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

Last updated date: 17th Sep 2024
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NCERT Solutions for Maths Class 7 Chapter 6 The Triangle and its Properties - FREE PDF Download

Class 7 Maths NCERT Solutions for Chapter 6 Triangle and Its Properties PDF is now available online on the official website of Vedantu. This chapter introduces the basic concepts of integration, including various techniques and applications. Students can download and practice the problems free of cost. With the help of these solutions, students can gain a good command of the chapter The Triangles and its Properties Class 7. Vedantu provides clear and updated solutions for CBSE Class 7 Maths Syllabus to help students understand these concepts easily. They can practice and clear their doubts online on the official website.

Table of Content
1. NCERT Solutions for Maths Class 7 Chapter 6 The Triangle and its Properties - FREE PDF Download
2. Glance on Maths Chapter 6 Class 7 - The Triangle and its Properties | Vedantu
3. Access Exercise wise NCERT Solutions for Chapter 6 Maths Class 7
4. Exercises Under NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties
5. Access NCERT Solutions for Class 7 Maths Chapter 6 – The Triangle and its Properties
5.1Exercise 6.1
5.2Exercise 6.2
5.3Exercise 6.3
5.4Exercise 6.4
5.5Exercise 6.5
6. Class 7 Maths Chapter 6: Exercises Breakdown
7. Other Study Material for CBSE Class 7 Maths Chapter 6
8. Chapter-Specific NCERT Solutions for Class 7 Maths
FAQs

Glance on Maths Chapter 6 Class 7 - The Triangle and its Properties | Vedantu

• Chapter 6 of Triangles and its Properties Class 7 Maths, focuses on the different properties and types of triangles, including the Pythagorean theorem and triangle inequalities.

• Types of Triangles are categorized by their sides (isosceles, equilateral, scalene) and angles (acute, obtuse, right-angled).

• An isosceles triangle has two equal sides, an equilateral triangle has three equal sides, and a scalene triangle has no equal sides are properties of Triangles .

• In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides then it is known as Pythagorean Theorem.

• Triangles are congruent if they meet criteria like SSS, SAS, ASA, or RHS, indicating they have the same shape and size then it is Congruence Criteria.

• Medians connect a vertex to the midpoint of the opposite side, altitudes are perpendicular lines from a vertex to the opposite side, and angle bisectors split an angle into two equal parts.

• This article contains chapter notes, important questions, exemplar solutions, exercises and video links for Chapter 6 The Triangles and its Properties Class 7, which you can download as PDFs.

• There are five exercises (21 fully solved questions) in class 7th maths chapter 6 The Triangles and its Properties Class 7.

Exercises Under NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties

6.1 Introduction

Students may learn about triangles in earlier classes while learning shapes. Here, students can understand what a triangle is, its types, angles and properties. Triangles have two different classifications based on sides and angles. These are Equilateral, Scalene and Isosceles. These are based on sides and Acute Angle, Obtuse Angle and Right Angle triangles.

6.2 Medians of Triangle

In this topic, the median of a triangle is defined as the line which connects the vertex and midpoint of the opposite side. It is explained more clearly in the PDF with an example and a detailed explanation. So, students need to have a glance at NCERT Solutions for Class 7 Maths Chapter 6 PDF.

6.3 Altitudes of a Triangle

Students need to find out the height of a triangle to learn about the altitudes of the triangle. The altitude can be defined as a line segment having two ends, one at a vertex and the other at the line on the opposite side. Students may have to solve some problems based on these to understand them more clearly.

6.4 Exterior Angle of a Triangle and its Property

Students need to remember a law that states that the exterior angle of a triangle is equal to the sum of opposite interior angles. This is the property of the exterior angle concerning the triangle. NCERT Solution for Class 7 Maths Chapter 6 PDF has explained and derived this law. It is beneficial to check out the PDF once or twice to understand the topic better.

6.5 Angle Sum Property of a Triangle

In this section of NCERT Class 7 Maths Chapter 6 Solutions, students have to learn a crucial property called Angle Sum Property. It deals with all the three angles of a triangle. The sum of all three angles is 180 °. To prove this, students need four activities and can use exterior angle property also.

