RD Sharma Solutions for Class 9 Maths Chapter 11–Co ...

RD Sharma Solutions for Class9 Maths Chapter 11–CoordinateGeometryClass 9: Maths Chapter 11 solutions. Complete Class 9 Maths Chapter 11 Notes.

RD Sharma Solutions for Class 9 Maths Chapter11–Coordinate GeometryRD Sharma 9th Maths Chapter 11, Class 9 Maths Chapter 11 solutions

https://www.indcareer.com/schools/rd-sharma-solutions-for-class-9-maths-chapter-11-co-ordinate-geometry/



Question 1.

In a ∆ABC, if ∠A = 55°, ∠B = 40°, find ∠C.

Solution:

∵ Sum of three angles of a triangle is 180°

∴ In ∆ABC, ∠A = 55°, ∠B = 40°

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

⇒ 55° + 40° + ∠C = 180°

⇒ 95° + ∠C = 180°

∴ ∠C= 180° -95° = 85°

Question 2.

If the angles of a triangle are in the ratio 1:2:3, determine three angles.

Solution:

Ratio in three angles of a triangle =1:2:3

Let first angle = x

Then second angle = 2x

and third angle = 3x




∴ x + 2x + 3x = 180° (Sum of angles of a triangle)

⇒6x = 180°

⇒x = 180∘6 = 30°

∴ First angle = x = 30°

Second angle = 2x = 2 x 30° = 60°

and third angle = 3x = 3 x 30° = 90°

∴ Angles are 30°, 60°, 90°

Question 3.

The angles of a triangle are (x – 40)°, (x – 20)° and (12 x – 10)°. Find the value of x.

Solution:

∵ Sum of three angles of a triangle = 180°

∴ (x – 40)° + (x – 20)° + (12x-10)0 = 180°

⇒ x – 40° + x – 20° + 12x – 10° = 180°

⇒ x + x+ 12x – 70° = 180°

⇒ 52x = 180° + 70° = 250°

⇒ x = 250∘x25 = 100°

∴ x = 100°

Question 4.

Two angles of a triangle are equal and the third angle is greater than each of those anglesby 30°. Determine all the angles of the triangle.

Solution:

Let each of the two equal angles = x

Then third angle = x + 30°




But sum of the three angles of a triangle is 180°

∴ x + x + x + 30° = 180°

⇒ 3x + 30° = 180°

⇒3x = 150° ⇒x = 150∘3 = 50°

∴ Each equal angle = 50°

and third angle = 50° + 30° = 80°

∴ Angles are 50°, 50° and 80°

Question 5.

If one angle of a triangle is equal to the sum of the other two, show that the triangle is aright triangle.

Solution:

In the triangle ABC,

∠B = ∠A + ∠C

But ∠A + ∠B + ∠C = 180°

⇒∠B + ∠A + ∠C = 180°

⇒∠B + ∠B = 180°




⇒2∠B = 180°

∴ ∠B = 180∘2 = 90°

∵ One angle of the triangle is 90°

∴ ∆ABC is a right triangle.

Question 6.

Can a triangle have:

(i) Two right angles?

(ii) Two obtuse angles?

(iii) Two acute angles?

(iv) All angles more than 60°?

(v) All angles less than 60°?

(vi) All angles equal to 60°?

Justify your answer in each case.

Solution:

(i) In a triangle, two right-angles cannot be possible. We know that sum of three angles is 180°and if there are two right-angles, then the third angle will be zero which is not possible.

(ii) In a triangle, two obtuse angle cannot be possible. We know that the sum of the three anglesof a triangle is 180° and if there are

two obtuse angle, then the third angle will be negative which is not possible.

(iii) In a triangle, two acute angles are possible as sum of three angles of a trianlge is 180°.

(iv) All angles more than 60°, they are also not possible as the sum will be more than 180°.

(v) All angles less than 60°. They are also not possible as the sum will be less than 180°.

(vi) All angles equal to 60°. This is possible as the sum will be 60° x 3 = 180°.




Question 7.

