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# NCERT Solutions for Class 7 Maths Chapter 6 - The Triangle and its Properties Exercise 6.5

Last updated date: 20th Jul 2024
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## NCERT Class 7 Maths Exercise 6.5 Solutions for Chapter 6 - FREE PDF Download

NCERT Solutions for Class 7 Maths Chapter 6 explores the fascinating world of triangles and their properties. This chapter delves into the fundamental aspects of triangles, including their types, properties, and the relationships between their angles and sides. Learn about the various kinds of triangles, such as equilateral, isosceles, and scalene, and understand important concepts like the angle sum property and exterior angle property in Class 7th Maths Chapter 6 Exercise 6.5.

Table of Content
1. NCERT Class 7 Maths Exercise 6.5 Solutions for Chapter 6 - FREE PDF Download
2. Glance on NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5 | Vedantu
3. Formulas Used in Class 7th Maths Exercise 6.5
4. Access Maths Class 7 NCERT Chapter 6 The Triangle and its Properties Exercise 6.5
5. Class 7 Maths Chapter 6: Exercises Breakdown
6. CBSE Class 7 Maths Chapter 6 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 7 Maths
FAQs

One of the key highlights of this chapter is the Pythagoras Theorem, a fundamental principle used to solve problems involving right-angled triangles. Through engaging exercises and practical examples, you'll gain a thorough understanding of how to apply these properties and theorems to solve geometric problems. Get ready to enhance your mathematical skills and deepen your understanding of one of the most essential shapes in geometry!

## Glance on NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5 | Vedantu

• Pythagorean Theorem is a fundamental theorem in geometry that states: in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

• Here's the formula for the Pythagorean Theorem: a² + b² = c², where: a, b are the lengths of the two shorter sides (legs) of the right triangle and c is the length of the hypotenuse of the right triangle.

• Important concepts covered in Exercise 6.5:

• Identifying right-angled triangles based on the Pythagorean Theorem.

• Applying the Pythagorean Theorem to find the missing side of a right-angled triangle.

• Understanding the converse of the Pythagorean Theorem: If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is right-angled.

• There are 8 questions in class 7 maths Ex 6.5 which are fully solved by experts at Vedantu.

## Formulas Used in Class 7th Maths Exercise 6.5

• Pythagorean Theorem: a² + b² = c²

## Access Maths Class 7 NCERT Chapter 6 The Triangle and its Properties Exercise 6.5

1. $PQR$ is a triangle, right angled at $P$. If $PQ = 10$cm and $PR = 24$cm, find $QR$.

Ans: Given: $PQ = 10$cm, $PR = 24$cm

Let $QR$ be $x$cm.

In right angled triangle $QPR$,

By Pythagoras Theorem, ${\text{Hypotenus}}{{\text{e}}^2} = {\text{ Bas}}{{\text{e}}^2} + {\text{ Perpendicula}}{{\text{r}}^2}$

${\left( {QR} \right)^2} = {\left( {PQ} \right)^2} + {\left( {PR} \right)^2}$

${\left( x \right)^2} = {\left( {10} \right)^2} + {\left( {24} \right)^2}$

${\left( x \right)^2} = 100 + 576 = 676$

$x = \sqrt {676} = 26cm$

Thus, the length of  $QR$ is $26cm$.

2. $ABC$ is a triangle, right angled at $C$. If $AB = 25$cm and $AC = 7$cm, find $BC$.

Ans: Given: $AB = 25$cm, $AC = 7$cm

Let $BC$ be $x$cm.

In right angled triangle $ACB$,

By Pythagoras Theorem, ${\text{Hypotenus}}{{\text{e}}^2} = {\text{ Bas}}{{\text{e}}^2} + {\text{ Perpendicula}}{{\text{r}}^2}$

${\left( {AB} \right)^2} = {\left( {AC} \right)^2} + {\left( {BC} \right)^2}$

${\left( {25} \right)^2} = {\left( 7 \right)^2} + {\left( x \right)^2}$

$625 = 49 + {x^2}$

${x^2} = 625 - 49 = 576$

$x = \sqrt {576} = 24cm$

Thus, the length of  $BC$is $24cm$.

3. A $15$m long ladder reached a window $12$m high from the ground on placing it against a wall at a distance $a$. Find the distance of the foot of the ladder from the wall.

Ans: Let $AC$ be the ladder and $A$ be the window.

Given: $AC = 15$m, $AB = 12$m

Let $CB$ be $a$m.

