## RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry

### RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry Ex 11.1

Question 1.

In a âˆ†ABC, if âˆ A = 55Â°, âˆ B = 40Â°, find âˆ C.

Solution:

âˆµ Sum of three angles of a triangle is 180Â°

âˆ´ In âˆ†ABC, âˆ A = 55Â°, âˆ B = 40Â°

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

â‡’ 55Â° + 40Â° + âˆ C = 180Â°

â‡’ 95Â° + âˆ C = 180Â°

âˆ´ âˆ C= 180Â° -95Â° = 85Â°

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

âˆ´ x + 2x + 3x = 180Â° (Sum of angles of a triangle)

â‡’6x = 180Â°

â‡’x = \(\frac { { 180 }^{ \circ } }{ 6 }\)Â = 30Â°

âˆ´ First angle = x = 30Â°

Second angle = 2x = 2 x 30Â° = 60Â°

and third angle = 3x = 3 x 30Â° = 90Â°

âˆ´ Angles are 30Â°, 60Â°, 90Â°

Question 3.

The angles of a triangle are (x – 40)Â°, (x – 20)Â° and (\(\frac { 1 }{ 2 }\) x – 10)Â°. Find the value of x.

Solution:

âˆµ Sum of three angles of a triangle = 180Â°

âˆ´ (x – 40)Â° + (x – 20)Â° + (\(\frac { 1 }{ 2 }\)x-10)0 = 180Â°

â‡’ x – 40Â° + x – 20Â° + \(\frac { 1 }{ 2 }\)x – 10Â° = 180Â°

â‡’ x + x+ \(\frac { 1 }{ 2 }\)x – 70Â° = 180Â°

â‡’ \(\frac { 5 }{ 2 }\)x = 180Â° + 70Â° = 250Â°

â‡’ x = \(\frac { { 250 }^{ \circ }x 2 }{ 5 }\)Â = 100Â°

âˆ´ x = 100Â°

Question 4.

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

Solution:

Let each of the two equal angles = x

Then third angle = x + 30Â°

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

âˆ´ x + x + x + 30Â° = 180Â°

â‡’ 3x + 30Â° = 180Â°

â‡’3x = 150Â° â‡’x = \(\frac { { 150 }^{ \circ } }{ 3 }\) = 50Â°

âˆ´ Each equal angle = 50Â°

and third angle = 50Â° + 30Â° = 80Â°

âˆ´ Angles are 50Â°, 50Â° and 80Â°

Question 5.

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

Solution:

In the triangle ABC,

âˆ B = âˆ A + âˆ C

But âˆ A + âˆ B + âˆ C = 180Â°

â‡’âˆ B + âˆ A + âˆ C = 180Â°

â‡’âˆ B + âˆ B = 180Â°

â‡’2âˆ B = 180Â°

âˆ´ âˆ B = \(\frac { { 180 }^{ \circ } }{ 2 }\) = 90Â°

âˆµ One angle of the triangle is 90Â°

âˆ´ âˆ†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 60Â°?

(v) All angles less than 60Â°?

(vi) All angles equal to 60Â°?

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 180Â° 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 angles of a triangle is 180Â° 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 180Â°.

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

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

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

Question 7.

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

Solution:

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

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

âˆ´ x + (x+ 10)Â° + (x + 20) = 180Â°

â‡’ x + x+10Â°+ x + 20 = 180Â°

â‡’ 3x + 30Â° = 180Â°

â‡’ 3x = 180Â° – 30Â° = 150Â°

âˆ´ x = \(\frac { { 180 }^{ \circ } }{ 2 }\) = 50Â°

âˆ´ Angle are 50Â°, 50 + 10, 50 + 20

i.e. 50Â°, 60Â°, 70Â°

Question 8.

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

Solution:

In âˆ†ABC, âˆ A = 12Â° and bisectors of âˆ B and âˆ C meet at O

Now âˆ B + âˆ C = 180Â° – 12Â° = 108Â°

âˆµ OB and OC are the bisectors of âˆ B and âˆ C respectively

âˆ´ âˆ OBC + âˆ OCB = \(\frac { 1 }{ 2 }\) (B + C)

= \(\frac { 1 }{ 2 }\) x 108Â° = 54Â°

But in âˆ†OBC,

âˆ´ âˆ OBC + âˆ OCB + âˆ BOC = 180Â°

â‡’ 54Â° + âˆ BOC = 180Â°

âˆ BOC = 180Â°-54Â°= 126Â°

OR

According to corollary,

âˆ BOC = 90Â°+ \(\frac { 1 }{ 2 }\) âˆ A

= 90+ \(\frac { 1 }{ 2 }\) x 72Â° = 90Â° + 36Â° = 126Â°

Question 9.

