# NCERT Solutions Class 7 Maths Chapter 6

## NCERT Solutions for Class 7 Mathematics Chapter 6 The Triangle and its Properties

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

You may have learnt about triangles in your earlier classes while learning about the different types of geometric shapes. In Chapter 6 Maths Class 7 you'll be able to develop a better understanding of its core concept:  A triangle, its- types,  angles , sides and its properties in more detail. NCERT Solutions of Class 7th Mathematics Chapter 6 offers you solutions to the textbook questions that will act as a reference guide while you solve the exercise questions of the chapters.

### NCERT Solutions for Class 7 Mathematics Chapter 6 The Triangle and Its Properties -

(include NCERT Solutions for Class 7 Mathematics Chapter 6 The Triangle and Its Properties)

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

## NCERT Solutions for Class 7 Mathematics Chapter 6

Go through the following concepts before referring to the solutions of NCERT Class 7 Mathematics Chapter 6. It will help you in understanding the solutions better:

1. A median of a triangle is a line segment drawn from a vertex of a triangle to the midpoint of the opposite side. A triangle can, thus, have three medians in all.
2. An altitude of a triangle is a perpendicular line drawn from a vertex of a triangle to the opposite side. In simpler words, a triangle can have three altitudes in total, just like the median.
3. The sum of the interior angles of a triangle is always equal to its opposite exterior angle.
4. Rule of angle sum property of a triangle: The sum of all the angles of a triangle is always equal to 180°.

### NCERT Solutions for Class 7 Mathematics Chapter 6

Chapter 6 of Mathematics NCERT Class 7 covers the following topics:

 UNIT TOPIC 6.1 Introduction 6.2 Medians of a Triangle 6.3 Altitudes of a Triangle 6.4 Exterior Angle of a Triangle and Its Properties 6.5 Angle Sum Property of a Triangle 6.6 Two Special Triangles - Equilateral and Isosceles 6.7 Sum of the Lengths of Two Sides of a Triangle 6.8 Right Angle Triangle and Pythagoras Property

6.1 Introduction

A triangle is a closed curve formed by the intersection of three line segments. The line segments are called sides, and the points of intersections are called the vertices while the angles formed at the vertices are known as the angles of the triangle. So, in a triangle, we can say that there are three sides, vertices and angles. .

Based on sides, the classification of triangles is of the following types:

1.  Equilateral triangle- A triangle in which all three sides are equal.
2. Isosceles triangle- A triangle in which only  two sides are equal.
3. Scalene triangle- A triangle in which no two sides are equal.

Based on the angles, the classification of triangles is of the following types:

1. Acute angled triangle- Any triangle in which all three interior angles of the triangle are less than 90°.
2. Obtuse angled triangle- Any triangle with one of the interior angles of the triangle is more than 90°.
3. Right-angled triangle- A triangle in which one of the interior angles of the triangle is 90°.

6.2 Medians of Triangle

This section deals with a line segment joining a vertex of a triangle to the midpoint of its opposite side, which is called a median. It bisects the opposite side of the triangle into two equal parts and further divides the triangle into two, the two of them having the same area.

6.3 Altitudes of a Triangle

This section deals with an altitude of a triangle which is a perpendicular line segment drawn from the vertex of the triangle to its opposite side. When meeting the opposite side, it makes an angle of 90°. A triangle has three altitudes and the point of intersection of all the three is known as the orthocenter.

6.4 Exterior Angle of a Triangle and its Property

According to this law the exterior angle of a triangle is always equal to the sum of its opposite interior angles as this is a property of every exterior angle concerning a triangle.

6.5 Angle Sum Property of a Triangle

In this section, students need to learn yet another important property of a triangle called the angle sum property. It states that the sum of all three angles is always equal to 180°. To prove this, the students can refer to the NCERT Solutions Class 7 Mathematics Chapter 6 and understand how it  can  be proved by using the exterior angle property rule.

6.6 Two Special Triangles- Equilateral and Isosceles

This section helps the students get a detailed understanding of the different types of triangles based on sides, with a prime focus on Equilateral (all sides equal implying all angles of the triangle are also equal) and Isosceles triangle (any two sides equal implying two angles of the triangle also equal).

