**Properties of Angles and Lines** ICSE Class-6th Concise Selina Mathematics Solutions Chapter-25 (Including Parallel Lines). We provide step by step Solutions of Exercise / lesson-25 **Properties of Angles and Lines** (Including Parallel Lines) for ICSE Class-6 Concise Selina Mathematics.

Our Solutions contain all type Questions of Exe-25 A, Exe-25 B, Exe-25 C, Exe-25 D and Revision Exercise to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-6.

**Properties of Angles and Lines** ICSE Class-6th Concise Selina Mathematics Solutions Chapter-25 (Including Parallel Lines)

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### Exercise – 25 A **Properties of Angles and Lines **(Including Parallel Lines) for ICSE Class-6th Concise Selina Mathematics Solutions

**Question -1-.**

Two straight lines AB and CD intersect each other at a point O and angle AOC = 50° ; find :

(i) angle BOD

(ii) ∠AOD

(iii) ∠BOC

**Answer-1**

**(i)**∠BOD = ∠AOC

(Vertically opposite angles are equal)

∴ ∠BOD =50°

**(ii)** ∠AOD

∠AOD + ∠BOD = 180°

∠AOD + 50° = 180° [From (i)]

∠AOD = 180°-50°

∠AOD = 130°

**(iii)** ∠BOC = ∠AOD

(Vertically opposite angles are equal)

∴ ∠BOC =130°

**Question- 2.**

The adjoining figure, shows two straight lines AB and CD intersecting at point P. If ∠BPC = 4x – 5° and ∠APD = 3x + 15° ; find :

(i) the value of x.

(ii) ∠APD

(iii) ∠BPD

(iv) ∠BPC

**Answer-2**

(i) 4x – 5 = 3x + 15 (∵ ∠CPB = ∠APD; opposite angles)

⇒ 4x – 3x = 15 + 5

⇒ x = 20

(ii) ∠APD = 3x + 15

= (3 × 20) + 15

= 60 + 15 = 75°

(iii) ∠BPD = 180 – ∠BPC

= 180 – (4x – 5)

= 180 – 4x + 5

= 185 – (4 × 20)

= 185 – 80 = 105°

(iv) ∠BPC = 4x – 5

= (4 × 20) – 5

= 80 – 5

= 75°

**Question- 3.**

**Answer-3**

Since, the exterior arms of the adjacent angles are in a straight line ; the adjacent angles are supplementary

∴ ∠AOB + ∠AOC = 180°

⇒ 68° + 3x – 20° = 180°

⇒ 3x = 180° + 20° – 68°

⇒ 3x = 200° – 68° ⇒ 3x =132°

x = ^{132}⁄_{3} = 44°

**Question -4.**

Each figure given below shows a pair of adjacent angles AOB and BOC. Find whether or not the exterior arms OA and OC are in the same straight line.

**Answer-4**

**(i)** ∠AOB + ∠COB = 180°

Since, the sum of adjacent angles AOB and COB = 180°

(90° -x) + (90°+ x) = 180°

⇒ 90°-x + 90° + x = 180°

⇒ 180° =180°

The exterior arms. OA and OC are in the same straight line.

**(ii)** ∠AOB + ∠BOC = 97° + 83° = 180°

⇒ The sum of adjacent angles AOB and BOC is 180°.

∴ The exterior arms OA and OC are in the same straight line.

**(iii)**∠COB + ∠AOB = 88° + 112° = 200° ; which is not 180°.

⇒ The exterior amis OA and OC are not in the same straight line.

**Question -5.**

A line segment AP stands at point P of a straight line BC such that ∠APB = 5x – 40° and ∠APC = .x+ 10°; find the value of x and angle APB.

**Answer-5**

AP stands on BC at P and

∠APB = 5x – 40°, ∠APC = x + 10°

**(i)** ∵APE is a straight line

∠APB + ∠APC = 180°

⇒ 5x – 40° + x + 10° = 180°

⇒ 6x-30°= 180°

⇒6x= 180° + 30° = 210°

x = 2106° = 35°

**(ii)** and ∠APB = 5x – 40° = 5 x 35° – 40°

= 175 ° – 140° = 135°

**Properties of Angles and Lines **(Including Parallel Lines) Exercise-25 B for ICSE Class-6th Concise Selina Solutions

**Question -1.**

Identify the pair of angles in each of the figure given below :

adjacent angles, vertically opposite angles, interior alternate angles, corresponding angles or exterior alternate angles.

