VMC VIQ Scholarship Test
ApplyRegister for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.
NCERT Solutions for Class 9 Maths Chapter 7 Triangles
Edited By Ramraj Saini | Updated on Sep 27, 2023 10:38 PM IST
NCERT Solutions for Class 9 Maths Chapter 7 Triangles
NCERT solutions for class 9 maths chapter 7 Triangles are provided here. These NCERT Solutions are developed by subject matter expert at careersers360 considering latest CBSE syllabus 2023. Also these provide step by step solutions to all NCERT problems in comprehensive and simple way therefore these are easy to understand and ultimately beneficial for exams. In Class 9 Maths NCERT Syllabus Triangles, you will learn something of a higher level. NCERT triangles class 9 questions and answers can be a good tool whenever you are stuck in any of the problems. This Class 9 NCERT book chapter will be covering the properties of triangles like congruence of triangles, isosceles triangle, etc in detail.
Scholarship Test: Vidyamandir Intellect Quest (VIQ)
Don't Miss: JEE Main 2027: Narayana Scholarship Test Preparation Kit for Class 9
This Story also Contains
- NCERT Solutions for Class 9 Maths Chapter 7 Triangles
- Triangles Class 9 Questions And Answers PDF Free Download
- Triangles Class 9 Solutions - Important Formulae And Points
- Triangles Class 9 NCERT Solutions (Intext Questions and Exercise)
- NCERT Solutions For Class 9 - Chapter Wise
- NCERT Solutions For Class 9 - Subject Wise
- NCERT Solutions for Class 9 Maths
- NCERT Books and NCERT Syllabus
By using NCERT class 9 maths chapter 7 question answer, you can prepare 360 degree for your school as well as for the competitive examinations. Here you will get NCERT solutions for class 9 Maths also.
Triangles Class 9 Questions And Answers PDF Free Download
Triangles Class 9 Solutions - Important Formulae And Points
Congruence:
Congruent refers to figures that are identical in all aspects, including their shapes and sizes. For example, two circles with the same radii or two squares with the same side lengths are considered congruent.
Congruent Triangles:
Two triangles are considered congruent if and only if one of them can be superimposed (placed or overlaid) over the other in such a way that they entirely cover each other.
>> Congruence Rules for Triangles:
Side-Angle-Side (SAS) Congruence:
Angle-Side-Angle (ASA) Congruence:
Angle-Angle-Side (AAS) Congruence:.
Side-Side-Side (SSS) Congruence:
Right-Angle Hypotenuse Side (RHS) Congruence:
Free download NCERT Solutions for Class 9 Maths Chapter 7 Triangles for CBSE Exam.
Triangles Class 9 NCERT Solutions (Intext Questions and Exercise)
NCERT solutions for class 10 maths chapter 7 Triangles - Excercise: 7.1
Q1 In quadrilateral ,
and
bisects
(see Fig.). Show that
. What can you say about
and
?
Answer:
In the given triangles we are given that:-
(i)
(ii) Further, it is given that AB bisects angle A. Thus BAC
BAD.
(iii) Side AB is common in both the triangles.
Hence by SAS congruence, we can say that :
By c.p.c.t. (corresponding parts of congruent triangles are equal) we can say that
Q2 (i) is a quadrilateral in which
and
(see Fig. ). Prove that
Answer:
It is given that :-
(i) AD = BC
(ii)
(iii) Side AB is common in both the triangles.
So, by SAS congruence, we can write :
Q2 (ii) is a quadrilateral in which
and
(see Fig.). Prove that
Answer:
In the previous part, we have proved that .
Thus by c.p.c.t. , we can write :
Q2 (iii) is a quadrilateral in which
and
(see Fig.). Prove that
.
Answer:
In the first part we have proved that .
Thus by c.p.c.t. , we can conclude :
Q3 and
are equal perpendiculars to a line segment
(see Fig.). Show that
bisects
.
