RD Chapter 10- Congruent Triangles Ex-10.1 |
RD Chapter 10- Congruent Triangles Ex-10.2 |
RD Chapter 10- Congruent Triangles Ex-10.3 |
RD Chapter 10- Congruent Triangles Ex-VSAQS |
RD Chapter 10- Congruent Triangles Ex-MCQS |

**Answer
1** :

AB || CD and l is transversal ∠1 : ∠2 = 3 : 2

Let ∠1 = 3x

Then ∠2 = 2x

But ∠1 + ∠2 = 180° (Linear pair)

∴ 3x + 2x = 180° ⇒ 5x = 180°

⇒ x = = 36°

∴ ∠1 = 3x = 3 x 36° = 108°

∠2 = 2x = 2 x 36° = 72°

Now ∠1 = ∠3 and ∠2 = ∠4 (Vertically opposite angles)

∴ ∠3 = 108° and ∠4 = 72°

∠1 = ∠5 and ∠2 = ∠6 (Corresponding angles)

∴ ∠5 = 108°, ∠6 = 72°

Similarly, ∠4 = ∠8 and

∠3 = ∠7

∴ ∠8 = 72° and ∠7 = 108°

Hence, ∠1 = 108°, ∠2= 72°

∠3 = 108°, ∠4 = 72°

∠5 = 108°, ∠6 = 72°

∠7 = 108°, ∠8 = 12°

**Answer
2** :

l || m || n and p is then transversal which intersects then at X, Y and Z respectively ∠4 = 120°

∠2 = ∠4 (Alternate angles)

∴ ∠2 = 120°

But ∠3 + ∠4 = 180° (Linear pair)

⇒ ∠3 + 120° = 180°

⇒ ∠3 = 180° – 120°

∴ ∠3 = 60°

But ∠l = ∠3 (Corresponding angles)

∴ ∠l = 60°

Hence ∠l = 60°, ∠2 = 120°, ∠3 = 60°

**Answer
3** :

Given : In the figure, AB || CD and CD || EF

∠BAC = 70°, ∠CEF = 130°

∵ EF || CD

∴ ∠ECD + ∠CEF = 180° (Co-interior angles)

⇒ ∠ECD + 130° = 180°

∴ ∠ECD = 180° – 130° = 50°

∵ BA || CD

∴ ∠BAC = ∠ACD (Alternate angles)

∴ ∠ACD = 70° (∵ ∠BAC = 70°)

∵ ∠ACE = ∠ACD – ∠ECD = 70° – 50° = 20°

**Answer
4** :

In the figure,

∵ ∠ACD = ∠CDE = 100°

But they are alternate angles

∴ AC || DE

**Answer
5** :

In the figure, l || m, n|| p and ∠1 = 85°

∵ n || p

∴ ∠1 = ∠3 (Corresponding anlges)

But ∠1 = 85°

∴ ∠3 = 85°

∵ m || 1

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

⇒ 85° + ∠2 = 180°

⇒ ∠2 = 180° – 85° = 95°

**Answer
6** :

**Answer
7** :

In ||gm ABCD,

∠A and ∠B are unequal

and ∠A : ∠B = 2 : 3

Let ∠A = 2x, then

∠B = 3x

But ∠A + ∠B = 180° (Co-interior angles)

∴ 2x + 3x = 180°

⇒ 5x = 180°

⇒ x == 36°

∴ ∠A = 2x = 2 x 36° = 72°

∠B = 3x = 3 x 36° = 108°

But ∠A = ∠C and ∠B = ∠D (Opposite angles of a ||gm)

∴ ∠C = 72° and ∠D = 108°

Hence ∠A = 72°, ∠B = 108°, ∠C = 72°, ∠D = 108°

**Answer
8** :

AB ⊥ line l and CD ⊥ line l

∴ ∠B = 90° and ∠D = 90°

∴ ∠B = ∠D

But there are corresponding angles

∴ AB || CD

**Answer
9** :

In the figure, a transversal n intersects two lines l and m

∠1 = 60° and

∠2 = rd of a right angle 2

= x 90° = 60°

∴ ∠1 = ∠2

But there are corresponding angles

∴ l || m

**Answer
10** :

In the figure,

l || m || n and a transversal p, intersects them at P, Q and R respectively

∠1 = 60°

∴ ∠1 = ∠3 (Corresponding angles)

∴ ∠3 = 60°

But ∠3 + ∠4 = 180° (Linear pair)

60° + ∠4 = 180° ⇒ ∠4 = 180° – 60°

∴ ∠4 = 120°

But ∠2 = ∠4 (Alternate angles)

∴ ∠2 = 120°

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