**Answer-1**

**(a)**

** (i)** Adjacent angles

**(ii)** Alternate exterior angles

**(iii)** Interior alternate angles

**(iv)** Corresponding angles

**(v)** Allied angles

**(b)**

** (i)** Alternate interior angles

**(ii)** Corresponding angles

**(iii)** Alternate exterior angles

**(iv)** Corresponding angles

**(v)** Allied angles.

**(c)**

** (i)** Corresponding

**(ii)** Alternate exterior

**(iii)** Alternate interior

**(iv)** Alternate interior

**(v)** Alternate exterior

**(vi)** Vertically opposite

**Question -2.**

Each figure given below shows a pair of parallel lines cut by a transversal For each case, find a and b, giving reasons.

**Answer-2**

(i) a + 140° = 180° (Linear pair)

∴ a = 180° – 140° = 40°

But b = a (alternate angles)

= 40°

∴ a = 40°, b = 40°

(ii) ∵ l || m and p intersects them

b + 60° = 180° (Linear pair)

∴ b = 180° – 60° = 120°

and a = 60° (corresponding angle)

∴ a = 60°, b = 120°

(iii) a = 110° (Vertically opposite angles)

b = 180° – a (Co-interior angles)

= 180° – 110° = **70°**

(iv) a = 60° (Alternate interior angles)

b = 180° – a (Co-interior angles)

= 180° – 60° = **120°**

**(v) **a = 72° (Alternate interior angles)

b = a (Vertically opposite angles)

i.e. b = 72°

(vi) b = 100° (Corresponding angles)

a = 180° – b (Linear pair of angles)

a = 180° – 100° = 80°

(vii) a = 180° – 130° = 50° (Co-interior angles)

b = 130° (Vertically opposite angles)

(viii) b = 62° (Corresponding angles)

a = 180° – b (Linear pairs of angles)

a = 180° – 62° = 118°

(viii) a = 180° – 90° (Linear pairs of angles)

= 90°

b = **90° ** (Corresponding angles)

**Question -3.**

**Answer-3**

l || m and p is their transversal and ∠1 = 120°

∠1 + ∠2 = 180° (Straight line angle)

∴ 120° + ∠2 = 180°

⇒ ∠2 = 180° – 120° = 60°

∴ ∠2 = 60°

But ∠1 = ∠3 (Vertically opposite angles)

∴ ∠3 = ∠1 = 120°

Similarly ∠4 = ∠2 (Vertically opposite angles)

∴ ∠4 = 60°

∠5 = ∠1 (Corresponding angles)

∴ ∠5 = 120°

Similarly ∠6 = ∠2 (Corresponding angles)

∴ ∠6 = 60°

∠7 = ∠5 (Vertically opposite angles)

∴ ∠7 = 120°

and ∠8 = ∠6 (Vertically opposite angles)

∴ ∠8 = 60°

Hence ∠2 = 60°, ∠3 = 120°, ∠4 = 60°,∠5 = 120°, ∠6 = 60°, ∠7 = 120° and ∠8 = 60°.

**Question- 4.**

In the figure given below, find the measure of the angles denoted by x,y, z,p,q and r.

**Answer-4**

x = 180 – 100 [L.P. of angles] = 80°

y = x [Alternate exterior angles]

= 80°

z = 100° [Corresponding angles]

p = x [Vertically opp. angles]

= 80°

q = 100° [Vertically opp. angles]

r = q [Corresponding angles]

= 100°

**Question -5.**

Using the given figure, fill in the blanks.

**Answer-5**

x = **60°** [Corresponding angles]

z = x [Corresponding angles]

= **60°**

p = z [Vertically opposite angle]

= **60°**

q = 180 – P [Linear Pair of angles]

= 180 – 60 = **120°**

r = 180 – x [Linear Pair of angles]

= 180 – 60 = **120°**

s = r [Vertically opposite angle] = **120°**

**Question- 6.**

In the given figure, find the angles shown by x,y, z and w. Give reasons.