Access NCERT Solutions for Class 7 Maths Chapter 6 – The Triangle and its Properties

Exercise 6.1

1. In $\vartriangle PQR$, $D$ is the mid-point of $\overline {QR}$.

Fill the following blanks.

$\overline {PM}$ is ___________

$PD$ is _______

Is $QM = MR$?

Ans: It is given a $\vartriangle PQR$, $D$ is the mid-point of $\ overline {QR}$. It means $QD = DR$. Since $\overline {PM}$ is perpendicular to the side $QR$ of triangle $\vartriangle PQR$, $\overline {PM}$ is the altitude of  $\vartriangle PQR$. Since $QD = DR$, $PD$ is the median of triangle $\vartriangle PQR$.

No, $QM \ne MR$ because $D$ is the mid-point of $\overline {QR}$. It means $QD = DR$.

2. Draw rough sketches for the following:

a) In $\vartriangle ABC$, $BE$ is a median.

Ans: A median of a triangle is a line segment that is drawn from a vertex to the opposite side of the vertex and it divides the opposite side into two equal parts. The rough sketch of $\vartriangle ABC$, where $BE$ is a median, is drawn below.

Here $BE$ is a median in $\vartriangle ABC$ and $AE = EC$.

(b) In $\vartriangle PQR$, $PQ$ and $PR$ are altitudes of the triangle.

Ans: An altitude of a triangle is defined as a perpendicular drawn from the vertex to the line containing the opposite side of the triangle. The rough sketch of $\vartriangle PQR$, where $PQ$ and $PR$ are the altitudes of the triangle is drawn below.

(c) In $\vartriangle XYZ$, $YL$ is an altitude in the exterior of the triangle

Ans: An altitude of a triangle is defined as a perpendicular drawn from the vertex to the line containing the opposite side of the triangle. The rough sketch of $\vartriangle XYZ$, $YL$ is an altitude in the exterior of the triangle is drawn below.

3. Verify by drawing a diagram if the median and altitude of an isosceles triangle can be the same.

Ans: It is given an isosceles triangle. It is required to verify if the median and altitude of the given triangle can be the same. To do this, construct an isosceles triangle. An isosceles triangle has two equal sides.

Construct an isosceles triangle $\vartriangle ABC$ with sides$AB = AC$ and draw a median $AL$ that divides the base of the triangle into two equal parts.

From the triangle, it can be seen that the median makes a ${90^ \circ }$ angle with the base $BC$. So, $AL$ is the altitude of the triangle $\vartriangle ABC$. Hence verified, $AL$ is the median and altitude of the given triangle $\vartriangle ABC$.

Exercise 6.2

1. Find the value of the unknown exterior angle $x$ in the following diagrams.

(i)

Ans: According to the exterior angle property, the sum of the opposite non-adjacent interior angles equals the exterior angle of a triangle.

Hence,

$x = 70^\circ + 50^\circ \\$

$= 120^\circ \\$

The value of $x$ is $120^\circ$.

(ii)

Ans: According to the exterior angle property, the sum of the opposite non-adjacent interior angles equals the exterior angle of a triangle.

Hence,

$x = 65^\circ + 45^\circ \\$

$= 110^\circ \\$

The value of $x$ is $110^\circ$.

(iii)

Ans: According to the exterior angle property, the sum of the opposite non-adjacent interior angles equals the exterior angle of a triangle.

Hence,

$x = 30^\circ + 40^\circ \\$

$= 70^\circ \\$

The value of $x$ is $70^\circ$.

(iv)

Ans: According to the exterior angle property, the sum of the opposite non-adjacent interior angles equals the exterior angle of a triangle.

Hence,

$x = 60^\circ + 60^\circ \\$

$= 120^\circ \\$

The value of $x$ is $120^\circ$.

(v)

Ans: According to the exterior angle property, the sum of the opposite non-adjacent interior angles equals the exterior angle of a triangle.

Hence,

$x = 50^\circ + 50^\circ \\$

$= 100^\circ \\$

The value of $x$ is ${100^ \circ }$.

(vi) Find the value of $x$ in the following diagram.

Ans: According to the exterior angle property, the sum of the opposite non-adjacent interior angles equals the exterior angle of a triangle.

