Class 9 Exam > Class 9 Notes > Mathematics (Maths) Class 9 > RS Aggarwal MCQs: Triangles
RS Aggarwal MCQs: Triangles | Mathematics (Maths) Class 9 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Q u e s t i o n : 3 0
In ?ABC, if 3 ?A = 4 ?B = 6 ?C then A : B : C = ?
a
3 : 4 : 6
b
4 : 3 : 2
c
Page 2
Q u e s t i o n : 3 0
In ?ABC, if 3 ?A = 4 ?B = 6 ?C then A : B : C = ?
a
3 : 4 : 6
b
4 : 3 : 2
c
2 : 3 : 4
d
6 : 4 : 3
S o l u t i o n :
LCM of 3, 4 and 6 = 12
3 ?A = 4 ?B = 6 ?C Given
Dividing throughout by 12, we get
3 ?A
12
=
4 ?B
12
=
6 ?C
12
?
?A
4
=
?B
3
=
?C
2
Let
?A
4
=
?B
3
=
?C
2
= k
, where k is some constant
Then, ?A = 4k, ?B = 3k, ?C = 2k
? ?A : ?B : ?C = 4k : 3k : 2k = 4 : 3 : 2
Hence, the correct answer is option b
.
Q u e s t i o n : 3 1
In a ?ABC, if ?A - ?B = 42° and ?B - ?C = 21° then ?B = ?
a 32°
b 63°
c 53°
d 95°
Figure
S o l u t i o n :
c 53°
Let
?A - ?B = 42° . . . (i) and ?B - ?C = 21° . . . (ii)Adding (i) and (ii), we get: ?A - ?C = 63° ?B = ?A -42° [Using (i)] ?C = ?A -63° [Using (iii)]
? ?A + ?B + ?C = 180° [Sum of the angles of a triangle] ? ?A + ?A -42°+ ?A -63° = 180° ? 3 ?A -105° = 180° ? 3 ?A = 285° ? ?A = 95° ? ?B = (95 -42)° ? ?B = 53°
Q u e s t i o n : 3 2
In ?ABC, side BC is produced to D. If ?ABC = 50° and ?ACD = 110° then ?A = ?
a
160°
b
60°
c
80°
d
30°
S o l u t i o n :
? ?A + ?B = ?ACD ? ?A +50° = 110° ? ?A = 60°
Hence, the correct answer is option b
.
Q u e s t i o n : 3 3
Side BC of ?ABC has been produced to D on left and to E on right hand side of BC such that ?ABD = 125° and ?ACE = 130°. Then ?A = ?
a
50°
b
55°
c
65°
d
75°
S o l u t i o n :
d 75°
We have :
? ?ABD + ?ABC = 180° [ ? DE is a straight line] ? 125°+ ?ABC = 180° ? ?ABC = 55°
Also,
Page 3
Q u e s t i o n : 3 0
In ?ABC, if 3 ?A = 4 ?B = 6 ?C then A : B : C = ?
a
3 : 4 : 6
b
4 : 3 : 2
c
2 : 3 : 4
d
6 : 4 : 3
S o l u t i o n :
LCM of 3, 4 and 6 = 12
3 ?A = 4 ?B = 6 ?C Given
Dividing throughout by 12, we get
3 ?A
12
=
4 ?B
12
=
6 ?C
12
?
?A
4
=
?B
3
=
?C
2
Let
?A
4
=
?B
3
=
?C
2
= k
, where k is some constant
Then, ?A = 4k, ?B = 3k, ?C = 2k
? ?A : ?B : ?C = 4k : 3k : 2k = 4 : 3 : 2
Hence, the correct answer is option b
.
Q u e s t i o n : 3 1
In a ?ABC, if ?A - ?B = 42° and ?B - ?C = 21° then ?B = ?
a 32°
b 63°
c 53°
d 95°
Figure
S o l u t i o n :
c 53°
Let
?A - ?B = 42° . . . (i) and ?B - ?C = 21° . . . (ii)Adding (i) and (ii), we get: ?A - ?C = 63° ?B = ?A -42° [Using (i)] ?C = ?A -63° [Using (iii)]
? ?A + ?B + ?C = 180° [Sum of the angles of a triangle] ? ?A + ?A -42°+ ?A -63° = 180° ? 3 ?A -105° = 180° ? 3 ?A = 285° ? ?A = 95° ? ?B = (95 -42)° ? ?B = 53°
Q u e s t i o n : 3 2
In ?ABC, side BC is produced to D. If ?ABC = 50° and ?ACD = 110° then ?A = ?
a
160°
b
60°
c
80°
d
30°
S o l u t i o n :
? ?A + ?B = ?ACD ? ?A +50° = 110° ? ?A = 60°
Hence, the correct answer is option b
.
