DP

∴ By SSS criterion of congruence, we have :

∆AQB = ∆CPD

(iv) AQ = CP (Corresponding parts of congruent triangles)

(v) In As AQC and PCA, we have AQ = CP, CQ = AP and AC = AC

∆AQC ≅ ∆PCA (By SSS Rule)

So, ∠ACQ = ∠CAD (CPCT)

i.e., QA ∥ CP

Since QA = CP and QA ∥ CP i.e., one pair of opposite sides are equal and parallel.

∴ APCQ is a parallelogram.

Question 10.

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

(i) ∆APB ≅ ∆CQD

(ii) AP = CQ

Answer:

(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, we have :

∆APB ≅ ∆CQD

(ii) Since ∆APB = ∆CQD, therefore

AP =CQ (∵ Corresponding parts of congruent triangles are equal)

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.

Answer:

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

To 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.

(i) Consider the quadrilateral ABED.

We have : AB = DE and AB ∥ DE.

That is, one pair of opposite sides are equal and parallel.

So, ABED is a parallelogram.

(ii) Now, consider quadrilateral BEFC. We have :

BC = EF and BC ∥ EF

That is, one pair of opposite sides are equal and parallel.

So, BEFC is a parallelogram.

(iii) Now, AD = BE and AD ∥ BE (∵ ABED is a ∥gm) (1)

and CF = BE and CF ∥BE (∵ BEFC is a ∥gm) (2)

From (1) and (2), we have :

AD = CF and AD ∥ CF.

(iv) Since AD = CF and AD ∥ CF, therefore one pair of opposite sides are equal and parallel So, ACFD is a parallelogram.

(v) Since ACFD is a parallelogram.

AC = DF (Opp. sides of a ∥gm ACFD)

(vi) In As ABC and DEF, we have :

AB = DE (Given)

BC = EF (Given)

and, CA = FD [Proved in (v)]

∴ By SSS criterion of congruence, we have :

∆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 [Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.

Answer:

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

To show that:

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) ∆ABC = ∆BAD

(iv) diagonal AC = diagonal BD

Construction :

Produce AB and draw a line CE ∥ DA. Also, join AC and BD.

(i) Since AD ∥ CE and transversal AB cuts them at A and E respectively, therefore

∠A + ∠E = 180° (1)

Since AB ∥ CD and AD ∥ CE, therefore AECD is a parallelogram.

So, AD = CE

∴ BC = CE [v AD = BC (given)]

Thus, in ABCE, we have : BC = CE

∴∠CBE = ∠CEB

So, 180° – ∠B = ∠E

180° – ∠E = ∠B (2)

From (1) and (2), we get ∠A = ∠B

(ii) Since ∠A = ∠B, therefore ∠BAD = ∠ABC

∴ 180° – ∠BAD = 180° – ∠ABC

So, ∠ADC = ∠BCD

or ∠D = ∠C, i.e., ∠C = ∠D.

(iii) In As ABC and BAD, we have :

BC = AD (Given)

AB = BA (Common)

∠B = ∠A (Shown above)

∴ By SAS criterion of congruence, we have :

∆ABC = ∆BAD

∆ABC = ∆BAD, therefore AC =BD

(Corresponding parts of congruent triangles are equal.)

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