RD Chapter 11- Co-ordinate Geometry Ex-11.1 |
RD Chapter 11- Co-ordinate Geometry Ex-VSAQS |
RD Chapter 11- Co-ordinate Geometry Ex-MCQS |

**Answer
1** :
In ∆ABC, base BC is produced both ways to D and E respectivley forming ∠ABE = 104° and ∠ACD = 136°

**Answer
2** :

In ∆ABC, sides BC, CA and BA are produced to D, E and F respectively.

∠ACD = 105° and ∠EAF = 45°

∠ACD + ∠ACB = 180° (Linear pair)

⇒ 105° + ∠ACB = 180°

⇒ ∠ACB = 180°- 105° = 75°

∠BAC = ∠EAF (Vertically opposite angles)

= 45°

But ∠BAC + ∠ABC + ∠ACB = 180°

⇒ 45° + ∠ABC + 75° = 180°

⇒ 120° +∠ABC = 180°

⇒ ∠ABC = 180°- 120°

∴ ∠ABC = 60°

Hence ∠ABC = 60°, ∠BCA = 75°

and ∠BAC = 45°

**Answer
3** :
(i) In ∆ABC, sides BC and CA are produced to D and E respectively

(ii) In ∆ABC, side BC is produced to either side to D and E respectively

∠ABE = 120° and ∠ACD =110°

∵ ∠ABE + ∠ABC = 180° (Linear pair)

(iii) In the figure, BA || DC

**Answer
4** :

In ∆ABC, ∠A : ∠B : ∠C = 3 : 2 : 1

BC is produced to D and CE ⊥ AC

∵ ∠A + ∠B + ∠C = 180° (Sum of angles of a triangles)

Let∠A = 3x, then ∠B = 2x and ∠C = x

∴ 3x + 2x + x = 180° ⇒ 6x = 180°

⇒ x = = 30°

∴ ∠A = 3x = 3 x 30° = 90°

∠B = 2x = 2 x 30° = 60°

∠C = x = 30°

In ∆ABC,

Ext. ∠ACD = ∠A + ∠B

⇒ 90° + ∠ECD = 90° + 60° = 150°

∴ ∠ECD = 150°-90° = 60°

**Answer
5** :

In the figure, AB || DE

AE and BD intersect each other at C ∠BAC = 30° and ∠CDE = 40°

∵ AB || DE

∴ ∠ABC = ∠CDE (Alternate angles)

⇒ ∠ABC = 40°

In ∆ABC, BC is produced

Ext. ∠ACD = Int. ∠A + ∠B

= 30° + 40° = 70°

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

(i) Sum of the three angles of a triangle is 180°.

(ii) A triangle can have two right angles.

(iii) All the angles of a triangle can be less than 60°.

(iv) All the angles of a triangle can be greater than 60°.

(v) All the angles of a triangle can be equal to 60°.

(vi) A triangle can have two obtuse angles.

(vii) A triangle can have at most one obtuse angles.

(viii) If one angle of a triangle is obtuse, then it cannot be a right angled triangle.

(ix) An exterior angle of a triangle is less than either of its interior opposite angles.

(x) An exterior angle of a triangle is equal to the sum of the two interior opposite angles.

(xi) An exterior angle of a triangle is greater than the opposite interior angles.

**Answer
6** :

(i) True.

(ii) False. A right triangle has only one right angle.

(iii) False. In this, the sum of three angles will be less than 180° which is not true.

(iv) False. In this, the sum of three angles will be more than 180° which is not true.

(v) True. As sum of three angles will be 180° which is true.

(vi) False. A triangle has only one obtuse angle.

(vii) True.

(viii)True.

(ix) False. Exterior angle of a triangle is always greater than its each interior opposite angles.

(x) True.

(xi) True.

Fill in the blanks to make the following statements true:

(i) Sum of the angles of a triangle is ………

(ii) An exterior angle of a triangle is equal to the two …….. opposite angles.

(iii) An exterior angle of a triangle is always …….. than either of the interior opposite angles.

(iv) A triangle cannot have more than ………. right angles.

(v) A triangles cannot have more than ……… obtuse angles.

**Answer
7** :

(i) Sum of the angles of a triangle is 180°.

(ii) An exterior angle of a triangle is equal to the two interior opposite angles.

(iii) An exterior angle of a triangle is always greater than either of the interior opposite angles.

(iv) A triangle cannot have more than one right angles.

(v) A triangles cannot have more than one obtuse angles.

**Answer
8** :

Given : In ∆ABC, sides AB and AC are produced to D and E respectively. Bisectors of interior ∠B and ∠C meet at P and bisectors of exterior angles B and C meet at Q.

To prove : ∠BPC + ∠BQC = 180°

Proof : ∵ PB and PC are the internal bisectors of ∠B and ∠C

∠BPC = 90°+ ∠A …(i)

Similarly, QB and QC are the bisectors of exterior angles B and C

∴ ∠BQC = 90° + ∠A …(ii)

Adding (i) and (ii),

∠BPC + ∠BQC = 90° + ∠A + 90° – ∠A

= 90° + 90° = 180°

Hence ∠BPC + ∠BQC = 180°

**Answer
9** :

In the figure,

∠ABC = 45°, ∠BAD = 35° and ∠BCD = 50° Join BD and produce it E