ML Aggarwal Solutions for Class 9 Maths Chapter 13 Rectilinear Figures

ML Aggarwal Solutions for Class 9 Maths Chapter 13 Rectilinear Figures provides precise solutions to the exercise problems present in the chapter. These solutions are created in a step-by-step method for easy understanding. These are effective resources for any doubt clearance and also for a quick reference to concepts present in the ML Aggarwal textbooks. Further, students can now refer to solutions of this chapter from the ML Aggarwal Solutions for Class 9 Maths Chapter 13 Rectilinear Figures PDF, which is available in the link given below.

Chapter 13 has problems based on the properties of parallelograms and angles sum property of a quadrilateral. ML Aggarwal solutions are available in PDF format, so that the students can use them to solve exercise problems effortlessly. Using this resource on a daily basis will improve the problem-solving skills of students and hence, boost their confidence to perform well in their examinations.

ML Aggarwal Solutions for Class 9 Maths Chapter 13 Rectilinear Figures Download PDF

 

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Access ML Aggarwal Solutions for Class 9 Maths Chapter 13 Rectilinear Figures

Exercise 13.1

1. If two angles of a quadrilateral are 40° and 110° and the other two are in the ratio 3 : 4, find these angles.

Solution:

We know that,

Sum of all four angles of a quadrilateral = 360o

Sum of two given angles = 40o + 110o = 150o

So, the sum of remaining two angles = 360o – 150o = 210o

Also given,

Ratio in these angles = 3 : 4

Hence,

Third angle = (210o x 3)/(3 + 4)

= (210o x 3)/7

= 90o

And,

Fourth angle = (210o x 4)/(3 + 4)

= (210o x 4)/7

= 120o

ML Aggarwal Solutions for Class 9 Chapter 13 - 12. If the angles of a quadrilateral, taken in order, are in the ratio 1 : 2 : 3 : 4, prove that it is a trapezium.

Solution:

Given,

In trapezium ABCD in which

∠A : ∠B : ∠C : ∠D = 1 : 2 : 3 : 4

We know,

The sum of angles of the quad. ABCD = 360o

∠A = (360o x 1)/10 = 36o

∠B = (360o x 2)/10 = 72o

∠C = (360o x 3)/10 = 108o

∠D = (360o x 4)/10 = 144o

Now,

∠A + ∠D = 36o + 114o = 180o

Since, the sum of angles ∠A and ∠D is 180o and these are co-interior angles

Thus, AB || DC

Therefore, ABCD is a trapezium.

3. If an angle of a parallelogram is two-thirds of its adjacent angle, find the angles of the parallelogram.

Solution:

Here ABCD is a parallelogram.

Let ∠A = xo

Then, ∠B = (2x/3)o (Given condition)

So,

∠A + ∠B = 180o (As the sum of adjacent angles in a parallelogram is 180o)

ML Aggarwal Solutions for Class 9 Chapter 13 - 2
ML Aggarwal Solutions for Class 9 Chapter 13 - 3

Hence, ∠A = 108o

∠B = 2/3 x 108o = 2 x 36o = 72o

∠B = ∠D = 72o (opposite angles in a parallelogram is same)

Also,

∠A = ∠C = 108o (opposite angles in a parallelogram is same)

Therefore, angles of parallelogram are 108o, 72o, 108o and 72o.

4. (a) In figure (1) given below, ABCD is a parallelogram in which ∠DAB = 70°, ∠DBC = 80°. Calculate angles CDB and ADB.

(b) In figure (2) given below, ABCD is a parallelogram. Find the angles of the AAOD.

(c) In figure (3) given below, ABCD is a rhombus. Find the value of x.

ML Aggarwal Solutions for Class 9 Chapter 13 - 4

Solution:

(a) Since, ABCD is a || gm

We have, AB || CD

∠ADB = ∠DBC (Alternate angles)

∠ADB = 80o (Given, ∠DBC = 80o)

Now,

In ∆ADB, we have

∠A + ∠ADB + ∠ABD = 180o (Angle sum property of a triangle)

70o + 80o + ∠ABD = 180o

150o + ∠ABD = 180o

∠ABD = 180o – 150o = 30o

Now, ∠CDB = ∠ABD (Since, AB || CD and alternate angles)

So,

∠CDB = 30o

Hence, ∠ADB = 80o and ∠CDB = 30o.

(b) Given, ∠BOC = 35o and ∠CBO = 77o

In ∆BOC, we have

∠BOC + ∠BCO + ∠CBO = 180o (Angle sum property of a triangle)

∠BOC = 180o – 112o = 68o

Now, in || gm ABCD

We have,

∠AOD = ∠BOC (Vertically opposite angles)

Hence, ∠AOD = 68o.

(c) ABCD is a rhombus

So, ∠A + ∠B = 180o (Sum of adjacent angles of a rhombus is 180o)

72o + ∠B = 180o (Given, ∠A = 72o)

∠B = 180o – 72o = 108o

Hence,

x = ½ B = ½ x 108o = 54o

5. (a) In figure (1) given below, ABCD is a parallelogram with perimeter 40. Find the values of x and y.

(b) In figure (2) given below. ABCD is a parallelogram. Find the values of x and y.

(c) In figure (3) given below. ABCD is a rhombus. Find x and y.

ML Aggarwal Solutions for Class 9 Chapter 13 - 5

Solution:

(a) Since, ABCD is a parallelogram

So, AB = CD and BC = AD

⇒ 3x = 2y + 2

3x – 2y = 2 … (i)

Also, AB + BC + CD + DA = 40

⇒ 3x + 2x + 2y + 2 + 2x = 40

7x + 2y = 40 – 2

7x + 2y = 38 … (ii)

Now, adding (i) and (ii) we get

3x – 2y = 2

7x + 2y = 38

——————

10x = 40

⇒ x = 40/10 = 4

On substituting the value of x in (i), we get

3(4) – 2y = 2

12 – 2y = 2

2y = 12 – 2

⇒ y = 10/2 = 5

Hence, x = 4 and y = 5

(b) In parallelogram ABCD, we have

∠A = ∠C (Opposite angles are same in || gm)

3x – 20o = x + 40o

3x – x = 40o + 20o

2x = 60o

x = 60o/2 = 30o … (i)

Also, ∠A + ∠B = 180o (Sum of adjacent angles in || gm is equal to 180o)

3x – 20o + y + 15o = 180o

3x + y = 180o + 20o – 15o

3x + y = 185o

3(30o) + y = 185o [Using (i)]

90o + y = 185o

y = 185o – 90o = 95o

Hence,

x = 30o and 95o

(c) ABCD is a rhombus

So, AB = CD

3x + 2 = 4x – 4

3x – 4x = -4 – 2

-x = -6

x = 6

Now, in ∆ABD we have

∠BAD = 60o and AB = AD

∠ADB = ∠ABD

So,

∠ADB = (180o – ∠BAD)/ 2

= (180o – 60o)/ 2

= 120o/2 = 60o

As ∆ABD is an equilateral triangle, all the angles of the triangle are 60o

Hence, AB = BD

3x + 2 = y – 1

3(6) + 2 = y – 1 (Substituting the value of x)

18 + 2 = y – 1

20 = y – 1

y = 20 + 1

y = 21

Thus, x = 6 and y = 21.

6. The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD = 50°, find ∠DPC.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 6

Given, ABCD is a rectangle

We know that the diagonals of rectangle are same and bisect each other

So, we have

AP = BP

∠PAB = ∠PBA (Equal sides have equal opposite angles)

∠PAB = 50o (Since, given ∠PBA = 50o)

Now, in ∆APB

∠APB + ∠ABP + ∠BAP = 180o

∠APB + 50o + 50o = 180o

∠APB = 180o – 100o

∠APB = 80o

Then,

∠DPB = ∠APB (Vertically opposite angles)

Hence,

∠DPB = 80o

7. (a) In figure (1) given below, equilateral triangle EBC surmounts square ABCD. Find angle BED represented by x.