Q u e s t i o n : 3 3
Side BC of ?ABC has been produced to D on left and to E on right hand side of BC such that ?ABD = 125° and ?ACE = 130°. Then ?A = ?
a
50°
b
55°
c
65°
d
75°
S o l u t i o n :
d 75°
We have :
? ?ABD + ?ABC = 180° [ ? DE is a straight line] ? 125°+ ?ABC = 180° ? ?ABC = 55°
Also,
?ACE + ?ACB = 180° ? 130°+ ?ACB = 180° ? ?ACB = 50° ? ?BAC + ?ABC + ?ACB = 180° [Sum of the angles of a triangle] ? ?BAC +55°+50° = 180° ? ?BAC = 75° ? ?A = 75°
Q u e s t i o n : 3 4
In the given figure, the sides CB and BA of ?ABC have been produced to D and E, respectively, such that ?ABD = 110° and ?CAE = 135°. Then ?ACB = ?
a
65°
b
45°
c
55°
d
35°
S o l u t i o n :
a 65°
We have :
?ABD + ?ABC = 180° [ ? CBD is a straight line] ? 100°+ ?ABC = 180° ? ?ABC = 70°
Side AB of triangle ABC is produced to E.
? ?CAE = ?ABC + ?ACB ? 135° = 70°+ ?ACB ? ?ACB = 65°
Q u e s t i o n : 3 5
The sides BC, CA and AB of ?ABC have been produced to D, E and F, respectively. ?BAE + ?CBF + ?ACD = ?
a 240°
b 300°
c 320°
d 360°
S o l u t i o n :
d 360°
We have :
?1 + ?BAE = 180° . . . (i) ?2 + ?CBF = 180° . . . (ii) and ?3 + ?ACD = 180° . . . (iii)
Adding (i), (ii) and (iii), we get:( ?1 + ?2 + ?3)+( ?BAE + ?CBF+ ?ACD) = 540°
? 180°+ ?BAE + ?CBF+ ?ACD = 540° [ ? ?1 + ?2 + ?3 = 180°] ? ?BAE + ?CBF+ ?ACD = 360°
Q u e s t i o n : 3 6
The the given figure, EAD ? BCD. Ray FAC cuts ray EAD at a point A such that ?EAF = 30°. Also, in ?BAC, ?BAC = x° and ?ABC = (x + 10)°. Then, the value of x is
a 20
b 25
c 30
d 35
S o l u t i o n :
In the given figure, ?CAD = ?EAF Verticallyoppositeangles
? ?CAD = 30°
In ?ABD,
?ABD + ?BAD + ?ADB = 180° Anglesumproperty
? (x + 10)° + (x° + 30°) + 90° = 180°
? 2x° + 130° = 180°
? 2x° = 180° - 130° = 50°
? x = 25
Thus, the value of x is 25.
Page 4
Q u e s t i o n : 3 0
In ?ABC, if 3 ?A = 4 ?B = 6 ?C then A : B : C = ?
a
3 : 4 : 6
b
4 : 3 : 2
c
2 : 3 : 4
d
6 : 4 : 3
S o l u t i o n :
LCM of 3, 4 and 6 = 12
3 ?A = 4 ?B = 6 ?C Given
Dividing throughout by 12, we get
3 ?A
12
=
4 ?B
12
=
6 ?C
12
?
?A
4
=
?B
3
=
?C
2
Let
?A
4
=
?B
3
=
?C
2
= k
, where k is some constant
Then, ?A = 4k, ?B = 3k, ?C = 2k
? ?A : ?B : ?C = 4k : 3k : 2k = 4 : 3 : 2
Hence, the correct answer is option b
.