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

Solution:

In right âˆ†ABC, âˆ A is the vertex angle and OB and OC are the bisectors of âˆ B and âˆ C respectively

To prove : âˆ BOC cannot be a right angle

Proof: âˆµ OB and OC are the bisectors of âˆ B and âˆ C respectively

âˆ´ âˆ BOC = 90Â° x \(\frac { 1 }{ 2 }\) âˆ A

Let âˆ BOC = 90Â°, then

\(\frac { 1 }{ 2 }\) âˆ A = O

â‡’âˆ A = O

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

Question 10.

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

Solution:

Given : In âˆ†ABC, OB and OC are the bisectors of âˆ B and âˆ C and âˆ BOC = 135Â°

To prove : âˆ†ABC is a right angled triangle

Proof: âˆµ Bisectors of base angles âˆ B and âˆ C of the âˆ†ABC meet at O

âˆ´ âˆ BOC = 90Â°+ \(\frac { 1 }{ 2 }\)âˆ A

But âˆ BOC =135Â°

âˆ´ 90Â°+ \(\frac { 1 }{ 2 }\) âˆ A = 135Â°

â‡’ \(\frac { 1 }{ 2 }\)âˆ A= 135Â° -90Â° = 45Â°

âˆ´ âˆ A = 45Â° x 2 = 90Â°

âˆ´ âˆ†ABC is a right angled triangle

Question 11.

In a âˆ†ABC, âˆ ABC = âˆ ACB and the bisectors of âˆ ABC and âˆ ACB intersect at O such that âˆ BOC = 120Â°. Show that âˆ A = âˆ B = âˆ C = 60Â°.

Solution:

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

To prove : âˆ A = âˆ B = âˆ C = 60Â°

Proof : âˆµ BO and CO are the bisectors of âˆ B and âˆ C

âˆ´ âˆ BOC = 90Â° + \(\frac { 1 }{ 2 }\)âˆ A

But âˆ BOC = 120Â°

âˆ´ 90Â°+ \(\frac { 1 }{ 2 }\) âˆ A = 120Â°

âˆ´ \(\frac { 1 }{ 2 }\) âˆ A = 120Â° – 90Â° = 30Â°

âˆ´ âˆ A = 60Â°

âˆµ âˆ A + âˆ B + âˆ C = 180Â° (Angles of a triangle)

âˆ B + âˆ C = 180Â° – 60Â° = 120Â° and âˆ B = âˆ C

âˆµ âˆ B = âˆ C = \(\frac { { 120 }^{ \circ } }{ 2 }\) = 60Â°

Hence âˆ A = âˆ B = âˆ C = 60Â°

Question 12.

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

Solution:

In a âˆ†ABC,

Let âˆ A < âˆ B + âˆ C

â‡’âˆ A + âˆ A < âˆ A + âˆ B + âˆ C

â‡’ 2âˆ A < 180Â°

â‡’ âˆ A < 90Â° (âˆµ Sum of angles of a triangle is 180Â°)

Similarly, we can prove that

âˆ B < 90Â° and âˆ C < 90Â°

âˆ´ Each angle of the triangle are acute angle.

### RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry Ex 11.2

Question 1.

The exterior angles obtained on producing the base of a triangle both ways are 104Â° and 136Â°. Find all the angles of the triangle.

Solution:

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

Question 2.

In the figure, the sides BC, CA and AB of a âˆ†ABC have been produced to D, E and F respectively. If âˆ ACD = 105Â° and âˆ EAF = 45Â°, find all the angles of the âˆ†ABC.

Solution:

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

âˆ ACD = 105Â° and âˆ EAF = 45Â°

âˆ ACD + âˆ ACB = 180Â° (Linear pair)

â‡’ 105Â° + âˆ ACB = 180Â°

â‡’ âˆ ACB = 180Â°- 105Â° = 75Â°

âˆ BAC = âˆ EAF (Vertically opposite angles)

= 45Â°

But âˆ BAC + âˆ ABC + âˆ ACB = 180Â°

â‡’ 45Â° + âˆ ABC + 75Â° = 180Â°

â‡’ 120Â° +âˆ ABC = 180Â°

â‡’ âˆ ABC = 180Â°- 120Â°

âˆ´ âˆ ABC = 60Â°

Hence âˆ ABC = 60Â°, âˆ BCA = 75Â°

and âˆ BAC = 45Â°

Question 3.

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

Solution:

(i) In âˆ†ABC, sides BC and CA are produced to D and E respectively

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

âˆ ABE = 120Â° and âˆ ACD =110Â°

âˆµ âˆ ABE + âˆ ABC = 180Â° (Linear pair)

(iii) In the figure, BA || DC

Question 4.