6.7 Sum of the Lengths of Two Sides of Triangle

This section explains  how the length of any two sides of  a triangle is always greater  than the length of its third side. To understand the concept clearer, students can refer to the NCERT Solutions Class 7 Mathematics Chapter 6 where this rule is proven and solved through many illustrations.

6.8 Right Angle Triangle and Pythagoras Property

A right-angle triangle is the one in which at least one angle of the triangle is 90°. Such triangles have a special property which is called the Pythagoras property or Pythagoras theorem. As per this property, the area of a triangle whose one side is a slanting line, called hypotenuse (it is the side opposite to the right angle) is always equal to the squares of the sum of the other two sides. From the solutions, students will be able to get a better understanding of this property.

## Key Features of NCERT Solutions for Class 7 Mathematics Chapter 6

The key points regarding the benefits that the students will receive through the solutions provided on this official website of Extramarks are:

• Our NCERT Solutions will provide you with a better understanding of all the topics in this chapter.
• It will enrich the young learners  as the solutions are easily available  and can be accessed anywhere, anytime as per your schedule and convenience.
• It will explain everything in a simple language,  so that the maximum number of students can benefit from it.

### NCERT Solutions for Class 7

The NCERT Solutions helps students in developing a better understanding of the complicated topics . It provides you with in-depth comprehension of the course and encourages you to develop deductive reasoning skills to solve the questions. It encourages the students to master the topic and increase their confidence in achieving a higher grade.

The NCERT Solutions of Extramarks for Class 7 cover all the subjects and concepts in a detailed manner. Students are assisted with  up-to-date solutions and studying Mathematics  couldn’t have  been easier.  So in case you are interested in looking for the solutions for all the subjects of Class 7, you can click on the link given below.

Q.1

$\mathrm{In}\mathrm{\Delta }\mathrm{PQR},\mathrm{Disthemid}–\mathrm{point}\mathrm{of}\overline{\mathrm{QR}.}$

$\begin{array}{l}\overline{\text{PM}}\text{is}_________________.\\ \overline{\text{PD}}\text{is}_________________.\\ \text{Is QM}=\text{MR}?\end{array}$

Ans

$\begin{array}{l}\overline{\text{PM}}\text{is}\overline{)\underset{_}{\text{Altitude}}}.\\ \overline{\text{PD}}\text{is}\overline{)\underset{_}{\text{Median}}}.\\ \text{No}\text{.}\end{array}$

Q.2

$\begin{array}{l}\mathrm{Draw}\mathrm{rough}\mathrm{sketches}\mathrm{for}\mathrm{the}\mathrm{following}:\\ \left(\mathrm{a}\right)\mathrm{In}\mathrm{\Delta ABC},\mathrm{BE}\mathrm{is}\mathrm{a}\mathrm{median}.\\ \left(\mathrm{b}\right)\mathrm{In}\mathrm{\Delta }\mathrm{PQR},\mathrm{PQ}\mathrm{and}\mathrm{PR}\mathrm{are}\mathrm{a}\mathrm{ltitudes}\mathrm{of}\mathrm{the}\mathrm{triangle}.\\ \left(\mathrm{c}\right)\mathrm{In}\mathrm{\Delta }\mathrm{XYZ},\mathrm{YL}\mathrm{is}\mathrm{an}\mathrm{altitude}\mathrm{in}\mathrm{the}\mathrm{exterior}\mathrm{of}\mathrm{the}\mathrm{triangle}.\end{array}$

Ans

(a)

(b)

(c)

For ΔXYZ, YL is an altitude drawn exterior to side XZ which is extended up to point L.

Q.3

$\begin{array}{l}\mathrm{Verify}\mathrm{by}\mathrm{drawing}\mathrm{a}\mathrm{diagram}\mathrm{if}\mathrm{the}\mathrm{median}\mathrm{and}\mathrm{altitude}\mathrm{of}\mathrm{an}\\ \mathrm{isosceles}\mathrm{triangle}\mathrm{can}\mathrm{be}\mathrm{same}.\end{array}$

Ans

Draw a line segment AD perpendicular to BC. It is an altitude for the triangle.
The length of BD and DC are same
.

