NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1

Mathematics is an integral part of human thought and logic and is integral to students’ attempts to understand the world and themselves. As well as building mental discipline, it promotes logical reasoning and problem-solving abilities. Moreover, Mathematics is crucial to understanding other school subjects, such as Science, Social Science, and even Music and Art.

In order to determine quantitative solutions to experimental laws, scientists use the fundamentals of Mathematics. It is used in a variety of fields, including Finance, Medicine, Computer Science, Social Sciences, Natural Sciences, and Engineering. So, the concepts of Mathematics are incorporated into the curriculum of all subjects in higher education. Class 9 Mathematics introduces students to many essential concepts of the subject, which form the basis for higher education. To perform well in the examinations and be able to apply those concepts in their future studies, students in Class 9 should develop strong concepts in mathematics.

Chapter 7: Class 9 Mathematics is Triangles. In lower grades, students have seen a glimpse of this chapter. There is no substitute for diligent practice when it comes to Mathematics. NCERT structures the fundamentals of all the subjects. Therefore, Extramarks provides students with r so that they do not have to look elsewhere for properly detailed and easy solutions. Class 9 is an essential academic year in the academic career of students. Besides preparing students for their examinations, the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.1 provide them with the foundation for academic success. Students must develop comprehensive learning habits in Class 9, which helps them perform better in the Class 10 board examinations. In order to improve students’ academic performance and have a bright academic career, Extramarks provides them with Exercise 7.1 Class 9 Maths NCERT Solutions.

Multiple public and private schools in India are managed and controlled by the Central Board of Secondary Education. Most of the schools in the country follow this board of education, making it one of the most important boards of education in the country. By resolution of the Indian government, the CBSE Board was established in 1929. Schools affiliated with this educational board follow the NCERT curriculum. NCERT promotes and coordinates research related to school education and prepares and publishes model textbooks, supplementary materials, etc. Additionally, it serves as a clearinghouse for ideas and information related to school education, as well as a government agency dedicated to achieving universal elementary education goals. Founded in 1961 by the Government of India, the National Council of Educational Research and Training (NCERT) assists the Central and State Governments with educational policies and programmes.

NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.1 Triangles

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Access NCERT Answers for Class-9 Maths Chapter 7 – Triangles

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles (Ex 7.1) Exercise 7.1

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Chapter wise NCERT Solutions for Class 9 Maths

The Extramarks website provides young learners with the Class 9 Maths Chapter 7 Exercise 7.1, that are curated by its in-house expert educators. These solutions provide step-by-step, detailed instructions for each and every question. To ensure that students have strong and clear concepts of Class 9 Maths Exercise 7.1, Extramarks walks them through the logic behind each step of the problems in the exercise. It is easy to download the Class 9 Maths Triangles Exercise 7.1 from the Extramarks website. The NCERT curriculum must be on the tips of the students in Class 9. They learn basic concepts of the subject through the NCERT Curriculum. The chapter entitled Triangles of Class 9 Mathematics can be one of the most scoring chapters in the Class 9 final examinations. Since the NCERT books are written by subject experts, students can understand all the concepts of every exercise if they thoroughly go through the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.1. Extramarks provides these solutions to help students improve their problem-solving skills and prepare for the examination. All the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.1, are explained step by step with all the details so that the students do not get confused. It is recommended that learners thoroughly review these solutions prior to their board examinations.

Class 9 Maths Chapter 7 Includes:

Students can achieve success in their examinations with Extramarks’ NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.1. Every topic which can appear in the examination is covered in the NCERT textbook. In addition, NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.1 are compiled by professionals with extensive experience. They are also a very helpful tool for students who find Mathematics challenging since they help them understand the basics of the chapter. Practising the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.1 helps students with a steady pace for solving the problems related to the curriculum, which is very critical for them to score well in the examinations. Once students go through the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.1, they will be able to solve any complicated problem that they can encounter in their examinations. Extramarks recommends learners practice the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.1 thoroughly prior to their examinations.

NCERT Solutions for Class 9

The subject of mathematics can be very intimidating for many students. The first and foremost step that learners must take for the preparation of their Class 9 Mathematics examination is to thoroughly practice the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.1. There are, however, no solutions to the questions in the NCERT textbooks.