(b) In figure (2) given below, ABCD is a rectangle and diagonals intersect at O. AC is produced to E. If ∠ECD = 146°, find the angles of the ∆ AOB.

(c) In figure (3) given below, ABCD is rhombus and diagonals intersect at O. If ∠OAB : ∠OBA = 3:2, find the angles of the ∆ AOD.

ML Aggarwal Solutions for Class 9 Chapter 13 - 7

Solution:

Since, EBC is an equilateral triangle, we have

EB = BC = EC … (i)

Also, ABCD is a square

So, AB = BC = CD = AD … (ii)

From (i) and (ii), we get

EB = EC = AB = BC = CD = AD … (iii)

Now, in ∆ECD

∠ECD = ∠BCD + ∠ECB

= 90o + 60o

= 150o … (iv)

Also, EC = CD [From (iii)]

So, ∠DEC = ∠CDE … (v)

∠ECD + ∠DEC + ∠CDE = 180o [Angles sum property of a triangle]

150o + ∠DEC + ∠DEC = 180o [Using (iv) and (v)]

2 ∠DEC = 180o – 150o = 30o

∠DEC = 30o/2

∠DEC = 15o … (vi)

Now, ∠BEC = 60o [BEC is an equilateral triangle]

∠BED + ∠DEC = 60o

xo + 15o = 60o [From (vi)]

x = 60o – 15o

x = 45o

Hence, the value of x is 45o.

(b) Given, ABCD is a rectangle

∠ECD = 146o

As ACE is a straight line, we have

146o + ∠ACD = 180o [Linear pair]

∠ACD = 180o – 146o = 34o … (i)

And, ∠CAB = ∠ACD [Alternate angles] … (ii)

From (i) and (ii), we have

∠CAB = 34o ⇒ ∠OAB = 34o … (iii)

In ∆AOB

AO = OB [Diagonals of a rectangle are equal and bisect each other]

∠OAB = ∠OBA … (iv) [Equal sides have equal angles opposite to them]

From (iii) and (iv),

∠OBA = 34o … (v)

Now,

∠AOB + ∠OBA + ∠OAB = 180o

∠AOB + 34o + 34o = 180o [Using (3) and (5)]

∠AOB + 68o = 180o

∠AOB = 180o – 68o = 112o

Hence, ∠AOB = 112o, ∠OAB = 34o and ∠OBA = 34o

(c) Here, ABCD is a rhombus and diagonals intersect at O and ∠OAB : ∠OBA = 3 : 2

Let ∠OAB = 2xo

Then,
∠OBA = 2xo

We know that diagonals of rhombus intersect at right angle,

So, ∠OAB = 90o

Now, in ∆AOB

∠OAB + ∠OBA = 180o

90o + 3xo + 2xo = 180o

90o + 5xo = 180o

5xo = 180o – 90o = 90o

xo = 90o/5 = 18o

Hence,

∠OAB = 3xo = 3 x 18o = 54o

OBA = 2xo = 2 x 18o = 36o and

∠AOB = 90o

8. (a) In figure (1) given below, ABCD is a trapezium. Find the values of x and y.

(b) In figure (2) given below, ABCD is an isosceles trapezium. Find the values of x and y.

(c) In figure (3) given below, ABCD is a kite and diagonals intersect at O. If ∠DAB = 112° and ∠DCB = 64°, find ∠ODC and ∠OBA.

ML Aggarwal Solutions for Class 9 Chapter 13 - 8

Solution:

(a) Given: ABCD is a trapezium

∠A = x + 20o, ∠B = y, ∠C = 92o, ∠D = 2x + 10o

We have,

∠B + ∠C = 180o [Since AB || DC]

y + 92o = 180o

y = 180o – 92o = 88o

Also, ∠A + ∠D = 180o

x + 20o + 2x + 10o = 180o

3x + 30o = 180o

3x = 180o – 30o = 150o

x = 150o/3 = 50o

Hence, the value of x = 50o and y = 88o.

(b) Given: ABCD is an isosceles trapezium BC = AD

∠A = 2x, ∠C = y and ∠D = 3x

Since, ABCD is a trapezium and AB || DC

⇒ ∠A + ∠D = 180o

2x + 3x = 180o

5x = 180o

x = 180o/5 = 36o … (i)

Also, AB = BC and AB || DC

So, ∠A + ∠C = 180o

2x + y = 180o

2 × 36o + y = 180o

72o + y = 180o

y = 180o – 72o = 108o

Hence, value of x = 72o and y = 108o.

(c) Given: ABCD is a kite and diagonal intersect at O.

∠DAB = 112o and ∠DCB = 64o

As AC is the diagonal of kite ABCD, we have

∠DCO = 64o/2 = 32o

And, ∠DOC = 90o [Diagonal of kites bisect at right angles]

In ∆OCD, we have

∠ODC = 180o – (∠DCO + ∠DOC)

= 180o – (32o + 90o)

= 180o – 122o

= 58o

In ∆DAB, we have

∠OAB = 112o/2 = 56o

∠AOB = 90o [Diagonal of kites bisect at right angles]

In ∆OAB, we have

∠OBA = 180o – (∠OAB + ∠AOB)

= 180o – (56o + 90o)

= 180o – 146o

= 34o

Hence, ∠ODC = 58o and ∠OBA = 34o.

9. (i) Prove that each angle of a rectangle is 90°.
(ii) If the angle of a quadrilateral are equal, prove that it is a rectangle.
(iii) If the diagonals of a rhombus are equal, prove that it is a square.
(iv) Prove that every diagonal of a rhombus bisects the angles at the vertices.

Solution:

(i) Given: ABCD is a rectangle

ML Aggarwal Solutions for Class 9 Chapter 13 - 9To prove: Each angle of rectangle = 90o

Proof:

In a rectangle opposite angles of a rectangle are equal

So, ∠A = ∠C and ∠B = ∠C

But, ∠A + ∠B + ∠C + ∠D = 360o [Sum of angles of a quadrilateral]

∠A + ∠B + ∠A + ∠B = 360o

2(∠A + ∠B) = 360o

(∠A + ∠B) = 360o/2

∠A + ∠B = 180o

But, ∠A = ∠B [Angles of a rectangle]

So, ∠A = ∠B = 90o

Thus,

∠A = ∠B = ∠C = ∠D = 90o

Hence, each angle of a rectangle is 90°.

ML Aggarwal Solutions for Class 9 Chapter 13 - 1(ii) Given: In quadrilateral ABCD, we have

∠A = ∠B = ∠C = ∠D

To prove: ABCD is a rectangle

Proof:

∠A = ∠B = ∠C = ∠D

⇒ ∠A = ∠C and ∠B = ∠D

But these are opposite angles of the quadrilateral.

So, ABCD is a parallelogram

And, as ∠A = ∠B = ∠C = ∠D

Therefore, ABCD is a rectangle.

(iii) Given: ABCD is a rhombus in which AC = BD

To prove: ABCD is a square

Proof:

ML Aggarwal Solutions for Class 9 Chapter 13 - 10Join AC and BD.

Now, in ∆ABC and ∆DCB we have

∠AB = ∠DC [Sides of a rhombus]

∠BC = ∠BC [Common]

∠AC = ∠BD [Given]

So, ∆ABC ≅ ∆DCB by S.S.S axiom of congruency

Thus,

∠ABC = ∠DBC [By C.P.C.T]

But these are made by transversal BC on the same side of parallel lines AB and CD.

So, ∠ABC + ∠DBC = 180o

∠ABC = 90o

Hence, ABCD is a square.

(iv) Given: ABCD is rhombus.

