Students can Download Class 9 Maths Chapter 7 Quadrilaterals Ex 7.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 9 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 9 Maths Chapter 7 Quadrilaterals Ex 7.1

Question 1.

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

Solution:

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

Let ∠A+ ∠B : ∠C : ∠D = 3 : 5 : 9 : 13.

Sum of ratio = 3x + 5x + 9x + 13x = 30x

Sum of 4 angles of quadrilateral is 360°

∴ ∠A + ∠B + ∠C + ∠D = 360°

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

30x = 360

∴ x = \(\frac{360}{30}\) = 12°

∠A = 3x = 3 × 12 = 36°

∠B = 5x = 5 × 12 = 60°

∠C = 9x = 9 × 12 = 108°

∠D = 13x = 13 × 12 = 156°

Question 2.

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

Solution:

Data: Diagonals of a parallelogram are equal.

To Prove: ABCD is a rectangle.

Proof: Now ABCD is a parallelogram and diagonal AC = Diagonal BD (Data)

In ∆ABC and ∆ABD,

BC = AD (Opposite sides of a quadrilateral)

AC = BD (Data)

AB common.

∴ ∆ABC ≅ ∆ABD (SSS postulate)

∠ABC = ∠BAD

But, ∠ABC + ∠BAD = 180°

∠ABC + ∠ABC = 180°

2 ∠ABC = 180°

∴ ∠ABC = 90°

If an angle of a parallelogram is a right angle, it is called a rectangle.

∴ 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:

Data: ABCD is a parallelogram and diagonals AC and BD bisect at right angles at O’.

To Prove: ABCD is a rhombus.

Proof: Here, AC and BD bisect each other at right angles.

∴ AO = OC

BO = OD

and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°

If sides are equal to each other, then ABCD is said to be a rhombus.

Now, ∆AOD and ∆COD,

AO = OC (Data)

∠AOD = ∠COD = 90° (Data)

OD is common.

∴ ∆AOD ≅ ∆COD (SAS Postulate)

∴ AD = CD …………… (i)

Similarly,

∆AOD = ∆AOB

AD = AB ………… (ii)

∆AOB ≅ ∆COB

∴ AB = BC ……….. (iii)

∆COB ≅ ∆COD

∴ BC = CD ……………. (iv)

From (i), (ii), (iii) and (iv),

AB = BC = CD = AD

All 4 sides of parallelogram ABCD are equal, then it is a rhombus.

Question 4.

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

Solution:

Data: ABCD is a parallelogram. Its diagonals are AC and BD. They meet at ‘O’.

To Prove: i) AO = OC

BO = OD

ii) ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°.

Proof: In ∆ABC and ∆ABD,

BC = AD (diagonals of square is equal)

∠ABC = ∠BAD = 90° (angles of square)

AB is common.

∴ ∆ABC ≅ ∆ABD (SAS Postulate.)

In ∆AOB and ∆COD,

AB = DC (sides of square)

∠OAB = ∠OCD (alternate angles)

∠OBA = ∠ODC (alternate angles)

∴ ∆AOB ≅ ∆COD (ASA Postulate)

AO = OC

BO = OD …………….. (i)

Similarly,

∆AOB ≅ ∆BOC.

∴ ∠AOB = ∠BOC = 90°

Now, ∆COD ≅ ∆AOD, then

∴ ∠COD = ∠DOA = 90°

∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90° …………. (ii)

from (i) and (ii)

∴ Sides of a square are equal and bisect 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:

Data: ABCD is a quadrilateral and bisects each other at right angles, then it is a square.

AO = OC and AC = BD

BO = OD

∠AOB = ∠BOC = ∠COD + ∠DOA = 90°.

To Prove: ABCD is a square.

Proof: In ∆AOB and ∆COD,

AO = OC

BO = OD (Data)

∠AOB = ∠COD (Vertically opposite angles)

∴ ∆AOB ≅ ∆COD (SAS Postulate)

AB = CD …………. (i)

∠ABO = ∠CDO

∴ AB || CD ………… (ii)

From (i) and (ii) ABCD is a parallelogram.