**Answer-6**

x = 115° [Vertically of angles]

y = 70° [vertically opposite Angles]

z = 70° [Alternate interior angles]

w = 115° [Alternate interior angles]

**Question -7.**

Find a, b, c and d in the figure given below :

**Answer-7**

a = 130° [Vertically of angles]

b = 150° [vertically opposite Angles]

c = 150° [Alternate interior angles]

d = 130° [Alternate interior angles]

**Question- 8.**

Find x, y and z in the figure given below :

**Answer-8**

x = 180 – 75 = **105°** [Co-interior angles]

y = 180 – x [Co-interior angles]

= 180 – 105 = **75°**

z = **75°** [Co-interior angles]

### Concise Maths Selina Solutions of Exercise 25 C **Properties of Angles and Lines **(Including Parallel Lines) for ICSE Class-6th

**Question- 1.**

In your note-book copy the following angles using ruler and a pair compass only.

**Answer-1**

(i) **Steps of Construction:**

1. At point Q, draw line QR = OB.

2. With O as a centre, draw an arc of any suitable radius, to cut the arms of the angle at C and D.

3. With Q as a centre, draw the arc of the same size as drawn for C and D. Let this arc cuts line QR at point T.

4. In your compasses, take the distance equal to distance between C and D; and then with T as a centre, draw an arc which cuts the earlier arc at S.

5. Join QS and produce up to a suitable point P. ∠PQR so obtained, is the angle equal to the given ∠AOB.

(ii) **Steps of Construction:**

1. A t point E, draw line EF.

2. With E as centre, draw an arc of any suitable radius, to cut the amis of the angle at C and D.

3. With Q as a centre, draw the arc of the same size as drawn for C and D. Let this arc cuts line QR at point T.

4. In your compasses, take the distance equal to the distance between C and D; and then with T as a centre, draw an arc which cuts the earlier arc at S.

5. Join QS and produce up to a suitable point R ∠PQR, so obtained, is the angle equal to the given ∠DEE

(iii) **Steps of Construction:**

1. At point A draw line AB = QP

2. With Q as centre, draw an arc of any suitable radius, to cut the arms of the angle A + C and D.

3. With A as centre, draw the arc of the same size as drawn for C and D. Let this arc cuts line AB at D.

4. In your compasses, take the distance equal to distance between 7 and 5; and then with D as centre, draw an arc which cuts the earlier arc at E.

5. Join AE and produced up to a suitable point C. ∠BAC, so obtained is the angle equal to the given ∠PQR.

**Question -2.**

Construct the following angles, using ruler and a pair of compass only

(i) 60°

(ii) 90°

(iii) 45°

(iv) 30°

(v) 120°

(vi) 135°

(vii) 15°

**Answer-2**

**(i) 60°** **Steps of Construction:**

To construct an angle of 60°.

- Draw a line OA of any suitable length.
- At O, draw an arc of any size to cut OA at B.
- With B as a centre, draw the same size arc, to cut the previous arc at C.
- Join OC and extend up to a suitable point D, Then, ∠DOA = 60°.

**(ii) 90° ****Steps of Construction:**

To construct an angle of 90°.

- With O as a center, draw an arc to cut OA at B.
- With B as a centre, draw the same size arc to cut the previous arc at C.
- Again with C as centre and with the same radius, draw one more arc to cut the first arc at D.
- With C and D as centres, draw two arcs of equal radii to cut each other at point E.
- Join O and E, Then, ∠AOE = 90°.

**(iii) 45°**

Draw an angle of 90° as in the previous question and bisects it. Each angle so obtained will be 45°.

**(iv) 30° ****Steps of construction:**

To construct an angle of 30°.

- Draw a line OB of any suitable length.
- At O, draw an arc of any size to cut OB at D.
- With D as a centre, draw the same size arc, to cut the previous arc at C.
- Join OC and extend up to a suitable point A. Then, ∠AOB = 60°.
- Bisect this angle of get two angles each of 30°. Thus, ∠EOB = 30°.

**(v) 120° ****Steps of construction:**

To construct an angle of 120°.

