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

NCERT solutions for class 7 maths chapter 6 The Triangle and Its Properties- The concepts studied in triangles and its properties are highly useful in higher classes. So practicing questions is important. This is where the solutions of NCERT class 7 maths chapter 6 the triangle and its properties are useful. In this chapter, we will discuss triangles and its properties like medians and altitudes of a triangle, exterior angle of a triangle, angle sum property of a triangle, equilateral and isosceles triangles, the sum of the lengths of two sides of the triangle, right-angled triangles, and Pythagoras theorem. There are a total of 5 exercises with 21 questions discussed in the CBSE NCERT solutions for class 7 maths chapter 6 the triangle and its properties. Topic-wise questions are also discussed in the solutions of NCERT for class 7 maths chapter 6 the triangle and its properties. A tool like NCERT solutions is extremely helpful for the students to understand the basics of each chapter and also help them to clear all their doubts easily. Here you will get solutions to five exercises of this chapter.

Exercise:6.1

Exercise:6.2

Exercise:6.3

Exercise:6.4

Exercise:6.5

## 6.1 Introduction

6.2 Medians of a Triangle

6.3 Altitudes of a Triangle

6.4 Exterior Angle of a Triangle and its Property

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-Angled Triangles and Pythagoras Property

## NCERT solutions for class 7 maths chapter 6 the triangle and its properties topic 6.1

The Triangle  $\small \Delta ABC$

The Elements of the triangle are:

Sides$\overline {AB},\overline{BC}\:\:and\:\:\overline{CA}$

Angles$\angle ABC, \angle BCA,\:\:\angle CAB$

(i) Side opposite to the vertex Q of $\small \Delta PQR$

(ii) Angle opposite to the side LM of $\small \Delta LMN$

(iii) Vertex opposite to the side RT of $\small \Delta RST$

(i) The side opposite to the vertex Q of $\small \Delta PQR$ $=\overline {PR}$

(ii) Angle opposite to the side LM of $\small \Delta LMN$ $=\angle MNL$

(iii) Vertex opposite to the side RT of $\small \Delta RST$  = Vertex S.

(a) SidesThe 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?

(b) Angles

i) The triangle ABC

Based on Side: In Triangle ABC, Since two sides (BC and AC ) are equal (= 8 cm ) The given triangle is an isosceles triangle.

Based on Angle: In Triangle ABC, Since all the triangles are less than 90 degrees, So the given triangle is Acute angled triangle.

ii) The Triangle PQR

Based on Side: In Triangle PQR, All the sides are different so, The given triangle is a scalene triangle.

Based on Angle: In Triangle PQR, Since angle QRP is a right angle, So the given triangle is Right-angled triangle.

iii)The Triangle LMN

Based on Side: In Triangle LMN, Since two sides (MN and NL ) are equal (=7 cm ), The given triangle is an isosceles triangle.

Based on Angle: In Triangle LMN, Since angle MNL is an Obtuse angle, So the given triangle is Obtuse angled triangle.

iv) The Triangle RST

Based on Side: In Triangle RST, All the sides are equal (=5.2 cm) so, The given triangle is an Equilateral triangle.

Based on Angle: In Triangle RST, Since all the triangles are less than 90 degrees, So the given triangle is Acute angled triangle.

v)The triangle ABC

Based on Side: In Triangle ABC, Since two sides (AB and BC ) are equal (= 3 cm ) The given triangle is an isosceles triangle.

Based on Angle: In Triangle ABC, Since angle ABC is greater than 90 degrees  So the given triangle is Obtuse angled triangle.

vi)The Triangle PQR

Based on Side: In Triangle PQR, Since two sides (PQ and QR ) are equal (= 6 cm ) The given triangle is an isosceles triangle.

Based on Angle: In Triangle PQR, Since angle PQR is a right angle, So the given triangle is Right-angled triangle.

## Solutions for NCERT class 7 maths chapter 6 the triangle and its properties topic 6.2

In geometry, a median of a triangle is a line segment joining a vertex to the midpoint of the opposite side, thus bisecting that side. Every triangle has exactly three medians, one from each vertex.

Yes, The median always lies in the interior of the triangle.

As we can see in all three cases the median lies inside the triangle.