Answer:
In the given figure consider AOD and
BOC.
(i) AD = BC (given)
(ii) A =
B (given that the line AB is perpendicular to AD and BC)
(iii) AOD =
BOC (vertically opposite angles).
Thus by AAS Postulate, we have
Hence by c.p.c.t. we can write :
And thus CD bisects AB.
Answer:
In the given figure, consider ABC and
CDA :
(i)
(ii)
(iii) Side AC is common in both the triangles.
Thus by ASA congruence, we have :
Answer:
In the given figure consider and
,
(i) (Right angle)
(ii) (Since it is given that I is bisector)
(iii) Side AB is common in both the triangle.
Thus AAS congruence, we can write :
Answer:
In the previous part we have proved that .
Thus by c.p.c.t. we can write :
Thus B is equidistant from arms of angle A.
Answer:
From the given figure following result can be drawn:-
Adding to the both sides, we get :
Now consider and
, :-
(i) (Given)
(ii) (proved above)
(iii) (Given)
Thus by SAS congruence we can say that :
Hence by c.p.c.t., we can say that :
Answer:
From the figure, it is clear that :
Adding both sides, we get :
or
Now, consider and
:
(i) (Proved above)
(ii) (Since P is the midpoint of line AB)
(iii) (Given)
Hence by ASA congruence, we can say that :
Answer:
In the previous part we have proved that .
Thus by c.p.c.t., we can say that :
Answer:
Consider and
,
(i) (Since M is the mid-point)
(ii) (Vertically opposite angles are equal)
(iii) (Given)
Thus by SAS congruency, we can conclude that :
Answer:
In the previous part, we have proved that .
By c.p.c.t. we can say that :
This implies side AC is parallel to BD.
Thus we can write : (Co-interior angles)
and,
or
Hence is a right angle.
Answer:
Consider and
,
(i) (Common in both the triangles)
(ii) (Right angle)
(iii) (By c.p.c.t. from the part (a) of the question.)
Thus SAS congruence we can conclude that :
Answer:
In the previous part we have proved that .
Thus by c.p.c.t., we can write :
or (Since M is midpoint.)
or .
Hence proved.
NCERT Class 9 Maths Chapter 7 Question Answer - Excercise: 7.2
Answer:
In the triangle ABC,
Since AB = AC, thus
or
or (Angles bisectors are equal)
Thus as sides opposite to equal are angles are also equal.
Answer:
Consider and
,
(i) (Given)
(ii) (Common in both the triangles)
(iii) (Proved in previous part)
Thus by SSS congruence rule, we can conclude that :
Now, by c.p.c.t.,
Hence AO bisects .
Q2 In , AD is the perpendicular bisector of BC (see Fig). Show that
is an isosceles triangle in which
.
Answer:
Consider ABD and
ADC,
(i) (Common in both the triangles)
(ii) (Right angle)
(iii) (Since AD is the bisector of BC)
Thus by SAS congruence axiom, we can state :
Hence by c.p.c.t., we can say that :
Thus is an isosceles triangle with AB and AC as equal sides.
Answer:
Consider and
,
(i) is common in both the triangles.
(ii) (Right angles)
(iii) (Given)
Thus by AAS congruence axiom, we can conclude that :
Now, by c.p.c.t. we can say :
Hence these altitudes are equal.
Q4 (i) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig). Show that
Answer:
Consider and
,
(i) is common in both the triangles.
(ii) (Right angles)
(iii) (Given)
Thus by AAS congruence, we can say that :
Answer:
From the prevoius part of the question we found out that :
Now, by c.p.c.t. we can say that :
Hence is an isosceles triangle.
Q5 ABC and DBC are two isosceles triangles on the same base BC (see Fig.). Show that .