In right angled triangle $ACB$,

By Pythagoras Theorem, ${\text{Hypotenus}}{{\text{e}}^2} = {\text{ Bas}}{{\text{e}}^2} + {\text{ Perpendicula}}{{\text{r}}^2}$

${\left( {AC} \right)^2} = {\left( {CB} \right)^2} + {\left( {AB} \right)^2}$

${\left( {15} \right)^2} = {\left( {12} \right)^2} + {\left( a \right)^2}$

$225 = {a^2} + 144$

${a^2} = 225 - 144 = 81$

$x = \sqrt {81} = 9m$

Hence, the distance of the foot of the ladder from the wall is $9m$.

4. Which of the following can be the sides of a right triangle? In the case of right angled triangles, identify the right angles.

(i) $2.5$cm, $6.5$cm, $6$cm

Ans: Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem,

${\text{Hypotenus}}{{\text{e}}^2} = {\text{ Bas}}{{\text{e}}^2} + {\text{ Perpendicula}}{{\text{r}}^2}$

In $\vartriangle ABC$,               ${\left( {AC} \right)^2} = {\left( {AB} \right)^2} + {\left( {BC} \right)^2}$

$L.H.S. = {\left( {6.5} \right)^2} = 42.25cm$

$R.H.S. = {\left( 6 \right)^2} + {\left( {2.5} \right)^2} = 36 + 6.25 = 42.25cm$

Since, $L.H.S. = R.H.S.$

Therefore, the given sides form a right angled triangle.

Right angle lies on the opposite to the greater side $6.5$cm, i.e., at $B$.

(ii) $2$cm, $2$cm, $5$cm

Ans: Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem,

${\text{Hypotenus}}{{\text{e}}^2} = {\text{ Bas}}{{\text{e}}^2} + {\text{ Perpendicula}}{{\text{r}}^2}$

${\left( 5 \right)^2} = {\left( 2 \right)^2} + {\left( 2 \right)^2}$

$L.H.S. = {\left( 5 \right)^2} = 25cm$

$R.H.S. = {\left( 2 \right)^2} + {\left( 2 \right)^2} = 4 + 4 = 8cm$

Since, $L.H.S. \ne R.H.S.$

Therefore, the given sides do not form a right angled triangle.

(iii) $1.5$cm, $2$cm, $2.5$cm

Ans: Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem,

${\text{Hypotenus}}{{\text{e}}^2} = {\text{ Bas}}{{\text{e}}^2} + {\text{ Perpendicula}}{{\text{r}}^2}$

In $\vartriangle PQR$,               ${\left( {\operatorname{P} R} \right)^2} = {\left( {PQ} \right)^2} + {\left( {RQ} \right)^2}$

$L.H.S. = {\left( {2.5} \right)^2} = 6.25cm$

$R.H.S. = {\left( 2 \right)^2} + {\left( {1.5} \right)^2} = 4 + 2.25 = 6.25cm$

Since, $L.H.S. = R.H.S.$

Therefore, the given sides form a right angled triangle.

Right angle lies on the opposite to the greater side $2.5$cm, i.e., at $Q$.

5. A tree is broken at a height of $5$m from the ground and its top touches the ground at a distance of $12$m from the base of the tree. Find the original height of the tree.

Ans: Let $A'CB$ represents the tree before it broken at the point $C$ and let the

top $A'$ touches the ground at $A$ after it broke. Then $\vartriangle ABC$ is a right angled triangle, right angled at $B$.

$AB = 12$m and $BC = 5$m

Using Pythagoras theorem, In $\vartriangle ABC$

${\left( {AC} \right)^2} = {\left( {AB} \right)^2} + {\left( {BC} \right)^2}$

${\left( {AC} \right)^2} = {\left( {12} \right)^2} + {\left( 5 \right)^2}$

${\left( {AC} \right)^2} = 144 + 25$

${\left( {AC} \right)^2} = 169$

$AC = 13m$

The total height of the tree is sum of sides $AC$ and $CB$ that is$AC + CB = 13 + 5 = 18m$.

6. Angles $Q$ and $R$ of a $\vartriangle PQR$ are ${25^ \circ }$ and ${65^ \circ }$.

Write which of the following is true:

(i) $P{Q^2} + Q{R^2} = R{P^2}$

(ii) $P{Q^2} + R{P^2} = Q{R^2}$

(iii) $R{P^2} + Q{R^2} = P{Q^2}$

Ans: In $\vartriangle PQR$,

$\angle PQR + \angle QRP + \angle RPQ = {180^ \circ }$                 By Angle sum property of a [$\vartriangle$]

${25^ \circ } + {65^ \circ } + \angle RPQ = {180^ \circ }$              $\Rightarrow$          ${90^ \circ } + \angle RPQ = {180^ \circ }$

$\angle RPQ = {180^ \circ } - {90^ \circ } = {90^ \circ }$

Thus, $\vartriangle PQR$ is a right angled triangle, right angled at $P$.