Q u e s t i o n : 3 1
In a ?ABC, if ?A - ?B = 42° and ?B - ?C = 21° then ?B = ?
a 32°
b 63°
c 53°
d 95°
Figure
S o l u t i o n :
c 53°
Let
?A - ?B = 42° . . . (i) and ?B - ?C = 21° . . . (ii)Adding (i) and (ii), we get: ?A - ?C = 63° ?B = ?A -42° [Using (i)] ?C = ?A -63° [Using (iii)]
? ?A + ?B + ?C = 180° [Sum of the angles of a triangle] ? ?A + ?A -42°+ ?A -63° = 180° ? 3 ?A -105° = 180° ? 3 ?A = 285° ? ?A = 95° ? ?B = (95 -42)° ? ?B = 53°
Q u e s t i o n : 3 2
In ?ABC, side BC is produced to D. If ?ABC = 50° and ?ACD = 110° then ?A = ?
a
160°
b
60°
c
80°
d
30°
S o l u t i o n :
? ?A + ?B = ?ACD ? ?A +50° = 110° ? ?A = 60°
Hence, the correct answer is option b
.
Q u e s t i o n : 3 3
Side BC of ?ABC has been produced to D on left and to E on right hand side of BC such that ?ABD = 125° and ?ACE = 130°. Then ?A = ?
a
50°
b
55°
c
65°
d
75°
S o l u t i o n :
d 75°
We have :
? ?ABD + ?ABC = 180° [ ? DE is a straight line] ? 125°+ ?ABC = 180° ? ?ABC = 55°
Also,
?ACE + ?ACB = 180° ? 130°+ ?ACB = 180° ? ?ACB = 50° ? ?BAC + ?ABC + ?ACB = 180° [Sum of the angles of a triangle] ? ?BAC +55°+50° = 180° ? ?BAC = 75° ? ?A = 75°
Q u e s t i o n : 3 4
In the given figure, the sides CB and BA of ?ABC have been produced to D and E, respectively, such that ?ABD = 110° and ?CAE = 135°. Then ?ACB = ?
a
65°
b
45°
c
55°
d
35°
S o l u t i o n :
a 65°
We have :
?ABD + ?ABC = 180° [ ? CBD is a straight line] ? 100°+ ?ABC = 180° ? ?ABC = 70°
Side AB of triangle ABC is produced to E.
? ?CAE = ?ABC + ?ACB ? 135° = 70°+ ?ACB ? ?ACB = 65°
Q u e s t i o n : 3 5
The sides BC, CA and AB of ?ABC have been produced to D, E and F, respectively. ?BAE + ?CBF + ?ACD = ?
a 240°
b 300°
c 320°
d 360°
S o l u t i o n :
d 360°
We have :
?1 + ?BAE = 180° . . . (i) ?2 + ?CBF = 180° . . . (ii) and ?3 + ?ACD = 180° . . . (iii)
Adding (i), (ii) and (iii), we get:( ?1 + ?2 + ?3)+( ?BAE + ?CBF+ ?ACD) = 540°
? 180°+ ?BAE + ?CBF+ ?ACD = 540° [ ? ?1 + ?2 + ?3 = 180°] ? ?BAE + ?CBF+ ?ACD = 360°
Q u e s t i o n : 3 6
The the given figure, EAD ? BCD. Ray FAC cuts ray EAD at a point A such that ?EAF = 30°. Also, in ?BAC, ?BAC = x° and ?ABC = (x + 10)°. Then, the value of x is
a 20
b 25
c 30
d 35
S o l u t i o n :
In the given figure, ?CAD = ?EAF Verticallyoppositeangles
? ?CAD = 30°
In ?ABD,
?ABD + ?BAD + ?ADB = 180° Anglesumproperty
? (x + 10)° + (x° + 30°) + 90° = 180°
? 2x° + 130° = 180°
? 2x° = 180° - 130° = 50°
? x = 25
Thus, the value of x is 25.
Hence, the correct answer is option b
.
Q u e s t i o n : 3 7
In the given figure, two rays BD and CE intersect at a point A. The side BC of ?ABC have been produced on both sides to points F and G respectively. If ?ABF = x°, ?ACG = y° and
?DAE = z° then z = ?
a x + y – 180
b x + y + 180
c 180 – (x + y)
d x + y + 360°
S o l u t i o n :
In the given figure, ?ABF + ?ABC = 180° Linearpairofangles
? x° + ?ABC = 180°
? ?ABC = 180° - x° .....1
Also, ?ACG + ?ACB = 180° Linearpairofangles
? y° + ?ACB = 180°
? ?ACB = 180° - y° .....2
Also, ?BAC = ?DAE = z° .....3
Verticallyoppositeangles
In ?ABC,
?BAC + ?ABC + ?ACB = 180° Anglesumproperty
? z° + 180° - x° + 180° - y° = 180°
Using(1), (2)and(3)
? z = x + y - 180
Hence, the correct answer is option a
.
