Gujarat Board GSEB Solutions Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1

Question 1.

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

Solution:

We have

∠A : n∠B : ∠C : ∠D = 3 : 5 : 9 : 13

Let ∠A = 3x,

∠B = 5x,

∠C = 9x,

and ∠D = 13x

∠A + ∠B + ∠C + ∠D

Sum of all angles of a quadrilateral

3x + 5x + 9x + 13x = 360°

⇒ 30 x = 360°

x = \(\frac {360°}{30}\)

⇒ x = 12°

Hence, ∠A = 3x = 3 x 12° = 36°

∠B = 5x = 5 x 12° = 60°

∠C = 9x = 9 x 12° = 108°

and ∠D = 13 = 13 x 12° = 156°

Question 2.

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

Solution:

Given: ABCD is a parallelogram in which AC = BD.

To prove: ABCD is a rectangle.

Proof: In ΔABC and ΔBAD,

AB = BA (Common)

BC – AD (Opposite sides of a parallelogram)

AC = BD (Given)

∴ ΔABC s ΔBAD (by SSS congruency)

Hence ∠ABC = ∠BAD (by CPCT)

∠ABC + ∠BAD = 180° (Consecutive interior

angles)

∠BAD + ∠BAD = 180°

⇒ 2∠BAD = \(\frac {180°}{2}\)

∴ ∠BAD = 90°

Therefore, 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:

Given: ABCD is a quadrilateral in which diagonals AC and BD bisect each other at right angle, i.e., ∠AOB = 90°.

To Prove: ABCD is a rhombus.

Proof: In ΔAOB and ΔAOD,

OB = OD (Given)

∠AOB = ∠AOD (each 90°)

OA = OA (Common)

∴ ΔAOB ≅ ΔAOD (by SAS congruency)

Hence, AB = AD …….(1) (by CPCT)

Similarly, we can prove

AB = BC ……(2)

and BC = CD …….(3)

AB = BC = CD = AD [From (1), (2) and (3)]

Hence, ABCD is a rhombus.

Question 4.

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

Solution:

Given: ABCD is a square in which AC and BD are diagonals.

To Prove: (i) AC = BD

(ii) AC and BD bisect each other at right angles.

Proof: (i) In ΔABC and ΔBAD,

AB = BA (Common)

BC = AD (Opposite sides of a square)

∠ABC =∠BAD (each 90°)

ΔABC – ΔBAD (by SAS congruency)

Therefore AC = BD (by CPCT)

(iii) Now in ΔAOB and ΔCOD,

∠AOB = ∠COD (Vertically opposite angles)

∠OAB = ∠OCD (Alternate interior angles)

AB = CD (Opposite side of a square)

∴ ΔAOB – ΔCOD (by AAS congruency)

Hence, OA = OC (by CPCT)

OB = OD (by CPCT)

Now in ΔAOB and ΔBOC,

OA = OC (proved above)

AB = BC (Side of a square)

∴ ΔAOB = ΔBOC (by SSS congruency)

∴ ∠AOB = ∠BOC (by CPCT)

∠AOB + ∠BOC = 180° (Linear pair axiom)

⇒ ∠BOC + ∠BOC = 180°

⇒ 2∠BOC = 180°

⇒∠BOC = \(\frac {180°}{2}\)

∴ ∠BOC = 90°

Hence AC and BD bisect each other 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: In Quadrilateral ABCD, AC = BD and OA = OC, OB = OD and ∠AOB = 90°.

To Prove: ABCD is a square.

Proof In ΔAOB and ΔCOÐ

OA = OC (given)

∠AOB = ∠COD (each 90°)

OB = OD (given)

ΔAOB-ΔCOD (by SAS congruency)

AB = CD (by CPCT)

∠OAB = ∠OCD (by CPCT)

(Alternate interior angles)

⇒ AB || CD

Hence ABCD is a parallelogram.