Now, in ∆AOD and ∆COD,

AO = OC (Data)

∠AOD = ∠COD = 90° (Data)

OD is common

∴ ∆AOD ≅ ∆COD (SAS Postulate)

AD = CD …………. (iii)

AD = BC ………….. (iv)

From (ii), (iii) and (iv)

AB = BC = CD = AD

Four sides of a quadrilateral are they are equal to each other and bisect each other at right angles, then it is a square.

∴ 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:

Data: Diagonal AC of a parallelogram ABCD bisects ∠A.

To Prove: (i) Diagonal AC also bisects ∠C.

(ii) ABCD is a rhombus.

Proof: i) Diagonal AC also bisects ∠A.

∴ ∠DAC = ∠BAC …………… (i)

But, ∠DAC= ∠BCA (alternate angles)(ii)

∠BAC=∠DCA (alternate angles)(iii)

From (i), (ii) and (iii),

∠BCA = ∠DCA

∴ AC bisects ∠C.

(ii) ∠DAC = ∠DCA

∴ AD = DC

But, AD = BC

∴ DC = AB

∴ AB = BC = CD = DA

∴ 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:

ABCD is a rhombus. Diagonal AC bisects ∠A and ∠C.

To Prove: Diagonal BD bisects ∠B and ∠D.

Proof: In a rhombus all sides are equal and opposite angles are equal to each other.

In ∆ABC,

AB = BC

∴ ∠BAC = ∠BCA

∠B AC = ∠DCA (alternate angles)

∴ ∠BCA = ∠DCA …………… (i)

∴ AC bisects ∠C.

Now, ∠BCA = ∠DAC (alternate angles)

∠BAC = ∠DAC

∴ AC bisects ∠A

In ∆ABD

AD = DB

∴ ∠ABD = ∠ADB

∠ABD = ∠CDB

∴ BD bisects ∠D

∠ADB = ∠CBD

∠ABD = ∠CBD

∴ BD bisects ∠B.

Question 8.

ABCD is a rectangle in which diagonal

(i) ABCD is a square,

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

Solution:

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

To Prove:

(i) ABCD is a square.

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

Proof: (i) ABCD is a rectangle. AC and BD diagonals intersect at O’.

AB = DC (Opposite sides of rectangle)

AD = BC

∠BAD = ∠BCD (Opposite angles)

∴ \(\frac{1}{2}\) ∠BAD = \(\frac{1}{2}\) ∠BCD

∴ ∠DAC = ∠DCA (AC bisects ∠A)

∴ DC = DA

But, AB = CD and DA = BC (Data)

∴ AB = BC = CD = AD

∴ ABCD is a square.

(ii) In ∆ABD.

AB = AD

∠ABD = ∠ADB

∠ABD = ∠CDB (alternate angles)

∠ADB = ∠CDB

∴ BD bisects ∠D.

∠ADB = ∠CDB (alternate angles)

∠ABD = ∠CBD (alternate angles)

∴ BD bisects ∠B.

Question 9.

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

(i) ∆APD ≅ ∆CQB

(ii) AP = CQ

(iii) ∆AQB ≅ ∆CPD

(iv) AQ = CP

(v) APCQ is a parallelogram.

Solution:

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

To Prove:

(i) ∆AAPD ≅ ∆CQB

(ii) AP = CQ

(iii) ∆AQB ≅ ∆CPD

(iv) AQ = CP

(v) APCQ is a parallelogram.

Proof: ABCD is a parallelogram.

