• AglaSem
  • Schools
  • Admission
  • Career
  • News
  • Hindi
  • Mock Test
  • Docs
  • ATSE
aglasem
  • CBSE
    • Date Sheet
    • Syllabus
    • Sample Papers
    • Question Papers
  • ICSE / ISC
  • State Boards
    • Date Sheet
    • Admit Card
    • Result
    • Sample Paper
    • Question Paper
  • NCERT
    • NCERT Books
    • NCERT Solutions
    • NCERT Exempler
    • Teaching Material
  • Mock Tests
  • Study Material
    • Notes
    • Solved Sample Papers
    • Maps
    • Writing Skill Format
  • Olympiads
    • ATSE
    • KVPY
    • NTSE
    • NMMS
  • School Admission
  • Entrance Exams
    • JEE Main
    • NEET
    • CLAT
  • Students Guide
    • Careers Opportunities
    • Courses & Career
    • Courses after 12th
  • Others
    • RD Sharma Solutions
    • HC Verma Solutions
    • Teaching Material
    • Classes Wise Resources
    • Videos
No Result
View All Result
aglasem
  • CBSE
    • Date Sheet
    • Syllabus
    • Sample Papers
    • Question Papers
  • ICSE / ISC
  • State Boards
    • Date Sheet
    • Admit Card
    • Result
    • Sample Paper
    • Question Paper
  • NCERT
    • NCERT Books
    • NCERT Solutions
    • NCERT Exempler
    • Teaching Material
  • Mock Tests
  • Study Material
    • Notes
    • Solved Sample Papers
    • Maps
    • Writing Skill Format
  • Olympiads
    • ATSE
    • KVPY
    • NTSE
    • NMMS
  • School Admission
  • Entrance Exams
    • JEE Main
    • NEET
    • CLAT
  • Students Guide
    • Careers Opportunities
    • Courses & Career
    • Courses after 12th
  • Others
    • RD Sharma Solutions
    • HC Verma Solutions
    • Teaching Material
    • Classes Wise Resources
    • Videos
No Result
View All Result
aglasem
No Result
View All Result

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

by aglasem
May 1, 2022
in 10th Class
Reading Time: 1 min read
0
NCERT Solutions

NCERT Solutions Class 10 Maths Chapter 11 Constructions – Here are all the NCERT solutions for Class 10 Maths Chapter 11. This solution contains questions, answers, images, explanations of the complete Chapter 11 titled Constructions of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 11 Constructions. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 11 Constructions in one place.

Go to Exercise

NCERT Solutions Class 10 Maths Chapter 11 Constructions

Subscribe For Latest Updates

Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 10 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 11 Constructions , Maths, Class 10.

Class 10
Subject Maths
Book Mathematics
Chapter Number 11
Chapter Name

Constructions

NCERT Solutions Class 10 Maths chapter 11 Constructions

  • LPU Admission 2022 Open - India's Top Ranked University
  • ATSE 2022, Olympiad Registration Open.
  • Rungta Institute Admission 2022 Open
  • REVA University Admission 2022 Open

Class 10, Maths chapter 11, Constructions solutions are given below in PDF format. You can view them online or download PDF file for future use.

Constructions Download

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Download

Did you find NCERT Solutions Class 10 Maths chapter 11 Constructions helpful? If yes, please comment below. Also please like, and share it with your friends!

NCERT Solutions Class 10 Maths chapter 11 Constructions- Video

You can also watch the video solutions of NCERT Class10 Maths chapter 11 Constructions here.

If you liked the video, please subscribe to our YouTube channel so that you can get more such interesting and useful study resources.

Download NCERT Solutions Class 10 Maths chapter 11 Constructions In PDF Format

You can also download here the NCERT Solutions Class 10 Maths chapter 11 Constructions in PDF format.

Click Here to download NCERT Solutions for Class 10 Maths chapter 11 Constructions

Question & Answer

Q.1: In the question, give the justification of the construction also Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.