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

Solution:

In âˆ†ABC, âˆ A : âˆ B : âˆ C = 3 : 2 : 1

BC is produced to D and CE âŠ¥ AC

âˆµ âˆ A + âˆ B + âˆ C = 180Â° (Sum of angles of a triangles)

Letâˆ A = 3x, then âˆ B = 2x and âˆ C = x

âˆ´ 3x + 2x + x = 180Â° â‡’ 6x = 180Â°

â‡’ x = \(\frac { { 180 }^{ \circ } }{ 6 }\)Â = 30Â°

âˆ´ âˆ A = 3x = 3 x 30Â° = 90Â°

âˆ B = 2x = 2 x 30Â° = 60Â°

âˆ C = x = 30Â°

In âˆ†ABC,

Ext. âˆ ACD = âˆ A + âˆ B

â‡’ 90Â° + âˆ ECD = 90Â° + 60Â° = 150Â°

âˆ´ âˆ ECD = 150Â°-90Â° = 60Â°

Question 5.

In the figure, AB || DE, find âˆ ACD.

Solution:

In the figure, AB || DE

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

âˆµ AB || DE

âˆ´ âˆ ABC = âˆ CDE (Alternate angles)

â‡’ âˆ ABC = 40Â°

In âˆ†ABC, BC is produced

Ext. âˆ ACD = Int. âˆ A + âˆ B

= 30Â° + 40Â° = 70Â°

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 180Â°.

(ii) A triangle can have two right angles.

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

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

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

(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 180Â° which is not true.

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

(v) True. As sum of three angles will be 180Â° 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 opposite angles.

(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 180Â°.

(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 âˆ†ABC, the internal bisectors of âˆ B and âˆ C meet at P and the external bisectors of âˆ B and âˆ C meet at Q. Prove that âˆ BPC + âˆ BQC = 180Â°.

Solution:

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

To prove : âˆ BPC + âˆ BQC = 180Â°

Proof : âˆµ PB and PC are the internal bisectors of âˆ B and âˆ C

âˆ BPC = 90Â°+ \(\frac { 1 }{ 2 }\) âˆ A …(i)

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

âˆ´ âˆ BQC = 90Â° + \(\frac { 1 }{ 2 }\) âˆ A …(ii)

Adding (i) and (ii),

âˆ BPC + âˆ BQC = 90Â° + \(\frac { 1 }{ 2 }\) âˆ A + 90Â° – \(\frac { 1 }{ 2 }\) âˆ A

= 90Â° + 90Â° = 180Â°

Hence âˆ BPC + âˆ BQC = 180Â°

Question 9.

In the figure, compute the value of x.

Solution:

In the figure,

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

Question 10.

In the figure, AB divides âˆ 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 âˆ C intersect each other at D. Prove that âˆ D = \(\frac { 1 }{ 2 }\) âˆ A.

Solution:

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

Question 12.

In the figure, AM âŠ¥ BC and AN is the bisector of âˆ A. If âˆ B = 65Â° and âˆ C = 33Â°, find âˆ MAN.

Solution:

Question 13.

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

Solution:

Given : In âˆ†ABC,

âˆ C > âˆ B and AD is the bisector of âˆ A

To prove : âˆ ADB > âˆ ADC

Proof: In âˆ†ABC, AD is the bisector of âˆ A

âˆ´ âˆ 1 = âˆ 2

In âˆ†ADC,

Ext. âˆ ADB = âˆ l+ âˆ C

â‡’ âˆ C = âˆ ADB – âˆ 1 …(i)

Similarly, in âˆ†ABD,

Ext. âˆ ADC = âˆ 2 + âˆ B

â‡’ âˆ B = âˆ ADC – âˆ 2 …(ii)

From (i) and (ii)

âˆµ âˆ C > âˆ B (Given)

âˆ´ (âˆ ADB – âˆ 1) > (âˆ ADC – âˆ 2)

But âˆ 1 = âˆ 2

âˆ´ âˆ ADB > âˆ ADC

Question 14.

In âˆ†ABC, BD âŠ¥ AC and CE âŠ¥ AB. If BD and CE intersect at O, prove that âˆ BOC = 180Â°-âˆ A.

Solution:

Given : In âˆ†ABC, BD âŠ¥ AC and CEâŠ¥ AB BD and CE intersect each other at O

To prove : âˆ BOC = 180Â° – âˆ A

Proof: In quadrilateral ADOE

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

â‡’ âˆ A + 90Â° + âˆ DOE + 90Â° = 360Â°

âˆ A + âˆ DOE = 360Â° – 90Â° – 90Â° = 180Â°

But âˆ BOC = âˆ DOE (Vertically opposite angles)

â‡’ âˆ A + âˆ BOC = 180Â°

âˆ´ âˆ BOC = 180Â° – âˆ A

Question 15.

