Selina Solutions Concise Mathematics Class 6 Chapter 25 Properties of Angles and Lines are well structured by highly experienced teachers at BYJUâ€™S, as per the understanding capacity of students. The solutions help students, to improve confidence and solve the problems with ease in the annual exam. Practising these solutions, on a daily basis, enable students to obtain expertise in Mathematics. Students who aspire to secure high marks, can refer to Selina Solutions Concise Mathematics Class 6 Chapter 25 Properties of Angles and Lines PDF links, which are given here with free download option.

## Selina Solutions Concise Mathematics Class 6 Chapter 25: Properties of Angles and Lines Download PDF

### Exercises of Selina Solutions Concise Mathematics Class 6 Chapter 25 Properties of Angles and Lines

## Access Selina Solutions Concise Mathematics Class 6 Chapter 25: Properties of Angles and Lines

Exercise 25(A)

**1. Two straight lines AB and CD intersect each other at a point O and angle AOC = 50 ^{0}; find: **

**(i) angle BOD**

**(ii) âˆ AOD **

**(iii) âˆ BOC**

**Solution:**

(i) âˆ BOD

Given âˆ AOC = 50^{0}

We know that,

Vertically opposite angles are equal

So, âˆ BOD = âˆ AOC

Therefore, âˆ BOD = 50^{0}

(ii) âˆ AOD

âˆ AOD + âˆ BOD = 180^{0}

âˆ AOD + 50^{0} = 180^{0} {From (i)}

âˆ AOD = 180^{0} â€“ 50^{0}

We get,

âˆ AOD = 130^{0}

(iii) âˆ BOC

We know that,

Vertically opposite angles are equal

So, âˆ BOC = âˆ AOD

Therefore, âˆ BOC = 130^{0}

**2. The adjoining figure, shows two straight lines AB and CD intersecting at P. If âˆ BPC = 4x â€“ 5 ^{0} and âˆ APD = 3x + 15^{0}; find:**

**(i) the value of x**

**(ii) âˆ APD**

**(iii) âˆ BPD**

**(iv) âˆ BPC**

**Solution:**

(i) The value of x is calculated as,

3x + 15^{0} = 4x â€“ 5^{0}

3x â€“ 4x = – 5^{0} – 15^{0}

– x =- 20^{0}

x = 20^{0}

(ii) the value of âˆ APD is calculated as,

âˆ APD = 3x + 15^{0}

= 3 Ã— 20^{0} + 15^{0}

We get,

= 60^{0} + 15^{0}

= 75^{0}

(iii) The value of âˆ BPD is calculated as,

âˆ BPD = 180 â€“ âˆ BPC

= 180^{0} â€“ (4x – 5^{0})

= 180^{0} â€“ (4 Ã— 20^{0} â€“ 5^{0})

= 180^{0} â€“ 80^{0} + 5^{0}

We get,

= 105^{0}

(iv) The value of âˆ BPC is calculated as,

âˆ BPC = (4x â€“ 5^{0})

= (4 Ã— 20^{0} â€“ 5^{0})

We get,

= 80^{0} – 5^{0}

= 75^{0}

**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:**

Here, the exterior arms of the adjacent angles are in a straight line,

Hence, the adjacent angles are supplementary

âˆ AOB + âˆ AOC =180^{0}

68^{0} + (3x – 20^{0}) = 180^{0}

68^{0} + 3x â€“ 20^{0} = 180^{0}

3x = 180^{0} + 20^{0} – 68^{0}

3x = 200^{0} – 68^{0}

We get,

3x = 132^{0}

x = 132^{0} / 3

x = 44^{0}

**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.**

**(i)**

** **

**(ii)**

** **

**(iii) **

**Solution:**

(i) We know that,

The sum of adjacent angles AOB and COB = 180^{0}

Hence,

âˆ AOB + âˆ COB = 180^{0}

(90^{0} â€“ x) + (90^{0} + x) = 180^{0}

90^{0} â€“ x + 90^{0} + x = 180^{0}

We get,

180^{0} = 180^{0}

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

(ii) âˆ AOB + âˆ BOC = 97^{0} + 83^{0}

= 180^{0}

The sum of adjacent angles AOB and BOC is 180^{0}

Hence, the exterior arms OA and OC are in the same straight line

(iii) âˆ COB + âˆ AOB = 88^{0} + 112^{0}

We get,

= 200^{0} [which is not equal to 180^{0}]

Hence, the exterior arms OA and OC are not in the same straight line

**5. A line segment AP stands at point P of a straight line BC such that âˆ APB = 5x â€“ 40 ^{0} and âˆ APC = x + 10^{0}; find the value of x and angle APB.**

