RD Sharma Solutions Class 7 Lines And Angles Exercise 14.2

RD Sharma Class 7 Solutions Chapter 14 Ex 14.2 PDF Free Download

Chapter 14 Exercise 14.2

Q1) In Figure, line n is a transversal to line l and m. Identify the following:

i) Alternate and corresponding angles in Fig. 58 (i)

RD Sharma Class 7 Maths Exercise 14.2 Qs 1 A

Solution: The alternate angles are:

∠EGA and ∠FHD

∠EGB and ∠CHF

∠AGH and ∠GHD

∠BGH and ∠CHG

Corresponding angles:

∠EGA and ∠CHG

∠AGH and ∠CHF

∠EGB and ∠DHG

∠BGH and ∠DHF

ii) Angles alternate to ∠d and ∠g and angles corresponding to angle ∠f and ∠h in Fig. 58 (ii)

RD Sharma Class 7 Maths Exercise 14.2 Qs 1 B

Solutions: The angle alternate to ∠d is ∠e.

The angle alternate to ∠g is ∠b.

The angle corresponding to ∠f is ∠c.

The angle corresponding to ∠h is ∠a.

(iii) Angle alternate to ∠PQR, angle corresponding to ∠RQF and angle alternate to

∠PQE in Fig. 58 (iii)

RD Sharma Class 7 Maths Exercise 14.2 Qs 1C

Solutions: Angle alternate to ∠PQR is ∠ARQ

Angle corresponding to ∠RQF is ∠BRA

Angle alternate to ∠PQE is ∠BRA

(iv) Pairs of interior and exterior angles on the same side of the transversal in Fig. 58 (ii)

RD Sharma Class 7 Maths Exercise 14.2 Qs 1D

Solution: Pairs of Interior angles are:

∠a , ∠e

∠d, ∠f

Pairs of Exterior angles are:

∠c, ∠g

∠b, ∠h

Q2) In Figure 59, AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If ∠CMQ = 600, find all other angles in the figure.

RD Sharma Class 7 Maths Exercise 14.2 Qs 2

Solutions: ∠CMQ = 600 is given in the figure.

∠CMQ = ∠PLB = 600 (Alternate exterior angle)

∠CMQ = ∠LMD = 600 (Vertically opposite angles)

∠CMQ = ∠ALM = 600 (Corresponding angles)

∠PLB and ∠ALP is a linear pair.

So ∠PLB + ∠ALP = 1800

600 + ∠ALP = 1800

∠ALP = 1200

∠LMD and ∠LMC is a linear pair

∠LMD + ∠LMC = 1800

600 + ∠LMC = 1800

∠LMC = 1200

∠CMQ and ∠DMQ is a linear pair

∠CMQ + ∠DMQ = 1800

600 + ∠DMQ = 1800

∠DMQ = 1200

∠DMQ = ∠MLB = 1200 (Corresponding angles)

Q3) In Fig. 60, AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If ∠LMD = 350 Find ∠ALM and ∠PLA.

RD Sharma Class 7 Maths Exercise 14.2 Qs 3

Solution: It is given that ∠LMD = 350

∠LMD and ∠LMC is a linear pair.

∠LMD + ∠LMC = 1800

350 + ∠LMC = 1800

∠LMC = 1450

∠LMC = ∠PLA = 1450 (Corresponding angles)

∠LMC = ∠MLB = 1450 (Alternate Interior angles)

∠ALM and ∠MLB is a linear pair

∠ALM + ∠MLB = 1800

∠ALM + 1450 = 1800

∠ALM = 350

Q4) The line n is transversal to line l and m in Fig. 61. Identify the angle alternate to ∠13, angle corresponding to ∠15, and angle alternate to ∠15.

RD Sharma Class 7 Maths Exercise 14.2 Qs 4

Solution: Angle alternate to ∠13 is ∠7.

Angle corresponding to ∠15 is 7.

Angle alternate to ∠15 is 5.

Q5) In Fig. 62, line l || m and n is transversal. If ∠1 = 400, find all the angles and check that all corresponding angles and alternate angles are equal.