Solutions of NCERT class 7 maths chapter 6 the triangle and its properties topic 6.3

Every triangle has three bases (any of its sides) and three altitudes (heights). Every altitude is the perpendicular segment from a vertex to its opposite side (or the extension of the opposite side.

altitudes from A to $\small \overline{BC}$ for the triangles are:

No, the altitude of a triangle might lie outside the triangle. for example in the obtuse-angled triangle, we have to extend the base side for making altitude angle.

4. Can you think of a triangle in which two altitudes of the triangle are two of its sides?

Yes, in a Right-angled triangle, the two altitudes of the sides are two sides as they make an angle of 90 degrees with one another.

(Hint: For Q.No.  4 and 5, investigate by drawing the altitudes for every type of triangle).

Yes, the altitude and median can be the same in a triangle. for example, consider an equilateral triangle, the median which divides the side in equal is also perpendicular to the side and hence the altitude and the median is the same.

## NCERT solutions for class 7 maths chapter 6 the triangle and its properties topic 6.4

There are three more ways of getting exterior angles. Try to produce those rough sketches.

In a triangle, there are a total of six exterior angles

Exterior Angles in a Triangle

No, The exterior angle formed at the vertices of a triangle are not equal. The exterior angle is equal to the sum of the two opposite interior angles.

As we can see they both angles forms a straight line, Hence the sum of an exterior angle of a triangle and its adjacent interior angle is always $180^0$.

(i) a right angle?             (ii) an obtuse angle?             (iii) an acute angle?

(i) a right angle

When the exterior angle is $90^0$, the sum of opposite internal angles is $90^0$

(ii) an obtuse angle

When the exterior angle is the obtuse angle, the opposite interior angles can be both acute, one right and one acute and one obtuse and one acute.

(iii) an acute angle?

When the exterior angle is an acute angle, both the internal angle has to be acute angles.

No, the exterior angle of a triangle cannot be a straight angle because if the exterior angle is straight then there won't be any triangle left that would be a straight line. (imagine it visually).

As we know that in a triangle, the exterior angle is equal to the sum of opposite interior angles.

So,

According to the question,

$70^0=25^0+interior \:\:angle$

$interior \:\:angle =70^0-25^0$

$interior \:\:angle =45^0$

Hence the other interior angle is $45^0$.

As we know that in a triangle, the exterior angle is equal to the sum of opposite interior angles.

So,

Exterior Angle =  $\small 60^{\circ}$$\small 80^{\circ}$.

= $140^0$

Hence the measure of the exterior is $140^0$.

Yes, The measure of the exterior angle is given wrong.

As we know that in a triangle, the exterior angle is equal to the sum of opposite interior angles.

So,

Exterior angle = $50^0+50^0$

= $100^0$

Hence the exterior angle should be equal be $100^0$ instead of $50^0$.

## 1. Two angles of a triangle are $\small 30^{\circ}$ and $\small 80^{\circ}$. Find the third angle.

Let the third angle be $x$

Now,  As we know the sum of internal angles of a triangle is 180. so,

$30^0+80^0+x=180^0$

$x=180^0-30^0-80^0$

$x=70^0$

Hence the Third angle is $70^0$.

Let the two same angles in triangle be $x$.

Now,  As we know the sum of internal angles of a triangle is 180. so,

$80^0+x+x=180^0$

$2x=180^0-80^0$

$2x=100^0$

$x=50^0$.

Hence both other angles are $50^0$ each.

Let the angles of the triangles be $x,2x\:\:and\:\:x$

So,

As we know the sum of internal angles of a triangle is 180. so,

$x+2x+x=180^0$

$4x=180^0$

$x=45^0$

$2x=90^0$

Hence the angles of the triangles are $45^0,90^0\:\:and\:\:45^0$.

On the Basis of sides, the triangle is isosceles triangle as two sides of the triangle are equal.

On the Basis of angle, the triangle is Right-Angled Triangle as it has one angle equal to 90 degrees.

No, we cannot have a triangle with two right angles. as the sum of the angles of a triangle is always 180 degrees, If there are two right angles then the sum would exceed 180 which is not possible.

No, we can not have two obtuse angles in a triangle because the sum of angles of the triangle is always 180 degrees and if there are two obtuse angles in the triangle then the sum would be more than 180 degrees which are not possible.