Therefore, AD is also the median of this triangle
.

Q.4

$\begin{array}{l}\text{Find the value of the unknown exterior angle x in the}\\ \text{following diagrams:}\end{array}$

Ans

$\begin{array}{c}\text{(i)}x=70°+50°\text{\hspace{0.17em}}\left(\text{exterior angle theorem}\right)\\ =\overline{)120°}\\ \text{(ii)}x=65°+45°\text{\hspace{0.17em}}\left(\text{exterior angle theorem}\right)\\ =\overline{)110°}\\ \text{(iii)}x=40°+30°\text{\hspace{0.17em}}\left(\text{exterior angle theorem}\right)\\ =\overline{)70°}\\ \text{(iv)}x=60°+60°\text{\hspace{0.17em}}\left(\text{exterior angle theorem}\right)\\ =\overline{)120°}\\ \text{(v)}x=50°+50°\text{\hspace{0.17em}}\left(\text{exterior angle theorem}\right)\\ =\overline{)100°}\\ \text{(vi)}x=60°+30°\text{\hspace{0.17em}}\left(\text{exterior angle theorem}\right)\\ =\overline{)90°}\end{array}$

Q.5

$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{unknown}\mathrm{x}\mathrm{in}\mathrm{the}\mathrm{following}\mathrm{diagrams}:$

Ans

$\begin{array}{l}\text{Since, the sum of all interior angles of a triangle is 180}°\text{.}\\ \text{So,}\\ \text{(i)}x+50°+60°=180°\\ x+110°=180°\\ x=180°-110°=\overline{)70°}\\ \text{(ii)}x+90°+30°=180°\\ x+120°=180°\\ x=180°-120°=\overline{)60°}\\ \left(iii\right)\text{\hspace{0.17em}}x+110°+30°=180°\\ x+140°=180°\\ x=180°-140°=\overline{)40°}\\ \left(iv\right)\text{\hspace{0.17em}}x+x+50°=180°\\ 2x+50°=180°\\ 2x=180°-50°\\ =130°\\ x=\frac{130°}{2}=\overline{)65°}\\ \left(v\right)\text{\hspace{0.17em}}x+x+x=180°\\ 3x=180°\\ x=\frac{180°}{3}=\overline{)60°}\\ \left(vi\right)\text{\hspace{0.17em}}x+2x+90°=180°\\ 3x+90°=180°\\ x=180°-90°\\ 3x=90°\\ x=\frac{90°}{3}=\overline{)30°}\end{array}$

Q.6

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{the}\mathrm{unknowns}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{in}\mathrm{the}\mathrm{following}\\ \mathrm{diagrams}:\end{array}$

Ans

$\begin{array}{l}\text{(i)}\text{\hspace{0.17em}}\text{y+120}°\text{=180}°\text{}\text{}\text{}\text{}\text{(Linear pair)}\\ y=\text{180}°-\text{120}°=60°\\ \text{Now,}\\ x+y+50°=180°\text{\hspace{0.17em}}\text{}\text{}\text{(angle sum property)}\\ x+60°+50°=180°\\ x=180°-110°=\overline{)70°}\\ \text{(ii)}y=80°\text{}\text{}\text{}\text{}\text{}\text{(vertically opposite angle)}\\ y+x+50°=180°\\ 80°+x+50°=180°\\ x+130°=180°\\ x=180°-130°=\overline{)50°}\\ \text{(iii)}y+50°+60°=180°\text{}\text{(angle sum property)}\\ \text{y+110}°\text{=180}°\\ y=180°–110°=\overline{)\text{70}°}\\ \text{(iv)}x=60°\text{}\text{}\text{}\text{}\text{(vertically opposite angle)}\\ 30°+x+y=180°\\ 30°+60°+y=180°\\ y=180°-90°=\overline{)90°}\\ \text{(v)}y=90°\text{}\text{}\text{}\text{}\text{(vertically opposite angle)}\\ \text{x+x+y=180}°\\ 2x+90°=180°\\ 2x=180°-90°=90°\\ 2x=90°\\ x=\frac{90°}{2}=\overline{)45°}\\ \text{(vi)}y=x\text{}\text{}\text{}\text{}\text{}\text{(vertically opposite angle)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}a=x\text{\hspace{0.17em}}\text{}\text{}\text{}\text{}\text{(vertically opposite angle)}\\ \text{}b=x\text{}\text{}\text{}\text{}\text{}\text{(vertically opposite angle)}\\ a+b+y=180°\\ x+x+x=180°\\ 3x=180°\\ x=\frac{180°}{3}=60°\\ \text{So, we get}\overline{)x=y=60°}\end{array}$