Ans : A line segment of length 7.6 cm can be divided in the ratio of 5 : 8 as follows Step 1 Draw line segment AB of 7.6 cm and draw a ray AX making an acute angle with line segment AB . Step 2 Locate L3=(= 5+8) points, \(A_{1}, A_{2}, A_{3}, A_{4}, …..A_{13},\) on AX such that \(AA_{1}=A_{1}A_{2}=A_{2}A_{3}\) and so on . Step 3 Join \(BA_{13}\). Step 4 Through the point \(A_{5}\), draw a line parallel to \(BA_{13}\) (by making an angle equal to \(\angle A A_{13} B\)) at \(A_{5}\) intersecting AB at point C. C is the point dividing line segment AB of 7.6 cm in the required ratio of 5 : 8. The lengths of AC and CB can be measured. It comes out to 2.9 cm and 4.7 cm respectively. Justification The construction can be justified by proving that \(\begin{array}{l}{\frac{\mathrm{AC}}{\mathrm{CB}}=\frac{5}{8}} \\ {\text { By construction, we have } \mathrm{A}_{5} \mathrm{C}\left\|\mathrm{A}_{13} \mathrm{B} \text { . By applying Basic proportionality theorem for }\right.} \\ {\frac{\mathrm{AC}}{\mathrm{CB}}=\frac{\mathrm{AA}_{5}}{\mathrm{A}_{5} \mathrm{A}_{13}}}\end{array}\) \(\begin{array}{l}{\text { From the figure, it can be observed that } A A_{5} \text { and } A_{5} A_{13} \text { contain } 5 \text { and } 8 \text { equal }} \\ {\text { divisions of line segments respectively. }} \\ {\therefore \frac{A A_{f}}{A_{3} A_{13}}=\frac{5}{8}} \\ {\text { On comparing equations }(1) \text { and }(2), \text { we obtain }} \\ {\frac{A C}{C B}=\frac{5}{8}} \\ {\text { This justifies the construction. }}\end{array}\)

Q.2: In the question, give the justification of the construction also , construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.

Ans : Step 1 Draw a line segment AB = 4 cm. Taking point A as centre, draw and arc of 5 cm radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius. These arcs will intersect each other at point C. Now, AC = 5 cm and BC=6 cm And \(\triangle ABC\) is the required triangle. Step 2 Draw a ray AX making an acute angle with line AB on the opposite site of vertex C. Step 3 Locate 3 points (as 3 is greater between 2 and 3) online AX such that \(AA_{1}=A_{1}A_{2}=A_{2}A_{3}\). Step 4 \(Join \mathrm{BA}_{3}\) and draw a line through \(\mathrm{A}_{2}\) parallel to \(\mathrm{B} \mathrm{A}_{3}\) to intersect \(\mathrm{AB}\) at point \(\mathrm{B}^{\prime} .\) Step 5 Draw a line through B\(\prime\) parallel to the line BC to intersect AC at \(C^{\prime}\) \(\triangle A B^{\prime} C^{\prime}\)is the required triangle. Justification The construction can be justified by proving that \(\mathrm{AB}^{\prime}=\frac{2}{3} \mathrm{AB}, \mathrm{B} \mathrm{C}^{\prime}=\frac{2}{3} \mathrm{BC}, \mathrm{AC}^{\prime}=\frac{2}{3} \mathrm{AC}\) \(\begin{array}{l}{\text { By construction, we have } B^{\prime} C^{\prime}\|B C} \\ {\therefore \angle A B C^{\prime}=\angle A B C(\text {Corresponding angles) }}\end{array}\) \(\triangle A B^{\prime} C^{\prime} \text { and } \triangle A B C\) \(\begin{aligned} \angle \mathrm{ABC}^{\prime} &=\angle A B C(\text { Proved above }) \\ \angle \mathrm{B}^{\prime} \mathrm{AC}^{\prime} &=\angle B A C(\text { Common }) \end{aligned}\) \(\therefore \triangle \mathrm{ABC}^{\prime}-\Delta \mathrm{ABC}(\text { AA similarity criterion })\) \(\Rightarrow \frac{A B^{\prime}}{A B}=\frac{B^{\prime} C^{\prime}}{B C}=\frac{A C^{\prime}}{A C}\) \(\triangle \mathrm{AA}_{2} \mathrm{B}^{\prime} \text { and } \triangle \mathrm{A} \mathrm{A}_{3} \mathrm{B}\) \(\begin{array}{l}{\angle A_{2} A B^{\prime}=\angle A_{3} A B(\text {Corresponding} \text { angles })} \\ {\angle A A_{2} B^{\prime}=\angle A A_{3} B(\text {Corresponding angles) }}\end{array}\) \(\begin{array}{l}{\therefore \triangle A A_{2} B^{\prime}-\Delta A A_{3} B(\text { AA similarity criterion) }} \\ {\Rightarrow \frac{A B^{\prime}}{A B}=\frac{A A_{2}}{A A_{3}}}\end{array}\) \(\Rightarrow \frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{2}{3}\) \(\begin{array}{l}{\text { From equations }(1) \text { and }(2), \text { we obtain }} \\ {\frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{\mathrm{B} \mathrm{C}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{AC}^{\prime}}{\mathrm{AC}}=\frac{2}{3}} \\ {\Rightarrow \mathrm{AB}^{\prime}=\frac{2}{3} \mathrm{AB}, \mathrm{B}^{\prime} \mathrm{C}^{\prime}=\frac{2}{3} \mathrm{BC}, \mathrm{AC}^{\prime}=\frac{2}{3} \mathrm{AC}} \\ {\text { This justifies the construction. }}\end{array}\)