RD Sharma Class 7 Maths Exercise 14.2 Qs 5

Solution: It is given that ∠1 = 400

As we can see in the figure that ∠1 and ∠2 is a linear pair.

∠1 + ∠2 = 1800

400 + ∠2 = 1800

∠2 = 1400

∠2 = ∠6 = 1400 (Corresponding angles)

∠6 and ∠5 is a linear pair.

∠6 + ∠5 = 1800

1400 + ∠5 = 1800

∠5 = 400

∠3 = ∠5 = 400 (Alternate interior angles)

∠3 = ∠7 = 400 (Corresponding angles)

∠3 and ∠4 is a linear pair.

∠3 + ∠4 = 1800

400 + ∠4 = 1800

∠4 = 1400

∠4 = ∠8 = 1400 (Corresponding angles)

Q6) In Fig. 63, line l || m and a transversal n cuts them P and Q respectively. If ∠1 = 750, find all other angles.

RD Sharma Class 7 Maths Exercise 14.2 Qs 6

Solution: It is given that ∠1 = 750

As we can see in the figure that ∠1 and ∠2 is a linear pair.

∠1 + ∠2 = 1800

750 + ∠2 = 1800

∠2 = 1050

∠2 = ∠6 = 1050 (Corresponding angles)

∠6 and ∠5 is a linear pair.

∠6 + ∠5 = 1800

1050 + ∠5 = 1800

∠5 = 750

∠3 = ∠5 = 750 (Alternate interior angles)

∠3 = ∠7 = 750 (Corresponding angles)

∠3 and ∠4 is a linear pair.

∠3 + ∠4 = 1800

750 + ∠4 = 1800

∠4 = 1050

∠4 = ∠8 = 1050 (Corresponding angles)

Q7) In Fig. 64, AB || CD and a transversal PQ cuts at L and M respectively. If ∠QMD = 1000, find all the other angles.

RD Sharma Class 7 Maths Exercise 14.2 Qs 7

Solution: It is given that ∠QMD = 1000

∠QMD = ∠MLB = 1000 (Corresponding angles)

∠MLB and ∠MLA is a linear angles.

∠MLB + ∠MLA = 1800

1000 + ∠MLA = 1800

∠MLA = 800

∠QMD = ∠PLA = 1000 (Alternate exterior angles)

We can see from the figure, ∠PLA and PLB is a linear pair.

∠PLA + ∠PLB = 1800

1000 + ∠PLB = 1800

∠PLB = 800

∠PLB = ∠LMD = 800 (Corresponding angles)

∠LMD and ∠LMC is a linear pair.

∠LMD + ∠LMC = 1800

800 + ∠LMC = 1800

∠LMC = 1000

As from the figure, ∠QMD and ∠QMC is a linear pair.

∠QMD + ∠QMC = 1800

1000 + ∠QMC = 1800

∠QMC = 800

Q8) In Fig. 65, l || m and p || q. Find the values of x,y,z,t.

RD Sharma Class 7 Maths Exercise 14.2 Qs 8

Solutions: The angle given the figure is 800

∠z and 800 are vertically opposite angles.

So, ∠z = 800

∠z = ∠y = 800 (Corresponding angles)

∠z = ∠t = 800 (Corresponding angles)

∠y = ∠x = 800 (Corresponding angles)

Q9) In Fig. 66, line l || m, ∠1 = 1200 and ∠2= 1000, find out ∠3 and ∠4.

RD Sharma Class 7 Maths Exercise 14.2 Qs 9

Solutions: It is given in that,

∠1 = 1200, ∠2 = 1000

RD Sharma Class 7 Maths Exercise 14.2 Ans 9

∠2 = ∠6 = 1000 (Corresponding angles)

∠6 and ∠3 is a linear pair.

∠6 + ∠3 = 1800

100 + ∠3 = 1800

∠3 = 800

∠1 and ∠5 is a linear pair.

∠1 +∠5 = 1800

1200 + ∠5 = 1800

∠5 = 600

By using angle sum property;

∠5 + ∠4 + ∠3 = 1800

600 + ∠4 + 800 = 1800

∠4 = 1800 – 1400

∠4 = 400.

Q10) In Fig. 67, l || m. Find the values of a, b, c, d. Give reasons.