- With centre O on the line OA, draw an arc to cut this line at C.
- With C as a centre, drawn the same size arc which cuts the first arc at point D.
- With D as a centre, draw one more arc of same size which cuts the first arc at E.
- Join OE and produce it up to point B. Then, ∠AOB = 120°

**(vi) 135° ****Steps of construction:**

To construct an angle of 135°.

- Draw an angle BOA = 90° at point O of given line AC.
- Bisect the angle BOC on the other side of OB, which is also 90°
- Thus, ∠BOD = ∠COD = 45°

And, ∠AOD = 90° + 45° = 135°

**(vii) 15° ****Steps of construction:**

To construct an angle of 15°.

- Draw an angle of 60° as drawn above.
- Bisect this angle of get two angles each of 30°. Thus, ∠EOB = 30°
- Bisect this angle ∠EOB to get two angles each of 15°. ∠EOB = 15°.

**Question -3.**

Draw line AB = 6cm. Construct angle ABC = 60°. Then draw the bisector of angle ABC.

**Answer-3**

**Steps of Construction:**

1. Draw a line segment AB = 6 cm.

2. With the help of compass construct ∠CBA = 60°.

3. Bisect ∠CBA, with the help of a compass, take any radius which meet line AB and BC at point E and F.

4. Now, with the help of compass take radius more than 12 of EF and draw two arcs from point E and F, which intersect both arcs at G, proceed BG toward D ∠DBA is bisector of ∠CBA.

**Question -4.**

Draw a line segment PQ = 8cm. Construct the perpendicular bisector of the line segment PQ. Let the perpendicular bisector drawn meet PQ at point R. Measure the lengths of PR and QR. Is PR = QR ?

**Answer-4**

**Steps of Construction:**

1. With P and Q as centers, draw arcs on both sides of PQ with equal radii. The radius should be more than half the length of PQ.

2. Let these arcs cut each other at points R and RS

3. Join RS which cuts PQ at D.

Then RS = PQ Also ∠POR = 90°.

Hence, the line segment RS is the perpendicular bisector of PQ as it bisects PQ at P and is also perpendicular to PQ. On measuring the lengths of PR = 4cm, QR = 4 cm Since PR = QR, both are 4cm each

∴ PR = QR.

**Question -5.**

Draw a line segment AB = 7cm. Mark a point Pon AB such that AP=3 cm. Draw perpendicular on to AB at point P.

**Answer-5**

1. Draw a line segment AB = 7 cm.

2. Out point from AB – AP =3cm

3. From point P, cut arc on out side of AB, E and F.

4. From point E & F cut arcs on both side intersection each other at C & D.

5. Join point P, CD.

6. Which is the required perpendicular.

**Question -6.**

Draw a line segment AB = 6.5 cm. Locate a point P that is 5 cm from A and 4.6 cm from B. Through the point P, draw a perpendicular on to the line segment AB.

**Answer-6**

**Steps of Construction :**

** **

(i) Draw a line segment AB =6.5cm

(ii) With centre A and radius 5 cm, draw an arc and with centre B and radius 4.6 cm, draw another arc which intersects the first arc at P.

Then P is the required point.

(iii) With centre A and a suitable radius, draw an arc which intersect AB at E and F.

(iv) With centres E and F and radius greater than half of EF, draw the arcs which intersect each other at Q.

(v) Join PQ which intersect AB at D.

Then PD is perpendicular to AB

**ICSE Class-6 Mathematics Exe-25 D Properties of Angles and Lines **(Including Parallel Lines)

**Question -1.**

Draw a line segment OA = 5 cm. Use set-square to construct angle AOB = 60°, such that OB = 3 cm. Join A and B ; then measure the length of AB.

**Answer-1**

Measuring the length of AB = 4.4cm. (approximately)

**Question- 2.**

Draw a line segment OP = 8cm. Use set-square to construct ∠POQ = 90°; such that OQ = 6 cm. Join P and Q; then measure the length of PQ.

**Answer-2**

Measuring PQ = 10 cm

**Question- 3.**

Draw ∠ABC = 120°. Bisect the angle using ruler and compasses. Measure each angle so obtained and check whether or not the new angles obtained on bisecting ∠ABC are equal.

