NCERT Solutions for Exercise 7.2 Class 9 Maths Chapter 7 - Triangles

NCERT Solutions for Exercise 7.2 Class 9 Maths Chapter 7 - Triangles

Edited By Vishal kumar | Updated on Oct 06, 2023 02:32 PM IST

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2- Download Free PDF

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2- Welcome to NCERT Solutions for Class 9 Maths, Chapter 7 - Triangles, Exercise 7.2. This exercise delves deeper into the subject, helping you master the intricacies of triangle geometry. In 9th class maths exercise 7.2 answers, we encounter a range of thought-provoking problems that challenge your understanding and problem-solving skills. Our expert-crafted solutions offer clear explanations and step-by-step guidance to assist you in tackling these challenges. Furthermore, you can easily download the PDF version of these class 9 maths chapter 7 exercise 7.2 for free, ensuring accessibility wherever and whenever you need them.

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  1. NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2- Download Free PDF
  2. NCERT Solutions for Class 9 Maths Chapter 7 – Triangles Exercise 7.2
  3. Access Triangles Class 9 Chapter 7 Exercise: 7.2
  4. More About NCERT Solutions for Class 9 Maths Exercise 7.2
  5. Benefits of NCERT Solutions for Class 9 Maths Exercise 7.2
  6. Key Features of Class 9 Maths Chapter 7 Exercise 7.2
  7. NCERT Solutions of Class 10 Subject Wise
  8. Subject Wise NCERT Exemplar Solutions
NCERT Solutions for Exercise 7.2 Class 9 Maths Chapter 7 - Triangles
NCERT Solutions for Exercise 7.2 Class 9 Maths Chapter 7 - Triangles

Along with NCERT book Class 9 Maths, chapter 9 exercise 7.2 the following exercises are also present.

NCERT Solutions for Class 9 Maths Chapter 7 – Triangles Exercise 7.2

Download PDF

Access Triangles Class 9 Chapter 7 Exercise: 7.2

Q1 (i) In an isosceles triangle ABC, with \small AB=AC , the bisectors of \small \angle B and \small \angle C intersect each other at O. Join A to O. Show that : \small OB=OC

Answer:

In the triangle ABC,

Since AB = AC, thus \angle B\ =\ \angle C

or \frac{1}{2}\angle B\ =\ \frac{1}{2}\angle C

or \angle OBC\ =\ \angle OCB (Angles bisectors are equal)

Thus \small OB=OC as sides opposite to equal are angles are also equal.

Q1 (ii) In an isosceles triangle ABC, with \small AB=AC , the bisectors of \small \angle B and \small \angle C intersect each other at O. Join A to O. Show that : AO bisects \small \angle A

Answer:

Consider \Delta AOB and \Delta AOC ,

(i) AB\ =\ AC (Given)

(ii) AO\ =\ AO (Common in both the triangles)

(iii) OB\ =\ OC (Proved in previous part)

Thus by SSS congruence rule, we can conclude that :

\Delta AOB\ \cong \ \Delta AOC

Now, by c.p.c.t.,

\angle BAO\ =\ \angle CAO

Hence AO bisects \angle A .

Q2 In \Delta ABC , AD is the perpendicular bisector of BC (see Fig). Show that \small \Delta ABC is an isosceles triangle in which \small AB=AC .

1640157417015

Answer:

Consider \Delta ABD and \Delta ADC,

(i) AD\ =\ AD (Common in both the triangles)

(ii) \angle ADB\ =\ \angle ADC (Right angle)

(iii) <BD=<CD

Q3 ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig.). Show that these altitudes are equal.

1640165152512

Answer:

Consider \Delta AEB and \Delta AFC ,

(i) \angle A is common in both the triangles.

(ii) \angle AEB\ =\ \angle AFC (Right angles)

(iii) AB\ =\ AC (Given)

Thus by AAS congruence axiom, we can conclude that :

\Delta AEB\ \cong \Delta AFC

Now, by c.p.c.t. we can say : BE\ =\ CF

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 \small \Delta ABE \cong \Delta ACF

1640165171491

Answer:

Consider \Delta ABE and \Delta ACF ,

(i) \angle A is common in both the triangles.