Q.7

$\begin{array}{l}\mathrm{Is}\mathrm{it}\mathrm{possible}\mathrm{to}\mathrm{have}\mathrm{a}\mathrm{triangle}\mathrm{with}\mathrm{the}\mathrm{following}\mathrm{sides}?\\ \left(\mathrm{i}\right)2\mathrm{cm},3\mathrm{cm},5\mathrm{cm}\\ \left(\mathrm{ii}\right)3\mathrm{cm},6\mathrm{cm},7\mathrm{cm}\\ \left(\mathrm{iii}\right)6\mathrm{cm},3\mathrm{cm},2\mathrm{cm}\end{array}$

Ans

$\begin{array}{l}\text{For a triangle, the sum of lengths of either two sides is always}\\ \text{greater than the third side}\text{.}\\ \text{(i) Given sides of the triangle are 2 cm, 3 cm and 5 cm}\text{.}\\ \text{2 cm + 3 cm = 5 cm}\\ \text{So, the sum of two side is not greater than the third side}\text{.}\\ \text{Therefore, this triangle is not possible}\text{.}\\ \text{(ii) Given sides of the triangle are 3 cm, 6 cm and 7 cm}\text{.}\\ \text{3 cm + 6 cm = 9 cm}\\ \text{So, the sum of two side is greater than the third side}\text{.}\\ \text{Therefore, this triangle is possible}\text{.}\\ \text{(iii) Given sides of the triangle are 6 cm, 3 cm and 2 cm}\text{.}\\ \text{2 cm + 3 cm = 5 cm}\\ \text{So, the sum of two side is not greater than the third side}\text{.}\\ \text{Therefore, this triangle is not possible}\text{.}\end{array}$

Q.8 Take any point O in the interior of a triangle PQR. Is
(i) OP + OQ > PQ
?
(II) OQ + OR > QR?
(
iii )
OR + OP > RP?

Ans

$\begin{array}{l}\text{If O is a point in the interior of a given triangle, then three}\\ \text{triangles}\Delta \text{OPQ,}\text{\hspace{0.17em}}\Delta \text{OQR,}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\Delta \text{ORP can be constructed}\text{.}\\ \text{In a triangle, the sum of the lengths of either two sides}\\ \text{is always greater than the third side}\text{.}\\ \text{(i) Yes, as}\Delta \text{OPQ is a triangle with side OP, OQ and PQ}\text{.}\\ \text{So, OP + OQ > PQ}\\ \text{(ii) Yes, as}\Delta \text{OQR is a triangle with side OP, OQ and PQ}\text{.}\\ \text{So, OQ + OR > QR}\\ \text{(iii)}\text{\hspace{0.17em}}\text{Yes, as}\Delta \text{ORP is a triangle with side OP, OQ and PQ}\text{.}\\ \text{So, OP + OR > PR}\end{array}$

Q.9

$\begin{array}{l}\mathrm{AM}\mathrm{is}\mathrm{a}\mathrm{median}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{ABC}.\\ \mathrm{Is}\mathrm{AB}+\mathrm{BC}+\mathrm{CA}> 2\mathrm{AM}?\end{array}$

(Consider the sides of triangles ΔABM and ΔAMC.)