Yes, of course! we can have a triangle with two acute angles. all the obtuse-angled triangles have two acute angles in them.

No, we can not have a triangle with all the three angles greater than  $\small 60^{\circ}$ because then the sum of the angles of the triangles would be greater than 180 degree which is no possible.

Yes, we can have a triangle with all the three angles equal to $\small 60^{\circ}$ as the sum of the angles of the triangle would be exactly 180 degrees. such triangles are called equilateral triangles.

No, we can not have a triangle with all angles less than $60^0$ because then the sum of the angles would be less than 180 degrees which are also not possible.

The sum of angles of a triangle is exactly $180^0$ neither more than that nor less than that.

## 1. Find angle x in each figure:

As we know, in an isosceles triangle, two sides and the angles they make with the third side are equal.

and

the sum of angles of the triangle is equal to $180^0$. So,

i) $x = 40^0$

ii) $x+45^0+45^0=180^0$

$x=180^0-90^0$

$x=90^0$

iii) $x=50^0$

iv) $100^0+x+x=180^0$

$2x=180^0-100^0$

$2x=80^0$

$x=40^0$

v)$x+x+90^0=180^0$

$2x=180^0-90^0$

$2x=90^0$

$x=45^0$

vi)$x+x+40^0=180^0$

$2x=180^0-40^0$

$2x=140^0$

$x=70^0$

vii)$x=180^0-120^0$

$x=60^0$

viii) As we know, the exterior angle is equal to the sum of opposite internal angles in a triangle. So,

$x+x=110^0$

$2x=110^0$

$x=55^0$

ix)As we know when two lines are intersecting, the opposite angles are equal. So

$x=30^0$

i) As we know, in an isosceles triangle, two sides and the angles they make with the third side are equal. So,

$y+120^0=180^0$

$y=180^0-120^0$

$y=60^0$

Now,  As we know the sum of internal angles of a triangle is 180. so,

$x+60^0+60^0=180^0$

$x=180^0-60^0-60^0$

$x=60^0$

Hence, $x=60^0\:\:and\:\:y=60^0$.

ii)  As we know, in an isosceles triangle, two sides and the angles they make with the third side are equal.

AND

the sum of internal angles of a triangle is 180. so,

$x+x+90^0=180^0$

$2x=90^0$

$x=45^0$

Also,

$y=180^0-x$

$y=180^0-45^0$

$y=135^0$.

Hence $x=45^0\:\:and\:\:y=135^0$.

iii) As we know when two lines are intersecting, the opposite angles are equal.

And

the sum of internal angles of a triangle is 180. so,

$x+x+92^0=180^0$

$2x=180^0-92^0$

$2x=88^0$

$x=44^0$

Now, As we know, the exterior angle is equal to the sum of opposite internal angles in a triangle

$y=x+92^0$

$y=44^0+92^0$

$y=136^0$

Hence $x=44^0\:\:and\:\:y=136^0$.

## 1. Is the sum of any two angles of a triangle always greater than the third angle?

No, we can not say that. because we can have triangles in which the sum of two angles is less than the third angle. For Example:

A triangle ABC with

$\angle A=150^o$

$\angle B=20^o$

$\angle C=10^o$

## Q. Find the unknown length x in the following figures (Fig $\small 6.29$):

As we know in a Right-angled Triangle: By Pythagoras Theorem,

$(Hypotenus)^2=(Base)^2+(Perpendicular)^2$

So, using this theorem,

i)

$x^2=3^2+4^2$

$x^2=9+16$

$x^2=25$

$x=5$.

ii)

$x^2=6^2+8^2$

$x^2=36+64$

$x^2=100$

$x=10$

iii)

$x^2=15^2+8^2$

$x^2=225+64$

$x^2=289$

$x=17$

iv)

$x^2=24^2+7^2$

$x^2=576+49$

$x^2=625$

$x=25$

v) in this question as we can see from the figure, it is making the right angle with the half-length of x, so

$\left ( \frac{x}{2} \right )^2+12^2=37^2$

$\frac{x^2}{4} +12^2=37^2$

$\frac{x^2}{4} =37^2-12^2$

.$\frac{x^2}{4} =1369-144$

$\frac{x^2}{4} =1225$

$x^2=4\times 1225$

$x=2\times35$

$x=70.$

vi)

$x^2=12^2+5^2$

$x^2=144+25$

$x^2=169$

$x=13$

The hypotenuse is the longest side in a triangle. So when the right angle is at P, the Longest side would be QR.