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CBSE Study Materials for Class 9

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CBSE Study Materials

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Q.1

$\begin{array}{l} In quadrilateral ACBD, AC = AD and AB bisects ∠ A \\ (see figure below) . Show that ΔABC ≅ ΔABD . What can \\ you say about BC and BD . \end{array}$

Ans

$\begin{array}{l} Given : ABCD is a quadrilateral, AC = BD, AB bisects ∠ A. \\ To prove: Δ ABC ≅ Δ ABD \\ Proof:In Δ ABC and Δ ABD \\ AC = AD [Given] \\ ∠ BAC = ∠ ABD [Given] \\ AB = AB [Common] \\ ∴ Δ ABC ≅ Δ ABD [By SAS] \\ BC = BD [By CPCT] \end{array}$

Q.2

$\begin{array}{l} ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA \\ (see figure below) . Prove that \\ (i) ΔABD ≅ ΔBAC \\ (ii) BD = AC \\ (iii) ∠ ABD = ∠ BAC . \end{array}$

Ans

$\begin{array}{l} Given : ABCD is a quadrilateral, AD = BC, ∠ D AB = ∠ C BA . \\ To prove: \\ (i) Δ ABD ≅ Δ B AC \\ (i i) B D = A C \\ (i i i) ∠ ABD = ∠ BAC \\ Proof: (i) In Δ ABD and Δ B AC \\ AB = B A [Common] \\ ∠ B A D = ∠ A B C [Given] \\ A D = B C [Given] \\ ∴ Δ ABD ≅ Δ BAC [B y SAS] \\ (i i) B D = A C [B y CPCT] \\ (i i i) ∠ ABD = ∠ BAC [B y CPCT] \end{array}$

Q.3 AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.

Ans

$\begin{array}{l} Given : AD and BC are equal perpendiculars to a line segment AB. \\ To Prove: CD bisects AB i.e., CO = OD \\ Proof : In ΔBOC ≅ ΔAOD \\ ∠ CBO = ∠ DAO [Given] \\ ∠ BOC = ∠ AOD [Vertical opposite angles] \\ BC = AD [Given] \\ ∴ ΔBOC ≅ ΔAOD [By AAS] \\ ∴ BO = OA [By C.P.C.T.] \\ \Rightarrow CD bisects line segment AB. \end{array}$

Q.4

$\begin{array}{l} l and m are two parallell ines intersected by another pair of \\ parallel lines p and q (see figure below) . Show that Δ ABC ≅ ΔCDA . \end{array}$

Ans

$\begin{array}{l} Given : line l and m are parallel and \overset{⇀}{p} ∥ \overset{⇀}{q} . \\ To prove: Δ ABC ≅ ΔCDA \\ Proof: In Δ ABCand ΔCDA \\ ∠ BAC = ∠ DCA [Alternate interior angles] \\ AC = CA [Common] \\ ∠ BCA = ∠ DAC [Alternate interior angles] \\ Δ ABC ≅ ΔCDA [By ASA] \end{array}$

Q.5

$\begin{array}{l} Line l is the bisector of an angle ∠A and ∠B is any point on l. \\ BP and BQ are perpendiculars from B to the arms of ∠A \\ (see figure below) . Show that : \\ (i) ΔAPB ≅ ΔAQB \\ (ii) BP = BQ or B is equidistant from the arms of ∠ A . \end{array}$

Ans

$\begin{array}{l} Given : Line l is bisector of ∠ A, BP and BQ are perpendiculars to \\ the arms of angle. \\ To prove: (i) Δ APB ≅ ΔAQB \\ (ii) BP = BQ or B is equidistant from the arms of∠A. \\ Proof: (i) In Δ ABQ and ΔAPB \\ ∠ BAQ = ∠ BAP [Given] \\ AB = AB [Common] \\ ∠ AQB = ∠ APB [each 90°] \\ Δ ABQ ≅ ΔAPB [By AAS] \\ (ii) BQ = BP [By C.P.C.T.] \\ \Rightarrow BP = BQ \end{array}$

Q.6 In figure below, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Ans

$\begin{array}{l} Given : In Δ ABC and Δ ADE, AC = AE, AB = AD and ∠ BAD = ∠ EAC \\ To prove:BC=DE \\ Proof: In Δ ABC and ΔADE \\ AB = AD [Given] \\ ∠ BAC = ∠ DAE [∠ BAD + ∠ DAC = ∠ DAC + ∠ EAC] \\ AC = AE [Given] \\ Δ ABC ≅ ΔADE [By SAS] \\ BC = DE [By C.P.C.T.] \end{array}$