Ans

$\begin{array}{l}\text{Since, in a triangle, the sum of either two sides is always}\\ \text{greater than the third side}\text{.}\\ \text{So, in}\Delta \text{ABM}\\ \text{AB}+\text{BM}>\text{AM}\text{}\dots \left(\text{i}\right)\\ \text{Similary in}\Delta \text{ACM, we get}\\ \text{AC}+\text{CM}>\text{AM}\text{}\dots \left(\text{ii}\right)\\ \text{Adding (i) and (ii) to get}\\ \text{AB}+\text{BM}+\text{AC}+\text{CM}>\text{2AM}\\ \text{AB}+\text{CA}+\text{(BM}+\text{CM)}>\text{2AM}\\ \text{AB}+\text{AC}+\text{BC}>\text{2AM}\\ \overline{)\text{AB}+\text{BC}+\text{CA}>\text{2AM}}\\ \text{Therefore, the given expression is true}\text{.}\end{array}$

Q.10 ABCD is a quadrilateral. Is AB  BC + CD + DA>AC + BD?

Ans

$\begin{array}{l}\text{Since, in a triangle, the sum of either two sides is always}\\ \text{greater than the third side}\text{.}\\ \text{So, in}\Delta \text{ABC}\\ \text{AB}+\text{BC}>\text{CA}\text{}\dots \left(\text{i}\right)\\ \text{Similary in}\Delta \text{BCD, we get}\\ \text{BC}+\text{CD}>\text{DB}\text{}\dots \left(\text{ii}\right)\\ \text{In}\Delta \text{CDA, we get}\\ \text{CD}+\text{DA}>\text{AC}\text{}\dots \left(\text{iii}\right)\\ \text{In}\Delta \text{DAB, we get}\\ \text{DA}+\text{AB}>\text{DB}\text{}\dots \left(\text{iv}\right)\\ \text{Adding (i), (ii), (iii) and (iv) to get}\\ \text{AB}+\text{BC}+\text{BC}+\text{CD}+\text{CD}+\text{DA}+\text{DA}+\text{AB}>\text{CA}+\text{DB}+\text{AC}+\text{DB}\\ \text{2}\left(\text{AB}+\text{BC}+\text{CD}+\text{DA}\right)+\text{2(AC}+\text{DB)}\\ \overline{)\text{AB}+\text{BC}+\text{CD}+\text{DA}>\text{AC+BD}}\\ \text{Therefore, the given expression is true}\text{.}\end{array}$

Q.11

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

Ans

Since, in a triangle, the sum of either two sides is always greater than the third side. So, in ΔOAB OA+OB>AB ( i )
Similary in
ΔOBC, we get
OC+OB>BC ( ii )
In ΔOCD, we get OD+OC>CD ( iii )
In
ΔODA, we get
OA+OD>DA ( iv )
Adding (i), (ii), (iii) and (iv) to get
OA+OB+OC+OB+OD+OC+OA+OD>AB+BC+CD+DA
2
( OA+OB+OC+OD )>2( AC+BD )
( OA+OB+OC+OD )>( AC+BD )
Therefore, the given expression is true
.

Q.12

$\begin{array}{l}\mathrm{The}\mathrm{lengths}\mathrm{of}\mathrm{two}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{are}12\mathrm{cm}\mathrm{and}15\mathrm{cm}.\\ \mathrm{Between}\mathrm{what}\mathrm{two}\mathrm{measures}\mathrm{should}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{third}\\ \mathrm{side}\mathrm{fall}?\end{array}$

Ans

$\begin{array}{l}\text{In a triangle, the sum of either two sides is always greater}\\ \text{than the third side and also, the difference of the lengths of}\\ \text{either two sides is always lesser than the third side}\text{.}\\ \text{Here, the third side will be lesser than the sum of these two}\\ \text{(i}\text{.e}\text{.,}12+15=27\text{) and also, it will be greater than the difference}\\ \text{of these two (i}\text{.e}\text{.,}15-12=3\text{)}\text{.}\\ \text{Therefore, those two measures are 27 cm and 3 cm}\text{.}\end{array}$