Q.3: In the question, give the justification of the construction also ,construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.

Ans : Step 1 Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 5 cm radius respectively. Let these arcs intersect each other at point C.\(\triangle ABC\) is the required triangle having length of sides as 5 cm , 6 cm , 7 cm respectively. Step 2 Draw a ray AX making acute angle with line AB on the opposite side of vertex C. Step 3 \(\begin{array}{l}{\text { Locate } 7 \text { points, } A_{1}, A_{2}, A_{3}, A_{4} A_{5}, A_{5}, A_{7}(\text { as } 7 \text { is greater between } 5 \text { and } 7), \text { on line } A X} \\ {\text { such that } A A_{1}=A_{1} A_{2}=A_{2} A_{3}=A_{3} A_{4}=A_{4} A_{5}=A_{5} A_{6}=A_{6} A_{7}}\end{array}\) Step 4 Join \(BA_{5}\) and draw a line through \(A_{7}\) parallel to \(BA_{5}\) to intersect extended line segment AB at point \(B^{\prime}\). Step 5 Draw a line through \(B^{\prime}\) parallel to BC intersecting the extended line segment AC at \(C^{\prime}\).\(\triangle\)A\(B^{\prime}\)\(C^{\prime}\) is the required triangle. Justification The construction can be justified by proving that \(\begin{array}{l}{\mathrm{AB}^{\prime}=\frac{7}{5} \mathrm{AB}, \mathrm{B} \mathrm{C}^{\prime}=\frac{7}{5} \mathrm{BC}, \mathrm{AC}^{\prime}=\frac{7}{5} \mathrm{AC}} \\ {\text { In } \triangle \mathrm{ABC} \text { and } \Delta \mathrm{AB}^{\prime} \mathrm{C}^{\prime}} \\ {\angle \mathrm{ABC}=\angle \mathrm{AB}^{\prime} \mathrm{C}^{\prime}(\text { Corresponding angles })} \\ {\angle \mathrm{BAC}=\angle \mathrm{B}^{\prime} \mathrm{AC}^{\prime}(\text { Common })}\end{array}\) \(\therefore \triangle \mathrm{ABC}-\triangle \mathrm{AB}^{\prime} \mathrm{C}^{\prime}\) \(\Rightarrow \frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}=\frac{\mathrm{AC}}{\mathrm{AC}^{\prime}} \ldots\) \(\begin{array}{l}{\text { In } \triangle A A_{5} B \text { and } \triangle A A_{7} B^{\prime}} \\ {\angle A_{5} A B=\angle A_{7} A B^{\prime}(\text { Common })}\end{array}\) \(\angle A A_{\mathrm{s}} B=\angle A A_{7} B^{\prime}(\text { Corresponding angles })\) \(\begin{array}{l}{\therefore \triangle \mathrm{AA}_{5} \mathrm{B}-\triangle \mathrm{AA}_{7} \mathrm{B}^{\prime} \text { (AA similarity criterion) }} \\ {\Rightarrow \frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{\mathrm{AA}_{5}}{\mathrm{AA}_{7}}} \\ {\Rightarrow \frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{5}{7}}\end{array}\) \(\begin{array}{l}{\text { On comparing equations }(1) \text { and }(2), \text { we obtain }} \\ {\frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{BC}^{\prime}}=\frac{\mathrm{AC}}{\mathrm{AC}^{\prime}}=\frac{5}{7}} \\ {\Rightarrow \mathrm{AB}^{\prime}=\frac{7}{5} \mathrm{AB}, \mathrm{B}^{\prime} \mathrm{C}=\frac{7}{5} \mathrm{BC}, \mathrm{AC}=\frac{7}{5} \mathrm{AC}} \\ {\text { This justifies the construction. }}\end{array}\)