To prove: Diagonals AC and BD bisects ∠A, ∠C, ∠B and ∠D respectively

Proof:

In ∆AOD and ∆COD, we have

AD = CD [sides of a rhombus are all equal]

OD = OD [Common]

AO = OC [Diagonal of rhombus bisect each other]

So, ∆AOD ≅ ∆COD by S.S.S axiom of congruency

Thus,

∠AOD = ∠COD [By C.P.C.T]

So, ∠AOD + ∠COD = 180o [Linear pair]

∠AOD = 180o

∠AOD = 90o

And, ∠COD = 90o

Thus,

OD ⊥ AC ⇒ BD ⊥ AC

Also, ∠ADO = ∠CDO [By C.P.C.T]

So,

OD bisect ∠D BD bisect ∠D

Similarly, we can prove that BD bisect ∠B and AC bisect the ∠A and ∠C.

10. ABCD is a parallelogram. If the diagonal AC bisects ∠A, then prove that:
(i) AC bisects ∠C
(ii) ABCD is a rhombus
(iii) AC ⊥ BD.

Solution:

Given: In parallelogram ABCD in which diagonal AC bisects ∠A

ML Aggarwal Solutions for Class 9 Chapter 13 - 11To prove: (i) AC bisects ∠C

(ii) ABCD is a rhombus

(iii) AC ⊥ BD

Proof:

(i) As AB || CD, we have [Opposite sides of a || gm]

∠DCA = ∠CAB

Similarly, ∠DAC = ∠DCB

But, ∠CAB = ∠DAC [Since, AC bisects ∠A]

Hence,

∠DCA = ∠ACB and AC bisects ∠C.

(ii) As AC bisects ∠A and ∠C

And, ∠A = ∠C

Hence, ABCD is a rhombus.

(iii) Since, AC and BD are the diagonals of a rhombus and

AC and BD bisect each other at right angles

Hence, AC ⊥ BD

11. (i) Prove that
bisectors of any two adjacent angles of a parallelogram are at right angles.

(ii) Prove that bisectors of any two opposite angles of a parallelogram are parallel.

(iii) If the diagonals of a quadrilateral are equal and bisect each other at right angles, then prove that it is a square.

Solution:

(i) Given AM bisect angle A and BM bisects angle of || gm ABCD.

ML Aggarwal Solutions for Class 9 Chapter 13 - 12To prove: ∠AMB = 90o

Proof:

We have,

∠A + ∠B = 180o [AD || BC and AB is the transversal]

⇒ ½ (∠A + ∠B) = 180o/2

½ ∠A + ½ ∠B = 90o

∠MAB + ∠MBA = 90o [Since, AM bisects ∠A and BM bisects ∠B]

Now, in ∆AMB

∠AMB + ∠MAB + ∠MBA = 180o [Angles sum property of a triangle]

∠AMB + 90o = 180o

∠AMB = 180o – 90o = 90o

Hence, bisectors of any two adjacent angles of a parallelogram are at right angles.

(ii) Given: A || gm ABCD in which bisector AR of ∠A meets DC in R and bisector CQ of ∠C meets AB in Q

ML Aggarwal Solutions for Class 9 Chapter 13 - 13To prove: AR || CQ

Proof:

In || gm ABCD, we have

∠A = ∠C [Opposite angles of || gm are equal]

½ ∠A = ½ ∠C

∠DAR = ∠BCQ [Since, AR is bisector of ½ ∠A and CQ is the bisector of ½ ∠C]

Now, in ∆ADR and ∆CBQ

∠DAR = ∠BCQ [Proved above]

AD = BC [Opposite sides of || gm ABCD are equal]

So, ∆ADR ≅ ∆CBQ, by A.S.A axiom of congruency

Then by C.P.C.T, we have

∠DRA = ∠BCQ

And,

∠DRA = ∠RAQ [Alternate angles since, DC || AB]

Thus, ∠RAQ = ∠BCQ

But these are corresponding angles,

Hence, AR || CQ.

(iii) Given: In quadrilateral ABCD, diagonals AC and BD are equal and bisect each other at right angles

ML Aggarwal Solutions for Class 9 Chapter 13 - 14To prove: ABCD is a square

Proof:

In ∆AOB and ∆COD, we have

AO = OC [Given]

BO = OD [Given]

∠AOB = ∠COD [Vertically opposite angles]

So, ∆AOB ≅ ∆COD, by S.A.S axiom of congruency

By C.P.C.T, we have

AB = CD

and ∠OAB = ∠OCD

But these are alternate angles

AB || CD

Thus, ABCD is a parallelogram

In a parallelogram, the diagonal bisect each other and are equal

Hence, ABCD is a square.

12. (i) If ABCD is a rectangle in which the diagonal BD bisect ∠B, then show that ABCD is a square.
(ii) Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 15

(i) ABCD is a rectangle and its diagonals AC bisects ∠A and ∠C

To prove: ABCD is a square

Proof:

We know that the opposite sides of a rectangle are equal and each angle is 90o

As AC bisects ∠A and ∠C

So, ∠1 = ∠2 and ∠3 = ∠4

But, ∠A = ∠C = 90o

∠2 = 45o and ∠4 = 45o

And, AB = BC [Opposite sides of equal angles]

But, AB = CD and BC = AD

So, AB = BC = CD = DA

Therefore, ABCD is a square.

(ii) In quadrilateral ABCD diagonals AC and BD are equal and bisect each other at right angle

To prove: ABCD is a square

Proof:

In ∆AOB and ∆BOC, we have

AO = CO [Diagonals bisect each other at right angles]

OB = OB [Common]

∠AOB = ∠COB [Each 90o]

So, ∆AOB ≅ ∆BOC, by S.A.S axiom

By C.P.C.T, we have

ML Aggarwal Solutions for Class 9 Chapter 13 - 16AB = BC … (i)

Similarly, in ∆BOC and ∆COD

OB = OD [Diagonals bisect each other at right angles]

OC = OC [Common]

∠BOC = ∠COD [Each 90o]

So, ∆BOC ≅ ∆COD, by S.A.S axiom

By C.P.C.T, we have

BC = CD … (ii)

From (i) and (ii), we have

AB = BC = CD = DA

Hence, ABCD is a square.

13. P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.

Solution:

Given: ABCD is a parallelogram, P and Q are the points on AB and DC. Diagonals AC and BD intersect each other at O.

ML Aggarwal Solutions for Class 9 Chapter 13 - 17To prove:

Diagonals of || gm ABCD bisect each other at O

So, AO = OC and BO = OD

Now, in ∆AOP and ∆COQ we have

AO = OC and BO = OD

Now, in ∆AOP and ∆COQ

AO = OC [Proved]

∠OAP = ∠OCQ [Alternate angles]

∠AOP = ∠COQ [Vertically opposite angles]

So, ∆AOP ≅ ∆COQ by S.A.S axiom

Thus, by C.P.C.T

OP = OQ

Hence, O bisects PQ.

14. (a) In figure (1) given below, ABCD is a parallelogram and X is mid-point of BC. The line AX produced meets DC produced at Q. The parallelogram ABPQ is completed.

Prove that:
(i) the triangles ABX and QCX are congruent;
(ii)DC = CQ = QP

(b) In figure (2) given below, points P and Q have been taken on opposite sides AB and CD respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other.

ML Aggarwal Solutions for Class 9 Chapter 13 - 18

Solution:

(a) Given: ABCD is parallelogram and X is mid-point of BC. The line AX produced meets DC produced at Q and ABPQ is a || gm.