Q.13

$\begin{array}{l}\text{PQR is a triangle right angled at P. If PQ =10 cm}\\ \text{and PR=24cm,find QR.}\end{array}$

Ans

By applying Pythagoras theorem in ΔPQR to get

(PQ)2 + (PR)2 = (QR)2
102 + 242 = QR2
100
+ 576 = QR2

676
= QR2

QR
=

$\sqrt{676}$

= 26cm

Q.14

$\begin{array}{l}\mathrm{ABC}\mathrm{is}\mathrm{a}\mathrm{triangle}\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{C}.\mathrm{If}\mathrm{AB}= 25\mathrm{cm}\mathrm{and}\\ \mathrm{AC}= 7\mathrm{cm},\mathrm{find}\mathrm{BC}.\end{array}$

Ans

$\begin{array}{l}\text{By Pythagoras theorem in}\Delta \text{ABC, we get}\\ {\left(\text{AC}\right)}^{2}+{\left(BC\right)}^{2}={\left(AB\right)}^{2}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(BC\right)}^{2}={\left(AB\right)}^{2}-{\left(AC\right)}^{2}\\ \text{}\text{}\text{}={24}^{2}-{7}^{2}\\ \text{}\text{}\text{}=625-49=576\\ \text{}\text{}\overline{)\text{BC}=\text{24}\text{\hspace{0.17em}}\text{cm}}\end{array}$

Q.15

$\begin{array}{l}\mathrm{A}15\mathrm{m}\mathrm{long}\mathrm{ladder}\mathrm{reached}\mathrm{a}\mathrm{window}12\mathrm{m}\mathrm{high}\mathrm{from}\mathrm{the}\\ \mathrm{ground}\mathrm{on}\mathrm{placing}\mathrm{it}\mathrm{against}\mathrm{a}\mathrm{wall}\mathrm{at}\mathrm{a}\mathrm{distance}‘\mathrm{a}‘.\mathrm{Find}\mathrm{the}\\ \mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{foot}\mathrm{of}\mathrm{the}\mathrm{ladder}\mathrm{from}\mathrm{the}\mathrm{wall}.\end{array}$

Ans

$\begin{array}{l}\text{By Pythagoras theorem}\\ {\left(15\right)}^{2}={\left(12\right)}^{2}+{a}^{2}\\ 225-144={a}^{2}\\ 81={a}^{2}\\ \overline{)a=9\text{\hspace{0.17em}}\text{cm}}\\ \text{Therefore, the distance of the foot of the ladder from the wall}\\ \text{is 9 cm}\text{.}\end{array}$

Q.16

$\begin{array}{l}\mathrm{Which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{can}\mathrm{be}\mathrm{the}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{right}\mathrm{triangle}?\\ \left(\mathrm{i}\right)\mathrm{ }2.5\mathrm{cm},6.5\mathrm{cm},6\mathrm{cm}.\\ \left(\mathrm{ii}\right)\mathrm{ }2\mathrm{cm},2\mathrm{cm},5\mathrm{cm}.\\ \left(\mathrm{iii}\right)\mathrm{ }1.5\mathrm{cm},2\mathrm{cm},2.5\mathrm{cm}\end{array}$

Ans

$\begin{array}{l}\left(\text{i}\right)\text{2}.\text{5 cm},\text{6}.\text{5 cm},\text{6 cm}.\\ {2.5}^{2}=6.25,{6.5}^{2}=42.25{\text{and 6}}^{2}=36\\ \text{Here, 2}{\text{.5}}^{2}+{6}^{2}={6.5}^{2}\\ \text{So, the square of the length of one side is the sum of the}\\ \text{squares of the lengths of remaining two sides}\text{.}\\ \text{Hence, these are the sides of a right}-\text{angled triangle}\text{.}\\ \left(\text{ii}\right)\text{2 cm},\text{2 cm},\text{5 cm}.\\ {2}^{2}=4,{2}^{2}=4{\text{and 5}}^{2}=25\\ {\text{Here, 2}}^{2}+{2}^{2}\ne {5}^{2}\\ \text{So, the square of the length of one side is not the sum of the}\\ \text{squares of the lengths of remaining two sides}\text{.}\\ \text{Hence, these are not the sides of a right}-\text{angled triangle}\text{.}\\ \left(\text{iii}\right)\text{1}.\text{5 cm},\text{2cm},\text{2}.\text{5 cm}\\ {1.5}^{2}=2.25,{2}^{2}=4\text{and 2}{\text{.5}}^{2}=6.25\\ \text{Here, 1}{\text{.5}}^{2}+{2}^{2}={2.5}^{2}\\ \text{So, the square of the length of one side is the sum of the}\\ \text{squares of the lengths of remaining two sides}\text{.}\\ \text{Hence, these are the sides of a right}-\text{angled triangle}\text{.}\end{array}$