The hypotenuse is the longest side in a triangle. So when the right angle is at B, the Longest side would be AC.

The hypotenuse is the longest side in a Right-angled triangle.

Baudhayan Theorem and Pythagoras theorem are basically the same. Baudhayan Theorem used geometry to intuitively prove the Pythagoras theorem.

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

$\small \overline{PM}$  is _________________.

PD is _________________.

Is $\small QM=MR$ ?

$\small \overline{PM}$  is the Altitude of the triangle.

PD is the Mediun of the triangle.

No, $\small QM\neq MR$. As QD = DR.

## 2.   Draw rough sketches for the following:

(a) In  $\small \Delta ABC$, BE is a median.

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

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

(a) In  $\small \Delta ABC$, BE is a median.

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

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

Yes, it is very much possible that the median and altitude of an isosceles triangle is the same. for example, The given triangle has the same median and altitude.

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

As we know that the exterior angle is equal to the sum of the opposite internal angles. So,

i) $x=50^0+70^0=120^0$

ii) $x=65^0+45^0=110^0$

iii) $x=30^0+40^0=70^0$

iv) $x=60^0+60^0=120^0$

v) $x=50^0+50^0=100^0$

vi) $x=30^0+60^0=90^0$

As we know that the exterior angle is equal to the sum of the opposite internal angles. So,

i) $60^0+x=120^0\Rightarrow x=120^0-60^0=60^0$

ii) $70^0+x=100^0\Rightarrow x=100^0-70^0=30^0$

iii) $x+90^0=125^0\Rightarrow x=125^0-90^0=35^0$

iv) $x+60^0=120^0\Rightarrow x=120^0-60^0=60^0$

v) $x+30^0=80^0\Rightarrow x=80^0-30^0=50^0$

## Solutions of NCERT class 7 maths chapter 6 the triangle and its properties exercise 6.3

As we know that the sum of the internal angles of the triangle is equal to $180^0$. So,

i) $x+50^0+60^0=180^0$

$x=180^0-50^0-60^0$

$x=70^0$

ii) $x+30^0+90^0=180^0$

$x=180^0-30^0-90^0$

$x=60^0$

iii) $x+30^0+110^0=180^0$

$x=180^0-30^0-110^0$

$x=40^0$

iv) $x+x+50^0=180^0$

$2x=180^0-50^0$

$2x=130^0$

$x=65^0$

v) $x+x+x=180^0$

$3x=180^0$

$x=60^0$

vi) $x+2x^0+90^0=180^0$

$3x=180^0-90^0$

$3x=90^0$

$x=30^0$

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

i) As we know, the exterior angle is equal to the sum of opposite internal angles in a triangle.

$50^0+x=120^0$

$x=120^0-50^0$

$x=70^0$

Now, As we know the sum of internal angles of a triangle is 180. so,

$50^0+x+y=180^0$

$50^0+70^0+y=180^0$

$y=180^0-50^0-70^0$

$y=60^0$

Hence, $x=70^0\:\:and\:\:y=60^0$.

ii)  As we know when two lines are intersecting, the opposite angles are equal. So

$y=80^0$

Now, As we know the sum of internal angles of a triangle is 180. so,

$50^0+y+x=180^0$

$50^0+80^0+x=180^0$

$x=180^0-50^0-80^0$

$x=50^0$

Hence, $x=50^0\:\:and\:\:y=80^0$.

iii) As we know, the exterior angle is equal to the sum of opposite internal angles in a triangle

$x=50^0+60^0$

$x=110^0$

Now, As we know the sum of internal angles of a triangle is 180. so,

$y+50^0+60^0=180^0$

$y=180^0-50^0-60^0$

$y=70^0$

Hence, $x=110^0\:\:and\:\:y=70^0$.

iv)

