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NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.1
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NCERT Solutions for Class 9 Maths Chapter 7: Triangles - Exercise 7.1
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Exercise-7.1
1. In quadrilateral ACBD, AC = AD and AB bisects \[\angle A\] (See the given figure). Show that \[\Delta ABC{\text{ }} \cong \Delta ABD\]. What can you say about BC and BD?
![In quadrilateral ACBD, AC = AD and AB bisects \[\angle A\]](https://www.vedantu.com/seo/content-images/022ae806-c622-436d-8fb9-de1994dcc4cd.png)
Ans: Given: In quadrilateral ACBD, AC = AD and AB is bisected by \[\angle A\]
To find: Show that \[\Delta ABC{\text{ }} \cong \Delta ABD\].
In \[\Delta ABC{\text{ , }}\Delta ABD\]
\[AC{\text{ }} = {\text{ }}AD\] (Given)
\[\angle CAB{\text{ }} = \angle DAB\] (AB bisects ∠A)
\[AB{\text{ }} = {\text{ }}AB\] (Common)
\[\therefore \Delta ABC{\text{ }} \cong \Delta ABD\] (By SAS congruence rule)
\[\therefore BC{\text{ }} = {\text{ }}BD\] (By CPCT)
Therefore, BC and BD are of equal lengths.
2. ABCD is a quadrilateral in which AD = BC and \[\angle DAB{\text{ }} = \angle CBA\] (See the given figure). Prove that
(i) \[\Delta ABD{\text{ }} \cong \Delta BAC\]
(ii) BD = AC
(iii) \[\angle ABD{\text{ }} = \angle BAC\].
![ABCD is a quadrilateral in which AD = BC and \[\angle DAB{\text{ }} = \angle CBA\]](https://www.vedantu.com/seo/content-images/49493929-facc-409e-b432-ac77b407d608.png)
Ans : Given: ABCD is a quadrilateral where AD = BC and \[\angle DAB{\text{ }} = \angle CBA\]
To prove:
(i) \[\Delta ABD{\text{ }} \cong \Delta BAC\]
(ii) BD = AC
(iii)\[\angle ABD{\text{ }} = \angle BAC\].
In \[\Delta ABD{\text{ , }}\Delta BAC\],
AD = BC (Given)
\[\angle DAB{\text{ }} = \angle CBA\] (Given)
AB = BA (Common)
\[\therefore \Delta ABD{\text{ }} \cong \Delta BAC\] (By SAS congruence rule)
\[\therefore BD{\text{ }} = {\text{ }}AC\] (By CPCT)
And, \[\angle ABD{\text{ }} = \angle BAC\] (By CPCT)
3. AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB.
Ans: Given: AD and BC are equal perpendiculars to a line segment AB
To prove: CD bisects AB.
In \[\Delta BOC{\text{ , }}\Delta AOD\],
\[\angle BOC{\text{ }} = \angle AOD\] (Vertically opposite angles)
\[\angle CBO{\text{ }} = \angle DAO\] (Each right angle )
BC = AD (Given)
\[\therefore \Delta BOC{\text{ }} \cong \Delta AOD\] (AAS congruence rule)
\[\therefore BO{\text{ }} = {\text{ }}AO\] (By CPCT)
\[ \Rightarrow \]CD bisects AB.
4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that \[\Delta ABC \cong \Delta CDA\].
Given: l and m are two parallel lines intersected by another pair of parallel lines p and q To prove: \[\Delta ABC \cong \Delta CDA\].
In \[\Delta ABC{\text{ , }}\Delta CDA\],
\[\angle BAC{\text{ }} = \angle DCA\] (Alternate interior angles, as\[p{\text{ }}||{\text{ }}q\])
2AC = CA (Common)
\[\angle BCA{\text{ }} = \angle DAC\] (Alternate interior angles, as \[l{\text{ }}||{\text{ }}m\])
\[\therefore \Delta ABC \cong \Delta CDA\] (By ASA congruence rule)
5. Line l is the bisector of an angle $\angle A$ and B is any point on l. BP and BQ are perpendiculars from B to the arms of $\angle A$ (see the given figure). Show that:
(i) \[\Delta APB \cong \Delta AQB\]
(ii) BP = BQ or B is equidistant from the arms of ∠A.