The angles of a triangle are arranged in ascending order of magnitude. If the differencebetween two consecutive angle is 10°, find the three angles.

Solution:

Let three angles of a triangle be x°, (x + 10)°, (x + 20)°

But sum of three angles of a triangle is 180°

∴ x + (x+ 10)° + (x + 20) = 180°

⇒ x + x+10°+ x + 20 = 180°

⇒ 3x + 30° = 180°

⇒ 3x = 180° – 30° = 150°

∴ x = 180∘2 = 50°

∴ Angle are 50°, 50 + 10, 50 + 20

i.e. 50°, 60°, 70°

Question 8.

ABC is a triangle is which ∠A = 72°, the internal bisectors of angles B and C meet in O.Find the magnitude of ∠BOC.

Solution:

In ∆ABC, ∠A = 12° and bisectors of ∠B and ∠C meet at O




Now ∠B + ∠C = 180° – 12° = 108°

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

∴ ∠OBC + ∠OCB = 12 (B + C)

= 12 x 108° = 54°

But in ∆OBC,

∴ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ 54° + ∠BOC = 180°

∠BOC = 180°-54°= 126°

OR

According to corollary,

∠BOC = 90°+ 12 ∠A

= 90+ 12 x 72° = 90° + 36° = 126°

Question 9.

The bisectors of base angles of a triangle cannot enclose a right angle in any case.

Solution:

In right ∆ABC, ∠A is the vertex angle and OB and OC are the bisectors of ∠B and ∠Crespectively

To prove : ∠BOC cannot be a right angle

Proof: ∵ OB and OC are the bisectors of ∠B and ∠C respectively




∴ ∠BOC = 90° x 12 ∠A

Let ∠BOC = 90°, then

12 ∠A = O

⇒∠A = O

Which is not possible because the points A, B and C will be on the same line Hence, ∠BOCcannot be a right angle.

Question 10.

If the bisectors of the base angles of a triangle enclose an angle of 135°. Prove that thetriangle is a right triangle.

Solution:

Given : In ∆ABC, OB and OC are the bisectors of ∠B and ∠C and ∠BOC = 135°




To prove : ∆ABC is a right angled triangle

Proof: ∵ Bisectors of base angles ∠B and ∠C of the ∆ABC meet at O

∴ ∠BOC = 90°+ 12∠A

But ∠BOC =135°

∴ 90°+ 12 ∠A = 135°

⇒ 12∠A= 135° -90° = 45°

∴ ∠A = 45° x 2 = 90°

∴ ∆ABC is a right angled triangle

Question 11.

In a ∆ABC, ∠ABC = ∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O suchthat ∠BOC = 120°. Show that ∠A = ∠B = ∠C = 60°.

Solution:

Given : In ∠ABC, BO and CO are the bisectors of ∠B and ∠C respectively and ∠BOC = 120°and ∠ABC = ∠ACB

To prove : ∠A = ∠B = ∠C = 60°

Proof : ∵ BO and CO are the bisectors of ∠B and ∠Chttps://www.indcareer.com/schools/rd-sharma-solutions-for-class-9-maths-chapter-11-co-ordinate-geometry/



∴ ∠BOC = 90° + 12∠A

But ∠BOC = 120°

∴ 90°+ 12 ∠A = 120°

∴ 12 ∠A = 120° – 90° = 30°

∴ ∠A = 60°

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

∠B + ∠C = 180° – 60° = 120° and ∠B = ∠C

∵ ∠B = ∠C = 120∘2 = 60°

Hence ∠A = ∠B = ∠C = 60°

Question 12.

If each angle of a triangle is less than the sum of the other two, show that the triangle isacute angled.

Solution:

In a ∆ABC,

Let ∠A < ∠B + ∠C

⇒∠A + ∠A < ∠A + ∠B + ∠Chttps://www.indcareer.com/schools/rd-sharma-solutions-for-class-9-maths-chapter-11-co-ordinate-geometry/



⇒ 2∠A < 180°

⇒ ∠A < 90° (∵ Sum of angles of a triangle is 180°)

Similarly, we can prove that

∠B < 90° and ∠C < 90°

∴ Each angle of the triangle are acute angle.