As we know when two lines are intersecting, the opposite angles are equal. So

$x=60^0$

Now, As we know the sum of internal angles of a triangle is 180. so,

$30^0+y+x=180^0$

$30^0+y+60^0=180^0$

$y=180^0-30^0-60^0$

$y=90^0$

Hence, $x=60^0\:\:and\:\:y=90^0$

v) As we know when two lines are intersecting, the opposite angles are equal. So

$y=90^0$

Now, As we know the sum of internal angles of a triangle is 180. so,

$y+x+x=180^0$

$90^0+2x=180^0$

$2x=90^0$

$x=45^0$

Hence, $x=45^0\:\:and\:\:y=90^0$

vi)As we know when two lines are intersecting, the opposite angles are equal. So

$y=x$

Now, As we know the sum of internal angles of a triangle is 180. so,

$x+x+x=180^0$

$3x=180^0$

$x=60^0$

Hence, $x=60^0\:\:and\:\:y=60^0$.

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

(i) 2 cm, 3 cm, 5 cm             (ii) 3 cm, 6 cm, 7 cm
(iii) 6 cm, 3 cm, 2 cm

As we know, According to the Triangle inequality law, the sum of lengths of any two sides of a triangle would always greater than the length of the third side. So

Verifying this inequality by taking all possible combinations, we have,

(i) 2 cm, 3 cm, 5 cm

3 + 5 > 2 ----> True

2 + 5 >  3---->  True

2 + 3 > 5 ---->False

Hence the triangle is not possible.

(ii) 3 cm, 6 cm, 7 cm
3 + 6 > 7 -----> True

3 + 7 > 6 ------>True

6 + 7 > 3 ------>True

Hence, The triangle is possible.

(iii) 6 cm, 3 cm, 2 cm

6 + 3 > 2 ------>True

6 + 2 > 3 ------> True

3 + 2 > 6 ------->False

Hence triangle is not possible.

(i)  $\small OP+OQ> PQ$ ?
(ii) $\small OQ+OR> QR$ ?
(iii) $\small OR+OP> RP$?

i) As POQ is a triangle, the sum of any two sides will always be greater than the third side. so

Yes,  $\small OP+OQ> PQ$ .

ii) As ROQ is a triangle, the sum of any two sides will always be greater than the third side. so

Yes,  $\small OQ+OR> QR$

iii) As ROQ is a triangle, the sum of any two sides will always be greater than the third side. so

Yes,$\small OR+OP> RP$

As we know that the sum of two sides of ANY triangle is always greater than the third side(Triangles Inequality Rule).

So,

In $\small \Delta ABM$ :

$\overline {AB}+\overline {BM}>\overline{AM}...........(1)$

In $\small \Delta AMC$ :

$\overline {AC}+\overline {CM}>\overline{AM}...........(2)$

Adding (1) and (2), we get

$\overline{AB}+\overline {AC}+\overline{BM}+\overline {CM}>\overline{AM}+\overline{AM}$

As we can see M is the point in line BC So, we can say

$\overline{BM}+\overline {CM}=\overline {BC}$

So our equation becomes

$\overline{AB}+\overline {AC}+\left (\overline{BM}+\overline {CM} \right )>\overline{AM}+\overline{AM}$

$\small AB+BC+CA> 2AM$.

Hence it is a True statement.

As we know that the sum of two sides of ANY triangle is always greater than the third side(Triangles Inequality Rule).

So,

In $\small \Delta ABC$ :

$\overline {AB}+\overline {BC}>\overline{AC}...........(1)$

In $\small \Delta CDA$ :

$\overline {CD}+\overline {DA}>\overline{BD}...........(2)$

Adding (1) and (2) we get,

$\small AB+BC+CD+DA> AC+BD$

Hence the given statement is True.

Let the intersection point of the two diagonals be O.

As we know that the sum of two sides of ANY triangle is always greater than the third side(Triangles Inequality Rule).

So,

In $\small \Delta AOB$ :

$\overline {AO}+\overline {OB}>\overline{AB}...........(1)$

In $\small \Delta BOC$ :

$\overline {BO}+\overline {OC}>\overline{BC}...........(2)$

In $\small \Delta COD$ :

$\overline {CO}+\overline {OD}>\overline{CD}...........(3)$

In $\small \Delta DOA$ :

$\overline {DO}+\overline {OA}>\overline{DA}...........(4)$

Now, Adding all four equations we, get

$\overline{AO}+\overline {OB}+\overline{BO}+\overline{OC}+\overline {CO}+\overline {OD}+\overline{DO}+\overline{OA}>\overline{AB}+\overline{BC}+\overline{CD}+\overline{DA}$