In the figure, AE bisects âˆ CAD and âˆ B = âˆ C. Prove that AE || BC.

Solution:

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

âˆ B = âˆ C

To prove : AE || BC

Proof: In âˆ†ABC, BA is produced

âˆ´ Ext. âˆ CAD = âˆ B + âˆ C

â‡’ 2âˆ EAC = âˆ C + âˆ C (âˆµ AE is the bisector of âˆ CAE) (âˆµ âˆ B = âˆ C)

â‡’ 2âˆ EAC = 2âˆ C

â‡’ âˆ EAC = âˆ C

But there are alternate angles

âˆ´ AE || BC

### RD Sharma Solutions Class 9 Chapter 11 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 180Â°.

Question 3.

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

Solution:

In âˆ†ABC, âˆ B = 60Â°, âˆ C = 80Â°

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

âˆµ âˆ A + âˆ B + âˆ C = 180Â° (Sum of angles of a triangle)

â‡’ âˆ A + 60Â° + 80Â° = 180Â°

â‡’ âˆ A + 140Â° = 180Â°

âˆ´ âˆ A = 180Â°- 140Â° = 40Â°

= 90Â° + – x 40Â° = 90Â° + 20Â° = 110Â°

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 = 180Â°

Ratio in the angles = 2 : 1 : 3

Let first angle = 2x

Second angle = x

and third angle = 3x

âˆ´ 2x + x + 3x = 180Â° â‡’ 6x = 180Â°

âˆ´ x = \(\frac { { 180 }^{ \circ } }{ 6 }\)Â = 30Â°

âˆ´ First angle = 2x = 2 x 30Â° = 60Â°

Second angle = x = 30Â°

and third angle = 3x = 3 x 30Â° = 90Â°

Hence angles are 60Â°, 30Â°, 90Â°

Question 5.

State exterior angle theorem.

Solution:

Given : In âˆ†ABC, side BC is produced to D

To prove : âˆ ACD = âˆ A + âˆ B

Proof: In âˆ†ABC,

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

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

From (i) and (ii)

âˆ ACD + âˆ ACB = âˆ A + âˆ B + âˆ ACB

âˆ ACD = âˆ A + âˆ 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 âˆ†ABC,

âˆ A + âˆ C = âˆ B

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

âˆ´ âˆ B + âˆ A + âˆ C = 180Â°

â‡’ âˆ B + âˆ B = 180Â°

â‡’ 2âˆ B = 180Â°

â‡’ âˆ B = \(\frac { { 180 }^{ \circ } }{ 2 }\)Â = 90Â°

âˆ´ Third angle = 90Â°

Question 7.

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

Solution:

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

âˆµ EF || BC

âˆ´ âˆ A = âˆ ACH (Alternate angle)

âˆ´ âˆ ACH = 65Â°

âˆµâˆ GHC = âˆ DHF

(Vertically opposite angles)

âˆ´ âˆ GHC = 35Â°

Now in âˆ†GCH,

Ext. âˆ AGH = âˆ GCH + âˆ GHC

= 65Â° + 35Â° = 100Â°

Question 8.

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

Solution:

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

âˆ FGH = 125Â° and âˆ B = 55Â°

âˆ FGH + FGE = 180Â° (Linear pair)

â‡’ 125Â° + y – 180Â°

â‡’ y= 180Â°- 125Â° = 55Â°

âˆµ BA || FD and BD || FG

âˆ B = âˆ F = 55Â°

Now in âˆ†EFG,

âˆ F + âˆ FEG + âˆ FGE = 180Â°

(Angles of a triangle)

â‡’ 55Â° + x + 55Â° = 180Â°

â‡’ x+ 110Â°= 180Â°

âˆ´ x= 180Â°- 110Â° = 70Â°

Hence x = 70, y = 55Â°

Question 9.

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

Solution:

In âˆ†ABC,

âˆ A + âˆ B + âˆ C= 180Â° …(i)

and B – A = C – B

â‡’ B + B = A + C â‡’ 2B = A + C

From (i),

B + 2B = 180Â° â‡’ 3B = 180Â°

âˆ B = \(\frac { { 180 }^{ \circ } }{ 3 }\) = 60Â°

Hence âˆ B = 60Â°

Question 10.

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

Solution:

In âˆ†ABC, bisectors of âˆ B and âˆ C intersect at O and âˆ BOC = 120Â°

But âˆ BOC = 90Â°+ \(\frac { 1 }{ 2 }\)

90Â°+ \(\frac { 1 }{ 2 }\) âˆ A= 120Â°

â‡’ \(\frac { 1 }{ 2 }\) âˆ A= 120Â°-90Â° = 30Â°

âˆ´ âˆ A = 2 x 30Â° = 60Â°

Question 11.

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

Solution:

In âˆ†ABC, side BC is produced on both sides forming exterior âˆ ABE and âˆ ACD

Ext. âˆ ABE = âˆ A + âˆ ACB

and Ext. âˆ ACD = âˆ ABC + âˆ A

Adding we get,

âˆ ABE + âˆ ACD = âˆ A + âˆ ACB + âˆ A + âˆ ABC

â‡’ âˆ ABE + âˆ ACD – âˆ A = âˆ A 4- âˆ ACB + âˆ A + âˆ ABC – âˆ A (Subtracting âˆ A from both sides)

= âˆ A + âˆ ABC + âˆ ACB = âˆ A + âˆ B + âˆ C = 180Â° (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 âˆ ACD: âˆ ADC.

Solution:

In âˆ†ABC, AB = AC

AB is produced to D such that BD = BC

DC are joined

In âˆ†ABC, AB = AC

âˆ´ âˆ ABC = âˆ ACB

In âˆ† BCD, BD = BC

âˆ´ âˆ BDC = âˆ BCD

and Ext. âˆ ABC = âˆ BDC + âˆ BCD = 2âˆ BDC (âˆµ âˆ BDC = âˆ BCD)

â‡’ âˆ ACB = 2âˆ BCD (âˆµ âˆ ABC = âˆ ACB)

Adding âˆ BDC to both sides

â‡’ âˆ ACB + âˆ BDC = 2âˆ BDC + âˆ BDC

â‡’ âˆ ACB + âˆ BCD = 3 âˆ BDC (âˆµ âˆ BDC = âˆ BCD)

â‡’ âˆ ACB = 3âˆ BDC

Question 13.

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

Solution:

In the figure,

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

âˆ BAC = 68Â°

In âˆ†ABC,

Ext. âˆ ACD = âˆ A + âˆ B

â‡’ \(\frac { 1 }{ 2 }\) âˆ ACD = \(\frac { 1 }{ 2 }\) âˆ A + \(\frac { 1 }{ 2 }\) âˆ B

â‡’ âˆ 2= \(\frac { 1 }{ 2 }\) âˆ A + âˆ 1 …(i)

But in âˆ†BCE,

Ext. âˆ 2 = âˆ E + âˆ l

â‡’ âˆ E + âˆ l = âˆ 2 = \(\frac { 1 }{ 2 }\) âˆ A + âˆ l [From (i)]

â‡’ âˆ E = \(\frac { 1 }{ 2 }\) âˆ A = \(\frac { { 68 }^{ \circ } }{ 2 }\)Â =34Â°

### Class 9 RD Sharma Solutions Chapter 11 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) 90Â°

(b) 45Â°

(c) 60Â°

(d) 30Â°

Solution:

âˆµ Sum of three angles of a triangle = 180Â°

âˆ´ Each angle = \(\frac { { 180 }^{ \circ } }{ 3 }\)Â = 60Â° (c)

Question 2.

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

(a) 30Â°

(b) 45Â°

(c) 60Â°

(d) 90Â°

Solution:

In a right triangle, one angle = 90Â°

âˆ´ Sum of other two acute angles = 180Â° – 90Â° = 90Â°

âˆµ Both angles are equal

âˆ´ Each angle will be = \(\frac { { 90 }^{ \circ } }{ 2 }\)Â = 45Â° (b)

Question 3.

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

(a) 75Â°

(b) 80Â°

(c) 40Â°

(d) 50Â°

Solution:

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

âˆ´ Sum of interior opposite angles = 100Â°

âˆµ Both angles are equal

âˆ´ Each angle will be = \(\frac { { 100 }^{ \circ } }{ 2 }\)Â = 50Â° (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 âˆ A, âˆ B, âˆ C be the angles of a âˆ†ABC and let âˆ A = âˆ B + âˆ C

But âˆ A + âˆ B + âˆ C = 180Â°

( Sum of angles of a triangle)

âˆ´ âˆ A + âˆ A = 180Â° â‡’ 2âˆ A = 180Â°

â‡’ âˆ A = \(\frac { { 180 }^{ \circ } }{ 2 }\)Â = 90Â°

âˆ´ âˆ† is a right triangle (d)

Question 5.

Side BC of a triangle ABC has been produced to a point D such that âˆ ACD = 120Â°. If âˆ B = \(\frac { 1 }{ 2 }\)âˆ A, then âˆ A is equal to

(a) 80Â°

(b) 75Â°

(c) 60Â°

(d) 90Â°

Solution:

Side BC of âˆ†ABC is produced to D, then

Ext. âˆ ACB = âˆ A + âˆ B

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

Question 6.

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

(a) 35Â°

(b) 90Â°

(c) 70Â°

(d) 55Â°

Solution:

In âˆ†ABC, âˆ B = âˆ C

AX is the bisector of ext. âˆ CAD

âˆ DAX = 70Â°

âˆ´ âˆ DAC = 70Â° x 2 = 140Â°

But Ext. âˆ DAC = âˆ B + âˆ C

= âˆ C + âˆ C (âˆµ âˆ B = âˆ C)

= 2âˆ C

âˆ´ 2âˆ C = 140Â° â‡’ âˆ C = \(\frac { { 140 }^{ \circ } }{ 2 }\) = 70Â°

âˆ´ âˆ ACB = 70Â° (c)

Question 7.

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

(a) 55Â°

(b) 85Â°

(c) 40Â°

(d) 9.0Â°

Solution:

In âˆ†ABC, BA is produced to D such that âˆ CAD = 95Â°

and let âˆ C = 55Â° and âˆ B = xÂ°

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

âˆ´ âˆ CAD = âˆ B + âˆ C â‡’ 95Â° = x + 55Â°

â‡’ x = 95Â° – 55Â° = 40Â°

âˆ´ Other interior angle = 40Â° (c)

Question 8.

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

(a) 90Â°

(b) 180Â°

(c) 270Â°

(d) 360Â°

Solution:

In âˆ†ABC, sides AB, BC and CA are produced in order, then exterior âˆ FAB, âˆ DBC and âˆ ACE are formed

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

âˆ´ âˆ FAB = âˆ B + âˆ C

âˆ DBC = âˆ C + âˆ A and

âˆ ACE = âˆ A + âˆ B Adding we get,

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

= 2(âˆ A + âˆ B + âˆ C)

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

= 360Â° (d)

Question 9.

In âˆ†ABC, if âˆ A = 100Â°, AD bisects âˆ A and ADâŠ¥ BC. Then, âˆ B =

(a) 50Â°

(b) 90Â°

(c) 40Â°

(d) 100Â°

Solution:

In âˆ†ABC, âˆ A = 100Â°

AD is bisector of âˆ A and AD âŠ¥ BC

Now, âˆ BAD = \(\frac { { 100 }^{ \circ } }{ 2 }\) = 50Â°

In âˆ†ABD,

âˆ BAD + âˆ B + âˆ D= 180Â°

(Sum of angles of a triangle)

â‡’ âˆ 50Â° + âˆ B + 90Â° = 180Â°

âˆ B + 140Â° = 180Â°

â‡’ âˆ B = 180Â° – 140Â° âˆ B = 40Â° (c)

Question 10.

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

(a) 48Â°, 60Â°, 72Â°

(b) 50Â°, 60Â°, 70Â°

(c) 52Â°, 56Â°, 72Â°

(d) 42Â°, 60Â°, 76Â°

Solution:

In âˆ†ABC, BC is produced to D and âˆ ACD = 108Â°

Ratio in âˆ A : âˆ B = 4:5

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

âˆ´ âˆ ACD = âˆ A + âˆ B = 108Â°

Ratio in âˆ A : âˆ B = 4:5

Question 11.

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

(a) 60Â°

(b) 120Â°

(c) 150Â°

(d) 30Â°

Solution:

In âˆ†ABC, âˆ A = 60Â°, âˆ B = 80Â°

âˆ´ âˆ C = 180Â° – (âˆ A + âˆ B)

= 180Â° – (60Â° + 80Â°)

= 180Â° – 140Â° = 40Â°

Bisectors of âˆ B and âˆ C meet at O

Question 12.

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

(a) 100Â°

(b) 80Â°

(c) 90Â°

(d) 135Â°

Solution:

In the figure,

AB and CD intersect at O

and AC || DB, âˆ CAB = 45Â°

and âˆ CDB = 55Â°

âˆµ AC || DB

âˆ´ âˆ CAB = âˆ ABD (Alternate angles)

In âˆ†OBD,

âˆ BOD = 180Â° – (âˆ CDB + âˆ ABD)

= 180Â° – (55Â° + 45Â°)

= 180Â° – 100Â° = 80Â° (b)

Question 13.

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

(a) 20Â°

(b) 50Â°

(c) 60Â°

(d) 70Â°

Solution:

In the figure, EC || AB

âˆ ECD = 70Â°, âˆ BDO = 20Â°

âˆµ EC || AB

âˆ AOD = âˆ ECD (Corresponding angles)

â‡’ âˆ AOD = 70Â°

In âˆ†OBD,

Ext. âˆ AOD = âˆ OBD + âˆ BDO

70Â° = âˆ OBD + 20Â°

â‡’ âˆ OBD = 70Â° – 20Â° = 50Â° (b)

Question 14.