Hence,

$x = 30^\circ + 60^\circ \\$

$= 90^\circ \\$

The value of $x$ is $90^\circ$.

2. Find the value of the unknown exterior angle $x$ in the following diagrams.

(i)

Ans: According to the exterior angle property, the sum of the opposite non-adjacent interior angles equals the exterior angle of a triangle.

Hence,

$x + 50^\circ = 115^\circ$

Subtract ${50^ \circ }$ from both sides of the equation.

$x = 115^\circ - 50^\circ \\$

$= 65^\circ \\$

The value of $x$ is $65^\circ$.

(ii)

Ans: According to the exterior angle property, the sum of the oppositenon-adjacent interior angles equals the exterior angle of a triangle.

$x + {70^ \circ } = 100^\circ$

Subtract $70^\circ$ from both sides of the equation.

$x = 100^\circ - 70^\circ \\$

$= 30^\circ \\$

The value of $x$ is $30^\circ$.

(iii)

Ans: According to the exterior angle property, the sum of the opposite non-adjacent interior angles equals the exterior angle of a triangle.

$x + 90^\circ = 125^\circ$

Subtract $90^\circ$ from both sides of the equation.

$x + 90^\circ - 90^\circ = 125^\circ - 90^\circ \\$

$x = 35^\circ \\$

The value of $x$ is $35^\circ$.

(iv)

Ans: According to the exterior angle property, the sum of the opposite non-adjacent interior angles equals the exterior angle of a triangle.

$x + 60^\circ = 120^\circ$

Subtract $60^\circ$ from both sides of the equation.

$x + 60^\circ - 60^\circ = 120^\circ - 60^\circ \\$

$x= 60^\circ \\$

The value of $x$ is $60^\circ$.

(v)

Ans: According to the exterior angle property, the sum of the opposite non-adjacent interior angles equals the exterior angle of a triangle.

$x + 30^\circ = 80^\circ$

Subtract $30^\circ$ from both sides of the equation.

$x + 30^\circ - 30^\circ = 80^\circ - 30^\circ \\$

$x= 50^\circ \\$

The value of $x$ is $50^\circ$.

(vi)

Ans: According to the exterior angle property, the sum of the opposite non-adjacent interior angles equals the exterior angle of a triangle.

$x + 35^\circ = 75^\circ$

Subtract $35^\circ$ from both sides of the equation.

$x + 35^\circ - 35^\circ = 75^\circ - 35^\circ \\$

$x= 40^\circ \\$

The value of $x$ is $40^\circ$.

Exercise 6.3

1. Find the value of the unknown $x$  in the following diagrams:

i.

Ans: The sum of the internal angles of a triangle is $180^\circ$. In $\vartriangle ABC$, $\angle BAC + \angle ACB + \angle ABC = 180^\circ$.

$\Rightarrow x + 60^\circ + 50^\circ = 180^\circ$

$\Rightarrow x + 110^\circ = 180^\circ \\$

Subtract $110^\circ$ from both sides and simplify.

$\Rightarrow x + 110^\circ - 110^\circ = 180^\circ - 110^\circ$

$\Rightarrow x = 70^\circ$

The value of $x$ is $70^\circ$.

ii.

Ans: The sum of the internal angles of a triangle is $180^\circ$. In $\vartriangle PQR$, $\angle RPQ + \angle PQR + \angle QRP = 180^\circ$.

$\Rightarrow 90^\circ + 30^\circ + x = 180^\circ$

$\Rightarrow x + 120^\circ = 180^\circ$

Subtract $120^\circ$ from both sides and simplify.

$\Rightarrow x + 120^\circ - 120^\circ = 180^\circ - 120^\circ$

$\Rightarrow x = 60^\circ$

The value of $x$ is $60^\circ$.

iii.

Ans: The sum of the internal angles of a triangle is $180^\circ$. In $\vartriangle XYZ$, $\angle ZXY + \angle XYZ + \angle YZX = 180^\circ$.

$\Rightarrow 30^\circ + 110^\circ + x = 180^\circ$

$\Rightarrow x + 140^\circ = 180^\circ$

Subtract $140^\circ$ from both sides and simplify.