Q u e s t i o n : 3 8
In the given figure, lines AB and CD intersect at a point O. The sides CA and OB have been produced to E and F respectively such that ?OAE = x° and ?DBF = y°.
If ?OCA = 80°, ?COA = 40° and ?BDO = 70° then x° + y° = ?
a 190°
b 230°
c 210°
d 270°
S o l u t i o n :
In the given figure, ?BOD = ?COA Verticallyoppositeangles
? ?BOD = 40° .....1
In ?ACO,
?OAE = ?OCA + ?COA Exteriorangleofatriangleisequaltothesumoftwooppositeinteriorangles
? x° = 80° + 40° = 120° .....2
In ?BDO,
Page 5
Q u e s t i o n : 3 0
In ?ABC, if 3 ?A = 4 ?B = 6 ?C then A : B : C = ?
a
3 : 4 : 6
b
4 : 3 : 2
c
2 : 3 : 4
d
6 : 4 : 3
S o l u t i o n :
LCM of 3, 4 and 6 = 12
3 ?A = 4 ?B = 6 ?C Given
Dividing throughout by 12, we get
3 ?A
12
=
4 ?B
12
=
6 ?C
12
?
?A
4
=
?B
3
=
?C
2
Let
?A
4
=
?B
3
=
?C
2
= k
, where k is some constant
Then, ?A = 4k, ?B = 3k, ?C = 2k
? ?A : ?B : ?C = 4k : 3k : 2k = 4 : 3 : 2
Hence, the correct answer is option b
.
Q u e s t i o n : 3 1
In a ?ABC, if ?A - ?B = 42° and ?B - ?C = 21° then ?B = ?
a 32°
b 63°
c 53°
d 95°
Figure
S o l u t i o n :
c 53°
Let
?A - ?B = 42° . . . (i) and ?B - ?C = 21° . . . (ii)Adding (i) and (ii), we get: ?A - ?C = 63° ?B = ?A -42° [Using (i)] ?C = ?A -63° [Using (iii)]
? ?A + ?B + ?C = 180° [Sum of the angles of a triangle] ? ?A + ?A -42°+ ?A -63° = 180° ? 3 ?A -105° = 180° ? 3 ?A = 285° ? ?A = 95° ? ?B = (95 -42)° ? ?B = 53°
Q u e s t i o n : 3 2
In ?ABC, side BC is produced to D. If ?ABC = 50° and ?ACD = 110° then ?A = ?
a
160°
b
60°
c
80°
d
30°
S o l u t i o n :
? ?A + ?B = ?ACD ? ?A +50° = 110° ? ?A = 60°
Hence, the correct answer is option b
.
Q u e s t i o n : 3 3
Side BC of ?ABC has been produced to D on left and to E on right hand side of BC such that ?ABD = 125° and ?ACE = 130°. Then ?A = ?
a
50°
b
55°
c
65°
d
75°
S o l u t i o n :
d 75°
We have :
? ?ABD + ?ABC = 180° [ ? DE is a straight line] ? 125°+ ?ABC = 180° ? ?ABC = 55°
Also,
?ACE + ?ACB = 180° ? 130°+ ?ACB = 180° ? ?ACB = 50° ? ?BAC + ?ABC + ?ACB = 180° [Sum of the angles of a triangle] ? ?BAC +55°+50° = 180° ? ?BAC = 75° ? ?A = 75°
Q u e s t i o n : 3 4
In the given figure, the sides CB and BA of ?ABC have been produced to D and E, respectively, such that ?ABD = 110° and ?CAE = 135°. Then ?ACB = ?
a
65°
b
45°
c
55°
d
35°
S o l u t i o n :
a 65°
We have :
?ABD + ?ABC = 180° [ ? CBD is a straight line] ? 100°+ ?ABC = 180° ? ?ABC = 70°
Side AB of triangle ABC is produced to E.