Q.4: \(\begin{array}{l}{\text { In the question, give the justification of the construction also , construct an isosceles triangle whose base is } 8 \mathrm{cm} \text { and altitude } 4 \mathrm{cm} \text { and then another }} \\ {\text { triangle whose sides are } 1 \frac{1}{2} \text { times the corresponding sides of the isosceles triangle. }}\end{array}\)

Ans : Let us assume that ∠ABC is an isosceles triangle having CA and CB of equal lengths, base AB of 8 crn, and AD is the altitude of 4 cm. \(A \triangle A B^{\prime} C^{\prime}\)whose sides are \(\frac{3}{2}\) times of can be drawn as follows. Step 1 Draw a line segment Aa of 8 cm. Draw arcs of same radius on both sides of the line segment while taking point A and as its centre. Let these arcs intersect each other at O and O'. Join OO' Let OO' intersect AB at D. Step 2 Taking D as centre, draw an arc of 4 cm radius which cuts the extended line segment OO' at point C. An isosceles \(AA_{3}C\) is formed, having CD (altitude) as 4 cm and AB (base) as 8 cm. Step 3 Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C. Step 4 Locate 3 points (as 3 is greater between 3 and 2) \(A_{1}, A_{2}, \text { and } A_{3}\) on AX such that \(AA_{1}= A_{1}A_{2}= A_{2}A_{3}\). Step 5 Join \(BA_{2}\) and draw a line through \(A_{3}\) parallel to \(BA_{2}\) to intersect extended line segment AB at point \(B^{\prime}\). Step 6 Draw a line through \(B^{\prime}\) parallel to BC intersecting the extended the segment AC at \(C^{\prime}\). \(\Delta \mathrm{AB}^{\prime} \mathrm{C}^{\prime}\) Is the required triangle. Justification The construction can be justified by proving that \(\mathrm{AB}^{\prime}=\frac{3}{2} \mathrm{AB}, \mathrm{BC}^{\prime}=\frac{3}{2} \mathrm{BC}, \mathrm{AC}^{\prime}=\frac{3}{2} \mathrm{AC}\) \(\begin{array}{l}{\text { In } \triangle A B C \text { and } \Delta A B^{\prime} C^{\prime}} \\ {\angle A B C=\angle A B^{\prime} C^{\prime} \text { (Corresponding angles) }}\end{array}\) \(\begin{array}{l}{\angle B A C=\angle B^{\prime} A C^{\prime}(\text { Common })} \\ {\therefore \triangle A B C-\Delta A B^{\prime} C^{\prime} \text { (AA similarity criterion) }}\end{array}\) \(\begin{array}{l}{\Rightarrow \frac{A B}{A B^{\prime}}=\frac{B C}{B C^{\prime}}=\frac{A C}{A C^{\prime}}} \\ {\text { In } \triangle A A_{2} B \text { and } \Delta A A_{3} B^{\prime}}\end{array}\) \(\begin{array}{l}{\angle A_{2} A B=\angle A_{3} A B^{\prime}(\text { Corresponding angles })} \\ {\angle A A_{2} B=\angle A A_{3} B^{\prime} \text { (Corresponding angles) }}\end{array}\) \(\begin{array}{l}{\therefore \triangle \mathrm{AA}_{2} \mathrm{B}-\triangle \mathrm{AA}_{3} \mathrm{B}^{\prime} \text { (AA similarity criterion) }} \\ {\Rightarrow \frac{\mathrm{AB}}{\mathrm{AB}}=\frac{\mathrm{AA}_{2}}{\mathrm{AA}_{3}}} \\ {\Rightarrow \frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{2}{3}}\end{array}\) \(\begin{array}{l}{\text { On comparing equations }(1) \text { and }(2), \text { we obtain }} \\ {\frac{A B}{A B}=\frac{B C}{B C^{\prime}}=\frac{A C}{A C^{\prime}}=\frac{2}{3}} \\ {\Rightarrow A B^{\prime}=\frac{3}{2} A B, B C=\frac{3}{2} B C, A C=\frac{3}{2} A C} \\ {\text { This justifies the construction. }}\end{array}\)