RD Sharma Class 7 Maths Exercise 14.2 Qs 10

Solution: As it is given in that l||m

∠a = 1100 (Vertically opposite angles)

∠d = 850 (Vertically opposite angles)

From the figure;

∠a = ∠b = 1100 (Corresponding angles)

∠c = ∠d = 850 (Corresponding angles)

Q11) In Fig. 68, AB || CD and ∠1 and ∠2 are in the ratio of 3 : 2. Determine all angles from 1 to 8.

RD Sharma Class 7 Maths Exercise 14.2 Qs 11

Solution: As given in the question that;

∠1 and ∠2 are in 3:2.

Let the ∠1 = 3a and ∠2 = 2a

∠1 and ∠2 is a linear pair.

∠1 + ∠2 = 1800

3a + 2a = 1800

5a = 1800

a = 360

So, ∠1 = 3a = 3 × 360 = 1080

∠2 = 2a = 2 × 360 = 720

As we can see in the figure;

∠1 = ∠5 = 1080 (Corresponding angles)

∠2 = ∠6 = 720 (Corresponding angles)

∠5 = ∠3 = 1080 (Alternate interior angles)

∠6 = ∠4 = 720 (Alternate interior angles)

∠2 = ∠8 = 720 (Alternate exterior angles)

∠1 = ∠7 = 1080 (Alternate exterior angles)

Thus the angles are:

∠1 = 1080, ∠2 = 720, ∠3 = 1080, ∠4 = 720, ∠5 = 1080, ∠6 = 720, ∠7 = 1080, ∠8 = 720

Q12) In Fig. 69 l, m and n are parallel lines intersected by a transversal p at X, Y and Z respectively. Find ∠1, ∠2 and ∠3.

RD Sharma Class 7 Maths Exercise 14.2 Qs 12

Solution: ∠4 and 600 is a linear pair.

∠4 + 600 = 1800

∠4 = 1200

As we can see in the figure;

∠4 = ∠1 = 1200 (Corresponding angles)

∠1 = ∠2 = 1200 (Corresponding angles)

∠2 = ∠3 = 1200 (Vertically opposite angles)

Thus; ∠1 = 1200, ∠2 = 1200, ∠3 = 1200

Q13) In Fig. 70, if l || m || n and ∠1 = 600, find ∠2.

RD Sharma Class 7 Maths Exercise 14.2 Qs 13

Solution: As we can see in the figure:

∠1 = ∠3 (Corresponding angles)

It is given that ∠1 = 600

So, ∠3 = 600

∠3 and ∠4 are linear pair

∠3 + ∠4 = 1800

600 + ∠4 = 1800

∠4 = 1200

∠4 and ∠2 are alternate interior angles.

So, ∠2 = 1200

Q14) In Fig. 71, if AB || CD and CD || EF, find ∠ACE.

RD Sharma Class 7 Maths Exercise 14.2 Qs 14

Solution: It is given in the figure that ∠CEF = 1300

As per the property of sum of interior angles:

∠CEF + ∠ECD = 1800

1300 + ∠ECD = 1800

∠ECD = 500

∠ACD and ∠BAC are Alternate angles.

So, ∠ACD = ∠BAC = 700

∠ACE = ∠ACD – ∠ECD

∠ACE = 700 – 500

∠ACE = 200

Q15) In Fig. 72, if l || m, n || p and ∠1 = 850, find ∠2.

RD Sharma Class 7 Maths Exercise 14.2 Qs 15

Solution: It is given that ∠1 = 850

∠1 = ∠3 = 850 (Corresponding angles)

∠3 and ∠2 are Interior angles.

∠3 + ∠2 = 1800 (Sum of interior angle)

850 + ∠2 = 1800

∠2 = 950

Q16) In Fig. 73, a transversal n cuts two lines l and m. If ∠1 = 700 and ∠7 = 800, is l || m?

RD Sharma Class 7 Maths Exercise 14.2 Qs 16

Solution: It is given that ∠1 =700 and ∠7 = 800

As we can see in the figure that;

∠1 and ∠7 are Alternate exterior angles.

So, ∠1 should be equal to ∠7. But both the angles have different values.