$\Rightarrow x + 140^\circ - 140^\circ = 180^\circ - 140^\circ$

$\Rightarrow x = 40^\circ$

The value of $x$ is $40^\circ$.

iv.

Ans: The sum of the internal angles of a triangle is $180^\circ$. In the given triangle, $50^\circ + x + x = 180^\circ$.

Hence,

$50^\circ + 2x = 180^\circ$

Subtract $50^\circ$ from both sides and simplify.

$\Rightarrow 50^\circ - 50^\circ + 2x = 180^\circ - 50^\circ$

$\Rightarrow 2x = 130^\circ$

Divide both sides by 2 and simplify.

$\Rightarrow \dfrac{{2x}}{2} = \dfrac{{130^\circ }}{2}$

$\Rightarrow x = 65^\circ$

The value of $x$ is $65^\circ$.

v.

Ans: The sum of the internal angles of a triangle is $180^\circ$. In the given triangle, $x + x + x = 180^\circ$.

Hence,

$3x = 180^\circ$

Divide both sides by 3 and simplify.

$\Rightarrow \dfrac{{3x}}{3} = \dfrac{{180^\circ }}{3}$

$\Rightarrow x = 60^\circ$

The value of $x$ is $60^\circ$.

vi.

Ans: The sum of the internal angles of a triangle is $180^\circ$. In the given triangle, $90^\circ + x + 2x = 180^\circ$.

$\Rightarrow 90^\circ + 3x = 180^\circ$

Subtract $90^\circ$ from both sides and simplify.

$\Rightarrow 90^\circ - 90^\circ + 3x = 180^\circ - 90^\circ$

$\Rightarrow 3x = 90^\circ$

Divide both sides by 3 and simplify.

$\Rightarrow \dfrac{{3x}}{3} = \dfrac{{90^\circ }}{3}$

$\Rightarrow x = 30^\circ$

The value of $x$ is $30^\circ$.

2. Find the values of the unknowns $x$ and $y$ in the following diagrams:

i.

Ans: The exterior angle property of a triangle states that the exterior angle is equal to the sum of the opposite non-adjacent interior angles.

$x + 50^\circ = 120^\circ$

Subtract $50^\circ$ from both sides of the equation.

$\Rightarrow x = 120^\circ - 50^\circ$

$\Rightarrow x = 70^\circ$

The value of $x$ is $70^\circ$.

The sum of the internal angles of a triangle is $180^\circ$. In the given triangle, $50^\circ + 70^\circ + y = 180^\circ$.

$\Rightarrow 120^\circ + y = 180^\circ$

Subtract $120^\circ$ from both sides and simplify.

$\Rightarrow 120^\circ + y - 120^\circ = 180^\circ - 120^\circ$

$\Rightarrow y = 60^\circ$

The value of $y$ is $60^\circ$.

ii.

Ans: Since the vertical opposite angles are equal, $y = 80^\circ$.

The value of $y$ is $80^\circ$.

The sum of the internal angles of a triangle is $180^\circ$. In the given triangle, $50^\circ + 80^\circ + x = 180^\circ$.

$\Rightarrow 130^\circ + x = 180^\circ$

Subtract $130^\circ$ from both sides and simplify.

$\Rightarrow 130^\circ - 130^\circ + x = 180^\circ - 130^\circ$

$\Rightarrow x = 50^\circ$

The value of $y$ is $50^\circ$.

iii.

Ans: The exterior angle property of a triangle states that the exterior angle is equal to the sum of the opposite non-adjacent interior angles.

$50^\circ + 60^\circ = x$.

Hence,

$x = 110^\circ$

The value of $x$ is $110^\circ$.

The sum of the internal angles of a triangle is $180^\circ$. In the given triangle, $50^\circ + 60^\circ + y = 180^\circ$.

Hence,

$110^\circ + y = 180^\circ$

Subtract $110^\circ$ from both sides and simplify.

$\Rightarrow 110^\circ + y - 110^\circ = 180^\circ - 110^\circ$

$\Rightarrow y = 70^\circ$

The value of $y$ is $70^\circ$.

iv.

Ans: Since the vertical opposite angles are equal, $x = 60^\circ$.