$\therefore {\text{ Hypotenus}}{{\text{e}}^2} = {\text{ Bas}}{{\text{e}}^2} + {\text{ Perpendicula}}{{\text{r}}^2}$       (By Pythagoras theorem)

$Q{R^2} = P{Q^2} + R{P^2}$

Hence, Option (ii) is correct.

7. Find the perimeter of the rectangle whose length is $40$cm and a diagonal is $41$cm.

Ans: Given diagonal $PR = 41$cm, length $PR = 40$cm

Let breadth $\left( {QR} \right)$ be $x$cm.

Now, in right angled triangle $PQR$,

${\left( {PR} \right)^2} = {\left( {RQ} \right)^2} + {\left( {PQ} \right)^2}$                         (By Pythagoras theorem)

${\left( {41} \right)^2} = {\left( x \right)^2} + {\left( {40} \right)^2}$

$1681 = {\left( x \right)^2} + 1600$

${x^2} = 1681 - 1600$

${x^2} = 81$

$x = \sqrt {81} = 9cm$

Therefore the breadth of the rectangle is $9$cm.

${\text{Perimeter of rectangle = 2}}\left( {{\text{length + breadth}}} \right)$

$= 2\left( {9 + 49} \right)$

$= 2 \times 49 = 98cm$

Hence the perimeter of the rectangle is $98$cm.

8. The diagonals of a rhombus measure $16$cm and $30$cm. Find its perimeter.

Ans: Given: Diagonals $AC = 30$cm and $DB = 16$cm.

Since the diagonals of the rhombus bisect at right angle to each other so it divides the diagonals into equal halves.

Therefore, $OD = \frac{{DB}}{2} = \frac{{16}}{2} = 8cm$

And $OC = \frac{{AC}}{2} = \frac{{30}}{2} = 15cm$

Now, In right angle triangle $DOC$,

${\left( {DC} \right)^2} = {\left( {OD} \right)^2} + {\left( {OC} \right)^2}$                                   (By Pythagoras theorem)

${\left( {DC} \right)^2} = {\left( 8 \right)^2} + {\left( {15} \right)^2}$

${\left( {DC} \right)^2} = 64 + 225 = 289$

$DC = \sqrt {289} = 17cm$

${\text{Perimeter of rhombus = }}4 \times {\text{Side}}$

$= 4 \times 17 = 68cm$

Thus, the perimeter of rhombus is $68$ cm.

## Conclusion

In conclusion, Exercise 6.5 of Chapter 6 in Class 7 Maths is crucial for mastering the properties of right-angled triangles, especially the Pythagoras Theorem. Through this exercise, students gain practical experience in applying the theorem to find missing sides and verify given triangles. By working through these problems, learners not only strengthen their understanding of geometric principles but also build a solid foundation for more advanced mathematical concepts. Mastery of these skills is essential for academic success in geometry and related fields.

From the analysis of previous exams, the number of questions asked from Class 7 Maths Chapter 6 Exercise 6.5 typically ranges from 2 to 4 questions per exam. These questions cover various aspects of the chapter, including types of triangles, properties of triangles, the Pythagoras theorem, and application-based problems involving angle sum property and exterior angle property​

## Class 7 Maths Chapter 6: Exercises Breakdown

 Exercise Number of Questions Exercise 6.1 3 Questions & Solutions Exercise 6.2 2 Questions & Solutions Exercise 6.3 2 Questions & Solutions Exercise 6.4 6 Questions & Solutions

## Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 7 Maths Chapter 6 - The Triangle and its Properties Exercise 6.5

1. What are the different types of triangles covered in Class 7 Maths Chapter 6.5?

Chapter 6 covers three main types of triangles:

• Equilateral Triangle: All sides and angles are equal.

• Isosceles Triangle: Two sides and two angles are equal.

• Scalene Triangle: All sides and angles are different.

2. What is the angle sum property of a triangle in Class 7 Maths Ex 6.5 Solutions?

The angle sum property of a triangle states that the sum of the interior angles of a triangle is always 180 degrees.

3. How is the Pythagoras Theorem applied in this Class 7 Math Exercise 6.5?

The Pythagoras Theorem is applied to right-angled triangles in this chapter. It states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides: a2+b2=c2a2+b2=c2.

4. What kind of questions can be expected from Class 7th Exercise 6.5 in exams?

Typical questions include:

• Finding unknown angles using the angle sum property.

• Verifying if given side lengths form a right-angled triangle using the Pythagoras Theorem.

• Problems related to the properties of different types of triangles.