**Answer-3**

Measuring PQ = 60°

Yes, angles obtained in ∠ABC bisecting are equal.

**Question -4.**

Draw ∠PQR = 75° by using set- squares. On PQ mark a point M such that MQ = 3 cm. On QR mark a point N such that QN = 4 cm. Join M and N. Measure the length of MN.

**Answer-4**

Length of MN = 4.3 cm

### Revision Exercise **Properties of Angles and Lines **(Including Parallel Lines) for ICSE Class-6th Concise Selina Mathematics Solutions

**Question -1.**

In the following figures, AB is parallel to CD; find the values of angles x, y and z :

**Answer-1**

**(i) In the given figure,**

AB || CD

and LM is its transversal

∴ ∠ALM = ∠LMN (Alternate angles)

⇒ ∠x = 105°

∴ x = 105°

Similarly AB || CD and LN is its transversal

∴ ∠BLN = ∠LNM (Alternate angles)

∴ ∠z = 60°

∴ z = 60°

But x + y + z = 180° (Straight line angles)

⇒ 105° + y + 60° = 180°

⇒ y + 165° = 180°

⇒ y = 180° – 165° = 15°

Hence x = 105°, y = 15° and z = 60°

**(ii) In the given figure,**

AB || CD

and MN is its transversal

∴ ∠LMN = ∠NMD (Alternate angles)

⇒ y = 45°

and AB || CD and LM is its transversal

∴ ∠ALM = ∠CMP (Corresponding angles)

⇒ 75° = x

∴ x = 75°

and ∠ALM = ∠LMD (Alternate angles)

⇒ 75° = z + 45°

⇒ z =75° – 45° = 30°

Hence x = 75°, y = 45° and z = 30°

**Question -2.**

In each of the following figures, BA is parallel to CD. Find the angles a, b and c:

**Answer-2**

(i)

In the given figure,

ABC is a triangle and CD || BA, BC is produced to E.

∠A = 60°, ∠B = 70°

∵ AB || DC and BE is its transversal

∴ ∠DCE = ∠ABC (Corresponding angles)

⇒ a = 70°

∴ a = 70°

Similarly, AB || DC and AC is its transversal

∴ ∠ACD = ∠BAC (Alternate angles)

⇒ b = 60°

∴ b = 60°

But a + b + c = 180° (Straight line angle)

⇒ 70° + 60° + c = 180°

⇒ 130° + c = 180°

⇒ c = 180° – 130° = 50°

Hence a = 70°, b = 60° and c = 50°

(ii) In the given figure,

ABC is a triangle and AB || DC and AC is its transversal.

∠BAC = ∠ACD (Alternate angles)

⇒ b = 65°

Again AB || DC and BCE is its transversal

∴ ∠ABC = ∠DCE

⇒ C = 70°

But ∠ACB + ∠ACD + ∠DCE = 180 (Straight line angle)

∴ a + 65° + 70° = 180°

⇒ a + 135° = 180°

⇒ a = 180° – 135° = 45°

Hence a = 45°, b = 65° and c = 70°

**Question -3.**

In each of the following figures, PQ is parallel to RS. Find the angles a, b and c:

**Answer-3**

**(i) In the given figure,**

PQ || RS, ∠B = 75°, ∠ACS = 140°

AB is its transversal

∴ ∠PAB = ∠ABC

⇒ a = 75°

Again PQ || RS and AC is its transversal

∴ ∠QAC + ∠ACS = 180° (Co-interior angles)

⇒ c + 140° = 180°

⇒ c = 180° – 140° = 40°

But a + b + c = 180° (Straight line angles)

∴ 75° + b + 40° = 180°

⇒ b + 115° = 180°

⇒ b = 180° – 115° = 65°

Hence a = 75°, b = 65°, c = 40°

**(ii) In the given figure,**

PQ || RS, ∠BAR = 63°, ∠CAS = 57°

AB is its transversal.

∴ ∠CBA = ∠BAR (Alternate angles)

⇒ a = 63°

∵ PQ || RS and CA is its transversal

∴ ∠QCA + ∠CAS = 180° (Co-interior angles)

⇒ b + 57° = 180°

⇒ b = 180° – 57° = 123°

But ∠CAS + ∠CAB + ∠BAR = 180° (Straight line angles)

∴ 57° + c + 63° = 180°

⇒ c + 120° = 180°

⇒ c = 180° – 120° = 60°

Hence a = 63°, b = 123°, c = 60°

**Question- 4.**

Two straight lines are cut by a transversal. Are the corresponding angles always equal?