**Solution:**

Given

A line segment AP stands at P and

âˆ APB = 5x â€“ 40^{0}

âˆ APC = x + 10^{0}

(i) BPC is a straight line

âˆ APB + âˆ APC = 180^{0}

5x â€“ 40^{0} + x + 10^{0}_{ }= 180^{0}

6x â€“ 30^{0} = 180^{0}

6x = 180^{0} + 30^{0}

We get,

6x = 210^{0}

x = 210^{0} / 6

x = 35^{0}

(ii) âˆ APB = 5x – 40^{0}

= 5 Ã— 35^{0} – 40^{0}

We get,

= 175^{0} â€“ 40^{0}

= 135^{0}

Exercise 25(B)

**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.**

**(a) (i) âˆ 2 and âˆ 4**

**(ii) âˆ 1 and âˆ 8**

**(iii) âˆ 4 and âˆ 5**

**(iv) âˆ 1 and âˆ 5**

**(v) âˆ 3 and âˆ 5**

**(b) (i) âˆ 2 and âˆ 7 **

**(ii) âˆ 4 and âˆ 8**

**(iii) âˆ 1 and âˆ 8**

**(iv) âˆ 1 and âˆ 5**

**(v) âˆ 4 and âˆ 7**

**(c) (i) âˆ 1 and âˆ 10**

**(ii) âˆ 6 and âˆ 12**

**(iii) âˆ 8 and âˆ 10**

**(iv) âˆ 4 and âˆ 11**

**(v) âˆ 2 and âˆ 8**

**(vi) âˆ 5 and âˆ 7**

**Solution:**

(a) (i) âˆ 2 and âˆ 4 = Adjacent angles

(ii) âˆ 1 and âˆ 8 = Alternate exterior angles

(iii) âˆ 4 and âˆ 5 = Alternate interior angles

(iv) âˆ 1 and âˆ 5 = Corresponding angles

(v) âˆ 3 and âˆ 5 = Allied angles

(b) (i) âˆ 2 and âˆ 7 = Alternate interior angles

(ii) âˆ 4 and âˆ 8 = Corresponding angles

(iii) âˆ 1 and âˆ 8 = Alternate exterior angles

(iv) âˆ 1 and âˆ 5 =Corresponding angles

(v) âˆ 4 and âˆ 7 = Allied angles

(c)(i) âˆ 1 and âˆ 10 = Corresponding angles

(ii) âˆ 6 and âˆ 12 = Alternate exterior angles

(iii) âˆ 8 and âˆ 10 = Alternate interior angles

(iv) âˆ 4 and âˆ 11 = Alternate interior angles

(v) âˆ 2 and âˆ 8 = Alternate exterior angles

(vi) âˆ 5 and âˆ 7 = Vertically opposite angles

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

**(i)**

** **

**(ii)**

** **

**(iii)**

** **

**(iv)**

** **

**(v)**

** **

**Solution:**

(i) a + 140^{0} = 180^{0} [Linear pair]

a = 180^{0} – 140^{0}

We get,

a = 40^{0}

Here, b = a {alternate angles}

Hence, a = b = 40^{0}

(ii) Given

l || m and p intersects them

b + 60^{0} = 180^{0} [Linear pair]

b = 180^{0} – 60^{0}

We get,

b = 120^{0}

and a = 60^{0} {corresponding angles}

Hence, a = 60^{0}

b = 120^{0}

(iii) a = 110^{0} [Vertically opposite angles]

b = 180^{0} â€“ a [Co-interior angles]

= 180^{0} â€“ a

= 180^{0} â€“ 110^{0}

We get,

= 70^{0}

(iv) a = 60^{0} [Alternate interior angles]

b = 180^{0} â€“ a [Co-interior angles]

= 180^{0} â€“ 60^{0}

We get,

= 120^{0}

(v) a = 72^{0} [Alternate interior angles]

b = a [Vertically opposite angles]

Hence, b = 72^{0}

**3. If âˆ 1 = 120 ^{0}, find the measures of: âˆ 2, âˆ 3, âˆ 4, âˆ 5, âˆ 6, âˆ 7 and âˆ 8. Give reasons.**

**Solution:**

Given

l || m and p is their transversal and

âˆ 1 = 120^{0}

âˆ 1 + âˆ 2 = 180^{0} [Straight line angle]

120^{0} + âˆ 2 = 180^{0}

âˆ 2 = 180^{0} – 120^{0}

We get,

âˆ 2 = 60^{0}

Therefore, âˆ 2 = 60^{0}

But âˆ 1 and âˆ 3 [Vertically opposite angles]

Hence, âˆ 3 = âˆ 1 = 120^{0}

Similarly,

âˆ 4 = âˆ 2 [Vertically opposite angles]

âˆ 4 = 60^{0}

âˆ 5 = âˆ 1 [Corresponding angles]

Hence, âˆ 5 = 120^{0}

Similarly,

âˆ 6 = âˆ 2 [Corresponding angles]

âˆ 6 = 60^{0}

âˆ 7 = âˆ 5 [Vertically opposite angles]

Hence, âˆ 7 = 120^{0}

and âˆ 8 = âˆ 6 [Vertically opposite angles]

Hence, âˆ 8 = 60^{0}

Therefore, the measures of angles are,

âˆ 2 = 60^{0}

âˆ 3 = 120^{0}

âˆ 4 = 60^{0}

âˆ 5 = 120^{0}

âˆ 6 = 60^{0}

âˆ 7 = 120^{0} and

âˆ 8 = 60^{0}

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

**Solution:**

x = 180^{0} â€“ 100^{0} (Linear pair of angles)

x = 80^{0}

y = x (Alternate exterior angles)

y = 80^{0}

z = 100^{0} (Corresponding angles)

p = x (Vertically opposite angles)

p = 80^{0}

q = 100^{0} (Vertically opposite angles)

r = q (Corresponding angles)

r = 100^{0}

Therefore, the measures of angles are,

x = y = p = 80^{0}

q = r = z = 100^{0}

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

**âˆ x = â€¦â€¦â€¦â€¦;**

**âˆ z = â€¦â€¦â€¦â€¦;**

**âˆ p = â€¦â€¦â€¦â€¦;**

**âˆ q = â€¦â€¦â€¦â€¦;**

**âˆ r = â€¦â€¦â€¦â€¦.;**

**âˆ s = â€¦â€¦â€¦â€¦.;**

**Solutions:**

âˆ x = 60^{0} (Corresponding angles)

z = x (Corresponding angles)

= 60^{0}

p = z (Vertically opposite angles)

= 60^{0}

q = 180^{0} â€“ p (Linear pair of angles)

= 180^{0} – 60^{0}

We get,

= 120^{0}

r = 180^{0} â€“ x (Linear pair of angles)

= 180^{0} â€“ 60^{0}

We get,

= 120^{0}

s = r (Vertically opposite angles)

s = r = 120^{0}

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

**Solution:**

x = 115^{0} (By vertically opposite angles)

y = 70^{0} (By vertically opposite angles)

z = 70^{0} (By alternate interior angles)

w = 115^{0} (By alternate interior angles)

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

**Solution:**

a = 130^{0} (By vertically opposite angles)

b = 150^{0} (By vertically opposite angles)

c = 150^{0} (By alternate interior angles)

d = 130^{0} (By alternate interior angles)

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

**Solution:**

x = 180^{0} – 75^{0} (Co-interior angles)

We get,

x = 105^{0}

y = 180^{0} â€“ x (Co-interior angles)

y = 180^{0} â€“ 105^{0}

We get,

y = 75^{0}

z = 75^{0} (Corresponding angles)

Therefore, the angles are,

x = 105^{0}, y = 75^{0} and z = 75^{0}

Exercise 25(C)

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

**(i)**

** **

**(ii)**

** **

**(iii)**

** **

**Solution:**

(i) Steps of Construction:

1. Draw line QR = OB at point Q

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

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

4. With the help of compasses, take the distance equal to distance between C and D; and then taking T as centre, draw an arc which cuts the earlier arc at point S.

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

(ii) Steps of Construction:

1. Draw a line EF, at a point E

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

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

4. With the help of compasses, take the distance equal to the distance between C and D; and then taking T as centre, draw an arc which cuts the earlier arc at point S

5. Join QS and produce up to a suitable point S. Now, the obtained âˆ PQR is the angle equal to the the given âˆ DEF

(iii) Steps of Construction:

1. Draw a line AB = QP at a point A

** **

2. Taking Q as centre, draw an arc of any suitable radius, to cut the arms of the angle T and S

3. Taking A as centre, draw an arc of the same size as drawn for points T and S. Let this arc cuts the line AB at point D

4. With the help of compasses, take the distance equal to the distance between T and S; and then taking D as centre, draw an arc which cuts the earlier arc at point E

5. Join AE produced up to a suitable point C. The obtained âˆ BAC is the angle equal to the given âˆ PQR

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

**(i) 60 ^{0}**

**(ii) 90 ^{0}**

**(iii) 45 ^{0}**

**(iv) 30 ^{0}**

**(v) 120 ^{0}**

**Solution:**

(i) Steps of Construction:

Constructing the angle of 60^{0}

1. Draw a line OA of any suitable length

2. Taking O as centre, draw an arc of any size to cut OA at point B

3. Now, taking B as centre, draw the same size arc, to cut the previous arc at point C

4. Join OC and produce up to a suitable point D. Then, the obtained âˆ DOA is the angle of 60^{0}

(ii) Steps of Construction:

Constructing angle of 90^{0}

Let OA be the line and at point O, the angle of 90^{0} is to be drawn

1. Taking O as centre, draw an arc to cut the arm OA at point B

2. Taking B as centre, draw the same size arc to cut the previous arc at point C

3. Again with C as centre and with the same radius, draw one more arc to cut the previous arc at point D

4. Now, taking C and D as centres, draw two arcs of equal radii to cut each other at point E.

5. Join O and E. Then âˆ AOE = 90^{0} is obtained

(iii) Draw an angle of 90^{0}, following the steps as in question (ii) and bisect it. Each angle so obtained will be 45^{0}

(iv) Steps of Construction:

Constructing an angle of 30^{0}

1. Draw an angle of 60^{0} following the steps as drawn in question no.(i)

2. Now, bisecting this angle, we get two angles each of 30^{0}. Therefore âˆ EOB = 30^{0}

(v) Steps of Construction:

Constructing an angle of 120^{0}

1. Taking centre as O on the line OA, draw an arc to cut this line at point C

2. Now, taking C as centre, draw a same size arc which cuts the first arc at point D

3. Taking D as centre, draw one more arc of same size which cuts the first arc at point E

4. Join OE and produce it up to point B. Now, the âˆ AOB is the obtained angle whose measure is 120^{0}

**3. Draw line AB = 6 cm. Construct angle ABC = 60 ^{0}. Then draw the bisector of angle ABC.**

**Solution:**

Steps of Construction:

1. Draw a line segment AB of length 6 cm

2. Using compass construct âˆ CBA = 60^{0}

3. Bisect âˆ CBA, using compass, take any radius which meet line AB and BC at points E and F

4. Now, with the help of compass take radius more than Â½ of EF and draw two arcs from point E and F, where both the arcs intersects at point G, proceed BG towards** **D.** **Now the âˆ DBA is bisector of âˆ CBA

**4. Draw a line segment PQ = 8 cm. 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. Taking 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 point R and RS

3. Now, join RS which cuts PQ at point D

Now, RS = PQ. Also âˆ POR = 90^{0}

Therefore, the line segment RS is the perpendicular bisector of PQ as it bisects PQ at point P and it also perpendicular to PQ. Now, on measuring the length

PR = 4 cm

QR = 4 cm

Since, PR = QR = 4 cm

Therefore, PR = QR

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

**Solution:**

1. Draw a line segment AB of length 7 cm

2. Mark a point on AB such that, AB â€“ AP = 3 cm

3. From point P, cut arc on outside of AB, and mark them as point E and F

4.Now, from point E and F cut arcs on both side intersecting each other at point C and D

5. Join point P, C and D

6. Which is the required perpendicular

**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 of length 6.5 cm

(ii) Taking radius as 5 cm and with centre A, draw an arc and taking radius as 4.6 cm and with centre B, draw another arc which intersects the first arc at point P

Now, P is the required point

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

(iv) Now, taking E and F as centres and radius greater than half of EF, draw the arcs which intersect each other at point Q

(v) Join PQ which intersect AB at point D

(vi) Now, PD is perpendicular to AB

Exercise 25(D)

**1. Draw a line segment OA = 5 cm. Use set-square to construct angle AOB = 60 ^{0}, such that OB = 3 cm. Join A and B; then measure the length of AB.**

**Solution:**

The length of AB = 4.4 cm (approximately)

**2. Draw a line segment OP = 8 cm. Use set-square to construct âˆ POQ = 90 ^{0}; such that OQ = 6 cm. Join P and Q; then measure the length of PQ.**

**Solution:**

Measuring the length of PQ = 10 cm

**3. Draw âˆ ABC = 120 ^{0}. Bisect the angle using ruler and compasses. Measure each angle so obtained and check whether or not the new angles obtained on bisecting âˆ ABC are equal.**

**Solution:**

Each angle measure = 60^{0}

Yes, the angles obtained on bisecting âˆ ABC are equal

**4. Draw âˆ PQR = 75 ^{0} 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:**

The length of MN = 4.3 cm

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