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

## 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 = 360^{o}

Sum of two given angles = 40^{o} + 110^{o} = 150^{o}

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

Also given,

Ratio in these angles = 3 : 4

Hence,

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

= (210^{o} x 3)/7

= 90^{o}

And,

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

= (210^{o} x 4)/7

= 120^{o}

**2. 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 = 360^{o}

∠A = (360^{o} x 1)/10 = 36^{o}

∠B = (360^{o} x 2)/10 = 72^{o}

∠C = (360^{o} x 3)/10 = 108^{o}

∠D = (360^{o} x 4)/10 = 144^{o}

Now,

∠A + ∠D = 36^{o} + 114^{o} = 180^{o}

Since, the sum of angles ∠A and ∠D is 180^{o} 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 = x^{o}

Then, ∠B = (2x/3)^{o} (Given condition)

So,

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

^{}

Hence, ∠A = 108^{o}

∠B = 2/3 x 108^{o} = 2 x 36^{o} = 72^{o}

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

Also,

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

Therefore, angles of parallelogram are 108^{o}, 72^{o}, 108^{o }and 72^{o}.

**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.**

**Solution: **

(a) Since, ABCD is a || gm

We have, AB || CD

∠ADB = ∠DBC (Alternate angles)

∠ADB = 80^{o} (Given, ∠DBC = 80^{o})

Now,

In ∆ADB, we have

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

70^{o} + 80^{o} + ∠ABD = 180^{o}

150^{o} + ∠ABD = 180^{o}

∠ABD = 180^{o} – 150^{o} = 30^{o}

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

So,

∠CDB = 30^{o}

Hence, ∠ADB = 80^{o} and ∠CDB = 30^{o}.

(b) Given, ∠BOC = 35^{o} and ∠CBO = 77^{o}

In ∆BOC, we have

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

∠BOC = 180^{o} – 112^{o} = 68^{o}

Now, in || gm ABCD

We have,

∠AOD = ∠BOC (Vertically opposite angles)

Hence, ∠AOD = 68^{o}.

(c) ABCD is a rhombus

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

72^{o} + ∠B = 180^{o} (Given, ∠A = 72^{o})

∠B = 180^{o} – 72^{o} = 108^{o}

Hence,

x = ½ B = ½ x 108^{o} = 54^{o}

**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.**

**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 – 20^{o} = x + 40^{o}

3x – x = 40^{o} + 20^{o}

2x = 60^{o}

x = 60^{o}/2 = 30^{o} … (i)

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

3x – 20^{o} + y + 15^{o} = 180^{o}

3x + y = 180^{o} + 20^{o} – 15^{o}

3x + y = 185^{o}

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

90^{o} + y = 185^{o}

y = 185^{o} – 90^{o} = 95^{o}

Hence,

x = 30^{o} and 95^{o}

(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 = 60^{o} and AB = AD

∠ADB = ∠ABD

So,

∠ADB = (180^{o} – ∠BAD)/ 2

= (180^{o} – 60^{o})/ 2

= 120^{o}/2 = 60^{o}

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

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: **

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 = 50^{o } (Since, given ∠PBA = 50^{o})

Now, in ∆APB

∠APB + ∠ABP + ∠BAP = 180^{o}

∠APB + 50^{o} + 50^{o} = 180^{o}

∠APB = 180^{o} – 100^{o}

∠APB = 80^{o}

Then,

∠DPB = ∠APB (Vertically opposite angles)

Hence,

∠DPB = 80^{o}

**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.**

**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

= 90^{o} + 60^{o}

= 150^{o} … (iv)

Also, EC = CD [From (iii)]

So, ∠DEC = ∠CDE … (v)

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

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

2 ∠DEC = 180^{o} – 150^{o} = 30^{o}

∠DEC = 30^{o}/2

∠DEC = 15^{o} … (vi)

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

∠BED + ∠DEC = 60^{o}

x^{o} + 15^{o} = 60^{o} [From (vi)]

x = 60^{o} – 15^{o}

x = 45^{o}

Hence, the value of x is 45^{o}.