**Answer-4**

If a transversal cuts two straight lines, their the corresponding angles are not equal unless the lines are not parallel. One in case of parallel lines, the corresponding angles are equal.

**Question -5.**

Two straight lines are cut by a transversal so that the co-interior angles are supplementary. Are the straight lines parallel?

**Answer-5**

A transversal intersects two straight lines and co-interior angles are supplementary

∴ By deflations, the lines will be parallel.

**Question -6.**

Two straight lines are cut by a transversal so that the co-interior angles are equal. What must be the measure of each interior angle to make the straight lines parallel to each other ?

**Answer-6**

A transversal intersects two straight lines and co-interior angles are equal to each other,

∵ The two straight lines are parallel Their sum of co-interior angles = 180°

But both angles are equal

∴ Each angle will be ..^{180}⁄_{2}…= 90…

**Question- 7**

In each case given below, find the value of x so that POQ is straight line

**Answer-7**

(i) POQ is a straight line.

In the given figure,

∵ POQ is a straight line

∴ ∠POL + ∠LOM + ∠MOQ = 180° (Straight line angles)

⇒ x + 20° + 2x – 30° + 3x – 50°= 180°

⇒ 6x + 20° – 80° = 180°

⇒ 6x – 60° = 180°

⇒ 6x = 180° + 60° = 240°

⇒ x = 240∘6

⇒ x = 40°

∴ x = 40°

(ii) ∵ POQ is a straight line

∴ ∠POL + ∠LOQ = 180°

⇒ ^{7x}⁄_{11 }+x=180∘

⇒ ** ^{7x+11x}⁄_{11 }**=180°

⇒ ^{18x}⁄_{11 }=180°

⇒ x = 180°× ^{11}⁄_{18} = 110°

∴ x = 110°

(iii) 5.5x + 15° = 180°

⇒ 5.5x = 180° – 15°

⇒ 5.5x = 165°

⇒ ^{18x}⁄_{11}=180°

⇒ x = ^{165}⁄_{5.5 }= ^{165×10}⁄_{55} = 30°

∴ x = 30°

**Question- 8.**

in each case, given below, draw perpendicular to AB from an exterior point P

**Answer-8**

(i) **Steps of Construction:**

- From point P, draw an arc CD at line AB
- From point C and D draw arcs that intersect each other at point E, now draw PE, perpendicular to AB

(ii) **Steps of construction:**

- From point P, draw an arc CD at line AB.
- From point C and D draw arcs that intersect each other at point E, now draw PE, perpendicular to AB.

**Question -9.**

Draw a line segment BC = 8 cm. Using set-squares, draw ∠CBA = 60° and ∠BCA = 75°. Measure the angle BAC. Also measure the lengths of AB and AC.

**Answer-9**

Length AB = 11 cm

Length AC = 9.8 cm

∠BAC = 45°.

**Question -10.**

Draw a line AB = 9 cm. Mark a point P in AB such that AP=5 cm. Through P draw (using set-square) perpendicular PQ = 3 cm. Measure BQ.

**Answer-10**

BQ = 5 cm

**Question- 11.**

Draw a line segment AB = 6 cm. Without using set squares, draw angle OAB = 60° and angle OBA = 90°. Measure angle AOB and write this measurement.

**Answer-11**

**Steps of construction:**

- Draw a line segment AB = 6 cm
- At A, draw a ray making an angle of 60° with the help of a compass.
- At B, draw another ray making an angle of 90° which meet each other at O.

Now on measuring ∠AOB, it is 30°

**Question- 12.**

Without using set squares, construct angle ABC = 60° in which AB = BC = 5 cm. Join A and C and measure the length of AC.

**Answer-12**

**Steps of Construction:**

- Draw an angle ABC = 60

Such that AB = BC = 5 cm - Join AC, on measuring, the length of AC = 5 cm.

End of **Properties of Angles and Lines** Solutions :–

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