Ans: Given: Line l is the bisector of an angle \[\angle A\] and B is any point on l.
To prove: (i) \[\Delta APB \cong \Delta AQB\]
(ii) \[BP{\text{ }} = {\text{ }}BQ\] or B is equidistant from the arms of \[\angle A\].
In \[\Delta APB{\text{ , }}\Delta AQB\],
\[\angle APB{\text{ }} = \angle AQB\] (Each right angle )
\[\angle PAB{\text{ }} = \angle QAB\] (l is the angle bisector of \[\angle A\])
AB = AB (Common)
\[\therefore \Delta APB \cong \Delta AQB\] (By AAS congruence rule)
\[\therefore BP{\text{ }} = {\text{ }}BQ\] (By CPCT)
Or, it can be said that B is equidistant from the arms of \[\angle A\].
6. In the given figure, \[AC{\text{ }} = {\text{ }}AE,{\text{ }}AB{\text{ }} = {\text{ }}AD\] and \[\angle BAD{\text{ }} = \angle EAC\]. Show that BC = DE.
Ans:
![\[AC{\text{ }} = {\text{ }}AE,{\text{ }}AB{\text{ }} = {\text{ }}AD\] and \[\angle BAD{\text{ }} = \angle EAC\]](https://www.vedantu.com/seo/content-images/08a10653-36c1-44ce-a830-9587a4b21c90.png)
Given: \[\angle BAD{\text{ }} = \angle EAC\]
To prove: BC = DE
It is given that \[\angle BAD{\text{ }} = \angle EAC\]
\[\angle BAD{\text{ }} + \angle DAC{\text{ }} = \angle EAC{\text{ }} + \angle DAC\]
\[\angle BAC{\text{ }} = \angle DAE\]
In \[\Delta BAC{\text{ , }}\Delta DAE\],
AB = AD (Given)
\[\angle BAC{\text{ }} = \angle DAE\] (Proved above)
AC = AE (Given)
\[\therefore \Delta BAC{\text{ }} \cong \Delta DAE\] (By SAS congruence rule)
\[\therefore BC{\text{ }} = {\text{ }}DE\] (By CPCT)
7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that \[\angle BAD{\text{ }} = \angle ABE\] and \[\angle EPA{\text{ }} = \angle DPB\] (See the given figure). Show that
(i) \[\Delta DAP \cong \Delta EBP\]
(ii) AD = BE
Ans: Given: AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that \[\angle BAD{\text{ }} = \angle ABE\] and \[\angle EPA{\text{ }} = \angle DPB\]
To prove: (i) \[\Delta DAP \cong \Delta EBP\]
(ii) AD = BE
It is given that \[\angle EPA{\text{ }} = \angle DPB\]
\[\angle EPA{\text{ }} + \angle DPE{\text{ }} = \angle DPB{\text{ }} + \angle DPE\]
\[\therefore \angle DPA{\text{ }} = \angle EPB\]
In \[\Delta DAP{\text{ , }}\Delta EBP\],
\[\angle DAP{\text{ }} = \angle EBP\] (Given)
AP = BP (P is mid-point of AB)
\[\angle DPA{\text{ }} = \angle EPB\] (From above)
\[\therefore \Delta DAP \cong \Delta EBP\] (ASA congruence rule)
\[\therefore AD{\text{ }} = {\text{ }}BE\] (By CPCT)
8. 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 the given figure). Show that:
(i) \[\Delta AMC \cong \Delta BMD\]
(ii) \[\angle DBC\] is a right angle.
(iii) \[\Delta DBC \cong \Delta ACB\]
(iv) \[CM{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AB\]
Ans: Given: M is the mid-point of hypotenuse AB. DM = CM
(i) In \[\Delta AMC{\text{ }},{\text{ }}\Delta BMD\],
AM = BM (M is the mid-point of AB)
\[\angle AMC{\text{ }} = \angle BMD\] (Vertically opposite angles)
CM = DM (Given)
\[\therefore \Delta AMC \cong \Delta BMD\] (By SAS congruence rule)
\[\therefore AC{\text{ }} = {\text{ }}BD\] (By CPCT)
And, \[\angle ACM{\text{ }} = \angle BDM\](By CPCT)
(ii) \[\angle ACM{\text{ }} = \angle BDM\]
However, \[\angle ACM{\text{ , }}\angle BDM\] are alternate interior angles.
