## RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 14 Properties of Parallel Lines Ex 14.

**Question 1.**

**Solution:**

A transversal t intersects two parallel lines l and m.

∠ 1 = ∠ 5 (corresponding angles)

But ∠ 5 = 70° (given)

∠ 1 = 70°

But ∠ 3 = ∠ 5 (Alternate angles)

∠ 3 = 70°

∠4 + ∠5 = 180° (Sum of co-interior angles)

⇒ ∠4 + 70° = 180°

⇒ ∠4 = 180° – 70°

⇒ ∠4 = 110°

But ∠ 4 = ∠ 8 (corresponding angles)

∠ 8 = 110°

Hence ∠ 1 = 70°, ∠3 = 70°, ∠4 = 110° and ∠ 8 = 110°

**Question 2.**

**Solution:**

A transversal t intersects two parallel lines l and m

∠1 : ∠2 = 5 : 7

But ∠ 1 + ∠ 2 = 180° (Linear pair)

But ∠ 3 = ∠ 1 (vertically opposite angles)

∠ 3 = 75°

∠ 8 = ∠ 4 (corresponding angles)

and ∠ 4 = ∠ 2 (vertically opposite angles)

∠8 = ∠2 = 105°

Hence ∠ 1 = 75°, ∠2 = 105°, ∠3 = 75° and ∠ 8 = 105°

**Question 3.**

**Solution:**

A transversal t intersects two parallel lines l and m interior angles of the same side of t are (2x – 8)° and (3x – 7)°

(2x – 8)° + (3x – 7)° = 180° (sum of co-interior angles)

⇒ 2x – 8 + 3x – 7 = 180°

⇒ 5x – 15° = 180°

⇒ 5x = 180° + 15°

⇒ 5x = 195°

⇒ x = \(\frac { 195 }{ 5 }\) = 39°

First angle = 2x – 8° = 2 x 39° – 8° = 78° – 8° = 70°

Second angle = 3x – 7 = 3 x 39° – 7° = 117° – 7° = 110°

**Question 4.**

**Solution:**

l || m and two transversals intersect these lines but s is not parallel to t.

∠ 5 = ∠ 1 (vertically opposite angles)

∠ 5 = 50°

But l || m and s the transversal

∠ 5 + ∠ 2 = 180° (sum of co-interior angles)

⇒ 50° + x = 180°

⇒ x = 180° – 50° – 130°

x = 130°

∠ 4 = ∠ 6 (vertically opposite angles)

∠ 6 = y

But l || m and t is the transversal

∠ 6 + ∠ 3 = 180° (sum of co-interior angles)

⇒ y + 65° = 180°

⇒ y = 180° – 65° = 115°

y = 115°

Hence x = 130° and y = 115°

**Question 5.**

**Solution:**

In the figure, ABC is a triangle, DAE || BC

∠B = 65°, ∠C = 45°

∠ DAB = x° and ∠ EAC = y°

DAE || BC and AB is transversal

∠ DAB = ∠ B (Alternate angles)

⇒ x° = 65°

Similarly ∠ EAC = ∠ C (Alternate angles)

y° = 45°

Hence x = 65° and y = 45°

**Question 6.**

**Solution:**

In ∆ABC, AB || CE

∠BAC = 80°, ∠ECD = 35°

AB || CE and BCD is the transversal

∠ABC = ∠ECD (corresponding angles)

⇒ ∠ABC = 35° (∠ECD = 35°)

Again AB || CE and AC is the transversal

∠ BAC = ∠ ACE (alternate angles)

∠ACE = 80° (∠BAC = 80°)

In ∆ABC

∠A + ∠B + ∠ACB = 180° (Sum of angles of a triangle)

∠ 80° + ∠ 35° + ∠ACB = 180°

⇒ ∠ACB + ∠ 115° = 180°

⇒ ∠ACB = 180° – 115° = 65°

Hence ∠ ACE = 80°, ∠ ACB = 65° and ∠ ABC = 35°

**Question 7.**

**Solution:**

In the figure,

AO || CD, DB || CE and ∠AOB = 50°

AO || CD and CD is the transversal

∠ AOB = ∠ CDB (corresponding angles)

∠ CDB = 50° (∠ AOB = 50°)

Similarly CE || OB and CD in transversal

∠ECD + ∠CEB = 180° (sum of co-interior angles)

⇒ ∠ECD + 50° = 180°

⇒ ∠ECD = 180° – 50° = 130°

∠ECD = 130°

**Question 8.**

**Solution:**

In the fig, AB || CD

∠ABO = 50° and ∠CDO = 40°

From O, draw EOF || AB or CD

AB || EF and BO is the transversal

∠ABO = ∠ 1 (Alternate angles) …(i)

∠ CDO = ∠ 2 (Alternate angles) …(ii)

Similarly, EF || CD and OD is the transversal

Adding (i) and (ii),

∠ 1 + ∠ 2 = ∠ABO + ∠CDO

⇒ ∠BOD = 50° + 40° = 90°

Hence ∠ BOD = 90°

**Question 9.**

**Solution:**

Given : In the figure, AB || CD and EF is a transversal which intersects them at G and H respectively

GL and HM are the angle bisectors or ∠ AGH and ∠ GHD respectively.

To prove : GL || HM.