Answer:
Consider and
,
(i) (Common in both the triangles)
(ii) (Sides of isosceles triangle)
(iii) (Sides of isosceles triangle)
Thus by SSS congruency, we can conclude that :
Answer:
Consider ABC,
It is given that AB = AC
So, (Since angles opposite to the equal sides are equal.)
Similarly in ACD,
We have AD = AB
and
So,
or ...........................(i)
And in ADC,
..............................(ii)
Adding (i) and (ii), we get :
or
and
Q7 ABC is a right angled triangle in which and
. Find
and
.
Answer:
In the triangle ABC, sides AB and AC are equal.
We know that angles opposite to equal sides are also equal.
Thus,
Also, the sum of the interior angles of a triangle is .
So, we have :
or
or
Hence
Q8 Show that the angles of an equilateral triangle are each.
Answer:
Consider a triangle ABC which has all sides equal.
We know that angles opposite to equal sides are equal.
Thus we can write :
Also, the sum of the interior angles of a triangle is .
Hence,
or
or
So, all the angles of the equilateral triangle are equal ( ).
Class 9 Triangles NCERT Solutions - Excercise: 7.3
Answer:
Consider and
,
(i) (Common)
(ii) (Isosceles triangle)
(iii) (Isosceles triangle)
Thus by SSS congruency we can conclude that :
Answer:
Consider and
,
(i) is common side in both the triangles.
(ii) (This is obtained from the c.p.c.t. as proved in the previous part.)
(iii) (Isosceles triangles)
Thus by SAS axiom, we can conclude that :
Answer:
In the first part, we have proved that .
So, by c.p.c.t. .
Hence AP bisects .
Now consider and
,
(i) (Common)
(ii) (Isosceles triangle)
(iii) (by c.p.c.t. from the part (b))
Thus by SSS congruency we have :
Hence by c.p.c.t. we have :
or AP bisects .
Answer:
In the previous part we have proved that .
Thus by c.p.c.t. we can say that :
Also,
SInce BC is a straight line, thus :
or
or
Hence it is clear that AP is a perpendicular bisector of line BC.
Q2 (i) AD is an altitude of an isosceles triangle ABC in which . Show that AD bisects BC
Answer:
Consider and
,
(i) (Given)
(ii) (Common in both triangles)
(iii)
Thus by RHS axiom we can conclude that :
Hence by c.p.c.t. we can say that : or AD bisects BC.
Q2 (ii) AD is an altitude of an isosceles triangle ABC in which . Show that AD bisects
.
Answer:
In the previous part of the question we have proved that
Thus by c.p.c.t., we can write :
Hence bisects
.
(i)
(ii)
Answer:
(i) From the figure we can say that :
or
or
Now, consider and
,
(a) (Given)
(b) (Given)
(c) (Prove above)
Thus by SSS congruence rule, we can conclude that :
(ii) Consider and
:
(a) (Given)
(b) (by c.p.c.t. from the above proof)
(c) (Given)
Thus by SAS congruence rule,
Answer:
Using the given conditions, consider and
,
(i) (Right angle)
(ii) (Common in both the triangles)
(iii) (Given that altitudes are of the same length. )
Thus by RHS axiom, we can say that :
Hence by c.p.c.t.,
And thus (sides opposite to equal angles are also equal). Thus ABC is an isosceles triangle.
Q5 ABC is an isosceles triangle with . Draw
to show that
.
Answer:
Consider and
,
(i) (Since it is given that AP is altitude.)
(ii) (Isosceles triangle)
(iii) (Common in both triangles)
Thus by RHS axiom we can conclude that :
Now, by c.p.c.t.we can say that :
Class 9 maths chapter 7 NCERT solutions - Excercise: 7.4
Q1 Show that in a right-angled triangle, the hypotenuse is the longest side.
Answer:
Consider a right-angled triangle ABC with right angle at A.
We know that the sum of the interior angles of a triangle is 180.
So,
or
or
Hence and
are less than
(
).
Also, the side opposite to the largest angle is also the largest.