Ex 11.2

Question 1.

The exterior angles obtained on producing the base of a triangle both ways are 104° and136°. Find all the angles of the triangle.

Solution:

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




Question 2.

In the figure, the sides BC, CA and AB of a ∆ABC have been produced to D, E and Frespectively. If ∠ACD = 105° and ∠EAF = 45°, find all the angles of the ∆ABC.




Solution:

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°




Question 3.

Compute the value of x in each of the following figures:

Solution:

(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




Question 4.

In the figure, AC ⊥ CE and ∠A: ∠B : ∠C = 3:2:1, find the value of ∠ECD.

Solution:




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 = 180∘6 = 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°

Question 5.

In the figure, AB || DE, find ∠ACD.




Solution:

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°

Question 6.

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.

Solution:

(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.

Question 7.

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 oppositeangles.

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

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

Solution:




(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.

Question 8.

In a ∆ABC, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of∠B and ∠C meet at Q. Prove that ∠BPC + ∠BQC = 180°.

Solution:

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°+ 12 ∠A …(i)




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

∴ ∠BQC = 90° + 12 ∠A …(ii)

Adding (i) and (ii),

∠BPC + ∠BQC = 90° + 12 ∠A + 90° – 12 ∠A

= 90° + 90° = 180°

Hence ∠BPC + ∠BQC = 180°

Question 9.

In the figure, compute the value of x.

Solution:

In the figure,

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




Question 10.

In the figure, AB divides ∠D AC in the ratio 1 : 3 and AB = DB. Determine the value of x.




Solution:

In the figure AB = DB

Question 11.




ABC is a triangle. The bisector of the exterior angle at B and the bisector of ∠C intersecteach other at D. Prove that ∠D = 12 ∠A.

Solution:

Given : In ∠ABC, CB is produced to E bisectors of ext. ∠ABE and into ∠ACB meet at D.







Question 12.

In the figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠B = 65° and ∠C = 33°, find∠MAN.

Solution:




Question 13.

In a AABC, AD bisects ∠A and ∠C > ∠B. Prove that ∠ADB > ∠ADC.

Solution:




Given : In ∆ABC,

∠C > ∠B and AD is the bisector of ∠A

To prove : ∠ADB > ∠ADC

Proof: In ∆ABC, AD is the bisector of ∠A

∴ ∠1 = ∠2

In ∆ADC,

Ext. ∠ADB = ∠l+ ∠C

⇒ ∠C = ∠ADB – ∠1 …(i)

Similarly, in ∆ABD,

Ext. ∠ADC = ∠2 + ∠B

⇒ ∠B = ∠ADC – ∠2 …(ii)

From (i) and (ii)

∵ ∠C > ∠B (Given)

∴ (∠ADB – ∠1) > (∠ADC – ∠2)

But ∠1 = ∠2




∴ ∠ADB > ∠ADC

Question 14.

In ∆ABC, BD ⊥ AC and CE ⊥ AB. If BD and CE intersect at O, prove that ∠BOC =180°-∠A.

Solution:

Given : In ∆ABC, BD ⊥ AC and CE⊥ AB BD and CE intersect each other at O

To prove : ∠BOC = 180° – ∠A

Proof: In quadrilateral ADOE

∠A + ∠D + ∠DOE + ∠E = 360° (Sum of angles of quadrilateral)

⇒ ∠A + 90° + ∠DOE + 90° = 360°

∠A + ∠DOE = 360° – 90° – 90° = 180°

But ∠BOC = ∠DOE (Vertically opposite angles)

⇒ ∠A + ∠BOC = 180°

∴ ∠BOC = 180° – ∠A

Question 15.