Now in ΔAOB and ΔBOC

OA = OC (given)

∠AOB = ∠BOC (each 90°)

OB = OB (common)

∴ ΔAOB – ΔBOC

(by SAS congruency)

Hence AB = BC (by CPCT)

Similarly, we can prove BC = CD

and CD = DA

Now in ΔABC and ΔBAD

AB = BA (common)

BC = AD

(Opposite sides of a parallelogram)

AC = BD (given)

∴ ΔABC = ΔBAD (by SSS congruency)

Hence ∠ABC = ∠BAD (by CPCT)

∠ABC + ∠BAD = 180° (Consecutive interior angles of ||gm)

⇒ ∠BAD + ∠BAD = 180°

2∠BAD = 180°

∴ ∠BAD = 90°

Thus, ABCD is a square.

Question 6.

Diagonal AC of a parallelogram ABCD bisects ∠A. Show that

(i) it bisects ∠C also,

(ii) ABCD is a rhombus

Solution:

Given: ABCD is a parallelogram.

∠1 = ∠2

To Prove: ∠3 = ∠4

Proof: (i) ∠1 = ∠2 …….(1) (given)

AB || CD

∴ ∠1 = ∠4 ……..(2)

(Alternate interior angles)

BC || AD

∴ ∠2 = ∠3 …….(3)

(Alternate interior angles)

From eqn. (1), (2) and (3)

∠3 = ∠4

Hence, AC bisects ∠C.

(ii) From eqn. (1) and (3), we get

∠1 = ∠3

∴ AB = BC

(Sides opposite to equal angles are equal)

Similarly, from eqn. (1) and (2),

∠2 = ∠4

∴ AD = DC

But AB = DC

Hence AB = BC = DC = AD

Therefore, 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: Rhombus ABCD, in which AC and BD are diagonals.

To Prove:

(i) Diagonal AC bisects ∠A as well as ∠C.

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

Proof: (i) ABCD is a rhombus.

AD = DC (Sides of rhombus)

∴ ∠DAC = ∠DCA ………(1)

(Angle opposite to equal sides are equal)

AB || DC (Opposite sides of rhombus)

∴ ∠BAC = ∠DCA …….(2)

(Alternate interior angles)

From eqn. (1) and (2),

∠BAC = ∠DAC

∴ Diagonal AC bisects ∠A.

Similarly, we can prove

∠BCA = ∠DCA

Hence diagonal AC also bisects ZC.

(ii) BC = DC (Sides of rhombus)

∠DBC = ∠BDC …….(3)

(Angles opposite to equal sides are equal)

AB || DC

(Opposite sides of rhombus)

∠ABD = ∠BDC …….(4)

(Alternate interior angles)

From eqn. (3) and (4)

∠DBC = ∠ABD Hence diagonal BD bisects ∠B.

Similarly, we can prove diagonal BD bisects ∠D.

Question 8.

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

(i) ABCD is a square.

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

Solution:

AB || DC

(Opposite sides of a rectangle) ∠BAC = ∠DCA ……(1)

(Alternate interior angle) But ∠BAC = ∠DAC …….(2) (given)

From eqn. (1) and (2)

∠DCA = ∠DAC

.-. AD = DC

(Sides opposite to equal angles are equal)

Hence ABCD is a square.

(ii) AD || BC

(Opposite sides of a square)

∠CBD = ∠ADB …….(3)

(Alternate interior angle)

But AB = AD (Side of square)

Hence ∠ABD = ∠ADB

(Angle opposite sides to equal are equal)

From (3) and (4)

∠ABD = ∠CBD …….(4)

Hence diagonal BD bisects ZB as well ZD.

Question 9.

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ. Show that

(i) AAPD = ACQB

(ii) AP = CQ

(iii) ΔAQB = ΔCPD

(iv) AQ = CP

(v) APCQ is a parallelogram.

Solution:

To Prove:

(i) In ΔAPD and ΔCQB,

AD =BC

(Opposite sides of a parallelogram)

BC ||AD

∠ADB = ∠CBD

(Alternate interior angles)

∴∠ADP = ∠CBQ

PD = BQ (given)

∴ ΔAPD = ΔCQB

(by SAS congruency)

(ii) ΔAPD = ΔCQB (Proved above)

Hence AP = CQ …….(1) (by CPCT)

(iii) In ΔAQB and ΔCPD,

AB = CD

(Opposite sides of a parallelogram)

AB|| CD

∴ ∠ABD = ∠CDB

(Alternate interior angles)

∠ABQ = ∠CDP

BQ = PD (given)

Hence, AAQB – ACPD

(by SAS congruency)

(iv) ΔABQ = ΔCPD (Proved above)

AQ = CP …….(2) (by CPCT)

(v) From eqn. (1) and (2)

APCQ is a prallelogram.