Hence the side BC is largest is the hypotenuse of the .
Hence it is proved that in a right-angled triangle, the hypotenuse is the longest side.
Q2 In Fig, sides AB and AC of are extended to points P and Q respectively. Also,
. Show that
.
Answer:
We are given that,
......................(i)
Also, (Linear pair of angles) .....................(ii)
and (Linear pair of angles) .....................(iii)
From (i), (ii) and (iii) we can say that :
Thus ( Sides opposite to the larger angle is larger.)
Answer:
In this question, we will use the property that sides opposite to larger angle are larger.
We are given and
.
Thus, ..............(i)
and ...............(ii)
Adding (i) and (ii), we get :
or
Hence proved.
Answer:
Consider in the above figure :
(Given)
Thus (as angle opposite to smaller side is smaller)
Now consider ,
We have :
and
Adding the above result we get,
or
Similarly, consider ,
we have
Therefore
and in we have,
and
from the above result we have,
or
Hence proved.
Q5 In Fig , and PS bisects
. Prove that
.
Answer:
We are given that .
Thus
Also, PS bisects , thus :
Now, consider ,
(Exterior angle)
Now, consider ,
Thus from the above the result we can conclude that :
Answer:
Consider a right-angled triangle ABC with right angle at B.
Then (Since
)
Thus the side opposite to largest angle is also largest.
Hence the given statement is proved that all line segments are drawn from a given point, not on it, the perpendicular line segment is the shortest.
Triangles class 9 NCERT solutions - Excercise: 7.5
Q1 ABC is a triangle. Locate a point in the interior of which is equidistant from all the vertices of
.
Answer:
We know that circumcenter of a triangle is equidistant from all the vertices. Also, circumcenter is the point of intersection of the perpendicular bisectors of the sides of a triangle.
Thus, draw perpendicular bisectors of each side of the triangle ABC. And let them meet at a point, say O.
Hence O is the required point which is equidistant from all the vertices.
Answer:
The required point is called in-centre of the triangle. This point is the intersection of the angle bisectors of the interior angles of a triangle.
Hence the point can be found out in this case just by drawing angle bisectors of all the angles of the triangle.
Q3 In a huge park, people are concentrated at three points (see Fig.):
A : where there are different slides and swings for children,
B : near which a man-made lake is situated,
C : which is near to a large parking and exit.
Where should an icecream parlour be set up so that maximum number of persons can approach it? ( Hint : The parlour should be equidistant from A, B and C)
Answer:
The three main points form a triangle ABC. Now we have to find a point which is equidistant from all the three points.
Thus we need to find the circumcenter of the .
We know that circumcenter is defined as the point as the intersection point of the perpendicular bisectors of the sides of the triangle.
Hence the required point can be found out by drawing perpendicular bisectors of .
Answer:
For finding the number of triangles we need to find the area of the figure.
Consider the hexagonal structure :
Area of hexagon = 6 Area of 1 equilateral
Thus area of the equilateral triangle :
or
or
So, the area of the hexagon is :
And the area of an equilateral triangle having 1cm as its side is :
or
Hence a number of equilateral triangles that can be filled in hexagon are :
Similarly for star-shaped rangoli :
Area :
Thus the number of equilateral triangles are :
Hence star-shaped rangoli has more equilateral triangles.
NCERT Triangles Class 9 Chapter 7 - Topics
- Congruence of Triangles
- Criteria for Congruence of Triangles
- Some Properties of a Triangle
- Some More Criteria for Congruence of Triangles
- Inequalities in a Triangle
More About NCERT Solutions For Class 9 Maths Chapter 7 Triangles
In ch 7 maths class 9, there are a total of 5 exercises with 31 questions in them. NCERT solutions for class 9 maths chapter 7 Triangles is covering the entire chapter including the optional exercises. The chapter is full of properties and theorems that's why the examples and theorems are as important as the practice exercises. There is another aspect to look at the importance of this chapter, apart from school exams this is an essential part of competitive examinations like- CAT, SSC, NTSE, INO, etc.