In the figure, AE bisects ∠CAD and ∠B = ∠C. Prove that AE || BC.https://www.indcareer.com/schools/rd-sharma-solutions-for-class-9-maths-chapter-11-co-ordinate-geometry/



Solution:

Given : In AABC, BA is produced and AE is the bisector of ∠CAD

∠B = ∠C

To prove : AE || BC

Proof: In ∆ABC, BA is produced

∴ Ext. ∠CAD = ∠B + ∠C

⇒ 2∠EAC = ∠C + ∠C (∵ AE is the bisector of ∠CAE) (∵ ∠B = ∠C)




⇒ 2∠EAC = 2∠C

⇒ ∠EAC = ∠C

But there are alternate angles

∴ AE || BC

Coordinate Geometry VSAQS

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 pointO, 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 ofthe 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 measureof ∠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 themeasure 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 thesum 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 bothsides)

= ∠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°

Coordinate Geometry MCQS

Mark the correct alternative in each of the following:

Question 1.

If all the three angles of a triangle are equal, then each one of them is equal to




(a) 90°

(b) 45°

(c) 60°

(d) 30°

Solution:

∵ Sum of three angles of a triangle = 180°

∴ Each angle = 180∘3 = 60° (c)

Question 2.

If two acute angles of a right triangle are equal, then each acute is equal to

(a) 30°

(b) 45°

(c) 60°

(d) 90°

Solution:

In a right triangle, one angle = 90°

∴ Sum of other two acute angles = 180° – 90° = 90°

∵ Both angles are equal

∴ Each angle will be = 90∘2 = 45° (b)

Question 3.

An exterior angle of a triangle is equal to 100° and two interior opposite angles are equal.Each of these angles is equal to

(a) 75°

(b) 80°




(c) 40°

(d) 50°

Solution:

In a triangle, exterior angles is equal to the sum of its interior opposite angles

∴ Sum of interior opposite angles = 100°

∵ Both angles are equal

∴ Each angle will be = 100∘2 = 50° (d)

Question 4.

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is

(a) an isosceles triangle

(b) an obtuse triangle

(c) an equilateral triangle

(d) a right triangle

Solution:

Let ∠A, ∠B, ∠C be the angles of a ∆ABC and let ∠A = ∠B + ∠C




But ∠A + ∠B + ∠C = 180°

( Sum of angles of a triangle)

∴ ∠A + ∠A = 180° ⇒ 2∠A = 180°

⇒ ∠A = 180∘2 = 90°

∴ ∆ is a right triangle (d)

Question 5.

Side BC of a triangle ABC has been produced to a point D such that ∠ACD = 120°. If ∠B= 12∠A, then ∠A is equal to

(a) 80°

(b) 75°

(c) 60°

(d) 90°

Solution:

Side BC of ∆ABC is produced to D, then

Ext. ∠ACB = ∠A + ∠B

(Exterior angle of a triangle is equal to the sum of its interior opposite angles)




Question 6.

In ∆ABC, ∠B = ∠C and ray AX bisects the exterior angle ∠DAC. If ∠DAX = 70°, then∠ACB =

(a) 35°

(b) 90°

(c) 70°

(d) 55°

Solution:

In ∆ABC, ∠B = ∠C

AX is the bisector of ext. ∠CAD

∠DAX = 70°




∴ ∠DAC = 70° x 2 = 140°

But Ext. ∠DAC = ∠B + ∠C

= ∠C + ∠C (∵ ∠B = ∠C)

= 2∠C

∴ 2∠C = 140° ⇒ ∠C = 140∘2 = 70°

∴ ∠ACB = 70° (c)

Question 7.

In a triangle, an exterior angle at a vertex is 95° and its one of the interior opposite angleis 55°, then the measure of the other interior angle is

(a) 55°

(b) 85°

(c) 40°

(d) 9.0°

Solution:




In ∆ABC, BA is produced to D such that ∠CAD = 95°

and let ∠C = 55° and ∠B = x°

∵ Exterior angle of a triangle is equal to the sum of its opposite interior angle

∴ ∠CAD = ∠B + ∠C ⇒ 95° = x + 55°

⇒ x = 95° – 55° = 40°

∴ Other interior angle = 40° (c)

Question 8.