The value of $x$ is $60^\circ$.

The sum of the internal angles of a triangle is $180^\circ$. In the given triangle,

$\Rightarrow 60^\circ + 30^\circ + y = 180^\circ$

$\Rightarrow 90^\circ + y = 180^\circ$.

Subtract ${90^\circ }$ from both sides and simplify.

$\Rightarrow 90^\circ - 90^\circ + y = 180^\circ - 90^\circ$

$\Rightarrow y = 90^\circ$

The value of $y$ is $90^\circ$.

v.

Ans: Since the vertical opposite angles are equal, $y = 90^\circ$.

The value of $y$ is $90^\circ$.

The sum of the internal angles of a triangle is $180^\circ$. In the given triangle, $x + x + 90^\circ = 180^\circ$.

Hence,

$90^\circ + 2x = 180^\circ$

Subtract $90^\circ$ from both sides and simplify.

$\Rightarrow 90^\circ + 2x - 90^\circ = 180^\circ - 90^\circ$

$\Rightarrow 2x = 90^\circ$

Divide both sides by 2 and simplify.

$\Rightarrow \dfrac{{2x}}{2} = \dfrac{{90^\circ }}{2}$

$\Rightarrow x = 45^\circ$

The value of $x$ is $45^\circ$.

vi.

Ans: Since the vertical opposite angles are equal, $y = x$.

The sum of the internal angles of a triangle is $180^\circ$.

In the given triangle, $x + x + x = 180^\circ$.

Hence,

$3x = 180^\circ$

Divide both sides by 3 and simplify.

$\Rightarrow \dfrac{{3x}}{3} = \dfrac{{180^\circ }}{3}$

$\Rightarrow x = 60^\circ$

The value of $x$ is $60^\circ$ and the value of $y$ is $60^\circ$.

Exercise 6.4

1. Is it possible to have a triangle with the following sides?

i) 2 cm, 3 cm, 5 cm?

Ans: A triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. Take the two sides as 2 cm and 3 cm. Since $2 + 3 = 5$ and the third side is also 5 cm, it is not possible to have a triangle with the sides 2 cm, 3 cm and 5 cm.

ii) 3 cm, 6 cm, 7 cm?

Ans: A triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. First, take the two sides as 3 cm and 6 cm. Since $3 + 6 = 9$ and $9 > 7$, the property of the triangle is satisfied. Now take the two sides as 6 cm and 7 cm. Since $6 + 7 = 13$ and $13 > 3$, the property of the triangle is satisfied.  Now, take the two sides as 7 cm and 3 cm. Since $7 + 3 = 10$ and $10 > 6$, the property of the triangle is satisfied. Hence, it is possible to have a triangle with the sides 3 cm, 6 cm and 7 cm.

iii) 6 cm, 3 cm, 2 cm?

Ans: A triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. Take the two sides as 2 cm and 3 cm. Since $2 + 3 = 5$ and 5 is not greater than 6, it is not possible to have a triangle with the sides 6 cm, 3 cm and 2 cm.

2. Take any point $O$ in the interior of a triangle $PQR$. Is

i) $OP + OQ > PQ$?

Ans: In the given triangle, join $OR$, $OQ$ and $OP$.

From the diagram, it can be seen that $OPQ$ is a triangle and a triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. So, $OP + OQ > PQ$. Yes, $OP + OQ > PQ$.

ii) $OQ + OR > QR$?

Ans: In the given triangle, join $OR$, $OQ$ and $OP$.

From the diagram, it can be seen that $OQR$ is a triangle and a triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. So, $OQ + OR > QR$. Yes, $OQ + OR > QR$.

iii) $OR + OP > RP$?

Ans: In the given triangle, join $OR$, $OQ$ and $OP$.

From the diagram, it can be seen that $OPR$ is a triangle and a triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. So, $OR + OP > RP$. Yes, $OR + OP > RP$.

3. AM is the median of a triangle $ABC$. Is $AB + BC + CA > 2AM$?

Ans: The sum of two sides of a triangle is always greater than the third side of the triangle.

In $\Delta ABM$, $AB + BM > AM$ and in $\Delta AMC$, $AC + MC > AM$.