(i) In ∆APD and ∆CQB,

AD = BC (opposite sides)

∠ADP = ∠CBQ (alternate angles)

DP = BQ (Data)

∴ ∆APD ≅ ∆CQB (ASA Postulate)

(ii) AP = CQ

(iii) In ∆AQB and ∆CPD,

AB = CD (opposite sides)

∠ABQ = ∠CDP (alternate angles)

BQ = DP (Data)

∴ ∆AQB ≅ ∆CPD (ASA Postulate)

(iv) ∴ AQ = CP

(v) In ∆AQPand ∆CPQ,

AP = CQ (proved)

AQ = PC PQ is common.

∴ ∆AQP ≅ ∆CPQ

∴ ∠APQ = ∠CQP

These are pair of alternate angles.

∴AP || PC

Similarly, AQ || PC

∴ APCQ is a parallelogram.

Question 10.

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that

(i) ∆APB ≅ ∆CQD

(ii) AP = CQ.

Solution:

Data: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD.

To Prove:

(i) ∆APB ≅ ∆CQD

(ii) AP = CQ.

Proof: ABCD is a parallelogram. BD is diagonal.

AP ⊥ BD and CQ ⊥ BD.

(i) In ∆APB and ∆CQD,

∠APB = ∠CQD = 90°

AB = CD (opposite sides)

∠ABP = ∠CQD (alternate angles)

∴ ∆APB ≅ ∆CQD (AAS postulate)

(ii) ∴ AP = CQ.

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. 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:

Data: 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

To Prove:

(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.

Proof: (i) AB = DE and AB||DE (Data)

∴ BE = AD and BE || AD

∴ ABCD is a parallelogram.

(ii) Similarly, BC = EF and BC || EF.

∴ BE = CF

BE || CF

∴ BEFC is a parallelogram.

(iii) ABED is a parallelogram.

∴ AD = BE

AD || BE ………. (i)

Similarly, BEFC is a parallelogram.

∴ CF = BE

CF || BE ……………. (ii)

Comparing (i) and (ii),

AD = CF and AD || CF,

(iv) In a quadrilateral ACFD,

AD = CF

AD || CF (proved)

∴ AC = DF

AC || DF

∴ ACFD is a parallelogram.

(v) ACFD is a parallelogram.

AC = DF (opposite sides).

(vi) In ∆ABC and ∆DEF,

AB = DE (Data)

BC = EF (opposite sides of a

parallelogram)

AC = DF (Opposite sides of a

parallelogram)

∴ ∆ABC ≅ ∆DEF (SSS postulate).

Question 12.

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

Solution:

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

To Prove:

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) ∆ABC ≅ ∆BAD

(iv) diagonal AC = diagonal BD.

Construction: Produce straight line AB.

Draw a parallel line DA, parallel to AB such that the produced line meets this at E.

Proof: (i) ABCD is a trapezium.

AB || CD and AD = BC (Data)

Now, AD || CD and AB || DC.

∴ ADCE is a parallelogram.

∴ AD = CE (Opposite sides)

AD = BC (Data)

CE = BC

∴ ∠CBE = ∠CEB

Now, ∠DAB + ∠CEB = 180° (Sum of consecutive angles.)

∴ ∠DAB + ∠CEB = 180°

(∵ ∠CEB = ∠CBE) ……………… (i)

Now, ∠ABC + ∠CDE = 180°

(linear pair) ……………….. (ii)

Comparing (i) and (ii),

∠DAB + ∠CBE = ∠ABC + ∠CBE

∴ ∠DAB = ∠ABC Or ∠A = ∠B.

(ii) AB || CD

∠DAB + ∠ADC =180° ……….. (iii)

∠ABC + ∠BCD = 180° ………… (iv)

Compairing (iii) and (iv),

∠DAB + ∠ADC = ∠ABC + ∠BCD

∴ ∠ADC = ∠BCD (∴∠A = ∠B)

Or ∠D = ∠C.

(iii) In ∆ABC and ∆ABD,

BC = AD (Data)

∠ABC = ∠BAD (proved)

AB is common.

∴ ∆ABC ≅ ∆ABD (SAS Postulate).

(iv) ∆ABC ≅ ∆BAD (Proved)

∴ Diagonal AC = Diagonal BD.