$2\left (\overline{AO}+\overline {OB}+\overline{OC}+\overline {OD} \right )>\overline{AB}+\overline{BC}+\overline{CD}+\overline{DA}$

$2\left (\left (\overline{AO}+\overline {OC} \right )+\left (\overline{CO}+\overline {OD} \right ) \right )>\overline{AB}+\overline{BC}+\overline{CD}+\overline{DA}$

$2\left (\overline{AC}+ \overline{BD} \right )>\overline{AB}+\overline{BC}+\overline{CD}+\overline{DA}$

which can also be expressed as

$\small AB+BC+CD+DA< 2(AC+BD)$

Hence this is true.

Let ABC be a triangle with AB = 12cm and BC = 15cm

Now

As we know that the sum of two sides of ANY triangle is always greater than the third side(Triangles Inequality Rule).

AB + BC > CA

12 + 15 > CA

CA < 27......(1)

Also, in a similar way

AB + CA  > BC

CA > BC -  AB

CA > 15 - 12

CA > 3............(2)

Hence from (1) and (2), we can say that the length of third side of the triangle must be between 3cm to 27 cm.

## 1.  PQR is a triangle, right-angled at P. If  $\small PQ = 10\hspace{1mm}cm$  and  $\small PR = 24\hspace{1mm}cm$, find QR.

As we know,

In a Right-angled Triangle: By Pythagoras Theorem,

$(Hypotenus)^2=(Base)^2+(Perpendicular)^2$

As PQR is a right-angled triangle with

Base = PQ = 10 cm.

Perpendicular = PR = 24 cm.

Hypotenuse = QR

So, By Pythagoras theorem,

$(QR)^2=(PQ)^2+(PR)^2$

$(QR)^2=(10)^2+(24)^2$

$(QR)^2=100+576$

$(QR)^2=676$

$QR=26 cm$

Hence, Length od QR is 26 cm.

As we know,

In a Right-angled Triangle: By Pythagoras Theorem,

$(Hypotenus)^2=(Base)^2+(Perpendicular)^2$

As ABC is a right-angled triangle with

Base = AC= 7 cm.

Perpendicular = BC

Hypotenuse = AB = 25 cm

So, By Pythagoras theorem,

$(AB)^2=(AC)^2+(BC)^2$

$(25)^2=(7)^2+(BC)^2$

$(BC)^2=(25)^2-(7)^2$

$(BC)^2=625-49$

$(BC)^2= 576$

$BC= 24$

Hence, Length od BC is 24 cm.

Here. As we can see, The ladder with wall forms a right-angled triangle with

the vertical height of the wall = perpendicular = 12 m

length of ladder = Hypotenuse = 15 m

Now, As we know

In a Right-angled Triangle: By Pythagoras Theorem,

$(Hypotenus)^2=(Base)^2+(Perpendicular)^2$

$(15)^2=(Base)^2+(12)^2$

$(Base)^2=(15)^2-(12)^2$

$(Base)^2=225-144$

$(Base)^2=81$

$Base=9 m$

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

(i) $\small 2.5\hspace{1mm} cm$, $\small 6.5\hspace{1mm} cm$, 6 cm.

(ii) 2 cm, 2 cm, 5 cm.

(iii) $\small 1.5\hspace{1mm} cm$, 2cm, $\small 2.5\hspace{1mm} cm$

In the case of right-angled triangles, identify the right angles.

As we know,

In a Right-angled Triangle: By Pythagoras Theorem,

$(Hypotenus)^2=(Base)^2+(Perpendicular)^2$

(i) $\small 2.5\hspace{1mm} cm$, $\small 6.5\hspace{1mm} cm$, 6 cm.

As we know the hypotenuse is the longest side of the triangle, So

Hypotenuse = 6.5 cm

Verifying the Pythagoras theorem,

$(6.5)^2=(6)^2+(2.5)^2$

$42.25=36+6.25$

$42.25=42.25$

Hence it is a right-angled triangle.

The Right-angle lies on the opposite of the longest side (hypotenuse) So the right angle is at the place where 2.5 cm side and 6 cm side meet.

(ii) 2 cm, 2 cm, 5 cm.