Q.5: In the question, give the justification of the construction also , draw a triangle ABC with side BC = 6 cm, AB=5 cm and \(\angle \mathrm{ABC}\)=\(60^{\circ}\) . Then construct a triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the triangle ABC.

Ans : A \(\triangle \mathrm{A}^{\prime} \mathrm{BC}^{\prime}\) whose sides \(\frac{3}{4}\) of the corresponding sides of can be drawn as follows. Step 1 Draw a \(\triangle\)ABC with side BC = 6 crn, AB = 5 cm and \(\angle\)ABC = 60\(^\circ\). Step 2 Draw a ray BX making an acute angle with BC on the opposite side of vertex A. Step 3 Locate 4 points (as 4 is greater in 3 and 4), \(B_{1}, B_{2}, B_{3}, B_{4}\), on line segment BX. Step 4 Join \(B_{4} C\)and draw a line through \(B_{3}\) , parallel to \(\mathrm{B}_{4} \mathrm{C}\) intersecting BC at \(C^{\prime}\) Step 5 Draw a line through C' parallel to AC intersecting AB at A'.\(\triangle \mathrm{A}^{\prime} \mathrm{BC}^{\prime}\) is the required triangle. Justification The construction can be justified by proving that, \(\mathrm{AB}=\frac{3}{4} \mathrm{AB}, \mathrm{BC}^{\prime}=\frac{3}{4} \mathrm{BC}, \mathrm{AC}^{\prime}=\frac{3}{4} \mathrm{AC}\) \(\begin{array}{l}{\text { In } \triangle A^{\prime} B C^{\prime} \text { and } \triangle A B C \text { , }} \\ {\angle A^{\prime} C^{\prime} B=\angle A C B \text { (Corresponding angles) }} \\ {\angle A^{\prime} B C^{\prime}=\angle A B C(\text { Common})}\end{array}\) \(\therefore \triangle A^{\prime} B C^{\prime} \sim \triangle A B C(\text { AA similarity criterion })\) \(\Rightarrow \frac{A B}{A B}=\frac{B C}{B C}=\frac{A C}{A C}\) \(\begin{array}{l}{\text { In } \triangle B B_{3} C^{\prime} \text { and } \triangle B B_{4} C} \\ {\angle B_{3} B C^{\prime}=\angle B_{4} B C(\text {Common})}\end{array}\) \(\begin{array}{l}{\angle B B_{3} C^{\prime}=\angle B B_{4} C(\text { Corresponding angles) }} \\ {\therefore \triangle B B_{3} C^{\prime} \sim \Delta B B_{4} C(\text { AA similarity criterion) }}\end{array}\) \(\begin{array}{l}{\Rightarrow \frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{BB}_{3}}{\mathrm{BB}_{4}}} \\ {\Rightarrow \frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{3}{4}} \\ {\text { From equations }(1) \text { and }(2), \text { we obtain }}\end{array}\) \(\begin{array}{l}{\frac{A B}{A B}=\frac{B C}{B C}=\frac{A^{\prime} C^{\prime}}{A C}=\frac{3}{4}} \\ {\Rightarrow A B=\frac{3}{4} A B, B C^{\prime}=\frac{3}{4} B C, A C^{\prime}=\frac{3}{4} A C} \\ {\text { This justifies the construction. }}\end{array}\)

NCERT / CBSE Book for Class 10 Maths

You can download the NCERT Book for Class 10 Maths in PDF format for free. Otherwise you can also buy it easily online.