Proof : AB || CD and EF is a transversal

∠ AGH = ∠ CHD (Alternate angles)

GL is the bisector of ∠ AGH

∠ 1 = ∠2 = \(\frac { 1 }{ 2 }\) ∠ AGH

Similarly, HM is the bisectors of ∠ GHD

∠3 = ∠4 = \(\frac { 1 }{ 2 }\) ∠ GHD

∠ AGH = ∠ GHD (proved)

∠ 1 = ∠3

But, these are alternate angles

BL || HM

Hence proved.

**Question 10.**

**Solution:**

In the given figure,

AB || CD

∠ ABE = 120° and ∠ECD = 100° ∠ BEC = x°

From E, draw FG || AB or CD.

AB || EF

∠ABE + ∠1 = 180° (sum of co-interior angles)

⇒ 120° + ∠1 = 180°

⇒ ∠1 = 180°- 120° = 60°

Similarly CD || EG

∠ECD + ∠2 = 180°

⇒ 100° + ∠2 = 180°

⇒ ∠2 = 180° – 100°

∠ 2 = 80°

But ∠1 + ∠x + ∠2 = 180° (Angles on one side of a straight line)

⇒ 60° + x + 80° = 180°

⇒ x + 140° = 180°

⇒ x = 180° – 140° = 40°

x = 40°

**Question 11.**

**Solution:**

Given : In the figure, ABCD is a quadrilateral in which AB || DC and AD || BC

To prove : ∠ADC = ∠ABC

Proof : AB || DC and DA is the transversal

∠ADC + ∠ DAB = 180° (co-interior angles)

Similarly, AD || BC and AB is the transversal

∠DAB + ∠ABC = 180° …(ii)

from (i) and (ii),

∠ ADC + ∠ DAB = ∠DAB + ∠ABC

⇒ ∠ADC = ∠ABC

Hence ∠ ADC = ∠ ABC

Hence proved.

**Question 12.**

**Solution:**

In the figure,

l || m and p || q.

∠1 = 65°

∠ 2 = ∠ 1 (vertically opposite angles)

∠ 2 = 65°

⇒ a = 65°

p || q and l is the transversal

∠ 2 + ∠ 3 = 180° (co-interior angles)

⇒ a + b= 180°

⇒ 65° + b = 180°

⇒ b = 180° – 65° = 115°

Again l || m and p is the transversal

∠ 3 + ∠4 = 180°

⇒ b + c = 180°

⇒ 115° + c = 180°

⇒ c = 180° – 115° = 65°

l || m and q is the transversal

∠ 2 + ∠ 5 = 180°

⇒ a + d = 180°

⇒ 65° + d = 180°

⇒ d = 180° – 65° = 115°

Hence a = 65°, b = 115°, c = 65° and d = 115°

**Question 13.**

**Solution:**

In the given figure, AB || DC and AD || BC and AC is the diagonal of parallelogram ABCD.

∠BAC = 35°, ∠CAD = 40°, ∠ACB = x° and ∠ ACD = y°. .

AB || DC and CA is the transversal

∠ DCA = ∠ CAB (Alternate angles)

⇒ y = 35°

and similarly AD || BC and AC is the transversal

∠ CAD = ∠ ACB (Alternate angles)

⇒ 40° = x°

x = 40° and y = 35°

**Question 14.**

**Solution:**

In the figure, AB || CD and CD has been produced to E so that

∠ BAE = 125° ∠ BAC = x°, ∠ ABD = x°, ∠ BDC = y° and ∠ ACD = z°

DAE is a straight line and AB stands on it.

∠ BAD + ∠ BAE = 180° (Linear pair)

⇒ x + 125° = 180°

⇒ x = 180° – 125° = 55°

But ∠ABC = x = 55°

DC || AB and CB is the transversal

∠ABC + ∠ BCD = 180° (co-interior angles)

⇒ x + y = 180°

⇒ 55° + y = 180°

⇒ y = 180° – 55° = 125°

Again DC || AB and DAE is its transversal

∠ CDA = ∠ BAE (corresponding angles).

z = 125°

Hence x = 55°, y = 125° and z = 125°

**Question 15.**

**Solution:**

Given : In each figure,

l and m are two lines and t is the transversal

To prove : l || m or not

Proof:

(i) fig. (i)

A transversal t intersects two lines l and m

and ∠ 1 = 40°, ∠2 = 130°

But ∠ 1 + ∠3 = 180° (Linear pair)

⇒ 40° + ∠ 3 = 180°

⇒ ∠3 = 180° – 40° = 140°

l || m,

If ∠ 3 = ∠ 2

⇒ 140° = 130°

Which is not possible.

l is not parallel to m.

(ii) fig. (ii)

Transversal t, intersects l and m and ∠ 1 = 35°, ∠2 = 145°

But ∠ 1 = ∠ 3 (vertically opposite angles).

∠3 = 35°

l || m,

if ∠3 + ∠2 = 180°

if 35° + 145° = 180°

if 180°= 180°

which is true

l || m

(iii) Transversal t, intersects l and m.

∠ 1 = 125°, ∠ 2 = 60°

But ∠ 1 = ∠ 3 (vertically opposite angles)

∠ 3 = 125°

l || m

If ∠3 + ∠2 = 180° (co-interior angles)

If 125° + 60° = 180°

If 185° =180°

which is not possible.

Hence l is not parallel to m.

Hope given RS Aggarwal Solutions Class 7 Chapter 14 Properties of Parallel Lines Ex 14 are helpful to complete your math homework.

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