## RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS

Other Exercises

- RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1
- RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.2
- RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry VSAQS
- RD Sharma Class 9 Solutions Chapter 11 Co-ordinate 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 = \(\frac { { 180 }^{ \circ } }{ 3 }\) = 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 = \(\frac { { 90 }^{ \circ } }{ 2 }\) = 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 = \(\frac { { 100 }^{ \circ } }{ 2 }\) = 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 = \(\frac { { 180 }^{ \circ } }{ 2 }\) = 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 = \(\frac { 1 }{ 2 }\)∠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 = \(\frac { { 140 }^{ \circ } }{ 2 }\) = 70°

∴ ∠ACB = 70° (c)

Question 7.

In a triangle, an exterior angle at a vertex is 95° and its one of the interior opposite angle is 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 angles so 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 ∠ACE are 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 = \(\frac { { 100 }^{ \circ } }{ 2 }\) = 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 + ∠DOB

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 of the 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 = \(\frac { 180 x 3x }{ 12x }\) = 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 = \(\frac { 44 }{ 2 }\) = 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 l_{1} || l_{2}?

(a) 37

(b) 43

(c) 45

(d) 47

Solution:

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

∴ ∠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)

= 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 = \(\frac { 3 }{ 2 }\)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

∴ ∠CBA + ∠CBD = 180° (Linear pair)

⇒ 7y + 5y = 180°

⇒ 12y = 180°

⇒ y = \(\frac { 180 }{ 12 }\) = 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 + ∠D

= 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 Q respectively. 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 angles formed 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 O between 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° + \(\frac { 1 }{ 2 }\) ∠A

= 90°+ \(\frac { 1 }{ 2 }\) 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 = \(\frac { 1 }{ 2 }\) ∠A = \(\frac { 1 }{ 2 }\) 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) 72\(\frac { 1 }{ 2 }\) °

(c) 145°

(d) none of these

Solution:

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

(b) 30

(c) 45

(d) 60

Solution:

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

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 = \(\frac { 90 }{ 2 }\) = 45 (c)

Question 30.

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

(a) 40°

(b) 90°

(c) 80°

(d) 100°

Solution:

Hope given RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS are helpful to complete your math homework.

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