Q.17

$\begin{array}{l}\text{A tree is broken at a height of 5 m from the ground and its}\\ \text{top touches the ground at a distance of 12 m from the}\\ \text{Base of the tree.Find the original height of the tree.}\end{array}$

Ans

$\begin{array}{l}\text{In above figure, BC represents the unbroken part of the tree}\text{.}\\ \text{Point C represents the point where the tree broke and CA}\\ \text{represents the broken part of the tree}\text{.}\\ \text{Triangle ABC thus formed is a right-angled triangle}\text{.}\\ \text{So, applying Pythagoras theorewm to get}\\ {\text{AC}}^{2}={\text{AB}}^{2}+{\text{BC}}^{2}\\ \text{}={12}^{2}+{5}^{2}\\ \text{}=144+25\\ \text{}=169=13\text{\hspace{0.17em}}\text{cm}\\ \text{Thus,the original height of the tree is}=\text{AC}+\text{CB}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{13 cm}+\text{5 cm}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\overline{)\text{18 cm}}\end{array}$

Q.18

$\begin{array}{l}\mathrm{Angles}\mathrm{Q}\mathrm{and}\mathrm{R}\mathrm{of}\mathrm{a}\mathrm{\Delta PQR}\mathrm{are}25º\mathrm{and}65º.\\ \mathrm{Write}\mathrm{which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{is}\mathrm{true}:\\ \left(\mathrm{i}\right){\mathrm{PQ}}^{2}+{\mathrm{QR}}^{2}={\mathrm{RP}}^{2}\\ \left(\mathrm{ii}\right){\mathrm{PQ}}^{2}+{\mathrm{RP}}^{2}={\mathrm{QR}}^{2}\\ \left(\mathrm{iii}\right){\mathrm{RP}}^{2}+{\mathrm{QR}}^{2}={\mathrm{PQ}}^{2}\end{array}$

Ans

$\begin{array}{l}\text{Since, the sum of all interior angle is 180}°\text{.}\\ \text{So, 25}°\text{+ 65}°+\angle \text{QPR}=180°\\ \angle \text{QPR}=180°-90°\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=90°\\ \text{Therefore,}\Delta \text{PQR is a rigfht angled triangle at P}\text{.}\\ \text{Thus,}\overline{){\text{PQ}}^{\text{2}}+{\text{RP}}^{\text{2}}={\text{QR}}^{\text{2}}}\\ \text{So, (ii) is true}\text{.}\end{array}$

Q.19

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{perimeter}\mathrm{of}\mathrm{the}\mathrm{rectangle}\mathrm{whose}\mathrm{length}\mathrm{is}40\mathrm{cm}\\ \mathrm{and}\mathrm{a}\mathrm{diagonal}\mathrm{is}41\mathrm{cm}.\end{array}$

Ans

$\begin{array}{l}\text{In a rectangle},\text{all interior angles are of 9}0º\text{measure}.\\ \text{Therefore},\text{Pythagoras theorem can be applied here}.\\ \text{}\text{}{\left(\text{41}\right)}^{\text{2}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left(\text{4}0\right)}^{\text{2}}+{x}^{\text{2}}\\ \text{}\text{}\text{1681}=\text{16}00\text{}+{x}^{\text{2}}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{\text{2}}=\text{1681}-\text{16}00\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{81}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\text{9 cm}\\ \text{So, Perimeter}=\text{2}\left(\text{Length}+\text{Breadth}\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}\left(x+\text{4}0\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}\left(\text{9}+\text{4}0\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{}\overline{)\text{98 cm}}\end{array}$