(If each pair of opposite sides of a quadrilat¬eral are equal then it 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) ΔAPB = ΔCQD

(ii) AP = CQ

Solution:

Given: ABCD is a parallelogram and AP ⊥ BD and CQ ⊥ BD respectively.

(i) In ΔAPB and ACQD,

AB = DC

(Opposite sides of a parallelogram)

AB || CD

∴ ∠ABP = ∠CDQ

(Alternate interior angles)

∠APB = ∠CQD (each 90°)

Hence, ΔAPB s ΔCQD

(by AAS congruency)

(ii) ΔAPB = ΔCQD (Proved above)

Hence, AP = CQ (by CPCT)

Question 11.

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF, vertices A, B and C are joined to verticals 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) ΔABC = ΔDEF

Solution:

(i) In ΔABC and ΔDEF

AB = DE (given)

AB || DE (given)

∴ ABED is a parallelogram.

(In a quadrilateral if one pair of opposite sides is equal and parallel then it is a parallelogram).

(ii) In ΔABC and ΔDEF

BC = EF (given)

BC || EF (given)

∴ BEFC is a parallelogram.

(If one pair of opposite sides of a quadrilateral is equal and parallel then, it is a parallelogram).

(iii) ABED is a parallelogram. (Proved above)

AD =BE

∴ AD || BE ……(1)

and BEFC is a parallelogram.

(Proved above)

BE =CF

BE || CF ……(2)

From eqn. (1) and (2)

AD || CF and AD = CF

(iv) AD || CF (Proved above)

and AD = CF

∴ ACFD is a parallelogram.

(If one pair of opposite side of a quadrilateral is equal and parallel, then it is a parallelogram)

(v) ACFD is a parallelogram

∴ AC = DF

(Opposite sides of a parallelogram)

(vi) Now in ΔABC and ΔDEF,

BC = EF (given)

AB = DE (given)

and AC = DF (Proved above)

∴ ΔABC = ΔDEF (by SSS congruency).

Question 12.

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

(i) ∠A = ZB

(ii) ∠C = ZD

(iii) ΔABC = ΔBAD

(iv) Diagonal AC = diagonal BD.

Solution:

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

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

Proof:

(i) AB || DC (Given)

AE || DC

and AD || CE (by construction)

Hence ADCE is a parallelogram.

(Both pair of opposite sides of a quadrilateral are parallel)

∴ ∠ADC = ∠AEC

(Opposite angles of a parallelogram)

AD = BC

∴ AD = CE

(Opposite sides of a parallelogram)

Hence BC = CE

∠CBE = ∠CEB

(Angle opposite to equal sides are equal)

∠CBE = ∠AEC ……(1) (∠AEC = ∠CEB) ∠DAB + ∠AEC = 180° …….(2)

(Consecutive interior angles)

∠ABC + ∠CBE = 180° …….(3) (Linear pair)

From eqn. (1) and (3),

∠ABC = ∠AEC = 180° ……(4)

Now from eqn. (2) and (4),

∠DAB + ∠AEC = ∠ABC + ∠AEC

∠DAB = ∠ABC

(ii) AB || DC ∠DAB + ∠ADC = 180°

(Consecutive interior angles)

∠DAB = ∠ABC

(Proved above)

∠ABC + ∠ADC = 180° ……(5)

and ∠ABC + ∠BCD = 180° …….(6)

(Consecutive interior angles) From eqn. (5) and (6),

∠ABC + ∠ADC = ∠ABC + ∠BCD

∴ ∠ADC = ∠BCD

(iii) In ΔABC and ΔBAD

AB = BA (common)

BC = AD (given)

ΔABC = ΔDAB (Proved above)

ΔABC = ΔBAD

(by SAS congruence rule)

(iv) ΔABC = ΔBAD (Proved above)

AC = BD (CBCT)