In this chapter, we provide RD Sharma Solutions for Chapter 11 Coordinate Geometry MCQs for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 11 Coordinate Geometry MCQs Maths pdf, free RD Sharma Solutions for Chapter 11 Coordinate Geometry MCQs Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 9 |

Subject | Maths |

Chapter | Chapter 11 |

Chapter Name | Coordinate Geometry |

Exercise | MCQs |

**RD Sharma Solutions for Class 9 Chapter 11 Coordinate Geometry MCQs Download PDF**

Question 1.

Define a triangle.

Solution:

A figure bounded by three lines segments in a plane is called a triangle.

Question 2.

Write the sum of the angles of an obtuse triangle.

Solution:

The sum of angles of an obtuse triangle is 180°.

Question 3.

In ∆ABC, if ∠B = 60°, ∠C = 80° and the bisectors of angles ∠ABC and ∠ACB meet at a point O, then find the measure of ∠BOC.

Solution:

In ∆ABC, ∠B = 60°, ∠C = 80°

OB and OC are the bisectors of ∠B and ∠C

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

⇒ ∠A + 60° + 80° = 180°

⇒ ∠A + 140° = 180°

∴ ∠A = 180°- 140° = 40°

= 90° + – x 40° = 90° + 20° = 110°

Question 4.

If the angles of a triangle are in the ratio 2:1:3. Then find the measure of smallest angle.

Solution:

Sum of angles of a triangle = 180°

Ratio in the angles = 2 : 1 : 3

Let first angle = 2x

Second angle = x

and third angle = 3x

∴ 2x + x + 3x = 180° ⇒ 6x = 180°

∴ x = 180∘6 = 30°

∴ First angle = 2x = 2 x 30° = 60°

Second angle = x = 30°

and third angle = 3x = 3 x 30° = 90°

Hence angles are 60°, 30°, 90°

Question 5.

State exterior angle theorem.

Solution:

Given : In ∆ABC, side BC is produced to D

To prove : ∠ACD = ∠A + ∠B

Proof: In ∆ABC,

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

and ∠ACD + ∠ACB = 180° …(ii) (Linear pair)

From (i) and (ii)

∠ACD + ∠ACB = ∠A + ∠B + ∠ACB

∠ACD = ∠A + ∠B

Hence proved.

Question 6.

The sum of two angles of a triangle is equal to its third angle. Determine the measure of the third angle.

Solution:

In ∆ABC,

∠A + ∠C = ∠B

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

∴ ∠B + ∠A + ∠C = 180°

⇒ ∠B + ∠B = 180°

⇒ 2∠B = 180°

⇒ ∠B = 180∘2 = 90°

∴ Third angle = 90°

Question 7.

In the figure, if AB || CD, EF || BC, ∠BAC = 65° and ∠DHF = 35°, find ∠AGH.

Solution:

Given : In figure, AB || CD, EF || BC ∠BAC = 65°, ∠DHF = 35°

∵ EF || BC

∴ ∠A = ∠ACH (Alternate angle)

∴ ∠ACH = 65°

∵∠GHC = ∠DHF

(Vertically opposite angles)

∴ ∠GHC = 35°

Now in ∆GCH,

Ext. ∠AGH = ∠GCH + ∠GHC

= 65° + 35° = 100°

Question 8.

In the figure, if AB || DE and BD || FG such that ∠FGH = 125° and ∠B = 55°, find x and y.

Solution:

In the figure, AB || DF, BD || FG

∠FGH = 125° and ∠B = 55°

∠FGH + FGE = 180° (Linear pair)

⇒ 125° + y – 180°

⇒ y= 180°- 125° = 55°

∵ BA || FD and BD || FG

∠B = ∠F = 55°

Now in ∆EFG,

∠F + ∠FEG + ∠FGE = 180°

(Angles of a triangle)

⇒ 55° + x + 55° = 180°

⇒ x+ 110°= 180°

∴ x= 180°- 110° = 70°

Hence x = 70, y = 55°

Question 9.

If the angles A, B and C of ∆ABC satisfy the relation B – A = C – B, then find the measure of ∠B.

Solution:

In ∆ABC,

∠A + ∠B + ∠C= 180° …(i)

and B – A = C – B

⇒ B + B = A + C ⇒ 2B = A + C

From (i),

B + 2B = 180° ⇒ 3B = 180°

∠B = 180∘3 = 60°

Hence ∠B = 60°

Question 10.

In ∆ABC, if bisectors of ∠ABC and ∠ACB intersect at O at angle of 120°, then find the measure of ∠A.

Solution:

In ∆ABC, bisectors of ∠B and ∠C intersect at O and ∠BOC = 120°

But ∠BOC = 90°+ 12

90°+ 12 ∠A= 120°

⇒ 12 ∠A= 120°-90° = 30°

∴ ∠A = 2 x 30° = 60°

Question 11.

If the side BC of ∆ABC is produced on both sides, then write the difference between the sum of the exterior angles so formed and ∠A.

Solution:

In ∆ABC, side BC is produced on both sides forming exterior ∠ABE and ∠ACD

Ext. ∠ABE = ∠A + ∠ACB

and Ext. ∠ACD = ∠ABC + ∠A

Adding we get,

∠ABE + ∠ACD = ∠A + ∠ACB + ∠A + ∠ABC

⇒ ∠ABE + ∠ACD – ∠A = ∠A 4- ∠ACB + ∠A + ∠ABC – ∠A (Subtracting ∠A from both sides)

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

Question 12.

In a triangle ABC, if AB = AC and AB is produced to D such that BD = BC, find ∠ACD: ∠ADC.

Solution:

In ∆ABC, AB = AC

AB is produced to D such that BD = BC

DC are joined

In ∆ABC, AB = AC

∴ ∠ABC = ∠ACB

In ∆ BCD, BD = BC

∴ ∠BDC = ∠BCD

and Ext. ∠ABC = ∠BDC + ∠BCD = 2∠BDC (∵ ∠BDC = ∠BCD)

⇒ ∠ACB = 2∠BCD (∵ ∠ABC = ∠ACB)

Adding ∠BDC to both sides

⇒ ∠ACB + ∠BDC = 2∠BDC + ∠BDC

⇒ ∠ACB + ∠BCD = 3 ∠BDC (∵ ∠BDC = ∠BCD)

⇒ ∠ACB = 3∠BDC

Question 13.

In the figure, side BC of AABC is produced to point D such that bisectors of ∠ABC and ∠ACD meet at a point E. If ∠BAC = 68°, find ∠BEC.

Solution:

In the figure,

side BC of ∆ABC is produced to D such that bisectors of ∠ABC and ∠ACD meet at E

∠BAC = 68°

In ∆ABC,

Ext. ∠ACD = ∠A + ∠B

⇒ 12 ∠ACD = 12 ∠A + 12 ∠B

⇒ ∠2= 12 ∠A + ∠1 …(i)

But in ∆BCE,

Ext. ∠2 = ∠E + ∠l

⇒ ∠E + ∠l = ∠2 = 12 ∠A + ∠l [From (i)]

⇒ ∠E = 12 ∠A = 68∘2 =34°

**All Chapter RD Sharma Solutions For Class 9 Maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class 9**

**All Subject NCERT Solutions For Class 9**

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.