Q.7

$\begin{array}{l} AB is a line segment and P is its mid - point . D and E are \\ points on the same side of AB such that ∠ BAD = ∠ ABE \\ and ∠ EPA = ∠ DPB (see figure below) . Show that \\ (i) ΔDAP ≅ ΔEBP \\ (ii) AD = BE \end{array}$

Ans

$\begin{array}{l} Given : P is mid-point of AB, ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB \\ To prove: (i) ΔDAP ≅ ΔEBP \\ (ii) AD = BE \\ Proof: (i) In Δ DAPand ΔEBP \\ ∠ DPA = ∠ EPB [∠ APE + ∠ EPD = ∠ BPD + ∠ DPE] \\ A P = PB [Given] \\ ∠ DAP = ∠ EBP [Given] \\ Δ DAP ≅ ΔEPB [By SAS] \\ (ii) AD = BE [By C.P.C.T.] \end{array}$

Q.8

$\begin{array}{l} In right triangle ABC, right angled at C, M is the mid - point \\ of hypotenuse AB . C is joined to M and produced to a point \\ D such that DM = CM . Point D is joined to point B \\ (see figure below) . Show that : \\ (i) ΔAMC ≅ ΔBMD \\ (ii) ΔDBC is a right angle . \\ (iii) ΔDBC ≅ ΔACB \\ (iv) CM = \frac{1}{2} AB \end{array}$

Ans

$\begin{array}{l} Given : In right Δ ABC, ∠ C = 90 ° and M is mid-point on \\ hypotenuse AB, CM = DM. \\ To Prove: (i) ΔAMC ≅ ΔBMD \\ (ii) ∠ DBC is a right angle. \\ (iii) ΔDBC ≅ ΔACB \\ (iv) CM = \frac{1}{2} AB \\ Proof : (i) In ΔAMC and ΔBMD \\ AM = BM [Given] \\ ∠ AMC = ∠ BMD [Vertica l opposite angles] \\ CM = DM [Given] \end{array}$

$\begin{array}{l} ∴ ΔAMC ≅ ΔBMD [S . A . S .] \\ AC = DB [By C.P.C.T.] \end{array}$

$\begin{array}{l} (ii) ∠ ACM = ∠ BDM [By C.P.C.T.] \\ \Rightarrow ∠ ACD = ∠ BDC \\ \Rightarrow AC ∥ DB [Alternate angles are equal.] \\ ∠ ACB + ∠ DBC = 180 ° [Co-interior angles] \\ 90 ° + ∠ DBC = 180 ° \\ ∠ DBC = 180 ° - 90 ° \\ = 90 ° \\ So, ∠ DBC is a right angle. \\ (iii) In Δ DCB and ΔACB \\ DB = AC [Proved] \\ ∠ ACB = ∠ DBC [Each 90°] \\ BC = BC [Common] \\ ∴ Δ DCB ≅ ΔACB [S . A . S .] \\ (iv) DC = AB [By C.P.C.T.] \\ 2 CM = AB [CM = DM] \\ CM = \frac{1}{2} AB \end{array}$

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FAQs (Frequently Asked Questions)

1. Are the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.1 available on the Extramarks website?

Yes, the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.1 are easily available on the Extramarks website. As these solutions are available in PDF format, they can be easily downloaded on any device, they are curated by the Extramarks subject experts. Along with the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.1, Extramarks also provides students with various learning modules that help them with an enjoyable and systematic learning process. Extramarks has proved to the students that e-learning is extremely beneficial. Students can subscribe to the Extramarks website for access to authentic and credible NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.1.

2. Is Class 9 Chapter 7 Triangles difficult?

No, Chapter 7 Triangles is not difficult or complicated for the students of Class 9. If they review the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.1 thoroughly, they will be able to grasp the concepts of the chapter easily and apply those concepts to any complicated problem that can appear in the examinations.

3. Is it essential to practice all the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.1?

Yes, students should practice all the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.1 as every question helps them in building strong concepts. Furthermore, by practising them, students will be able to increase their speed and accuracy, which will help them avoid small calculation errors so that they may not lose marks during examinations.

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1

NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.1 Triangles

Access NCERT Answers for Class-9 Maths Chapter 7 – Triangles

NCERT Solutions for Class 9 Maths Chapter 7 Triangles (Ex 7.1) Exercise 7.1