? ?CAE = ?ABC + ?ACB ? 135° = 70°+ ?ACB ? ?ACB = 65°
Q u e s t i o n : 3 5
The sides BC, CA and AB of ?ABC have been produced to D, E and F, respectively. ?BAE + ?CBF + ?ACD = ?
a 240°
b 300°
c 320°
d 360°
S o l u t i o n :
d 360°
We have :
?1 + ?BAE = 180° . . . (i) ?2 + ?CBF = 180° . . . (ii) and ?3 + ?ACD = 180° . . . (iii)
Adding (i), (ii) and (iii), we get:( ?1 + ?2 + ?3)+( ?BAE + ?CBF+ ?ACD) = 540°
? 180°+ ?BAE + ?CBF+ ?ACD = 540° [ ? ?1 + ?2 + ?3 = 180°] ? ?BAE + ?CBF+ ?ACD = 360°
Q u e s t i o n : 3 6
The the given figure, EAD ? BCD. Ray FAC cuts ray EAD at a point A such that ?EAF = 30°. Also, in ?BAC, ?BAC = x° and ?ABC = (x + 10)°. Then, the value of x is
a 20
b 25
c 30
d 35
S o l u t i o n :
In the given figure, ?CAD = ?EAF Verticallyoppositeangles
? ?CAD = 30°
In ?ABD,
?ABD + ?BAD + ?ADB = 180° Anglesumproperty
? (x + 10)° + (x° + 30°) + 90° = 180°
? 2x° + 130° = 180°
? 2x° = 180° - 130° = 50°
? x = 25
Thus, the value of x is 25.
Hence, the correct answer is option b
.
Q u e s t i o n : 3 7
In the given figure, two rays BD and CE intersect at a point A. The side BC of ?ABC have been produced on both sides to points F and G respectively. If ?ABF = x°, ?ACG = y° and
?DAE = z° then z = ?
a x + y – 180
b x + y + 180
c 180 – (x + y)
d x + y + 360°
S o l u t i o n :
In the given figure, ?ABF + ?ABC = 180° Linearpairofangles
? x° + ?ABC = 180°
? ?ABC = 180° - x° .....1
Also, ?ACG + ?ACB = 180° Linearpairofangles
? y° + ?ACB = 180°
? ?ACB = 180° - y° .....2
Also, ?BAC = ?DAE = z° .....3
Verticallyoppositeangles
In ?ABC,
?BAC + ?ABC + ?ACB = 180° Anglesumproperty
? z° + 180° - x° + 180° - y° = 180°
Using(1), (2)and(3)
? z = x + y - 180
Hence, the correct answer is option a
.
Q u e s t i o n : 3 8
In the given figure, lines AB and CD intersect at a point O. The sides CA and OB have been produced to E and F respectively such that ?OAE = x° and ?DBF = y°.
If ?OCA = 80°, ?COA = 40° and ?BDO = 70° then x° + y° = ?
a 190°
b 230°
c 210°
d 270°
S o l u t i o n :
In the given figure, ?BOD = ?COA Verticallyoppositeangles
? ?BOD = 40° .....1
In ?ACO,
?OAE = ?OCA + ?COA Exteriorangleofatriangleisequaltothesumoftwooppositeinteriorangles
? x° = 80° + 40° = 120° .....2
In ?BDO,
?DBF = ?BDO + ?BOD Exteriorangleofatriangleisequaltothesumoftwooppositeinteriorangles
? y° = 70° + 40° = 110°
Using(1)
.....3
Adding 2
and 3
, we get
x° + y° = 120° + 110° = 230°
Hence, the correct answer is option b
.
Q u e s t i o n : 3 9
In a ?ABC, it is given that ?A : ?B : ?C = 3 : 2 : 1 and ?ACD = 90°. If BC is produced to E then ?ECD = ?
a 60°
b 50°
c 40°
d 25°
S o l u t i o n :
Let ?A = (3x)°, ?B = (2x)° and ?C = x°
Then,
3x +2x +x = 180° [Sum of the angles of a triangle] ? 6x = 180° ? x = 30°
Hence, the angles are
?A = 3 ×30° = 90°, ?B = 2 ×30° = 60° and ?C = 30°
Side BC of triangle ABC is produced to E.
? ?ACE = ?A + ?B ? ?ACD + ?ECD = 90°+60° ? 90°+ ?ECD = 150° ? ?ECD = 60°
Hence, the correct answer is option a
.
Q u e s t i o n : 4 0
In the given figure, BO and CO are the bisectors of ?B and ?C respectively. If ?A = 50° then ?BOC = ?
a 130°
b 100°
c 115°
d 120°
S o l u t i o n :
c 115°
In ? ABC
, we have:
?A + ?B + ?C = 180° [Sum of the angles of a triangle] ? 50°+ ?B + ?C = 180° ? ?B + ?C = 130° ?