Therefore, line l is not parallel to m (Lines are parallel only if alternate exterior angles are equal).

Q17) In Fig. 74, a transversal n cuts two lines l and m such that ∠2 = 650 and ∠8 = 650. Are the lines parallel?

RD Sharma Class 7 Maths Exercise 14.2 Qs 17

Solution: It is given that;

∠2 = 650 and ∠8 = 650

From the figure we can see that

∠2 and ∠8 are Alternate Exterior angles.

So, ∠2 = ∠8 = 1800

As the value for both angles are given equal in the question itself. Therefore, line l and m are parallel to each other (Two lines are parallel if the alternate angles formed with the transverse are equal).

Q18) In Fig. 75, Show that AB || EF.

RD Sharma Class 7 Maths Exercise 14.2 Qs 18

Solution: It is given that;

∠BAC = 570, ∠ACE = 220, ∠DCE = 350, ∠CEF = 1450

As we can see;

∠ACE + ∠DCE = ∠ACD

220 + 350 = ∠ACD

∠ACD = 570

As the ∠BAC and ∠ACD are alternate angles and both are equal to 570.

Therefore, line AB || CD (Two lines are parallel if the alternate angles are equal)……….(i)

Now,

∠DCE + ∠CEF

= 350 + 1450

= 1800

This shows that the sum of the alternate interior angles on the same side of the transverse line CE is 1800. So, the ∠DCE and ∠CEF are supplementary angles.

Thus, line CD || EF……………..(ii)

From equations i and ii

AB || CD || EF

Therefore, AB || EF.

Q19) In Fig. 76, AB || CD. Find the values of x, y, z.

RD Sharma Class 7 Maths Exercise 14.2 Qs 19

Solution: ∠x and 1250 are linear pairs.

∠x + 1250 = 1800

∠x = 550

The sum of interior angles on the same side of the transverse line AC is 1800.

∠x + ∠z = 180o

550 + ∠z = 1800

∠z = 1250

The sum of interior angles is 1800.

∠x + ∠y = 180o

550 + ∠y = 1800

∠y = 1250

Q20) In Fig. 77, find out ∠PXR, if PQ || RS.

RD Sharma Class 7 Maths Exercise 14.2 Qs 20

Solution: Look at the diagram below;

RD Sharma Class 7 Maths Exercise 14.2 Ans 20

∠QPX = ∠1 = 700 (Alternate angles)

∠SRX = ∠2 = 500 (Alternate angles)

∠PXR = ∠1 + ∠2

∠PXR = 700 + 500

∠PXR = 1200

Q21) In Fig. 78, we have

RD Sharma Class 7 Maths Exercise 14.2 Qs 21

i) ∠MLY = 2∠LMQ, find ∠LMQ

Solution: ∠MLY and ∠LMQ are interior angles.

∠MLY + ∠LMQ = 1800

2∠LMQ + ∠LMQ = 1800

3∠LMQ = 1800

∠LMQ = 600

ii) ∠XLM = (2x − 10)0 and ∠LMQ= x + 300, find x.

Solution: ∠XLM = ∠LMQ (Alternate angles)

2x – 100 = x + 300

X = 400

iii) ∠XLM = ∠PML, find ∠ALY

Solution: ∠XLM and ∠PML are interior angles.

∠XLM + ∠PML = 1800

∠XLM + ∠XML = 1800

2∠XML = 1800

∠XML = 900

∠ALY = ∠XML (Vertically opposite angles)

So, ∠ALY = 900

iv) ∠ALY = (2x − 15)0, ∠LMQ = (x + 40)0, find x

Solution: ∠ALY = ∠LMQ (Corresponding angles)

2x – 150 = x + 400

X = 550

Q22) In Fig. 79, DE || BC. Find the values of x and y.

RD Sharma Class 7 Maths Exercise 14.2 Qs 22

Solution: As we can see from the figure that;

∠DAB = ∠ABC (Alternate interior angles)

x = 400

∠EAC = CAD (Alternate interior angles)

y = 550

Q23) In Fig. 80, line AC || line DE and ∠ABD= 320, Find out the angles x and y if ∠E = 1220.