To prove: (i) ∆ABX ≅ ∆QCX

(ii) DC = CQ = QP

Proof:

In ∆ABX and ∆QCX, we have

BX = XC [X is the mid-point of BC]

∠AXB = ∠CXQ [Vertically opposite angles]

∠XCQ = ∠XBA [Alternate angle, since AB || CQ]

So, ABX ≅ ∆QCX by A.S.A axiom of congruence

Now, by C.P.C.T

CQ = AB

But,

AB = DC and AB = QP [As ABCD and ABPQ are || gms]

Hence,

DC = CQ = QP

(b) In || gm ABCD, P and Q are points on AB and CD respectively, PQ and AC intersect each other at O and AP = CQ

To prove: AC and PQ bisect each other i.e. AO = OC and PO = OQ

Proof:

In ∆AOP and ∆COQ

AP = CQ [Given]

∠AOP = ∠COQ [Vertically opposite angles]

∠OAP = ∠OCP [Alternate angles]

So, ∆AOP ≅ ∆COQ by A.A.S axiom of congruence

Now, by C.P.C.T

OP = OQ and OA = OC

Hence proved.

15. ABCD is a square. A is joined to a point P on BC and D is joined to a point Q on AB. If AP = DQ, prove that AP and DQ are perpendicular to each other.

Solution:

Given: ABCD is a square. P is any point on BC and Q is any point on AB and these points are taken such that AP = DQ

ML Aggarwal Solutions for Class 9 Chapter 13 - 19To prove: AP ⊥ DQ

Proof:

In ∆ABP and ∆ADQ, we have

AP = DQ [Given]

AD = AB [Sides of square ABCD]

∠DAQ = ∠ABP [Each 90o]

So, ∆ABP ≅ ∆ADQ by R.H.S axiom of congruency

Now, by C.P.C.T

∠BAP = ∠ADQ

But, ∠BAD = 90o

∠BAD = ∠BAP + ∠PAD … (i)

90o = ∠BAP + ∠PAD

∠BAP + ∠PAD = 90o

∠BAP + ∠ADQ = 90o

Now, in ∆ADM we have

(∠MAD + ∠ADM) + ∠AMD = 180o [Angles sum property of a triangle]

90o + ∠AMD = 180o [From (i)]

∠AMD = 180o – 90o = 90o

So, DM ⊥ AP

⇒ DQ ⊥ AP

Hence, AP ⊥ DQ

16. If P and Q are points of trisection of the diagonal BD of a parallelogram ABCD, prove that CQ || AP.

Solution:

Given: ABCD is a || gm is which BP = PQ = QD

To prove: CQ || AP

ML Aggarwal Solutions for Class 9 Chapter 13 - 20

Proof:

In || gm ABCD, we have

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

And BD is the transversal

So, ∠1 = ∠2 [Alternate interior angles] … (i)

Now, in ∆ABP and ∆DCQ

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

∠1 = ∠2 [From (i)]

BP = QD [Given]

So, ∆ABP ≅ ∆DCQ by S.A.S axiom of congruency

Then by C.P.C.T, we have

AP = QC

Also, ∠APB = ∠DQC [By C.P.C.T]

-∠APB = -∠DQC [Multiplying both sides by -1]

180o – ∠APB = 180o – ∠DQC [Adding 180o both sides]

∠APQ = ∠CQP

But, these are alternate angles

Hence, AP || QC ⇒ CQ || AP.

17. A transversal cuts two parallel lines at A and B. The two interior angles at A are bisected and so are the two interior angles at B ; the four bisectors form a quadrilateral ABCD. Prove that
(i) ABCD is a rectangle.
(ii) CD is parallel to the original parallel lines.

18. In a parallelogram ABCD, the bisector of ∠A meets DC in E and AB = 2 AD. Prove that
(i) BE bisects ∠B
(ii) ∠AEB = a right angle.

ML Aggarwal Solutions for Class 9 Chapter 13 - 21

Solution:

Given: LM || PQ and AB is the transversal line cutting ∠M at A and PQ at B

AC, AD, BC and BD is the bisector of ∠LAB, ∠BAM, ∠PAB and ∠ABQ respectively.

AC and BC intersect at C and AD and BD intersect at D.

A quadrilateral ABCD is formed.

To prove: (i) ABCD is a rectangle

(ii) CD || LM and PQ

Proof:

(1) ∠LAB + ∠BAM = 180o [LAM is a straight line]

½ (∠LAB + ∠BAM) = 90o

½ ∠LAB + ½ ∠BAM = 90o

∠2 + ∠3 = 90o [Since, AC and AD is bisector of ∠LAB & ∠BAM respectively]

∠CAD = 90o

∠A = 90o

(2) Similarly, ∠PBA + ∠QBA = 180o [PBQ is a straight line]

½ (∠PBA + ∠QBA) = 90o

½ ∠PBA + ½ ∠QBA = 90o

∠6 + ∠7 = 90o [Since, BC and BD is bisector of ∠PAB & ∠QBA respectively]

∠CBD = 90o

∠B = 90o

(3) ∠LAB + ∠ABP = 180o [Sum of co-interior angles is 180o and given LM || PQ]

½ ∠LAB + ½ ∠ABP = 90o

∠2 + ∠6 = 90o [Since, AC and BC is bisector of ∠LAB & ∠PBA respectively]

(4) In ∆ACB,

∠2 + ∠6 + ∠C = 180o [Angles sum property of a triangle]

(∠2 + ∠6) + ∠C = 180o

90o + ∠C = 180o [using (3)]

∠C = 180o – 90o

∠C = 90o

(5) ∠MAB + ∠ABQ = 180o [Sum of co-interior angles is 180o and given LM || PQ]

½ ∠MAB + ½ ∠ABQ = 90o

∠3 + ∠7 = 90o [Since, AD and BD is bisector of ∠MAB & ∠ABQ respectively]

(6) In ∆ADB,

∠3 + ∠7 + ∠D = 180o [Angles sum property of a triangle]

(∠3 + ∠7) + ∠D = 180o

90o + ∠D = 180o [using (5)]

∠D = 180o – 90o

∠D = 90o

(7) ∠LAB + ∠BAM = 180o

∠BAM = ∠ABP [From (1) and (2)]

½ ∠BAM = ½ ∠ABP

∠3 = ∠6 [Since, AD and BC is bisector of ∠BAM and ∠ABP respectively]

Similarly, ∠2 = ∠7

(8) In ∆ABC and ∆ABD,

∠2 = ∠7 [From (7)]

AB = AB [Common]

∠6 = ∠3 [From (7)]

So, ∆ABC ≅ ∆ABD by A.S.A axiom of congruency

Then, by C.P.C.T we have

AC = DB

Also, CB = AD

(9) ∠A = ∠B = ∠C =∠D = 90o [From (1), (2), (3) and (4)]

AC = DB [Proved in (8)]

CB = AD [Proved in (8)]

Hence, ABCD is a rectangle.

(10) Since, ABCD is a rectangle [From (9)]

OA = OD [Diagonals of rectangle bisect each other]

(11) In ∆AOD, we have

OA = OD [From (10)]

∠9 = ∠3 [Angles opposite to equal sides are equal]

(12) ∠3 = ∠4 [AD bisects ∠MAB]

(13) ∠9 = ∠4 [From (11) and (12)]

But these are alternate angles.

OD || LM ⇒ CD || LM

Similarly, we can prove that

∠10 = ∠8

But these are alternate angles,

So, OD || PQ ⇒ CD || PQ

(14) CD || LM [Proved in (13)]

CD || PQ [Proved in (13)]

18. In a parallelogram ABCD, the bisector of ∠A meets DC in E and AB = 2 AD. Prove that:

(i) BE bisects ∠B

(ii) ∠AEB is a right angle

Solution:

Given: ABCD is a || gm in which bisectors of angle A and B meets in E and AB = 2 AD

To prove: (i) BE bisects ∠B

(ii) ∠AEB = 90o

Proof:

(1) In || gm ABCD

∠1 = ∠2 [AD bisects angles ∠A]

(2) AB || DC and AE is the transversal

∠2 = ∠3 [Alternate angles]

(3) ∠1 = ∠2 [From (1) and (2)]

ML Aggarwal Solutions for Class 9 Chapter 13 - 22

(4) In ∆ADE, we have

∠1 = ∠3 [Proved in (3)]

DE = AD [Sides opposite to equal angles are equal]

⇒ AD = DE

(5) AB = 2 AD [Given]

AB/2 = AD

AB/2 = DE [using (4)]

DC/2 = DE [AB = DC, opposite sides of a || gm are equal]

So, E is the mid-point of D.

⇒ DE = EC

(6) AD = BC [Opposite sides of a || gm are equal]

(7) DE = BC [From (4) and (6)]

(8) EC = BC [From (5) and (7)]

(9) In ∆BCE, we have

EC = BC [Proved in (8)]

∠6 = ∠5 [Angles opposite to equal sides are equal]

(10) AB || DC and BE is the transerval

∠4 = ∠5 [Alternate angles]

(11) ∠4 = ∠6 [From (9) and (10)]

So, BE is bisector of ∠B

(12) ∠A + ∠B = 180o [Sum of co-interior angles is equal to 180o, AD || BC]

½ ∠A + ½ ∠B = 180o/ 2

∠2 + ∠4 = 90o [AE is bisector of ∠A and BE is bisector of ∠B]

(13) In ∆APB,

∠AEB + ∠2 + ∠4 = 180o

∠AEB + 90o = 180o

Hence, ∠AEB = 90o

19. ABCD is a parallelogram, bisectors of angles A and B meet at E which lie on DC. Prove that AB.

Solution:

Given: ABCD is a parallelogram in which bisector of ∠A and ∠B meets DC in E

To prove: AB = 2 AD

Proof:

In parallelogram ABCD, we have

AB || DC

∠1 = ∠5 [Alternate angles, AE is transversal]

ML Aggarwal Solutions for Class 9 Chapter 13 - 23∠1 = ∠2 [AE is bisector of ∠A, given]

Thus,
∠2 = ∠5 … (i)

Now, in ∆AED

DE = AD [Sides opposite to equal angles are equal]

∠3 = ∠6 [Alternate angles]

∠3 = ∠4 [Since, BE is bisector of ∠B (given)]

Thus, ∠4 = ∠6 … (ii)

In ∆BCE, we have

BC = EC [Sides opposite to equal angles are equal]

AD = BC [Opposite sides of || gm are equal]

AD = DE = EC [From (i) and (ii)]

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

AB = DE + EC

= AD + AD

Hence,

AB = 2 AD

20. ABCD is a square and the diagonals intersect at O. If P is a point on AB such that AO =AP, prove that 3 ∠POB = ∠AOP.

Solution:

Given: ABCD is a square and the diagonals intersect at O. P is the point on AB such that AO = AP

To prove: 3 ∠POB = ∠AOP

Proof:

(1) In square ABCD, AC is a diagonal

So, ∠CAB = 45o

∠OAP = 45o

(2) In ∆AOP,

∠OAP = 45o [From (1)]

AO = AP [Sides opposite to equal angles are equal]

Now,

∠AOP + ∠APO + ∠OAP = 180o [Angles sum property of a triangle]

∠AOP + ∠AOP + 45o = 180o

2 ∠AOP = 180o – 45o

∠AOP = 135o/2

(3) ∠AOB = 90o [Diagonals of a square bisect at right angles]

So, ∠AOP + ∠POB = 90o

135o/2 + ∠POB = 90o [From (2)]

∠POB = 90o – 135o/2

= (180o – 135o)/2

= 45o/2

3 ∠POB = 135o/2 [Multiplying both sides by 3]

Hence,

∠AOP = 3 ∠POB [From (2) and (3)]

21. ABCD is a square. E, F, G and H are points on the sides AB, BC, CD and DA respectively such that AE = BF = CG = DH. Prove that EFGH is a square.

Solution:

Given: ABCD is a square in which E, F, G and H are points on AB, BC, CD and DA

Such that AE = BF = CG = DH

ML Aggarwal Solutions for Class 9 Chapter 13 - 24EF, FG, GH and HE are joined

To prove: EFGH is a square

Proof:

Since, AE = BF = CG = DH

So, EB = FC = GD = HA

Now, in ∆AEH and ∆BFE

AE = BF [Given]

AH = EB [Proved]

∠A = ∠B [Each 90o]

So, ∆AEH ≅ ∆BFE by S.A.S axiom of congruency

Then, by C.P.C.T we have

EH = EF

And ∠4 = ∠2

But ∠1 + ∠4 = 90o

∠1 + ∠2 = 90o

Thus, ∠HEF = 90o

Hence, EFGH is a square.

22. (a) In the Figure (1) given below, ABCD and ABEF are parallelograms. Prove that
(i) CDFE is a parallelogram
(ii) FD = EC
(iii) Δ AFD = ΔBEC.
(b) In the figure (2) given below, ABCD is a parallelogram, ADEF and AGHB are two squares. Prove that FG = AC.

ML Aggarwal Solutions for Class 9 Chapter 13 - 25

Solution:

Given: ABCD and ABEF are || gms

To prove: (i) CDFE is a parallelogram

(ii) FD = EC

(iii) Δ AFD = ΔBEC

Proof:

(1) DC || AB and DC = AB [ABCD is a || gm]

(2) FE || AB and FE = AB [ABEF is a || gm]

(3) DC || FE and DC = FE [From (1) and (2)]

Thus, CDFE is a || gm

(4) CDEF is a || gm

So, FD = EC

(5) In ∆AFD and ∆BEC, we have

AD = BC [Opposite sides of || gm ABCD are equal]

AF = BE [Opposite sides of || gm ABEF are equal]

FD = BE [From (4)]

Hence, ∆AFD ≅ ∆BEC by S.S.S axiom of congruency

(b) Given: ABCD is a || gm, ADEF and AGHB are two squares

To prove: FG = AC

Proof:

(1) ∠FAG + 90o + 90o + ∠BAD = 360o [At a point total angle is 360o]

∠FAG = 360o – 90o – 90o – ∠BAD

∠FAG = 180o – ∠BAD

(2) ∠B + ∠BAD = 180o [Adjacent angle in || gm is equal to 180o]

∠B = 180o – ∠BAD

(3) ∠FAG = ∠B [From (1) and (2)]

(4) In ∆AFG and ∆ABC, we have

AF = BC [FADE and ABCD both are squares on the same base]

Similarly, AG = AB

∠FAG = ∠B [From (3)]

So, ∆AFG ≅ ∆ABC by S.A.S axiom of congruency

Hence, by C.P.C.T

FG = AC

23. ABCD is a rhombus in which ∠A = 60°. Find the ratio AC : BD.

Solution:

Let each side of the rhombus ABCD be a

∠A = 60o

So, ABD is an equilateral triangle

⇒ BD = AB = a

ML Aggarwal Solutions for Class 9 Chapter 13 - 26We know that, the diagonals of a rhombus bisect each other at right angles

So, in right triangle AOB, we have

AO2 + OB2 = AB2 [By Pythagoras Theorem]

AO2 = AB2 – OB2

= a2 – (½ a)2

= a2 – a2/4

= 3a2/4

AO = √(3a2/4) = √3a/2

But, AC = 2 AO = 2 x 3a/2 = 3a

Hence,

AC : BD = √3a : a = √3 : 1

Exercise 13.2

1. Using ruler and compasses only, construct the quadrilateral ABCD in which ∠ BAD = 45°, AD = AB = 6cm, BC = 3.6cm, CD = 5cm. Measure ∠ BCD.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 27

Steps of construction:

(i) Draw a line segment AB = 6cm

(ii) At A, draw a ray AX making an angle of 45o and cut off AD = 6cm

(iii) With centre B and radius 3.6 cm and with centre D and radius 5 cm, draw two arcs intersecting each other at C.

(iv) Join BC and DC.

Thus, ABCD is the required quadrilateral.

On measuring ∠BCD, it is 60o.

2. Draw a quadrilateral ABCD with AB = 6cm, BC = 4cm, CD = 4 cm and ∠ BC = ∠ BCD = 90°.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 28

Steps of construction:

(i) Draw a line segment BC = 4 cm

(ii) At B and C draw rays BX and CY making an angle of 90o each

(iii) From BX, cut off BA = 6 cm and from CY, cut off CD = 4cm

(iv) Join AD.

Thus, ABCD is the required quadrilateral.

3. Using ruler and compasses only, construct the quadrilateral ABCD given that AB = 5 cm, BC = 2.5 cm, CD = 6 cm, ∠BAD = 90° and the diagonal AC = 5.5 cm.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 29

Steps of construction:

(i) Draw a line segment AB = 5cm

(ii) With centre A and radius 5.5cm and with centre B and radius 2.5cm draw arcs which intersect each other at C.

(iii) Join AC and BC.

(iv) At A, draw a ray AX making an angle of 90o.

(v) With centre C and radius 6cm, draw an arc intersecting AX at D

(vi) Join CD.

Thus, ABCD is the required quadrilateral.

4. Construct a quadrilateral ABCD in which AB = 3.3 cm, BC = 4.9 cm, CD = 5.8 cm, DA = 4 cm and BD = 5.3 cm.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 30

Steps of construction:

(i) Draw a line segment AB = 3.3cm

(ii) with centre A and radius 4cm, and with centre B and radius 5.3cm, draw arcs intersecting each other at D.

(iii) Join AD and BD.

(iv) With centre B and radius 4.9 cm and with centre D and radius 5.8cm, draw arcs intersecting each other at C.

(v) Join BC and DC.

Thus, ABCD is the required quadrilateral.

5. Construct a trapezium ABCD in which AD || BC, AB = CD = 3 cm, BC = 5.2cm and AD = 4 cm.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 31

Steps of construction:

(i) Draw a line segment BC = 5.2cm

(ii) From BC, cut off BE = AD = 4cm

(iii) With centre E and C, and radius 3 cm, draw arcs intersecting each other at D.

(iv) Join ED and CD.

(v) With centre D and radius 4cm and with centre B and radius 3cm, draw arcs intersecting each other at A.

(vi) Join BA and DA

Thus, ABCD is the required trapezium.

6. Construct a trapezium ABCD in which AD || BC, ∠B= 60°, AB = 5 cm. BC = 6.2 cm and CD = 4.8 cm.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 32

Steps of construction:

(i) Draw a line segment BC = 6.2cm

(ii) At B, draw a ray BX making an angle of 60o and cut off AB = 5cm

(iii) From A, draw a line AY parallel to BC.

(iv) With centre C and radius 4.8cm, draw an arc which intersects AY at D and D’.

(v) Join CD and CD’

Thus, ABCD and ABCD’ are the required two trapeziums.

7. Using ruler and compasses only, construct a parallelogram ABCD with AB = 5.1 cm, BC = 7 cm and ∠ABC = 75°.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 33

Steps of construction:

(i) Draw a line segment BC = 7cm

(ii) A to B, draw a ray Bx making an angle of 75o and cut off AB = 5.1cm

(iii) With centre A and radius 7cm with centre C and radius 5.1cm, draw arcs intersecting each other at D.

(iv) Join AD and CD.

Thus, ABCD is the required parallelogram.

8. Using ruler and compasses only, construct a parallelogram ABCD in which AB = 4.6 cm, BC = 3.2 cm and AC = 6.1 cm.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 34

Steps of construction:

(i) Draw a line segment AB = 4.6cm

(ii) With centre A and radius 6.1cm and with centre B and radius 3.2cm, draw arcs intersecting each other at C.

(iii) Join AC and BC.

(iv) Again, with centre A and radius 3.2cm and with centre C and radius 4.6cm, draw arcs intersecting each other at C.

(v) Join AD and CD.

Thus, ABCD is the required parallelogram.

9. Using ruler and compasses, construct a parallelogram ABCD give that AB = 4 cm, AC = 10 cm, BD = 6 cm. Measure BC.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 35

Steps of construction:

(i) Construct triangle OAB such that

OA = ½ x AC = ½ x 10cm = 5cm

OB = ½ x BD = ½ x 6cm = 3cm

As, diagonals of || gm bisect each other and AB = 4cm

(ii) Produce AO to C such that OA = OC = 5cm

(iii) Produce BO to D such that OB = OD = 3cm

(iv) Join AD, BC and CD

Thus, ABCD is the required parallelogram

(v) Measure BC which is equal to 7.2cm

10. Using ruler and compasses only, construct a parallelogram ABCD such that BC = 4 cm, diagonal AC = 8.6 cm and diagonal BD = 4.4 cm. Measure the side AB.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 36

Steps of construction:

(i) Construct triangle OBC such that

OB = ½ x BD = ½ x 4.4cm = 2.2cm

OC = ½ x AC = ½ x 8.6cm = 4.3cm

Since, diagonals of || gm bisect each other and BC = 4cm

(ii) Produce BO to D such that BO = OD = 2.2cm

(iii) Produce CO to A such that CO = OA = 4.3cm

(iv) Join AB, AD and CD

Thus, ABCD is the required parallelogram.

(v) Measure the side AB, AB = 5.6cm

11. Use ruler and compasses to construct a parallelogram with diagonals 6 cm and 8 cm in length having given the acute angle between them is 60°. Measure one of the longer sides.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 37

Steps of construction:

(i) Draw AC = 6cm

(ii) Find the mid-point O of AC. [Since, diagonals of || gm bisect each other]

(iii) Draw line POQ such that POC = 60o and OB = OD = ½ BD = ½ x 8cm = 4cm

So, from OP cut OD = 4cm and from OQ cut OB = 4cm

(iv) Join AB, BC, CD and DA.

Thus, ABCD is the required parallelogram.

(v) Measure the length of side AD = 6.1cm

12. Using ruler and compasses only, draw a parallelogram whose diagonals are 4 cm and 6 cm long and contain an angle of 75°. Measure and write down the length of one of the shorter sides of the parallelogram.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 38

Steps of construction:

(i) Dra a line segment AC = 6cm

(ii) Bisect AC at O.

(iii) At O, draw a ray XY making an angle of 75o at O.

(iv) From OX and OY, cut off OD = OB = 4/2 = 2cm

(v) Join AB, BC, CD and DA.

Thus, ABCD is the required parallelogram.

On measuring one of the shorter sides, we get

AB = CD = 3cm

13. Using ruler and compasses only, construct a parallelogram ABCD with AB = 6 cm, altitude = 3.5 cm and side BC = 4 cm. Measure the acute angles of the parallelogram.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 39

Steps of construction:

(i) Draw AB = 6cm

(ii) At B, draw BP ⊥ AB

(iii) From BP, cut BE = 3.5cm = height of || gm

(iv) Through E draw QR parallel to AB

(v) With B as centre and radius BC = 4cm draw an arc which cuts QR at C.

(vi) Since, opposite sides of || gm are equal

So, AD = BC = 4cm

(vii)With A as centre and radius = 4cm draw an arc which cuts QR at D.

Thus, ABCD is the required parallelogram.

(viii) To measure the acute angle of parallelogram which is equal to 61o.

14. The perpendicular distances between the pairs of opposite sides of a parallelogram ABCD are 3 cm and 4 cm and one of its angles measures 60°. Using ruler and compasses only, construct ABCD.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 40

Steps of construction:

(i) Draw a straight-line PQ, take a point A on it.

(ii) At A, construct ∠QAF = 60o

(iii) At A, draw AE ⊥ PQ from AE cut off AN = 3cm

(iv) Through N draw a straight line to PQ to meet AF at D.

(v) At D, draw AG ⊥ AD, from AG cut off AM = 4cm

(vi) Through M, draw at straight line parallel to AD to meet AQ in B and ND in C.

Then, ABCD is the required parallelogram

15. Using ruler and compasses, construct a rectangle ABCD with AB = 5cm and AD = 3 cm.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 41

Steps of construction:

(i) Draw a straight-line AB = 5cm

(ii) At A and B construct ∠XAB and ∠YBA = 90o

(iii) From A and B cut off AC and BD = 3cm each

(iv) Join CD

Thus, ABCD is the required rectangle.

16. Using ruler and compasses only, construct a rectangle each of whose diagonals measures 6cm and the diagonals intersect at an angle of 45°.

Solution:

Steps of construction:

(i) Draw a line segment AC = 6cm

(ii) Bisect AC at O.

(iii) At O, draw a ray XY making an angle of 45o at O.

(iv) From XY, cut off OB = OD = 6/2 = 3cm each

(v) Join AB, BC CD and DA.

Thus, ABCD is the required rectangle.

17. Using ruler and compasses only, construct a square having a diagonal of length 5cm. Measure its sides correct to the nearest millimeter.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 42

Steps of construction:

(i) Draw a line segment AC = 5cm

(ii) Draw its percentage bisector XY bisecting it at O

(iii) From XY, cut off

OB = OD = 5/2 = 2.5cm

(iv) Join AB, BC, CD and DA.

Thus, ABCD is the required square

On measuring its sides, each = 3.6cm (approximately)

18. Using ruler and compasses only construct A rhombus ABCD given that AB 5cm, AC = 6cm measure ∠BAD.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 43

Steps of construction:

(i) Draw a line segment AB = 5cm

(ii) With centre A and radius 6cm, with centre B and radius 5cm, draw arcs intersecting each other at C.

(iii) Join AC and BC

(iv) With centre A and C and radius 5cm, draw arc intersecting each other 5cm, draw arcs intersecting each other at D

(v) Join AD and CD.

Thus, ABCD is a rhombus

On measuring, ∠BAD = 106o.

19. Using ruler and compasses only, construct rhombus ABCD with sides of length 4cm and diagonal AC of length 5 cm. Measure ∠ABC.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 44

Steps of construction:

(i) Draw a line segment AC = 5cm

(ii) With centre A and C and radius 4cm, draw arcs intersecting each other above and below AC at D and B.

(iii) Join AB, BC, CD and DA.

Thus, ABCD is the required rhombus.

20. Construct a rhombus PQRS whose diagonals PR and QS are 8cip and 6cm respectively.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 45

Steps of construction:

(i) Draw a line segment PR = 8cm

(ii) Draw its perpendicular bisector XY intersecting it at O.

(iii) From XY, cut off OQ = OS = 6/2 = 3cm each

(iv) Join PQ, QR, RS and SP

Thus, PQRS is the required rhombus.

21. Construct a rhombus ABCD of side 4.6 cm and ∠BCD = 135°, by using ruler and compasses only.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 46

Steps of construction:

(i) Draw a line segment BC = 4.6cm

(ii) At C, draw a ray CX making an angle of 135o and cut off CD = 4.6cm

(iii) With centres B and D, and radius 4.6cm draw arcs intersecting each other at A.

(iv) Join BA and DA.

Thus, ABCD is the required rhombus.

22. Construct a trapezium in which AB || CD, AB = 4.6 cm, ∠ ABC = 90°, ∠ DAB = 120° and the distance between parallel sides is 2.9 cm.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 47

Steps of construction:

(i) Draw a line segment AB = 4.6cm

(ii) At B, draw a ray BZ making an angle of 90o and cut off BC = 2.9cm (distance between AB and CD)

(iii) At C, draw a parallel line XY to AB.

(iv) At A, draw a ray making an angle of 120o meeting XY at D.

Thus, ABCD is the required trapezium.

23. Construct a trapezium ABCD when one of parallel sides AB = 4.8 cm, height = 2.6cm, BC = 3.1 cm and AD = 3.6 cm.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 48

Step construction:

(i) Draw a line segment AB = 4.8cm

(ii) At A, draw a ray AZ making an angle of 90o cut off AL = 2.6cm

(iii) At L, draw a line XY parallel to AB.

(iv) With centre A and radius 3.6cm and with centre B and radius 3.1cm, draw arcs intersecting XY at D and C respectively.

(v) Join AD and BC

Thus, ABCD is the required trapezium.

24. Construct a regular hexagon of side 2.5 cm.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 49

Steps of construction:

(i) With O as centre and radius = 2.5cm, draw a circle

(ii) take any point A on the circumference of circle.

(iii) With A as centre and radius = 2.5cm, draw an arc which cuts the circumference of circle at B.

(iv) With B as centre and radius = 2.5cm, draw an arc which cuts the circumference of circle at C.

(v) With C as centre and radius = 2.5cm, draw an arc which cuts the circumference of circle at D.

(vi) With D as centre and radius = 2.5cm, draw an arc which cuts the circumference of circle at E.

(vii) With E as centre and radius = 2.5cm, draw an arc which cuts the circumference of circle at F.

(viii) Join AB, BC, CD, DE, EF and FA.

(ix) ABCDEF is the required Hexagon.

Chapter Test

1. In the given figure, ABCD is a parallelogram. CB is produced to E such that BE=BC. Prove that AEBD is a parallelogram.

ML Aggarwal Solutions for Class 9 Chapter 13 - 50Solution:

Given ABCD is a || gm in which CB is produced to E such that BE = BC

BD and AE are joined

To prove: AEBD is a parallelogram

Proof:

In ∆AEB and ∆BDC

EB = BC [Given]

∠ABE = ∠DCB [Corresponding angles]

AB = DC [Opposite sides of || gm]

Thus, ∆AEB ≅ ∆BDC by S.A.S axiom

So, by C.P.C.T

But, AD = CB = BE [Given]

As the opposite sides are equal and ∠AEB = ∠DBC

But these are corresponding angles

Hence, AEBD is a parallelogram.

2. In the given figure, ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle PAC and CD || BA. Show that (i) ∠DAC=∠BCA (ii) ABCD is a parallelogram.

Solution:

Given: In isosceles triangle ABC, AB = AC. AD is the bisector of ext. ∠PAC and CD || BA

To prove: (i) ∠DAC = ∠BCA

ML Aggarwal Solutions for Class 9 Chapter 13 - 51(ii) ABCD is a || gm

Proof:

In ∆ABC

AB = AC [Given]

∠C = ∠B [Angles opposite to equal sides]

Since, ext. ∠PAC = ∠B + ∠C

= ∠C + ∠C

= 2 ∠C

= 2 ∠BCA

So, ∠DAC = 2 ∠BCA

∠DAC = ∠BCA

But these are alternate angles

Thus, AD || BC

But, AB || AC

Hence, ABCD is a || gm.

3. Prove that the quadrilateral obtained by joining the mid-points of an isosceles trapezium is a rhombus.

Solution:

Given: ABCD is an isosceles trapezium in which AB || DC and AD = BC

P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively PQ, QR, RS and SP are joined.

ML Aggarwal Solutions for Class 9 Chapter 13 - 52To prove: PQRS is a rhombus

Construction: Join AC and BD

Proof:

Since, ABCD is an isosceles trapezium

Its diagonals are equal

AC = BD

Now, in ∆ABC

P and Q are the mid-points of AB and BC

So, PQ || AC and PQ = ½ AC … (i)

Similarly, in ∆ADC

S and R mid-point of CD and AD

So, SR || AC and SR = ½ AC … (ii)

From (i) and (ii), we have

PQ || SR and PQ = SR

Thus, PQRS is a parallelogram.

Now, in ∆APS and ∆BPQ

AP = BP [P is the mid-point]

AS = BQ [Half of equal sides]

∠A = ∠B [As ABCD is an isosceles trapezium]

So, ∆APS ≅ ∆BPQ by SAS Axiom of congruency

Thus, by C.P.C.T we have

PS = PQ

But there are the adjacent sides of a parallelogram

So, sides of PQRS are equal

Hence, PQRS is a rhombus

– Hence proved

4. Find the size of each lettered angle in the following figures. 

ML Aggarwal Solutions for Class 9 Chapter 13 - 53

Solution:

(i) As CDE is a straight line

∠ADE + ∠ADC = 180o

122o + ∠ADC = 180o

∠ADC = 180o – 122o = 58o … (i)

∠ABC = 360o – 140o = 220o … (ii)

[At any point the angle is 360o]

Now, in quadrilateral ABCD we have

∠ADC + ∠BCD + ∠BAD + ∠ABC = 360o

58o + 53o + x + 220o = 360o [Using (i) and (ii)]

331o + x = 360o

x = 360o – 331o

x = 29o

(ii) As DE || AB [Given]

∠ECB = ∠CBA [Alternate angles]

75o = ∠CBA

⇒ ∠CBA = 75o

Since, AD || BC we have

(x + 66o) + 75o = 180o

x + 141o = 180o

x = 180o – 141o

x = 39o … (i)

Now, in ∆AMB

x + 30o + ∠AMB = 180o [Angles sum property of a triangle]

39o + 30o + ∠AMB = 180o [From (i)]

69o + ∠AMB + 180o

∠AMB = 180o – 69o = 111o … (ii)

Since, ∠AMB = y [Vertically opposite angles]

⇒ y = 111o

Hence, x = 39o and y = 111o

(iii) In ∆ABD

AB = AD [Given]

∠ABD = ∠ADB [Angles opposite to equal sides are equal]

∠ABD = 42o [Since, given ∠ADB = 42o]

And,

∠ABD + ∠ADB + ∠BAD = 180o [Angles sum property of a triangle]

42o + 42o + y = 180o

84o + y = 180o

y = 180o – 84o

y = 96o

∠BCD = 2 x 26o = 52o

In ∆BCD,

As BC = CD [Given]

∠CBD = ∠CDB = x [Angles opposite to equal sides are equal]

∠CBD + ∠CDB + ∠BCD = 180o

x + x + 52o = 180o

2x + 52o = 180o

2x = 180o – 52o

x = 128o/2

x = 64o

Hence, x = 64o and y = 90o.

5. Find the size of each lettered angle in the following figures: 

ML Aggarwal Solutions for Class 9 Chapter 13 - 54

Solution:

(i) Here, AB || CD and BC || AD

So, ABCD is a || gm

y = 2 x ∠ABD

y = 2 x 53o = 106o … (1)

Also, y + ∠DAB = 180o

∠DAB = 180o – 106o

= 74o

Thus, x = ½ ∠DAB [As AC bisects ∠DAB]

x = ½ x 74o = 37o

and ∠DAC = x = 37o … (ii)

Also, ∠DAC = z … (iii) [Alternate angles]

From (ii) and (iii),

z = 37o

Hence, x = 37o, y = 106o and z = 37o

(ii) As ED is a straight line, we have

60o + ∠AED = 180o [Linear pair]

∠AED = 180o – 60o

∠AED = 120o … (i)

Also, as CD is a straight line

50o + ∠BCD = 180o [Linear pair]

∠BCD = 180o – 50o

∠BCD = 130o … (ii)

In pentagon ABCDE, we have

∠A + ∠B+ ∠AED + ∠BCD + ∠x = 540o [Sum of interior angles in pentagon is 540o]

90o + 90o 120o + 130o + x = 540o

430o + x = 540o

x = 540o – 430o

x = 110o

Hence, value of x = 110o

(iii) In given figure, AD || BC [Given]

60o + y = 180o and x + 110o = 180o

y = 180o – 60o and x = 180o – 110o

y = 120o and x = 70o

Since, CD || AF [Given]

∠FAD = 70o … (i)

In quadrilateral ADEF,

∠FAD + 75o + z + 130o = 360o

70o + 75o + z + 130o = 360o

275o + z = 360o

z = 360o – 275o = 85o

Hence,

x = 70o, y = 120o and z = 85o

6. In the adjoining figure, ABCD is a rhombus and DCFE is a square. If ∠ABC = 56°, find (i) ∠DAG (ii) ∠FEG (iii) ∠GAC (iv) ∠AGC.

Solution:

Here ABCD and DCFE is a rhombus and square respectively.

ML Aggarwal Solutions for Class 9 Chapter 13 - 55So, AB = BC = DC = AD … (i)

Also, DC = EF = FC = EF … (ii)

From (i) and (ii), we have

AB = BC = DC = AD = EF = FC = EF … (iii)

∠ABC = 56o [Given]

∠ADC = 56o [Opposite angle in rhombus are equal]

So, ∠EDA = ∠EDC + ∠ADC = 90o + 56o = 146o

In ∆ADE,

DE = AD [From (iii)]

∠DEA = ∠DAE [Equal sides have equal opposite angles]

∠DEA = ∠DAG = (180o – ∠EDA)/ 2

= (180o – 146o)/ 2

= 34o/ 2 = 17o

⇒ ∠DAG = 17o

Also,
∠DEG = 17o

∠FEG = ∠E – ∠DEG

= 90o – 17o

= 73o

In rhombus ABCD,

∠DAB = 180o – 56o = 124o

∠DAC = 124o/ 2 [Since, AC diagonals bisect the ∠A]

∠DAC = 62o

∠GAC = ∠DAC – ∠DAG

= 62o – 17o

= 45o

In ∆EDG,

∠D + ∠DEG + ∠DGE = 180o [Angles sum property of a triangle]

90o + 17o + ∠DGE = 180o

∠DGE = 180o – 107o = 73o … (iv)

Thus, ∠AGC = ∠DGE … (v) [Vertically opposite angles]

Hence from (iv) and (v), we have

∠AGC = 73o

7. If one angle of a rhombus is 60° and the length of a side is 8 cm, find the lengths of its diagonals.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 56

Each side of rhombus ABCD is 8cm

So, AB = BC = CD = DA = 8cm

Let ∠A = 60o

So, ∆ABD is an equilateral triangle

Then,

AB = BD = AD = 8cm

As we know, the diagonals of a rhombus bisect each other at right angles

AO = OC, BO = OD = 4cm and ∠AOB = 90o

Now, in right ∆AOB

By Pythagoras theorem

AB2 = AO2 + OB2

82 = AO2 + 42

64 = AO2 + 16

AO2 = 64 – 16 = 48

AO = √48 = 4√3cm

But, AC = 2 AO

Hence, AC = 2 × 4√3 = 8√3cm.

8. Using ruler and compasses only, construct a parallelogram ABCD with AB = 5 cm, AD = 2.5 cm and ∠BAD = 45°. If the bisector of ∠BAD meets DC at E, prove that ∠AEB is a right angle.

Solution:

ML Aggarwal Solutions for Class 9 Chapter 13 - 57

Steps of construction:

(i) Draw AB = 5.0cm

(ii) Draw BAP = 45o on side AB

(iii) Take A as centre and radius 2.5cm cut the line AP at D

(iv) Take D as centre and radius 5.0cm draw an arc

(v) Take B as centre and radius equal to 2.5cm cut the arc of step (iv) at C

(vi) Join BC and CD

(vii) ABCD is the required parallelogram

(viii) Draw the bisector of ∠BAD, which cuts the DC at E

(ix) Join EB

(x) Measure ∠AEB which is equal to 90o.

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