If the sides of a triangle are produced in order, then the sum of the three exterior anglesso formed is

(a) 90°

(b) 180°

(c) 270°

(d) 360°

Solution:




In ∆ABC, sides AB, BC and CA are produced in order, then exterior ∠FAB, ∠DBC and ∠ACEare formed

We know an exterior angles of a triangle is equal to the sum of its interior opposite angles

∴ ∠FAB = ∠B + ∠C

∠DBC = ∠C + ∠A and

∠ACE = ∠A + ∠B Adding we get,

∠FAB + ∠DBC + ∠ACE = ∠B + ∠C + ∠C + ∠A + ∠A + ∠B

= 2(∠A + ∠B + ∠C)

= 2 x 180° (Sum of angles of a triangle)

= 360° (d)

Question 9.

In ∆ABC, if ∠A = 100°, AD bisects ∠A and AD⊥ BC. Then, ∠B =

(a) 50°

(b) 90°




(c) 40°

(d) 100°

Solution:

In ∆ABC, ∠A = 100°

AD is bisector of ∠A and AD ⊥ BC

Now, ∠BAD = 100∘2 = 50°

In ∆ABD,

∠BAD + ∠B + ∠D= 180°

(Sum of angles of a triangle)

⇒ ∠50° + ∠B + 90° = 180°

∠B + 140° = 180°

⇒ ∠B = 180° – 140° ∠B = 40° (c)

Question 10.

An exterior angle of a triangle is 108° and its interior opposite angles are in the ratio 4:5.The angles of the triangle are




(a) 48°, 60°, 72°

(b) 50°, 60°, 70°

(c) 52°, 56°, 72°

(d) 42°, 60°, 76°

Solution:

In ∆ABC, BC is produced to D and ∠ACD = 108°

Ratio in ∠A : ∠B = 4:5

∵ Exterior angle of a triangle is equal to the sum of its opposite interior angles

∴ ∠ACD = ∠A + ∠B = 108°

Ratio in ∠A : ∠B = 4:5




Question 11.

In a ∆ABC, if ∠A = 60°, ∠B = 80° and the bisectors of ∠B and ∠C meet at O, then ∠BOC=

(a) 60°

(b) 120°

(c) 150°

(d) 30°

Solution:

In ∆ABC, ∠A = 60°, ∠B = 80°

∴ ∠C = 180° – (∠A + ∠B)

= 180° – (60° + 80°)

= 180° – 140° = 40°

Bisectors of ∠B and ∠C meet at O




Question 12.

Line segments AB and CD intersect at O such that AC || DB. If ∠CAB = 45° and ∠CDB =55°, then ∠BOD =

(a) 100°

(b) 80°

(c) 90°

(d) 135°

Solution:

In the figure,




AB and CD intersect at O

and AC || DB, ∠CAB = 45°

and ∠CDB = 55°

∵ AC || DB

∴ ∠CAB = ∠ABD (Alternate angles)

In ∆OBD,

∠BOD = 180° – (∠CDB + ∠ABD)

= 180° – (55° + 45°)

= 180° – 100° = 80° (b)

Question 13.

In the figure, if EC || AB, ∠ECD = 70° and ∠BDO = 20°, then ∠OBD is

(a) 20°

(b) 50°

(c) 60°




(d) 70°

Solution:

In the figure, EC || AB

∠ECD = 70°, ∠BDO = 20°

∵ EC || AB

∠AOD = ∠ECD (Corresponding angles)

⇒ ∠AOD = 70°

In ∆OBD,

Ext. ∠AOD = ∠OBD + ∠BDO

70° = ∠OBD + 20°

⇒ ∠OBD = 70° – 20° = 50° (b)

Question 14.