1
2
?B +
1
2
?C = 65° . . . (i)
In ? OBC
, we have:
?OBC + ?OCB + ?BOC = 180° ?
1
2
?B +
1
2
?C + ?BOC = 180° [Using (i)] ? 65°+ ?BOC = 180° ? ?BOC = 115°
Q u e s t i o n : 4 1
In the given figure, side BC of ?ABC has been produced to a point D. If ?A = 3y°, ?B = x°, ?C = 5y° and ?CBD = 7y°. Then, the value of x is
a 60
b 50
c 45
d 35
S o l u t i o n :
D i s c l a i m e r: In the question ?ACD should be 7y°.
In the given figure, ?ACB + ?ACD = 180° Linearpairofangles
? 5y° + 7y° = 180°
Read More
|
44 videos|412 docs|54 tests
|
Related Exams
About this Document
|
4.66/5 Rating |
|
Dec 18, 2024 Last updated |
Document Description: RS Aggarwal MCQs: Triangles for Class 9 2024 is part of Mathematics (Maths) Class 9 preparation.
The notes and questions for RS Aggarwal MCQs: Triangles have been prepared according to the Class 9 exam syllabus. Information about RS Aggarwal MCQs: Triangles covers topics
like and RS Aggarwal MCQs: Triangles Example, for Class 9 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RS Aggarwal MCQs: Triangles.
Introduction of RS Aggarwal MCQs: Triangles in English is available as part of our Mathematics (Maths) Class 9
for Class 9 & RS Aggarwal MCQs: Triangles in Hindi for Mathematics (Maths) Class 9 course.
Download more important topics related with notes, lectures and mock test series for Class 9
Exam by signing up for free. Class 9: RS Aggarwal MCQs: Triangles | Mathematics (Maths) Class 9
Description
Full syllabus notes, lecture & questions for RS Aggarwal MCQs: Triangles | Mathematics (Maths) Class 9 - Class 9 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 9 | Best notes, free PDF download
Information about RS Aggarwal MCQs: Triangles
In this doc you can find the meaning of RS Aggarwal MCQs: Triangles defined & explained in the simplest way possible. Besides explaining types of
RS Aggarwal MCQs: Triangles theory, EduRev gives you an ample number of questions to practice RS Aggarwal MCQs: Triangles tests, examples and also practice Class 9
tests
|
Explore Courses for Class 9 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
MCQs
,RS Aggarwal MCQs: Triangles | Mathematics (Maths) Class 9
,Objective type Questions
,video lectures
,Sample Paper
,Exam
,practice quizzes
,RS Aggarwal MCQs: Triangles | Mathematics (Maths) Class 9
,Free
,study material
,Important questions
,Viva Questions
,past year papers
,ppt
,shortcuts and tricks
,Extra Questions
,RS Aggarwal MCQs: Triangles | Mathematics (Maths) Class 9
,Previous Year Questions with Solutions
,Semester Notes
,Summary
,mock tests for examination
;
Additional Information about RS Aggarwal MCQs: Triangles for Class 9 Preparation
RS Aggarwal MCQs: Triangles Free PDF Download
The RS Aggarwal MCQs: Triangles is an invaluable resource that delves deep into the core of the Class 9 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the RS Aggarwal MCQs: Triangles now and kickstart your journey towards success in the Class 9 exam.
Importance of RS Aggarwal MCQs: Triangles
The importance of RS Aggarwal MCQs: Triangles cannot be overstated, especially for Class 9 aspirants.
This document holds the key to success in the Class 9 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
RS Aggarwal MCQs: Triangles Notes
RS Aggarwal MCQs: Triangles Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RS Aggarwal MCQs: Triangles.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, RS Aggarwal MCQs: Triangles Notes on EduRev are your ultimate resource for success.
RS Aggarwal MCQs: Triangles Class 9 Questions
The "RS Aggarwal MCQs: Triangles Class 9 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 9 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study RS Aggarwal MCQs: Triangles on the App
Students of Class 9 can study RS Aggarwal MCQs: Triangles alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RS Aggarwal MCQs: Triangles,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of RS Aggarwal MCQs: Triangles is prepared as per the latest Class 9 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup