RD Chapter 10- Congruent Triangles Ex-10.1 |
RD Chapter 10- Congruent Triangles Ex-10.2 |
RD Chapter 10- Congruent Triangles Ex-10.4 |
RD Chapter 10- Congruent Triangles Ex-VSAQS |
RD Chapter 10- Congruent Triangles Ex-MCQS |

**Answer
1** :

Two lines l1 and l2 intersect each other at O ∠x = 45°

∵ ∠z = ∠x (Vertically opposite angles)

= 45°

But x + y = 180° (Linear pair)

⇒45° + y= 180°

⇒ y= 180°-45°= 135°

But u = y (Vertically opposite angles)

∴ u = 135°

Hence y = 135°, z = 45° and u = 135°

**Answer
2** :

Three lines AB, CD and EF intersect at O

∠BOD = 90°, ∠DOF = 50°

∵ AB is a line

∴ ∠BOD + ∠DOF + FOA = 180°

⇒ 90° + 50° + u = 180°

⇒ 140° + w = 180°

∴ u= 180°- 140° = 40°

But x = u (Vertically opposite angles)

∴ x = 40°

Similarly, y = 50° and z = 90°

Hence x = 40°, y = 50°, z = 90° and u = 40°

**Answer
3** :

Two lines l_{1} andl_{2} intersect eachother at O

∴Vertically opposite angles are equal,

∴ y =25° and x = z

Now 25° + x = 180° (Linear pair)

⇒ x=180°-25°= 155°

∴ z = x= 155°

Hence x = 155°, y = 25°, z = 155°

**Answer
4** :

∵ EF and CD intersect each other at O

∴ Vertically opposite angles are equal,

∴ ∠1 = 2x

AB is a line

3x + ∠1 + 5x = 180° (Angles on the same side of a line)

⇒ 3x + 2x + 5x = 180°

⇒ 10x = 180° ⇒ x == 18°

Hence x = 18°

**Answer
5** :

Given : Two lines AB and CD intersect each other at O. ∠AOC = 90°

To prove: ∠AOD = ∠BOC = ∠BOD = 90°

Proof : ∵ AB and CD intersect each other at O

∴ ∠AOC = ∠BOD and ∠BOC = ∠AOD (Vertically opposite angles)

∴ But ∠AOC = 90°

∴ ∠BOD = 90°

∴ ∠AOC + ∠BOC = 180° (Linear pair)

⇒ 90° + ∠BOC = 180°

∴ ∠BOC = 180° -90° = 90°

∴ ∠AOD = ∠BOC = 90°

∴ ∠AOD = ∠BOC = ∠BOD = 90°

In the figure, rays AB and CD intersect at O.

(i) Determine y when x = 60°

(ii) Determine x when y = 40°

**Answer
6** :

In the figure,

AB is a line

∴ 2x + y = 180° (Linear pair)

(i) If x = 60°, then

2 x 60° + y = 180°

⇒ 120° +y= 180°

∴ y= 180°- 120° = 60°

(ii) If y = 40°, then

2x + 40° = 180°

⇒ 2x = 180° – 40° = 140°

⇒ x==70°

∴ x = 70°

**Answer
7** :

Three lines AB, CD and EF intersect each other at O

∠AOE = 40° and ∠BOD = 35°

(i) ∠AOC = ∠BOD (Vertically opposite angles)

= 35°

AB is a line

∴ ∠AOE + ∠DOE + ∠BOD = 180°

⇒ 40° + ∠DOE + 35° = 180°

⇒ 75° + ∠DOE = 180°

⇒ ∠DOE = 180°-75° = 105°

But ∠COF = ∠DOE (Vertically opposite angles)

∴ ∠COF = 105°

Similarly, ∠BOF = ∠AOE (Vertically opposite angles)

⇒ ∠BOF = 40°

Hence ∠AOC = 35°, ∠COF = 105°, ∠DOE = 105° and ∠BOF = 40°

**Answer
8** :

AB, CD and EF intersect at O. Such that OF is the bisector of

∠BOD ∠BOF = 35°

∵ OF bisects ∠BOD,

∴ ∠DOF = ∠BOF = 35° (Vertically opposite angles)

∴ ∠BOD = 35° + 35° = 70°

But ∠BOC + ∠BOD = 180° (Linear pair)

⇒ ∠BOC + 70° = 180°

⇒ ∠BOC = 180°-70°= 110°

But ∠AOD = ∠BOC (Vertically opposite angles)

= 110°

Hence ∠BOC = 110° and ∠AOD =110°

**Answer
9** :

In the figure, AB and CD intersect each other at O

∠AOC + ∠BOE = 70°

∠BOD = 40°

AB is a line

∴ ∠AOC + ∠BOE + ∠COE = 180° (Angles on one side of a line)

⇒ 70° + ∠COE = 180°

⇒ ∠COE = 180°-70°= 110°

and ∠AOC = ∠BOD (Vertically opposite angles)

⇒ ∠AOC = 40°

∴ ∠BOE = 70° – 40° = 30°

and reflex ∠COE = 360° – ∠COE

= 360°- 110° = 250°

Which of the following statements are true (T) and which are false (F)?

(i) Angles forming a linear pair are supplementary.

(ii) If two adjacent angles are equal, then each angle measures 90°.

(iii) Angles forming a linear pair can both be acute angles.

(iv) If angles forming a linear pair are equal, then each of these angles is of measure 90°.

**Answer
10** :

(i) True.

(ii) False. It can be possible if they are a linear pair.

(iii) False. In a linear pair, if one is acute, then the other will be obtuse.

(iv) True.

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