In the figure, x + y =

(a) 270




(b) 230

(c) 210

(d) 190°

Solution:

In the figure

Ext. ∠OAE = ∠AOC + ∠ACO

⇒ x = 40° + 80° = 120°

Similarly,

Ext. ∠DBF = ∠ODB + ∠DOBhttps://www.indcareer.com/schools/rd-sharma-solutions-for-class-9-maths-chapter-11-co-ordinate-geometry/



y = 70° + ∠DOB

[(∵ ∠AOC = ∠DOB) (vertically opp. angles)]

= 70° + 40° = 110°

∴ x+y= 120°+ 110° = 230° (b)

Question 15.

If the measures of angles of a triangle are in the ratio of 3 : 4 : 5, what is the measure ofthe smallest angle of the triangle?

(a) 25°

(b) 30°

(c) 45°

(d) 60°

Solution:

Ratio in the measures of the triangle =3:4:5

Sum of angles of a triangle = 180°

Let angles be 3x, 4x, 5x

Sum of angles = 3x + 4x + 5x = 12x

∴ Smallest angle = 180x3x12x = 45° (c)

Question 16.

In the figure, if AB ⊥ BC, then x =

(a) 18

(b) 22

(c) 25

(d) 32




Solution:

In the figure, AB ⊥ BC

∠AGF = 32°

∴ ∠CGB = ∠AGF (Vertically opposite angles)

= 32°

In ∆GCB, ∠B = 90°




∴ ∠CGB + ∠GCB = 90°

⇒ 32° + ∠GCB = 90°

⇒ ∠GCB = 90° – 32° = 58°

Now in ∆GDC,

Ext. ∠GCB = ∠CDG + ∠DGC

⇒ 58° = x + 14° + x

⇒ 2x + 14° = 58°

⇒ 2x = 58 – 14° = 44

⇒ x = 442 = 22°

∴ x = 22° (b)

Question 17.

In the figure, what is ∠ in terms of x and y?

(a) x + y + 180

(b) x + y – 180

(c) 180° -(x+y)

(d) x+y + 360°




Solution:

In the figure, BC is produced both sides CA and BA are also produced

In ∆ABC,

∠B = 180° -y

and ∠C 180° – x

∴ z = ∠A = 180° – (B + C)

= 180° – (180 – y + 180 -x)

= 180° – (360° – x – y)

= 180° – 360° + x + y = x + y – 180° (b)

Question 18.

In the figure, for which value of x is l1 || l2?

(a) 37

(b) 43

(c) 45




(d) 47

Solution:

In the figure, l1 || l2

∴ ∠EBA = ∠BAH (Alternate angles)

∴ ∠BAH = 78°

⇒ ∠BAC + ∠CAH = 78°

⇒ ∠BAC + 35° = 78°

⇒ ∠BAC = 78° – 35° = 43°

In ∆ABC, ∠C = 90°




∴ ∠ABC + ∠BAC = 90°

⇒ x + 43° = 90° ⇒ x = 90° – 43°

∴ x = 47° (d)

Question 19.

In the figure, what is y in terms of x?

Solution:

In ∆ABC,

∠ACB = 180° – (x + 2x)

= 180° – 3x …(i)

and in ∆BDG,

∠BED = 180° – (2x + y) …(ii)

∠EGC = ∠AGD (Vertically opposite angles)https://www.indcareer.com/schools/rd-sharma-solutions-for-class-9-maths-chapter-11-co-ordinate-geometry/



= 3y

In quad. BCGE,

∠B + ∠ACB + ∠CGE + ∠BED = 360° (Sum of angles of a quadrilateral)

⇒ 2x+ 180° – 3x + 3y + 180°- 2x-y = 360°

⇒ -3x + 2y = 0

⇒ 3x = 2y ⇒ y = 32x (a)

Question 20.

In the figure, what is the value of x?

(a) 35

(b) 45

(c) 50

(d) 60




Solution:

In the figure, side AB is produced to D

∴ ∠CBA + ∠CBD = 180° (Linear pair)

⇒ 7y + 5y = 180°

⇒ 12y = 180°

⇒ y = 18012 = 15

and Ext. ∠CBD = ∠A + ∠C

⇒ 7y = 3y + x

⇒ 7y -3y = x

⇒ 4y = x

∴ x = 4 x 15 = 60 (d)

Question 21.