(ii) \angle AEB\ =\ \angle AFC (Right angles)

(iii) BE\ =\ CF (Given)

Thus by AAS congruence, we can say that :

\small \Delta ABE \cong \Delta ACF

Q4 (ii) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig.). Show that \small AB=AC , i.e., ABC is an isosceles triangle.

1640157444769

Answer:

From the prevoius part of the question we found out that : \Delta ABE\ \cong \Delta ACF

Now, by c.p.c.t. we can say that : AB\ =\ AC

Hence \Delta \ ABC is an isosceles triangle.

Q5 ABC and DBC are two isosceles triangles on the same base BC (see Fig.). Show that \small \angle ABD\ \cong \ \angle ACD .

1640166831546

Answer:

Consider \Delta ABD and \Delta ACD ,

(i) AD\ =\ AD (Common in both the triangles)

(ii) AB\ =\ AC (Sides of isosceles triangle)

(iii) BD\ =\ CD (Sides of isosceles triangle)

Thus by SSS congruency, we can conclude that :

\small \angle ABD\ \cong \ \angle ACD

Q6 \Delta ABC is an isosceles triangle in which AB=AC . Side BA is produced to D such that AD=AB (see Fig.). Show that \angle BCD is a right angle.

1640157465047

Answer:

Consider \Delta ABC,
It is given that AB = AC


So, \angle ACB = \angle ABC (Since angles opposite to the equal sides are equal.)

Similarly in \Delta ACD,

We have AD = AB
and \angle ADC = \angle ACD
So,

\angle CAB + \angle ACB + \angle ABC = 180^{\circ}

\angle CAB\ +\ 2\angle ACB = 180^{\circ}
or \angle CAB\ = 180^{\circ}\ -\ 2\angle ACB ...........................(i)

And in \Delta ADC,
\angle CAD\ = 180^{\circ}\ -\ 2\angle ACD ..............................(ii)

Adding (i) and (ii), we get :
\angle CAB\ +\ \angle CAD\ = 360^{\circ}\ -\ 2\angle ACD\ -\ 2\angle ACB

or 180^{\circ}\ = 360^{\circ}\ -\ 2\angle ACD\ -\ 2\angle ACB

and \angle BCD\ =\ 90^{\circ}

Q7 ABC is a right angled triangle in which \small \angle A =90^{\circ} and \small AB=AC . Find \small \angle B and \small \angle C .

Answer:

In the triangle ABC, sides AB and AC are equal.

We know that angles opposite to equal sides are also equal.

Thus, \angle B\ =\ \angle C

Also, the sum of the interior angles of a triangle is 180^{\circ} .

So, we have :

\angle A\ +\ \angle B\ +\ \angle C\ =\ 180^{\circ}

or 90^{\circ} +\ 2\angle B\ =\ 180^{\circ}

or \angle B\ =\ 45^{\circ}

Hence \angle B\ =\ \angle C\ =\ 45^{\circ}

Q8 Show that the angles of an equilateral triangle are \small 60^{\circ} 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 : \angle A\ =\ \angle B\ =\ \angle C

Also, the sum of the interior angles of a triangle is 180 ^{\circ} .

Hence, \angle A\ +\ \angle B\ +\ \angle C\ =\ 180^{\circ}

or 3\angle A\ =\ 180^{\circ}

or \angle A\ =\ 60^{\circ}

So, all the angles of the equilateral triangle are equal ( 60^{\circ} ).

More About NCERT Solutions for Class 9 Maths Exercise 7.2

In NCERT Solutions for exercise 7.2 class 9 maths, we will begin by understanding the definitions and important terms related to triangles. It's essential to revise the basics before diving into any topic. A triangle is a closed figure formed by three intersecting lines (the prefix "tri" signifies "three"). A triangle consists of three sides, three angles, and three vertices.