Add both the inequalities and simplify.

$AB + BM + AC + MC > AM + AM$

$\Rightarrow AB + AC + \left( {BM + MC} \right) > 2AM$

Substitute $BC$ for $BM + MC$.

$\Rightarrow AB + AC + BC > 2AM$

Hence, $AB + BC + CA > 2AM$ is true.

4. $ABCD$ is a quadrilateral. Is $AB + BC + CD + DA > AC + BD$?

Ans: The sum of two sides of a triangle is always greater than the third side of the triangle.

In $\vartriangle ABC$, $AB + BC > AC$.

In $\vartriangle ADC$, $AD + DC > AC$.

In $\vartriangle DCB$, $DC + CB > DB$.

In $\vartriangle ADB$, $AD + AB > DB$.

Add all the four inequalities and simplify.

$AB + BC + AD + DC + DC + CB + AD + AB > AC + AC + DB + DB$

$\Rightarrow \left( {AB + AB} \right) + \left( {BC + BC} \right) + \left( {AD + AD} \right) + \left( {DC + DC} \right) > 2AC + 2DB$

$\Rightarrow 2AB + 2BC + 2AD + 2DC > 2\left( {AC + DB} \right)$

$\Rightarrow 2\left( {AB + BC + AD + DC} \right) > 2\left( {AC + DB} \right)$

Divide both sides by 2 and simplify.

$\Rightarrow \dfrac{2}{2}\left( {AB + BC + AD + DC} \right) > \dfrac{2}{2}\left( {AC + DB} \right)$

$\Rightarrow AB + BC + AD + DC > AC + DB$

Hence, $AB + BC + CD + DA > AC + BD$ is true.

5. $ABCD$ is a quadrilateral. Is $AB + BC + CD + DA < 2\left( {AC + BD} \right)$?

Ans: Draw a quadrilateral $ABCD$. Join $AC$ and $BD$.

The sum of two sides of a triangle is always greater than the third side of the triangle.

In $\vartriangle AOB$, $AB < OA + OB$.

In $\vartriangle BOC$, $BC < OB + OC$.

In $\vartriangle COD$, $DC < OC + OD$.

In $\vartriangle AOD$, $DA < OD + OA$.

Add all the four inequalities and simplify.

$AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OD + OA$

$\Rightarrow AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD$

$\Rightarrow AB + BC + CD + DA < 2\left[ {\left( {AO + OC} \right) + \left( {DO + OB} \right)} \right]$

Substitute $AC$ for $AO + OC$ and $BD$ for $DO + OB$.

$\Rightarrow AB + BC + CD + DA < 2\left[ {AC + BD} \right]$

Hence, $AB + BC + CD + DA < 2\left( {AC + BD} \right)$ is true.

6. The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?

Ans: It is given that two sides of a triangle are 12 cm and 15 cm. The sum of two sides of a triangle is always greater than the third side of the triangle.

Therefore, the third side of the triangle should be less than $12 + 15 = 27$ cm.

Also, the third side cannot be less than the difference of two sides of a triangle. Therefore, The third side should be more than $15 - 12 = 3$ cm.

Therefore, the length of the third side should be between 3 cm and 27 cm.

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.

Class 7 Maths Chapter 6: Exercises Breakdown

 Exercise Number of Questions Exercise 6.1 3 Problems (1 is Long Answer, and 2 are Short Answer Type Questions) Exercise 6.2 2 Sums (Both are Short Answers) Exercise 6.3 2 Sums (2 Sums are Short Answers) Exercise 6.4 6 Problems (1 is Long Answer Type, and 5 are Short Answer Type Questions) Exercise 6.5 8 Sums (6 are Short Answers, and 2 are Long Answers)

Conclusion

The NCERT Solutions for Class 7 Maths Chapter 6 - The Triangle and Its Properties offer a detailed understanding of various types of triangles, their properties, and important theorems like the Pythagorean theorem. It is essential to focus on triangle classifications, congruence criteria, and the triangle inequality theorem. In previous exams, around 3–4 questions were asked from this chapter class 7 triangle and its properties, highlighting its significance. Practising the solutions provided by Vedantu will help students master these concepts in class 7 triangle and its properties. Regular practice and a solid understanding of this chapter will prepare students well for their exams.