In the figure, x + y =

(a) 270

(b) 230

(c) 210

(d) 190Â°

Solution:

In the figure

Ext. âˆ OAE = âˆ AOC + âˆ ACO

â‡’ x = 40Â° + 80Â° = 120Â°

Similarly,

Ext. âˆ DBF = âˆ ODB + âˆ DOB

y = 70Â° + âˆ DOB

[(âˆµ âˆ AOC = âˆ DOB) (vertically opp. angles)]

= 70Â° + 40Â° = 110Â°

âˆ´ x+y= 120Â°+ 110Â° = 230Â° (b)

Question 15.

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

(a) 25Â°

(b) 30Â°

(c) 45Â°

(d) 60Â°

Solution:

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

Sum of angles of a triangle = 180Â°

Let angles be 3x, 4x, 5x

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

âˆ´ Smallest angle = \(\frac { 180 x 3x }{ 12x }\) = 45Â° (c)

Question 16.

In the figure, if AB âŠ¥ BC, then x =

(a) 18

(b) 22

(c) 25

(d) 32

Solution:

In the figure, AB âŠ¥ BC

âˆ AGF = 32Â°

âˆ´ âˆ CGB = âˆ AGF (Vertically opposite angles)

= 32Â°

In âˆ†GCB, âˆ B = 90Â°

âˆ´ âˆ CGB + âˆ GCB = 90Â°

â‡’ 32Â° + âˆ GCB = 90Â°

â‡’ âˆ GCB = 90Â° – 32Â° = 58Â°

Now in âˆ†GDC,

Ext. âˆ GCB = âˆ CDG + âˆ DGC

â‡’ 58Â° = x + 14Â° + x

â‡’ 2x + 14Â° = 58Â°

â‡’ 2x = 58 – 14Â° = 44

â‡’ x = \(\frac { 44 }{ 2 }\) = 22Â°

âˆ´ x = 22Â° (b)

Question 17.

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

(a) x + y + 180

(b) x + y – 180

(c) 180Â° -(x+y)

(d) x+y + 360Â°

Solution:

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

In âˆ†ABC,

âˆ B = 180Â° -y

and âˆ C 180Â° – x

âˆ´ z = âˆ A = 180Â° – (B + C)

= 180Â° – (180 – y + 180 -x)

= 180Â° – (360Â° – x – y)

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

Question 18.

In the figure, for which value of x is l_{1} || l_{2}?

(a) 37

(b) 43

(c) 45

(d) 47

Solution:

In the figure, l_{1} || l_{2}

âˆ´ âˆ EBA = âˆ BAH (Alternate angles)

âˆ´ âˆ BAH = 78Â°

â‡’ âˆ BAC + âˆ CAH = 78Â°

â‡’ âˆ BAC + 35Â° = 78Â°

â‡’ âˆ BAC = 78Â° – 35Â° = 43Â°

In âˆ†ABC, âˆ C = 90Â°

âˆ´ âˆ ABC + âˆ BAC = 90Â°

â‡’ x + 43Â° = 90Â° â‡’ x = 90Â° – 43Â°

âˆ´ x = 47Â° (d)

Question 19.

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

Solution:

In âˆ†ABC,

âˆ ACB = 180Â° – (x + 2x)

= 180Â° – 3x …(i)

and in âˆ†BDG,

âˆ BED = 180Â° – (2x + y) …(ii)

âˆ EGC = âˆ AGD (Vertically opposite angles)

= 3y

In quad. BCGE,

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

â‡’ 2x+ 180Â° – 3x + 3y + 180Â°- 2x-y = 360Â°

â‡’ -3x + 2y = 0

â‡’ 3x = 2y â‡’ y = \(\frac { 3 }{ 2 }\)x (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

âˆ´ âˆ CBA + âˆ CBD = 180Â° (Linear pair)

â‡’ 7y + 5y = 180Â°

â‡’ 12y = 180Â°

â‡’ y = \(\frac { 180 }{ 12 }\) = 15

and Ext. âˆ CBD = âˆ A + âˆ C

â‡’ 7y = 3y + x

â‡’ 7y -3y = x

â‡’ 4y = x

âˆ´ x = 4 x 15 = 60 (d)

Question 21.

In the figure, the value of x is

(a) 65Â°

(b) 80Â°

(c) 95Â°

(d) 120Â°

Solution:

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

Now in âˆ†ABD,

Ext. âˆ DBC = âˆ A + âˆ D

= 55Â° + 25Â° = 80Â°

Similarly, in âˆ†BCE,

Ext. âˆ DEC = âˆ EBC + âˆ ECB

= 80Â° + 40Â° = 120Â° (d)

Question 22.