Triangle Congruence: Two triangles are considered congruent if the sides and angles of one triangle are equal to the corresponding sides and angles of the other. When PQR and ABC are congruent, we denote it as PQR = ABC.

The NCERT syllabus for Class 9 Maths Exercise 7.2 covers different criteria for establishing the congruence of triangles, which include: Side-Angle-Side, Angle-Side-Angle, Side-Side-Side, and Right-angle-Hypotenuse-Side.

Theorem 1: In an isosceles triangle, the angles opposite the equal sides are equal.

Theorem 2: In a triangle, the sides opposite equal angles are equal.

Some very important properties of triangles which we should know and are helpful in solving this exercise are:

  • The entire angles of a triangle sum up to 180o.
  • The sum of two inside opposite angles forms the external angle formed by a triangle's side.
  • Angles opposing equal sides of a triangle are equal.
  • The sides of a triangle that have equal angles are equal.
  • Each of an equilateral triangle's angles is 60.

Also Read| Triangles Class 9 Notes

Benefits of NCERT Solutions for Class 9 Maths Exercise 7.2

  • Exercise 7.2 Class 9 Maths, is related to triangles.

  • From Class 9 Maths chapter 7 exercise 7.2 introduces us to theorems related to isosceles triangles.

  • Understanding the principles from Class 9 Maths Chapter 7 Exercise 7.2 will assist us in understanding the notions of congruency and isosceles triangle characteristics.

Key Features of Class 9 Maths Chapter 7 Exercise 7.2

  1. Comprehensive Coverage: The 9th class maths exercise 7.2 answers covers various aspects of triangles, including congruence criteria and theorems related to triangle properties.

  2. Clear Explanations: The class 9 maths ex 7.2 solutions provide clear and concise explanations, making it easier for students to understand and apply the concepts.

  3. Step-by-Step Approach: Each ex 7.2 class 9 problem is solved step by step, allowing students to follow the thought process and replicate it in similar problems.

  4. Variety of Problems: The class 9 ex 7.2 includes a range of problems based on different congruence criteria, ensuring that students get a well-rounded practice.

  5. Application of Theorems: The exercise 7.2 class 9 maths helps students apply theorems related to triangles, enhancing their problem-solving skills.

  6. Conceptual Clarity: By solving these problems, students gain a deeper understanding of the properties and congruence criteria of triangles.

  7. Aligned with Curriculum: The content aligns with the Class 9 mathematics curriculum, preparing students for examinations and future topics in geometry.

  8. Free Resource: These solutions are available to students at no cost, making quality math resources accessible to all.

Also, See

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. What is the definition of an isosceles triangle?

An isosceles triangle is one with two equal sides. It has two equal angles as well.

2. How many sides and angles of an isosceles triangle are equal?

An isosceles triangle has two sides and two angles equal to each other.

3. Can we consider an equilateral triangle as an isosceles triangle?

Yes, every equilateral triangle is an isosceles triangle, but every isosceles triangle is not an equilateral triangle.

4. Find the unequal side of a right-angled triangle with equal sides of 10cm?

Yes, every equilateral triangle is an isosceles triangle, but every isosceles triangle is not an equilateral triangle.

5. The ratio in which the angle bisector of the vertex angle divides the base is?

The angle bisector of the vertex angle divides the base into 1:1 ratio which means into two equal parts i.e., it passes through the midpoint of the base.

6. Find the equal angels of the isosceles triangle if the vertex angle is 80 degrees?

The angle bisector of the vertex angle divides the base into 1:1 ratio which means into two equal parts i.e., it passes through the midpoint of the base.

7. Find the angles of the triangles if they are in a ratio of 1 : 2 : 2 ?

The angle bisector of the vertex angle divides the base into 1:1 ratio which means into two equal parts i.e., it passes through the midpoint of the base.

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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)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

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Option 1)

2.45×10−3 kg

Option 2)

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Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

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2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

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K/2\,

Option 2)

\; K\;

Option 3)

zero\;

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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

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33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

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67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

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Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

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be a function of the molecular mass of the substance.

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Option 2)

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