In the figure, the value of x is

(a) 65°




(b) 80°

(c) 95°

(d) 120°

Solution:

In the figure, ∠A = 55°, ∠D = 25° and ∠C = 40°

Now in ∆ABD,

Ext. ∠DBC = ∠A + ∠Dhttps://www.indcareer.com/schools/rd-sharma-solutions-for-class-9-maths-chapter-11-co-ordinate-geometry/



= 55° + 25° = 80°

Similarly, in ∆BCE,

Ext. ∠DEC = ∠EBC + ∠ECB

= 80° + 40° = 120° (d)

Question 22.

In the figure, if BP || CQ and AC = BC, then the measure of x is

(a) 20°

(b) 25°

(c) 30°

(d) 35°

Solution:

In the figure, AC = BC, BP || CQ




∵ BP || CQ

∴ ∠PBC – ∠QCD

⇒ 20° + ∠ABC = 70°

⇒ ∠ABC = 70° – 20° = 50°

∵ BC = AC

∴ ∠ACB = ∠ABC (Angles opposite to equal sides)

= 50°

Now in ∆ABC,

Ext. ∠ACD = ∠B + ∠A

⇒ x + 70° = 50° + 50°

⇒ x + 70° = 100°

∴ x = 100° – 70° = 30° (c)

Question 23.

In the figure, AB and CD are parallel lines and transversal EF intersects them at P and Qrespectively. If ∠APR = 25°, ∠RQC = 30° and ∠CQF = 65°, then




(a) x = 55°, y = 40°

(b) x = 50°, y = 45°

(c) x = 60°, y = 35°

(d) x = 35°, y = 60°

Solution:

In the figure,

∵ AB || CD, EF intersects them at P and Q respectively,

∠APR = 25°, ∠RQC = 30°, ∠CQF = 65°

∵ AB || CD

∴ ∠APQ = ∠CQF (Corresponding anlges)

⇒ y + 25° = 65°

⇒ y = 65° – 25° = 40°

and APQ + PQC = 180° (Co-interior angles)




y + 25° + ∠1 +30°= 180°

40° + 25° + ∠1 + 30° = 180°

⇒ ∠1 + 95° = 180°

∴ ∠1 = 180° – 95° = 85°

Now, ∆PQR,

∠RPQ + ∠PQR + ∠PRQ = 180° (Sum of angles of a triangle)

⇒ 40° + x + 85° = 180°

⇒ 125° + x = 180°

⇒ x = 180° – 125° = 55°

∴ x = 55°, y = 40° (a)

Question 24.

The base BC of triangle ABC is produced both ways and the measure of exterior anglesformed are 94° and 126°. Then, ∠BAC = ?




(a) 94°

(b) 54°

(c) 40°

(d) 44°

Solution:

In ∆ABC, base BC is produced both ways and ∠ACD = 94°, ∠ABE = 126°

Ext. ∠ACD = ∠BAC + ∠ABC

⇒ 94° = ∠BAC + ∠ABC

Similarly, ∠ABE = ∠BAC + ∠ACB

⇒ 126° = ∠BAC + ∠ACB

Adding,

94° + 126° = ∠BAC + ∠ABC + ∠ACB + ∠BAC

220° = 180° + ∠BAC

∴ ∠BAC = 220° -180° = 40° (c)

Question 25.




If the bisectors of the acute angles of a right triangle meet at O, then the angle at Obetween the two bisectors is

(a) 45°

(b) 95°

(c) 135°

(d) 90°

Solution:

In right ∆ABC, ∠A = 90°

Bisectors of ∠B and ∠C meet at O, then 1

∠BOC = 90° + 12 ∠A

= 90°+ 12 x 90° = 90° + 45°= 135° (c)

Question 26.

The bisects of exterior angles at B and C of ∆ABC, meet at O. If ∠A = .x°, then ∠BOC=




Solution:

In ∆ABC, ∠A = x°

and bisectors of ∠B and ∠C meet at O.

Question 27.