  • Click here for NCERT Book for Class 10 Maths
  • Click here to buy NCERT Book for Class 10 Maths

All NCERT Solutions Class 10

  • NCERT Solutions for Class 10 English
  • NCERT Solutions for Class 10 Hindi
  • NCERT Solutions for Class 10 Maths
  • NCERT Solutions for Class 10 Science
  • NCERT Solutions for Class 10 Social Science
  • NCERT Solutions for Class 10 Sanskrit

All NCERT Solutions

You can also check out NCERT Solutions of other classes here. Click on the class number below to go to relevant NCERT Solutions of Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

Class 1 Class 2 Class 3
Class 4 Class 5 Class 6
Class 7 Class 8 Class 9
Class 10 Class 11 Class 12

Download the NCERT Solutions app for quick access to NCERT Solutions Class 10 Maths Chapter 11 Constructions. It will help you stay updated with relevant study material to help you top your class!


Previous Next

Exercise Covered under Maths Book Class 10 Maths Chapter 11 NCERT Solutions

Here on this page we have covered following exercises from NCERT Maths Book for Class 10 Maths Chapter 11 Constructions

  • Exercise: 11.1 Class 10 Maths Solutions
  • Exercise: 11.2 Class 10 Maths Solutions

To get fastest exam alerts and government job alerts in India, join our Telegram channel.

Tags: MathsNCERT Maths SolutionsNCERT SolutionsNCERT Solutions Class 10NCERT Solutions PDF

aglasem

Related Posts

CBSE Class 10 Answer Key 2022 for Term 2
10th Class

CBSE Class 10 Answer Key 2022 for Term 2

CBSE Class 10 Answer Keys
10th Class

Class 10 French Answer Key / Solution for Term 2 CBSE Board Exam 2022 (Code 018) – Download PDF

CBSE Class 10 Answer Keys
10th Class

Class 10 Hindi B Answer Key / Solution for Term 2 CBSE Board Exam 2022 (Code 085) – Download PDF

ICSE Class 10 Answer Key
10th Class

ICSE Biology Answer Key 2022 Class 10 Sem 2 (Out) – Check Solved Paper

Next Post
NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 10 Circles

Discussion about this post

Registration Open!!

  • REVA University Admission 2022 Open
  • Rungta Institute Admission 2022 Open
  • ATSE 2022, Olympiad Registration Open.
  • LPU Admission 2022 Open - India's Top Ranked University

Term 2 Solved Sample Paper 2022

  • Class 8 Sample Paper Term 2
  • Class 9 Sample Paper Term 2
  • Class 10 Sample Paper Term 2
  • Class 11 Sample Paper Term 2
  • Class 12 Sample Paper Term 2

CBSE Board Quick Links

  • CBSE Date Sheet
  • CBSE Result
  • CBSE Syllabus
  • CBSE Sample Papers
  • CBSE Question Papers
  • CBSE Notes
  • CBSE Practice Papers
  • CBSE Mock Tests

Class Wise Study Material

  • Class 1
  • Class 2
  • Class 3
  • Class 4
  • Class 5
  • Class 6
  • Class 7
  • Class 8
  • Class 9
  • Class 10
  • Class 11
  • Class 12

For 2022

  • Solved Sample Papers
  • Maps
  • Revision Notes
  • CBSE
  • State Board

Study Material

  • Class Notes
  • NCERT Solutions
  • NCERT Books
  • HC Verma Solutions
  • Courses After Class 12th

Exam Zone

  • JEE Main 2022
  • NEET 2022
  • CLAT 2022
  • Fashion & Design
  • Latest
  • Disclaimer
  • Terms of Use
  • Privacy Policy
  • Contact

© 2019 aglasem.com

No Result
View All Result
  • CBSE
  • ICSE / ISC
  • State Board
  • NCERT
  • Mock Test
  • Study Material
  • Olympiads
  • Schools Admission
  • Entrance Exams
  • Student Guide
  • HC Verma Solutions
  • Videos

© 2019 aglasem.com

REVA University 2022 Admission Open Apply Now!!