(b) Given, ABCD is a rectangle

∠ECD = 146^{o}

As ACE is a straight line, we have

146^{o} + ∠ACD = 180^{o} [Linear pair]

∠ACD = 180^{o} – 146^{o} = 34^{o }… (i)

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

From (i) and (ii), we have

∠CAB = 34^{o} ⇒ ∠OAB = 34^{o} … (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 = 34^{o} … (v)

Now,

∠AOB + ∠OBA + ∠OAB = 180^{o}

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

∠AOB + 68^{o} = 180^{o}

∠AOB = 180^{o} – 68^{o} = 112^{o}

Hence, ∠AOB = 112^{o}, ∠OAB = 34^{o} and ∠OBA = 34^{o}

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

Let ∠OAB = 2x^{o}

We know that diagonals of rhombus intersect at right angle,

So, ∠OAB = 90^{o}

Now, in ∆AOB

∠OAB + ∠OBA = 180^{o}

90^{o} + 3x^{o }+ 2x^{o} = 180^{o}

90^{o} + 5x^{o} = 180^{o}

5x^{o} = 180^{o} – 90^{o} = 90^{o}

x^{o} = 90^{o}/5 = 18^{o}

Hence,

∠OAB = 3x^{o} = 3 x 18^{o} = 54^{o}

OBA = 2x^{o} = 2 x 18^{o} = 36^{o} and

∠AOB = 90^{o}

**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.**

**Solution: **

(a) Given: ABCD is a trapezium

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

We have,

∠B + ∠C = 180^{o} [Since AB || DC]

y + 92^{o} = 180^{o}

y = 180^{o} – 92^{o} = 88^{o}

Also, ∠A + ∠D = 180^{o}

x + 20^{o} + 2x + 10^{o} = 180^{o}

3x + 30^{o} = 180^{o}

3x = 180^{o} – 30^{o} = 150^{o}

x = 150^{o}/3 = 50^{o}

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

(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 = 180^{o}

2x + 3x = 180^{o}

5x = 180^{o}

x = 180^{o}/5 = 36^{o} … (i)

Also, AB = BC and AB || DC

So, ∠A + ∠C = 180^{o}

2x + y = 180^{o}

2 × 36^{o} + y = 180^{o}

72^{o} + y = 180^{o}

y = 180^{o} – 72^{o} = 108^{o}

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

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

∠DAB = 112^{o} and ∠DCB = 64^{o}

As AC is the diagonal of kite ABCD, we have

∠DCO = 64^{o}/2 = 32^{o}

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

In ∆OCD, we have

∠ODC = 180^{o }– (∠DCO + ∠DOC)

= 180^{o} – (32^{o} + 90^{o})

= 180^{o} – 122^{o}

= 58^{o}

In ∆DAB, we have

∠OAB = 112^{o}/2 = 56^{o}

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

In ∆OAB, we have

∠OBA = 180^{o }– (∠OAB + ∠AOB)

= 180^{o} – (56^{o} + 90^{o})

= 180^{o} – 146^{o}

= 34^{o}

Hence, ∠ODC = 58^{o} and ∠OBA = 34^{o}.

**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

To prove: Each angle of rectangle = 90^{o}

Proof:

In a rectangle opposite angles of a rectangle are equal

So, ∠A = ∠C and ∠B = ∠C

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

∠A + ∠B + ∠A + ∠B = 360^{o}

2(∠A + ∠B) = 360^{o}

(∠A + ∠B) = 360^{o}/2

∠A + ∠B = 180^{o}

But, ∠A = ∠B [Angles of a rectangle]

So, ∠A = ∠B = 90^{o}

Thus,

∠A = ∠B = ∠C = ∠D = 90^{o}

Hence, each angle of a rectangle is 90°.

(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:

Join 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 = 180^{o}

∠ABC = 90^{o}

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 = 180^{o} [Linear pair]

∠AOD = 90^{o}

And, ∠COD = 90^{o}

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

To 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.

To prove: ∠AMB = 90^{o}

Proof:

We have,

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

⇒ ½ (∠A + ∠B) = 180^{o}/2

½ ∠A + ½ ∠B = 90^{o}

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

Now, in ∆AMB

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

∠AMB + 90^{o} = 180^{o}

∠AMB = 180^{o} – 90^{o} = 90^{o}

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

To 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

To 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: **

(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 90^{o}

As AC bisects ∠A and ∠C

So, ∠1 = ∠2 and ∠3 = ∠4

But, ∠A = ∠C = 90^{o}

∠2 = 45^{o} and ∠4 = 45^{o}

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 90^{o}]

So, ∆AOB ≅ ∆BOC, by S.A.S axiom

By C.P.C.T, we have

AB = BC … (i)

Similarly, in ∆BOC and ∆COD

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

OC = OC [Common]

∠BOC = ∠COD [Each 90^{o}]

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.

To 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.**

**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

To prove: AP ⊥ DQ

Proof:

In ∆ABP and ∆ADQ, we have

AP = DQ [Given]

AD = AB [Sides of square ABCD]

∠DAQ = ∠ABP [Each 90^{o}]

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

Now, by C.P.C.T

∠BAP = ∠ADQ

But, ∠BAD = 90^{o}

∠BAD = ∠BAP + ∠PAD … (i)

90^{o} = ∠BAP + ∠PAD

∠BAP + ∠PAD = 90^{o}

∠BAP + ∠ADQ = 90^{o}

Now, in ∆ADM we have

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

90^{o} + ∠AMD = 180^{o} [From (i)]

∠AMD = 180^{o} – 90^{o} = 90^{o}

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

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]

180^{o} – ∠APB = 180^{o} – ∠DQC [Adding 180^{o} 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.**

**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 = 180^{o} [LAM is a straight line]

½ (∠LAB + ∠BAM) = 90^{o}

½ ∠LAB + ½ ∠BAM = 90^{o}

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

∠CAD = 90^{o}

∠A = 90^{o}

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

½ (∠PBA + ∠QBA) = 90^{o}

½ ∠PBA + ½ ∠QBA = 90^{o}

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

∠CBD = 90^{o}

∠B = 90^{o}

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

½ ∠LAB + ½ ∠ABP = 90^{o}

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

(4) In ∆ACB,

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

(∠2 + ∠6) + ∠C = 180^{o}

90^{o} + ∠C = 180^{o} [using (3)]

∠C = 180^{o} – 90^{o}

∠C = 90^{o}

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

½ ∠MAB + ½ ∠ABQ = 90^{o}

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

(6) In ∆ADB,

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

(∠3 + ∠7) + ∠D = 180^{o}

90^{o} + ∠D = 180^{o} [using (5)]

∠D = 90^{o}

(7) ∠LAB + ∠BAM = 180^{o}

∠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 = 90^{o} [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 = 90^{o}

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)]

(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 = 180^{o} [Sum of co-interior angles is equal to 180^{o}, AD || BC]

½ ∠A + ½ ∠B = 180^{o}/ 2

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

(13) In ∆APB,

∠AEB + ∠2 + ∠4 = 180^{o}

∠AEB + 90^{o} = 180^{o}

^{‑}Hence, ∠AEB = 90^{o}

**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]

∠1 = ∠2 [AE is bisector of ∠A, given]

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 = 45^{o}

∠OAP = 45^{o}

(2) In ∆AOP,

∠OAP = 45^{o} [From (1)]

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

Now,

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

∠AOP + ∠AOP + 45^{o} = 180^{o}

2 ∠AOP = 180^{o} – 45^{o}

∠AOP = 135^{o}/2

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

So, ∠AOP + ∠POB = 90^{o}

135^{o}/2 + ∠POB = 90^{o} [From (2)]

∠POB = 90^{o} – 135^{o}/2

= (180^{o} – 135^{o})/2

= 45^{o}/2

3 ∠POB = 135^{o}/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

EF, 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]

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 = 90^{o}

∠1 + ∠2 = 90^{o}

Thus, ∠HEF = 90^{o}

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.**

**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 + 90^{o} + 90^{o} + ∠BAD = 360^{o} [At a point total angle is 360^{o}]

∠FAG = 360^{o} – 90^{o} – 90^{o }– ∠BAD

∠FAG = 180^{o} – ∠BAD

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

∠B = 180^{o} – ∠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 = 60^{o}

So, ABD is an equilateral triangle

⇒ BD = AB = a

We know that, the diagonals of a rhombus bisect each other at right angles

So, in right triangle AOB, we have

AO^{2} + OB^{2} = AB^{2} [By Pythagoras Theorem]

AO^{2} = AB^{2} – OB^{2}

= a^{2} – (½ a)^{2}

= a^{2} – a^{2}/4

= 3a^{2}/4

AO = √(3a^{2}/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: **

Steps of construction:

(i) Draw a line segment AB = 6cm

(ii) At A, draw a ray AX making an angle of 45^{o} 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 60^{o}.

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

**Solution: **

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 90^{o} 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: **

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 90^{o}.

(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: **

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: **

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: **

Steps of construction:

(i) Draw a line segment BC = 6.2cm

(ii) At B, draw a ray BX making an angle of 60^{o} 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: **

Steps of construction:

(i) Draw a line segment BC = 7cm

(ii) A to B, draw a ray Bx making an angle of 75^{o} 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: **

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: **

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: **

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: **

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 = 60^{o} 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: **

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 75^{o }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: **

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 61^{o}.

**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: **

Steps of construction:

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

(ii) At A, construct ∠QAF = 60^{o}

(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: **

Steps of construction:

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

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

(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 45^{o} 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: **

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: **

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 = 106^{o}.

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

**Solution: **

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 8 cm and 6cm respectively.**

**Solution: **

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: **

Steps of construction:

(i) Draw a line segment BC = 4.6cm

(ii) At C, draw a ray CX making an angle of 135^{o} 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: **

Steps of construction:

(i) Draw a line segment AB = 4.6cm

(ii) At B, draw a ray BZ making an angle of 90^{o} 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 120^{o} 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: **

Step construction:

(i) Draw a line segment AB = 4.8cm

(ii) At A, draw a ray AZ making an angle of 90^{o} 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: **

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.**

**Solution: **

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

(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.

To 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. **

**Solution: **

(i) As CDE is a straight line

∠ADE + ∠ADC = 180^{o}

122^{o} + ∠ADC = 180^{o}

∠ADC = 180^{o} – 122^{o} = 58^{o} … (i)

∠ABC = 360^{o} – 140^{o} = 220^{o} … (ii)

^{o}]

Now, in quadrilateral ABCD we have

∠ADC + ∠BCD + ∠BAD + ∠ABC = 360^{o}

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

331^{o} + x = 360^{o}

x = 360^{o} – 331^{o}

x = 29^{o}

(ii) As DE || AB [Given]

∠ECB = ∠CBA [Alternate angles]

75^{o} = ∠CBA

⇒ ∠CBA = 75^{o}

Since, AD || BC we have

(x + 66^{o}) + 75^{o} = 180^{o}

x + 141^{o} = 180^{o}

x = 180^{o }– 141^{o}

x = 39^{o} … (i)

Now, in ∆AMB

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

39^{o} + 30^{o} + ∠AMB = 180^{o} [From (i)]

69^{o} + ∠AMB + 180^{o}

∠AMB = 180^{o} – 69^{o} = 111^{o} … (ii)

Since, ∠AMB = y [Vertically opposite angles]

⇒ y = 111^{o}

Hence, x = 39^{o} and y = 111^{o}

(iii) In ∆ABD

AB = AD [Given]