In a ∆ABC, ∠A = 50° and BC is produced to a point D. If the bisectors of ∠ABC and∠ACD meet at E, then ∠E =

(a) 25°




(b) 50°

(c) 100°

(d) 75°

Solution:

In ∆ABC, ∠A = 50°

BC is produced

Bisectors of ∠ABC and ∠ACD meet at ∠E

∴ ∠E = 12 ∠A = 12 x 50° = 25° (a)

Question 28.

The side BC of AABC is produced to a point D. The bisector of ∠A meets side BC in L. If∠ABC = 30° and ∠ACD =115°,then ∠ALC =

(a) 85°

(b) 7212 °

(c) 145°

(d) none of these

Solution:https://www.indcareer.com/schools/rd-sharma-solutions-for-class-9-maths-chapter-11-co-ordinate-geometry/



In ∆ABC, BC is produced to D

∠B = 30°, ∠ACD = 115°

Question 29.




In the figure , if l1 || l2, the value of x is

(a) 22 12

(b) 30

(c) 45

(d) 60

Solution:

In the figure, l1 || l2

EC, EB are the bisectors of ∠DCB and ∠CBA respectively EF is the bisector of ∠GEB




∵ EC and EB are the bisectors of ∠DCB and ∠CBA respectively

∴ ∠CEB = 90°

∴ a + b = 90° ,

and ∠GEB = 90° (∵ ∠CEB = 90°)

2x = 90° ⇒ x = 902 = 45 (c)

Question 30.

In ∆RST (in the figure), what is the value of x?

(a) 40°

(b) 90°

(c) 80°

(d) 100°




Solution:




Chapterwise RD SharmaSolutions for Class 9 Maths :

● Chapter 1–Number System

● Chapter 2–Exponents of Real

Numbers

● Chapter 3–Rationalisation

● Chapter 4–Algebraic Identities

● Chapter 5–Factorization of

Algebraic Expressions

● Chapter 6–Factorization Of

Polynomials

● Chapter 7–Introduction to

Euclid’s Geometry

● Chapter 8–Lines and Angles

● Chapter 9–Triangle and its

Angles

● Chapter 10–Congruent Triangles

● Chapter 11–Coordinate Geometry

● Chapter 12–Heron’s Formula

● Chapter 13–Linear Equations in

Two Variables

● Chapter 14–Quadrilaterals

● Chapter 15–Area of

Parallelograms and Triangles

● Chapter 16–Circles

● Chapter 17–Construction

● Chapter 18–Surface Area and

Volume of Cuboid and Cube


Volume of A Right Circular

Cylinder


Volume of A Right Circular Cone

● Chapter 21–Surface Area And

Volume Of Sphere

● Chapter 22–Tabular

Representation of Statistical Data

● Chapter 23–Graphical

Representation of Statistical Data

● Chapter 24–Measure of Central

Tendency

● Chapter 25–Probability


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About RD Sharma

RD Sharma isn't the kind of author you'd bump into at lit fests. But his

bestselling books have helped many CBSE students lose their dread of

maths. Sunday Times profiles the tutor turned internet star

He dreams of algorithms that would give most people nightmares. And,

spends every waking hour thinking of ways to explain concepts like 'series

solution of linear differential equations'. Meet Dr Ravi Dutt Sharma —

mathematics teacher and author of 25 reference books — whose name

evokes as much awe as the subject he teaches. And though students have

used his thick tomes for the last 31 years to ace the dreaded maths exam,

it's only recently that a spoof video turned the tutor into a YouTube star.

R D Sharma had a good laugh but said he shared little with his on-screen

persona except for the love for maths. "I like to spend all my time thinking

and writing about maths problems. I find it relaxing," he says. When he is

not writing books explaining mathematical concepts for classes 6 to 12 and

engineering students, Sharma is busy dispensing his duty as vice-principal

and head of department of science and humanities at Delhi government's

Guru Nanak Dev Institute of Technology.




RD Sharma Solutions for Class 9 Maths Chapter 11–Co ...

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