Since alternate angles are equal,
It can be said that \[DB{\text{ }}||{\text{ }}AC\]
(Co-interior angles)
(iii) In \[\Delta DBC{\text{ , }}\Delta ACB\],
DB = AC (Already proved)
\[\angle DBC{\text{ }} = \angle ACB\] (Each \[{90^o}\])
BC = CB (Common)
\[\therefore \Delta DBC \cong \Delta ACB\] (SAS congruence rule)
(iv) \[\therefore \Delta DBC \cong \Delta ACB\]
AB = DC (By CPCT)
AB = 2 CM
\[CM{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AB\]
NCERT Solutions for Class 9 Maths Chapter 7 Triangles (Ex 7.1) Exercise 7.1
Opting for the NCERT solutions for Ex 7.1 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 7.1 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.
Chapter wise NCERT Solutions for Class 9 Maths
Class 9 Maths Chapter 7 Includes:
Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 9 students who are thorough with all the concepts from the Subject Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 9 Maths Chapter 7 Exercise 7.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.
Besides these NCERT solutions for Class 9 Maths Chapter 7 Exercise 7.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.
Do not delay any more. Download the NCERT solutions for Class 9 Maths Chapter 7 Exercise 7.1 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.
FAQs on NCERT Solutions for Class 9 Maths Chapter 7: Triangles - Exercise 7.1
1. What are the important concepts students learn in chapter 7 Triangles?
Class 9 is one of those few chapters that teach you concepts that will be of great importance and necessity in further higher classes. The basic important concepts that students learn in Class 9 Maths Chapter 7 are as follows:
Congruence of triangles.
Criteria for congruence of triangles - SAS congruence rule, ASA congruence rule, SSS congruence rule, RHS congruence rule.
Properties of triangles.
Inequalities of the triangle.
2. What are available in the NCERT Solutions Class 9 Maths Chapter 7 Exercise 7.1?
Chapter 7 has a total of 5 exercises. The first one - Exercise 7.1 has a total of 8 Questions - 6 Short Answer Questions, 2 Long Answer Question. In this exercise, students will learn the concepts of Congruence of Triangles, Criteria for Congruence and theorems related to the same. Also, most of the questions in this exercise are related to the proof. Therefore while solving the same, students must ensure that they mention the theorem every single time and also not miss any step while solving the same. Once the students finish solving this exercise, they can verify their answers with the NCERT Solutions Class 9 Maths Chapter 7 Exercise 7.1 provided by Vedantu. Apart from this, Vedantu also provides accurate solutions to all the problems in the following exercises as well:
Exercise 7.2 - 8 Questions - 6 Short Answer Questions, 2 Long Answer Question
Exercise 7.3 - 5 Questions - 3 Short Answer Questions, 2 Long Answer Question
Exercise 7.4 - 6 Questions - 5 Short Answer Questions, 1 Long Answer Question
Exercise 7.5 - Optional- 4 Questions
3. What are the basic properties of a triangle?
Some of the basic important properties of triangles that you must be aware of while studying this chapter is as follows:
The sum of all the angles of a triangle(of all types) is equal to 1800.
The sum of the length of the two sides of a triangle is greater than the length of the third side.
In the same way, the difference between the two sides of a triangle is less than the length of the third side.
The side opposite the greater angle is the longest side of all the three sides of a triangle.
The exterior angle of a triangle is always equal to the sum of the interior opposite angles. This property of a triangle is called an exterior angle property
Two triangles are said to be similar if their corresponding angles of both triangles are congruent and lengths of their sides are proportional.
Area of a triangle = ½ × Base × Height
4. What are the various types of triangles?
There are numerous types of triangles based on their sides and angles. Students must ensure that they are aware of them all in order to be able to solve triangle related sums with ease:
Scalene Triangle: All the sides and angles are unequal.
