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

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

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**IMPORTANT POINTS**

**1. Property :** When two straight lines intersect:

(i) sum of each pair of adjacent angles is always 180°.

(ii) vertically opposite angles are always equal. .

**2. Property :** If the sum of two adjacent angles is 180°, their exterior arms are always in the same straight line.

Conversely, if the exterior arms of two adjacent angles are in the same straight line ; the sum of angles is always 180°

**3. Parallel Lines :** Two straight lines are said to be parallel, if they do not meet anywhere, no matter how much they are produced in either direction.

**4. Concepts of Transversal Lines :** When a line cuts two or more lines (parallel or non-parallel); it is called a transversal line or simply, a transversal. In each of the following figures : PQ is a transversal line.

**5. Angles formed by two lines and their transversal line :** When a transversal cuts two parallel or nonparallel lines; eight (8) angles are formed which are marked 1 to 8 in the adjoining diagram.

These angles can further he distinguished, as given below:

**(i) Exterior Angles :** Angles marked 1, 2, 7 and 8 are exterior angles.

**(ii) Interior Angles** : Angles marked 3, 4, 5 and 6 are interior angles.

**(iii) Exterior Alternates Angles :** Two pairs of exterior alternate angles are marked as : 2 and 8 ; and, 1 and 7.

**(iv) Interior Alternate Angles :** Two pairs of interior alternate are marked as : 3 and 5 ; and 4 and 6. In general, interior alternate angles are simply called as alternate angles only.

**(v) Corresponding Angles :** Four pairs of corresponding angles are marked as : 1 and 5 ; 2 and 6 ; 3 and 7 ; and 4 and 8.

**(vi) Co-interior or Conjoined or Allied Angles :** Two pairs of co-interior or allied angles are marked as : 3 and 6 ; and 4 and 5.

**(vii) Exterior Allied Angles :** Two pairs of exterior allied angles are marked as : 2 and 7 ; and 1 and 8.

### Properties of Angles and Lines Exercise 25A – Selina Concise Mathematics Class 6 ICSE 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**

**Solution:**

**(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**

**Solution:**

**Question 3.**

**The given diagram, shows two adjacent angles AOB and AOC, whose exterior sides are along the same straight line. Find the value of x.**

**Solution:**

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 = \(\frac { 132 }{ 3 }\)° = 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.**

**Solution:**

**(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.**

**Solution:**

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 = \(\frac { 210 }{ 6 }\)° = 35°

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

= 175 ° – 140° = 135°

### Properties of Angles and Lines Exercise 25B – Selina Concise Mathematics Class 6 ICSE 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.**

**Solution:**

**(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.**

**Solution:**

**Question 3.**

**If ∠1 = 120°, find the measures of : ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8. Give reasons.**

**Solution:**

**Question 4.**

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

**Solution:**

**Question 5.**

**Using the given figure, fill in the blanks.**

**Solution:**

**Question 6.**

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

**Solution:**

**Question 7.**

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

**Solution:**

**Question 8.**

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

**Solution:**

### Properties of Angles and Lines Exercise 25C – Selina Concise Mathematics Class 6 ICSE Solutions

**Question 1.**

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

**Solution:**

**(i) Steps of Construction :**

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

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

3. With Q as 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 centre, draw an arc which cuts the earlier arc at S.

5. Join QS and produce upto 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 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 centre, draw an arc which cuts the earlier arc at S.

5. Join QS and produce upto 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 upto 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°**

**Solution:**

**Question 3.**

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

**Solution:**

**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 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 \(\frac { 1 }{ 2 }\) 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 ?**

**Solution:**

**Steps of Construction :**

1. With P and Q as centres, 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.**

**Solution:**

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 pont 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.**

**Solution:**

**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.

### Properties of Angles and Lines Exercise 25D – Selina Concise Mathematics Class 6 ICSE Solutions

**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 ofAB.**

**Solution:**

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.**

**Solution:**

**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.**

**Solution:**

**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.**

**Solution:**

### Properties of Angles and Lines Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

**Question 1.**

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

**Solution:**

**Question 2.**

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

**Solution:**

**Question 3.**

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

**Solution:**

**Question 4.**

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

**Solution:**

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 ?**

**Solution:**

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 ?**

**Solution:**

A transveral 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 \(\frac { 180 }{ 2 }\)° = 90°

**Question 7.**

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

**Solution:**

**Question 8.**

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

**Solution:**

**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.**

**Solution:**

**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.**

**Solution:**

**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.**

**Solution:**

**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.**

**Solution:**