Q.20

$\begin{array}{l}\mathrm{The}\mathrm{diagonals}\mathrm{of}\mathrm{a}\mathrm{rhombus}\mathrm{measure}16\mathrm{cm}\mathrm{and}30\mathrm{cm}.\\ \mathrm{Find}\mathrm{its}\mathrm{perimeter}.\end{array}$

Ans

$\begin{array}{l}\text{Let ABCD be a rhombus}\left(\text{all sides are of equal length}\right)\text{and its}\\ \text{diagonals},\text{AC and BD},\text{are intersecting each other at point O}.\\ \text{Since, diagonals of a rhombus bisect each other at 9}0°.\\ \text{So,}\\ \text{By applying Pythagoras theorem in}\Delta \text{AOB},\\ {\text{OA}}^{\text{2}}+{\text{OB}}^{\text{2}}={\text{AB}}^{\text{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{8}}^{\text{2}}+{\text{15}}^{\text{2}}={\text{AB}}^{\text{2}}\\ \text{\hspace{0.17em}}\text{64}+\text{225}={\text{AB}}^{\text{2}}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{289}={\text{AB}}^{\text{2}}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{AB}=\text{17 cm}\\ \text{Therefore},\text{the length of the side of rhombus is 17 cm}.\\ \text{Perimeter of rhombus}=\text{4}×\text{Side of the rhombus}\\ \text{}\text{}\text{}\text{}\text{}\text{}=\text{4}×\text{17}\\ \text{}\text{}\text{}\text{}\text{}\text{}=\text{}\overline{)\text{68 cm}}\end{array}$

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2. List out some of the important points to remember in Chapter 6 of Class 7 Mathematics.

Following are some key points to remember while solving exercise questions:

• Sum of all interior angles of a triangle is 180°.
• Kinds of angles including alternate, vertically opposite, adjacent and corresponding angles.
• Sum of two angles lying on the same plane.
• Angles based on degrees are of three kinds- acute, obtuse and right-angle.
• Parallel lines and transversal in a triangle.
• Properties of interior and exterior angles of a triangle.
• Sum of vertically opposite angles in a triangle.

3. How to score full marks in Class 7 Mathematics Chapter 6?

To score full marks in any topic of Mathematics, revision and practice is the key. Practice will not only help in understanding all the concepts of Chapter 6  at a deeper level, but it will also help you avoid silly mistakes while solving related problems. The more you practice , the easier it gets. Practice day in and day out, to experience exceptional results over time.

4. What is the difference between a triangle and a triangular region?

A triangle is a closed curve made by the intersection of three line segments. The line segments are called sides, and the points of intersections are called the vertices. The angles formed at the vertices are called the angles of the triangle. Whereas a triangular region includes the interior of the triangle along with the triangle itself.

5. Should the examples of Chapter 6 also be solved after solving the unsolved questions?

There is no hard and fast rule. However, it’s good to make a habit of first going through all the examples in the chapter and then solve the unsolved questions. The more you practice, it will be easier for you to grasp all the concepts of this chapter. And in case you fail to understand any of the concepts or problems, the solutions offered by Extramarks for NCERT solutions Class 7 Mathematics will come handy. You  won’t require any other assistance once you begin using solutions and it will save time for other subjects as well.

6. Explain types of triangles based on sides.

Based on sides, the classification of triangles is of the following types:

•  Equilateral triangle- A triangle in which all three sides are equal.
• Isosceles triangle- A triangle in which  any two sides are equal.
• Scalene triangle- A triangle in which no two sides are equal.

7. Explain types of triangles based on angles.

Based on the angles, the classification of triangles is of the following types:

• Acute angled triangle- Any triangle in which all three interior angles of the triangle are less than 90°.
• Obtuse angled triangle- Any triangle with one of the interior angles of the triangle is more than 90°.
• Right-angled triangle- A triangle in which one of the interior angles of the triangle is 90°.