## RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4C.

**Other Exercises**

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4A

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4B

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4C

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D

**Question 1.**

**Solution:**

AB || CD and a line t intersects them at E and F forming angles ∠1, ∠ 2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8.

**Question 2.**

**Solution:**

AB || CD and a transversal t intersects them at E and F respectively forming angles ∠l, ∠2, ∠3, ∠4, ∠5, ∠ 6, ∠ 7 and ∠ 8

**Question 3.**

**Solution:**

Given. In quadrilateral ABCD, AB || DC and AD || BC

To Prove : ∠ADC = ∠ABC

Proof : AB || DC and AD intersects their

∠DAB + ∠ADC = 180°

(sum of co-interior angles)

Similarly ∴ AD || BC

∠DAB + ∠ABC = 180° …(ii)

from (i) and (ii),

∠ DAB + ∠ ADC = ∠DAB + ∠ABC

∴∠ADC = ∠ABC. Hence proved.

**Question 4.**

**Solution:**

(i) In the figure, AB || CD

∠ABE = 35° and ∠EDC = 65°

Draw FEG || AB or CD

∴ AS || FG (const.)

**Question 5.**

**Solution:**

In the figure, AB || CD || EF,

∠ ABC = 70° and ∠ CEF = 130°

∴EF || CD (given)

∴∠ CEF +∠1 – 180°

(sum of Co-interior angles)

=> 130° + ∠ 1 = 180°

=> ∠ 1 = 180° – 130° = 50°

Again, AB || CD (given)

∴ ∠ ABC = ∠ BCD (Alternate angles)

=> 70° = ∠ BCD = x + ∠ 1

=> x + 50° – 70°

=> x = 70° – 50° = 20°

Hence x = 20° Ans.

**Question 6.**

**Solution:**

In the figure, AB || CD.

∠DCE = 130°

and ∠ CEA – 20°

From E, draw EF || AB or CD.

**Question 7.**

**Solution:**

Given. In the given figure, AB || CD.

**Question 8.**

**Solution:**

In the figure, AB || CD

**Question 9.**

**Solution:**

In the figure, AB || CD, ∠AEF = p

**Question 10.**

**Solution:**

In the figure, AB || PQ.

A transversal LM cuts them at E and F

∠ LEB = 75°

∠BEG = 20°

∠ EFG = 25°

∠ EGF = x° and ∠ GFD = y°

∴∠ LEB + ∠ BEF = 180° (Linear pair)

**Question 11.**

**Solution:**

In the figure

**Question 12.**

**Solution:**

In the figure, AB || CD

∠PEF = 85°, ∠QHC = 115°

∴ ∠ GHF = ∠ QHC

(Vertically opposite angles)

∠ GHF = 115°

∴AB || CD

∴∠ PEF = ∠ EGH

(Corresponding angles)

∴∠EGH = 85°

But ∠ QGH + ∠ EGH = 180°, (Linear pair)

=> ∠QGH + 85° = 180°

=> QGH = 180° – 85° = 95°

In ∆ GHQ,

Ext. ∠GHF = ∠QGH + ∠GQH

=> 115° = 95° + x

=> x = 115° – 95°

Hence, x = 20° Ans.

**Question 13.**

**Solution:**

In the figure, AB || CD

∠BAD = 75°, ∠ BCD = 35°

∴AB || CD

∴∠ABC = ∠BCD (Alternate angles)

=> x = 35°

and ∠ BAD = ∠ ADC (Alternate angles)

=> 75° = z

**Question 14.**

**Solution:**

In the figure, AB || CD

∠APQ = 75°, ∠ PRD = 125°

∴AB || CD.

∠ APQ = ∠ PQR (Alternate angles)

∴75° = y°

=> y° = 75°

**Question 15.**

**Solution:**

In the figure, AB || CD and EF || GH

∠APR = 110°, ∠LRF = 60°

∴∠PRQ = ∠LRF

(vertically opposite angles)

**Question 16.**

**Solution:**

(i) l is parallel to m

if 3x – 20° = 2x +10°

(Alternate angles are equal)

**Question 17.**

**Solution:**

**Given.** Two lines AB and CD are perpendiculars on EF

**To Prove :** AB ⊥ CD.

Hope given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4C are helpful to complete your math homework.

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