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

1. What will I learn in Class 7 Maths Chapter 6?

You will get to learn about the specific important properties of triangles, such as

• Sum of all interior angles of a triangle.

• Types of angles.

• Sum of two angles lying on the same plane.

• Types of angles based on degrees.

• Parallel lines and transversal.

• Comparison between interior and exterior angles.

• Properties of angles in a triangle.

• Sum of vertically opposite angles.

• Concept of area and perimeter of any given two-dimensional figure.

• Difference between a triangle and a quadrilateral.

• Calculation of area and other regular figures and perimeter of both triangles.

• Angle sum property of triangles.

• Isosceles and equilateral triangles.

• Right-angle triangles.

• Pythagoras Theorem.

2. Explain the basic properties of the triangles.

While solving the figure-based problems, you need to know the below-given points:

• The summation of all the angles in a triangle equals 180°.

• The difference between any two sides of a triangle is less than the length of its third side.

• The summation of the length of any two sides of a triangle is greater than the length of its third side.

• The side of a triangle that is opposite to the greater angle is the longest side of the triangle.

• The exterior angle property – herein, the exterior angle property of a triangle is equal to the summation of the interior opposite angles.

• If the corresponding angles of two triangles are congruent and the lengths of their respective sides are proportional, the two triangles are similar.

• The perimeter of a triangle equals the summation of all its three sides.

• The area of a triangle equals (½ x base X height).

3. How many exercises and questions are there in Chapter 6 of Class 7 Maths?

Class 7 Maths Chapter 6 consists of 6 exercises:

• Exercise 6.1 consists of 3 questions. This exercise mainly deals with Isosceles and equilateral triangles.

• Exercise 6.2 consists of 2 questions. This exercise focuses on interior and exterior angles, properties of angles in a triangle and the sum of all interior angles of a triangle.

• Exercise 6.3 consists of 2 questions. This exercise deals with the Isosceles triangle, equilateral triangles, Right-angled triangles and Pythagoras property.

• Exercise 6.4 consists of 6 questions. This exercise comprises questions related to the sum of vertically opposite angles, parallel lines and transversal.

• Exercise 6.5 consists of 8 questions. These particular exercise questions mainly focus on the Pythagoras Theorem, a triangle and a quadrilateral.

4. How to reduce the fear of Maths with NCERT Solutions by Vedantu?

Maths is a subject which is always a fear for many students. It is seen as a phobia among the majority of students. This is because they have never seen this subject in the way it actually had to be seen. The best way to get rid of this phobia is to take a smart forward step.

Our NCERT Solutions for Class 7 Maths Chapter 6 are one of the most essential study materials for class 7 students. Our expert teachers have framed these solutions with the utmost care in a bit-by-bit manner to make their preparation process for the exam way easier.

5. What is the Pythagorean theorem in class 7 maths chapter 6 PDF solutions?

In class 7 maths chapter 6 PDF solutions the Pythagorean theorem is a fundamental principle in geometry. 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. This theorem helps in calculating distances and is widely used in various mathematical applications.

6. How does the triangle inequality theorem help in understanding class 7 maths chapter triangle and its properties?

The triangle inequality theorem is a crucial concept in geometry. It states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. In class 7 maths chapter triangle and its properties, this theorem is essential for determining the possible dimensions of a triangle and understanding the properties of triangle formation.

7. What are the criteria for the congruence of triangles in class 7 ch 6 maths?

In class 7 ch 6 maths  triangles are congruent if they have the same shape and size. The criteria for congruence include Side-Side-Side (SSS), where all three sides are equal; Side-Angle-Side (SAS), where two sides and the included angle are equal; Angle-Side-Angle (ASA), where two angles and the included side are equal; and Right-angle-Hypotenuse-Side (RHS) for right-angled triangles.

8. Why is it important to practice the exercises in class 7th chapter 6?

Practising the class 7th chapter 6 is essential for mastering the concepts of triangle properties and theorems. It reinforces students' understanding and helps them apply these principles to solve problems. Regular practice ensures that students are well-prepared for exams and can tackle questions related to triangles confidently.