Also practice class 9 maths ch 7 question answer using the exercise given below.
- Triangles Class 9 NCERT Exercise 7.1
- Triangles Class 9 NCERT Exercise 7.2
- Triangles Class 9 NCERT Exercise 7.3
- Triangles Class 9 NCERT Exercise 7.4
NCERT Solutions For Class 9 - Chapter Wise
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | Triangles |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
NCERT Solutions For Class 9 - Subject Wise
How To Use NCERT Solutions For Class 9 Maths Chapter 7 Triangles
- Learn some basic properties of triangles using some books of previous classes.
- Go through all the theorems and examples given in the chapter.
- Once you have memorized all the theorems and properties, then you can practice the questions from practice exercises
- During the practice, you can use NCERT solutions for class 9 maths chapter 7 triangles as a helpmate.
- After doing all the exercises you can do some questions from past papers.
NCERT Books and NCERT Syllabus
Keep Working Hard and Happy Learning!
Frequently Asked Questions (FAQs)
1. What are the important topics in chapter 7 maths class 9 ?
Congruence of triangles, Criteria for congruence of triangles, Properties of triangles, Inequalities of triangles are the important topics covered in this chapter. Students can practice NCERT solutions for class 9 maths to get command in these concepts that ultimately help during the exams.
2. In NCERT Solutions for class 9 chapter 7 maths, what does the concept of "congruence of triangles" signify?
NCERT Solutions for maths chapter 7 class 9 explain that "congruence of triangles" refers to the condition where two triangles are identical copies of each other and overlap perfectly when superimposed. In simpler terms, two triangles are considered congruent when the angles and sides of one triangle are equivalent to the corresponding angles and sides of the other triangle.
3. Where can I find the complete triangles class 9 solutions ?
Here you will get the detailed NCERT solutions for class 9 maths by clicking on the link. you can practice these solutions to command the concepts.
4. In what ways can NCERT Solutions for Class 9 Maths Chapter 7 assist in preparing for CBSE exams?
NCERT Solutions for Class 9 Maths Chapter 7 can assist students in achieving a high score and excelling in the subject in their CBSE exams. These solutions are designed based on the latest CBSE syllabus and cover all the essential topics in the respective subject. By practicing these solutions, students can gain confidence and be better prepared to face the board exams. The topics covered in these solutions are fundamental and contribute significantly to obtaining top scores. Moreover, solving problems of varying difficulty levels helps students get accustomed to answering questions of all types. Thus, these solutions are highly recommended for students as a reference and for practice in preparation for their CBSE exams.
5. How many chapters are there in the CBSE class 9 maths ?
There are 15 chapters starting from the number system to probability in the CBSE class 9 maths.
Also Read
Articles
Upcoming School Exams
Admit Card Date:04 October,2024 - 29 November,2024
Admit Card Date:04 October,2024 - 29 November,2024
Application Date:07 October,2024 - 22 November,2024
Application Correction Date:08 October,2024 - 27 November,2024
Get answers from students and experts
Popular Questions
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
In the reaction,
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
| Option 1)
0.02 |
Option 2)
3.125 × 10-2 |
| Option 3)
1.25 × 10-2 |
Option 4)
2.5 × 10-2 |
Colleges After 12th
Popular Course After 12th
Applications for Admissions are open.
JEE Main Important Physics formulas
ApplyAs per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
JEE Main Important Chemistry formulas
ApplyAs per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
TOEFL ® Registrations 2024
ApplyAccepted by more than 11,000 universities in over 150 countries worldwide
Pearson | PTE
ApplyRegister now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 30th NOV'24! Trusted by 3,500+ universities globally
JEE Main high scoring chapters and topics
ApplyAs per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE
News and Notifications
Back to top