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

(a) 20Â°

(b) 25Â°

(c) 30Â°

(d) 35Â°

Solution:

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

âˆµ BP || CQ

âˆ´ âˆ PBC – âˆ QCD

â‡’ 20Â° + âˆ ABC = 70Â°

â‡’ âˆ ABC = 70Â° – 20Â° = 50Â°

âˆµ BC = AC

âˆ´ âˆ ACB = âˆ ABC (Angles opposite to equal sides)

= 50Â°

Now in âˆ†ABC,

Ext. âˆ ACD = âˆ B + âˆ A

â‡’ x + 70Â° = 50Â° + 50Â°

â‡’ x + 70Â° = 100Â°

âˆ´ x = 100Â° – 70Â° = 30Â° (c)

Question 23.

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

(a) x = 55Â°, y = 40Â°

(b) x = 50Â°, y = 45Â°

(c) x = 60Â°, y = 35Â°

(d) x = 35Â°, y = 60Â°

Solution:

In the figure,

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

âˆ APR = 25Â°, âˆ RQC = 30Â°, âˆ CQF = 65Â°

âˆµ AB || CD

âˆ´ âˆ APQ = âˆ CQF (Corresponding anlges)

â‡’ y + 25Â° = 65Â°

â‡’ y = 65Â° – 25Â° = 40Â°

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

y + 25Â° + âˆ 1 +30Â°= 180Â°

40Â° + 25Â° + âˆ 1 + 30Â° = 180Â°

â‡’ âˆ 1 + 95Â° = 180Â°

âˆ´ âˆ 1 = 180Â° – 95Â° = 85Â°

Now, âˆ†PQR,

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

â‡’ 40Â° + x + 85Â° = 180Â°

â‡’ 125Â° + x = 180Â°

â‡’ x = 180Â° – 125Â° = 55Â°

âˆ´ x = 55Â°, y = 40Â° (a)

Question 24.

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

(a) 94Â°

(b) 54Â°

(c) 40Â°

(d) 44Â°

Solution:

In âˆ†ABC, base BC is produced both ways and âˆ ACD = 94Â°, âˆ ABE = 126Â°

Ext. âˆ ACD = âˆ BAC + âˆ ABC

â‡’ 94Â° = âˆ BAC + âˆ ABC

Similarly, âˆ ABE = âˆ BAC + âˆ ACB

â‡’ 126Â° = âˆ BAC + âˆ ACB

Adding,

94Â° + 126Â° = âˆ BAC + âˆ ABC + âˆ ACB + âˆ BAC

220Â° = 180Â° + âˆ BAC

âˆ´ âˆ BAC = 220Â° -180Â° = 40Â° (c)

Question 25.

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

(a) 45Â°

(b) 95Â°

(c) 135Â°

(d) 90Â°

Solution:

In right âˆ†ABC, âˆ A = 90Â°

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

âˆ BOC = 90Â° + \(\frac { 1 }{ 2 }\) âˆ A

= 90Â°+ \(\frac { 1 }{ 2 }\) x 90Â° = 90Â° + 45Â°= 135Â° (c)

Question 26.

The bisects of exterior angles at B and C of âˆ†ABC, meet at O. If âˆ A = .xÂ°, then âˆ BOC=

Solution:

In âˆ†ABC, âˆ A = xÂ°

and bisectors of âˆ B and âˆ C meet at O.

Question 27.

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

(a) 25Â°

(b) 50Â°

(c) 100Â°

(d) 75Â°

Solution:

In âˆ†ABC, âˆ A = 50Â°

BC is produced

Bisectors of âˆ ABC and âˆ ACD meet at âˆ E

âˆ´ âˆ E = \(\frac { 1 }{ 2 }\) âˆ A = \(\frac { 1 }{ 2 }\) x 50Â° = 25Â° (a)

Question 28.

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

(a) 85Â°

(b) 72\(\frac { 1 }{ 2 }\) Â°

(c) 145Â°

(d) none of these

Solution:

In âˆ†ABC, BC is produced to D

âˆ B = 30Â°, âˆ ACD = 115Â°

Question 29.

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

(a) 22 \(\frac { 1 }{ 2 }\)

(b) 30

(c) 45

(d) 60

Solution:

In the figure, l_{1} || l_{2}

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

âˆµ EC and EB are the bisectors of âˆ DCB and âˆ CBA respectively

âˆ´ âˆ CEB = 90Â°

âˆ´ a + b = 90Â° ,

and âˆ GEB = 90Â° (âˆµ âˆ CEB = 90Â°)

2x = 90Â° â‡’ x = \(\frac { 90 }{ 2 }\) = 45 (c)

Question 30.

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

(a) 40Â°

(b) 90Â°

(c) 80Â°

(d) 100Â°

Solution:

NCERT Solutions for maths in Video https://www.youtube.com/user/cbsepapers/videos