Isosceles Triangle: It has two equal sides. Also, the angles opposite these equal sides are equal.
Equilateral Triangle: All the sides are equal and all the three angles equal to 60°.
Acute Angled Triangle: A triangle having all its angles less than 90°.
Right Angled Triangle: A triangle having one of the three angles is 90°.
Obtuse Angled Triangle: A triangle having one of the three angles more than 90°.
5. Which is the most critical question in Exercise 7.1 of Chapter 7 of Class 9 Mathematics?
Question Number 8 is the most critical question in Exercise 7.1 of Class 9 Mathematics as it is among some complex problems in the chapter. It also has multiple concepts involved. The solution to the question involves using rules like the SAS congruence rule, CPCT rule and also the property of vertically opposite angles and co-interior angles. It means the question is testing students on some basic properties of triangles as well as the concept of congruence of triangles. Vedantu’s in-house subject matter specialists address the exercise's problems/questions with great care and precision, adhering to all CBSE criteria. Students in Class 9 who are thoroughly familiar with all of the topics included in the Subject Maths textbook and are well-versed in all of the problems presented in the exercises will easily achieve the greatest possible score on the final exam. Students can easily grasp the pattern of questions that may be asked in the exam from this chapter and also study the marks weightage of the chapter with the assistance of this Exercise 7.1 of Chapter 7 of Class 9 Mathematics solutions. So that they can adequately prepare for the final examination.
6. Is Exercise 7.1 of Chapter 7 of Class 9 Mathematics easy to understand or difficult?
Exercise 7.1 of Chapter 7 of Class 9 Mathematics is composed of a total of 8 questions, 6 of which are short answer questions and the rest two are of long-form answers. The difficulty level of the exercise is moderate, meaning it is not too easy yet not too difficult. A student who has understood the concepts involved can easily solve the questions. However, if someone finds it difficult, one can refer to the easy to understand solutions provided by Vedantu free of cost. Apart from these NCERT solutions for Exercise 7.1 of Chapter 7 of Class 9 Mathematics, this chapter contains numerous exercises with numerous questions. As previously said, our in-house subject specialists solve/answer all of these questions. As a result, all of these are guaranteed to be of high quality, and anyone can use them to study for exams. It is critical to grasp all of the topics in the textbooks and solve the questions from the exercises supplied next to them in order to achieve the highest possible grade in the class.
7. What does Exercise 7.1 of Chapter 7 of Class 9 Mathematics solutions talk about?
The Congruence of Triangles is discussed in NCERT Solutions for Exercise 7.1 of Chapter 7 ‘Triangles’ of Class 9 Mathematics. The congruence of triangles is a key math issue that is used in construction and design. As a result, students must have a thorough understanding of this subject because it is important in real-life situations, such as the construction of huge buildings or architectural designs. Triangle congruences can also be seen in geometric art, stepping stone designs, carpet designs, and a variety of other places.
8. What kind of problems are discussed in Exercise 7.1 of Chapter 7 of Class 9 Mathematics?
NCERT Solutions for Exercise 7.1 of Chapter 7 ‘Triangle’ of Class 9 Mathematics problems are based on proving triangle congruence criteria. As a result, students should thoroughly understand all elements of triangle congruences. It will enable students to confidently answer all of the questions in this exercise.
Triangle congruences will also aid children in understanding the concepts of even forms. In the first exercise of the Class 9 maths NCERT solutions Chapter 7 Exercise 7.1, there are eight questions about proofs based on congruency norms.
9. Why should students go through the NCERT solutions of Chapter 7 of Class 9 Mathematics?
NCERT Solutions for Exercise 7.1 of Chapter 7 ‘Triangle’ of Class 9 Mathematics Triangles is an easy activity that just requires a rudimentary awareness of triangle congruence parameters. To improve their conceptual knowledge of congruence in triangles, students must also solve example problems based on practical circumstances or activities provided in between the exercises. In addition, the solved examples in NCERT Solutions for Exercise 7.1 of Chapter 7 ‘Triangle’ of Class 9 Mathematics will assist students in gaining the proper technique and reasoning to easily tackle the exercise questions.