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

∠ABD = 42^{o} [Since, given ∠ADB = 42^{o}]

And,

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

42^{o} + 42^{o} + y = 180^{o}

84^{o} + y = 180^{o}

y = 180^{o} – 84^{o}

y = 96^{o}

∠BCD = 2 x 26^{o} = 52^{o}

In ∆BCD,

As BC = CD [Given]

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

∠CBD + ∠CDB + ∠BCD = 180^{o}

x + x + 52^{o} = 180^{o}

2x + 52^{o} = 180^{o}

2x = 180^{o} – 52^{o}

x = 128^{o}/2

x = 64^{o}

Hence, x = 64^{o} and y = 90^{o}.

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

**Solution: **

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

So, ABCD is a || gm

y = 2 x ∠ABD

y = 2 x 53^{o} = 106^{o} … (1)

Also, y + ∠DAB = 180^{o}

∠DAB = 180^{o} – 106^{o}

= 74^{o}

Thus, x = ½ ∠DAB [As AC bisects ∠DAB]

x = ½ x 74^{o} = 37^{o}

and ∠DAC = x = 37^{o} … (ii)

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

From (ii) and (iii),

z = 37^{o}

Hence, x = 37^{o}, y = 106^{o} and z = 37^{o}

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

60^{o} + ∠AED = 180^{o} [Linear pair]

∠AED = 180^{o} – 60^{o}

∠AED = 120^{o} … (i)

Also, as CD is a straight line

50^{o} + ∠BCD = 180^{o} [Linear pair]

∠BCD = 180^{o} – 50^{o}

∠BCD = 130^{o} … (ii)

In pentagon ABCDE, we have

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

90^{o} + 90^{o} 120^{o} + 130^{o} + x = 540^{o}

430^{o} + x = 540^{o}

x = 540^{o} – 430^{o}

x = 110^{o}

Hence, value of x = 110^{o}

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

60^{o} + y = 180^{o} and x + 110^{o} = 180^{o}

y = 180^{o} – 60^{o} and x = 180^{o} – 110^{o}

y = 120^{o} and x = 70^{o}

Since, CD || AF [Given]

∠FAD = 70^{o} … (i)

In quadrilateral ADEF,

∠FAD + 75^{o} + z + 130^{o} = 360^{o}

70^{o} + 75^{o} + z + 130^{o} = 360^{o}

275^{o} + z = 360^{o}

z = 360^{o} – 275^{o} = 85^{o}

Hence,

x = 70^{o}, y = 120^{o} and z = 85^{o}

**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.

So, 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 = 56^{o} [Given]

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

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

In ∆ADE,

DE = AD [From (iii)]

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

∠DEA = ∠DAG = (180^{o} – ∠EDA)/ 2

= (180^{o} – 146^{o})/ 2

= 34^{o}/ 2 = 17^{o}

⇒ ∠DAG = 17^{o}

∠FEG = ∠E – ∠DEG

= 90^{o} – 17^{o}

= 73^{o}

In rhombus ABCD,

∠DAB = 180^{o} – 56^{o} = 124^{o}

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

∠DAC = 62^{o}

∠GAC = ∠DAC – ∠DAG

= 62^{o} – 17^{o}

= 45^{o}

In ∆EDG,

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

90^{o} + 17^{o} + ∠DGE = 180^{o}

∠DGE = 180^{o} – 107^{o} = 73^{o} … (iv)

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

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

∠AGC = 73^{o}

**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: **

Each side of rhombus ABCD is 8cm

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

Let ∠A = 60^{o}

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 = 90^{o}

Now, in right ∆AOB

By Pythagoras theorem

AB^{2} = AO^{2} + OB^{2}

8^{2} = AO^{2} + 4^{2}

64 = AO^{2} + 16

AO^{2} = 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: **

Steps of construction:

(i) Draw AB = 5.0cm

(ii) Draw BAP = 45^{o} 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 90^{o}.