RD Sharma Class 7 Maths Exercise 14.2 Qs 23

Solution: As we can see in the figure.

y + 320 = 180o (Sum of interior angles)

y = 1480

Now,

y and ∠BDE are linear pairs

1480 + ∠BDE = 1800

∠BDE = 320.

∠BED + 1220 =1800 (Linear pair angles)

∠BED = 580

By using angle sum property of a triangle;

∠BDE + ∠BED + x= 1800

320 + 580 + x = 1800

x = 1800 – 900

x = 900

Q24) In Fig. 81, side BC of ΔABC has been produced to D and CE || BA. If

∠ABC = 650, ∠BAC = 550, find ∠ACE, ∠ECD, ∠ACD.

RD Sharma Class 7 Maths Exercise 14.2 Qs 24

Solution: By using angle sum property of a triangle:

650 + 550 + ∠ACB = 1800

∠ACB = 1800 – 1200

∠ACB = 600

∠BAC = ∠ACE = 550 (Alternate interior angles)

∠ACB + ∠ACE + ∠ECD = 1800 (Linear pair angles)

600 + 550 + ∠ECD = 1800

∠ECD = 650

∠ACD = ∠ACE +∠ECD

∠ACD = 550 + 650

∠ACD = 1200

Q25) In Fig. 82, line CA ⊥ AB || line CR and line PR || line BD. Find ∠x, ∠y, ∠z.

RD Sharma Class 7 Maths Exercise 14.2 Qs 25

Solution: As it is given that

CA ⊥ AB

So, x = 900

By using angle sum property in triangle APQ;

x + ∠AQP + ∠APQ = 1800

900 + 200 + ∠APQ = 1800

∠APQ = 700

∠APQ = y = 700 (Corresponding angles)

∠APQ + z = 1800 (Interior angles)

700 + z = 1800

Z = 1100

Q26) In Fig. 83, PQ || RS. Find the value of x.

RD Sharma Class 7 Maths Exercise 14.2 Qs 26

Solution: ∠RCB + 1300 = 1800 (Linear pairs)

∠RCB = 500

Applying angle sum property in triangle BRC.

∠RCB + 550 + x = 1800

500 + 550 + x = 1800

x = 750

Q27) In Fig. 84, AB || CD and AE || CF, ∠FCG = 900 and ∠BAC = 1200. Find the value of x, y and z.

RD Sharma Class 7 Maths Exercise 14.2 Qs 27

Solution: ∠BAC and ∠ACG are Alternate interior angles.

So, ∠BAC = ∠ACG = 1200

∠ACG = y + 900

1200 = y + 900

y = 300

x + ∠ACG = 1800 (Linear pairs)

x + 1200 = 1800

x = 600

∠DCF = x + y

∠DCF = 600 + 300

∠DCF = 900

∠BAE = ∠DCF = 900

∠BAE + z + 1200 = 3600

900 + z + 1200 = 3600

z = 1500

Q28) In Fig. 85, AB || CD and AC || BD. Find the values of x, y, z.

RD Sharma Class 7 Maths Exercise 14.2 Qs 28

Solution: (i) As it is given AB || CD and AC || BD.

So, ABCD is a parallelogram.

Sum of the adjacent angles of a parallelogram is equal to 1800.

650 + z = 1800

Z = 1150

z = x = 1150 (Opposite angles of a parallelogram are equal)

y = 650 (Opposite angles of a parallelogram are equal)

(ii) x = 400 (Alternate interior angles)

y = 350 (Alternate interior angles)

Q29) In Fig. 86, state which lines are parallel and why?

RD Sharma Class 7 Maths Exercise 14.2 Qs 29

Solution: As the alternate angles are equals;

∠ACD = ∠EDC = 1000

Therefore the line AC || DE

Q30) In Fig. 87, the corresponding arms of ∠ABC and ∠DEF are parallel. If ∠ABC = 750, find ∠DEF.

RD Sharma Class 7 Maths Exercise 14.2 Qs 30

Solution: The line BC and DE intersect each other at point G as shown in the figure below.

RD Sharma Class 7 Maths Exercise 14.2 Ans 30

So, AB || DE and BC || EF

Thus, ∠ABC = ∠DEF = 750 (Corresponding angles)

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