As we know the hypotenuse is the longest side of the triangle, So

Hypotenuse = 5 cm

Verifying the Pythagoras theorem,

$(5)^2=(2)^2+(2)^2$

$25=4+4$

$25\neq8$

Hence it is Not a right-angled triangle.

(iii) $\small 1.5\hspace{1mm} cm$, 2cm, $\small 2.5\hspace{1mm} cm$.

As we know the hypotenuse is the longest side of the triangle, So

Hypotenuse = 2.5 cm

Verifying the Pythagoras theorem,

$(2.5)^2=(2)^2+(1.5)^2$

$6.25=4+2.25$

$6.25=6.25$

Hence it is a Right-angled triangle.

The right angle is the point where the base and perpendicular meet.

As we can see the tree makes a right angle with

Perpendicular = 5 m

Base = 12 m

As we know,

In a Right-angled Triangle: By Pythagoras Theorem,

$(Hypotenus)^2=(Base)^2+(Perpendicular)^2$

$(Hypotenus)^2=(12)^2+(5)^2$

$(Hypotenus)^2=144+25$

$(Hypotenus)^2=169$

$Hypotenus=13$

Here, The Hypotenuse of the triangle was also a part of the tree originally, So

The Original height of the tree = height + hypotenuse

= 5 m + 13 m

= 18 m.

Hence the original height of the tree was 18 m.

(i)  $\small PQ^2+QR^2=RP^2$

(ii) $\small PQ^2+RP^2=QR^2$

(iii) $\small RP^2+QR^2=PQ^2$

As we know the sum of the angles of any triangle is always 180. So,

$\angle P + \angle Q + \angle R = 180^0$

$\angle P + 25^0 + 65^0 = 180^0$

$\angle P = 180^0- 25^0 - 65^0$

$\angle P = 90^0$

Now. Since PQR is a right-angled triangle with right angle at P. So

$(Hypotenus)^2=(Base)^2+(Perpendicular)^2$

$(QR)^2=(PQ)^2+(RP)^2$

$\small PQ^2+RP^2=QR^2$

Hence option (ii) is correct.

As we can see in the rectangle,

By Pythagoras theorem,

$(Diagonal)^2=(Length)^2+(Width)^2$

Now as given in the question,

Diagonal = 41 cm.

Length = 40 cm.

So, Putting these value we get,

$(41)^2=(40)^2+(Width)^2$

$(Width)^2=(41)^2-(40)^2$

$(Width)^2=1681-1600$

$(Width)^2=81$

$Width=9cm$

Hence the width of the rectangle is 9 cm.

So

The perimeter of the rectangle = 2 ( Length + Width )

= 2 ( 40 cm + 9 cm )

= 2 x 49 cm

= 98 cm

Hence the perimeter of the rectangle is 98 cm.

As we know that the diagonals of the rhombus are perpendicular to each other and intersect at a point which is mid of both the diagonal.

So. By Pythagoras Theorem we can say that

$(Hypotenus)^2=(Base)^2+(Perpendicular)^2$

$(Side)^2=\left(\frac{16}{2}\right)^2+\left ( \frac{30}{2} \right )^2$

$(Side)^2= 8 ^2 + 15 ^2$

$(Side)^2= 64 + 225$

$(Side)^2= 289$

$Side= 17\:cm$

Hence Side of the rhombus is 17 cm.

So,

The Perimeter of the rhombus = 4 x 17 cm

= 68 cm.

Hence, the perimeter of the rhombus is 68 cm.

## NCERT Solutions for Class 7 Maths - Chapter-wise

 Chapter No. Chapter Name Chapter 1 Solutions of NCERT for class 7 maths chapter 1 Integers Chapter 2 CBSE NCERT solutions for class 7 maths chapter 2 Fractions and Decimals Chapter 3 NCERT solutions for class 7 maths chapter 3 Data Handling Chapter 4 Solutions of NCERT for class 7 maths chapter 4 Simple Equations Chapter 5 CBSE NCERT solutions for class 7 maths chapter 5 Lines and Angles Chapter 6 NCERT solutions for class 7 maths chapter 6 The Triangle and its Properties Chapter 7 Solutions of NCERT for class 7 maths chapter 7 Congruence of Triangles Chapter 8 NCERT solutions for class 7 maths chapter 8 comparing quantities Chapter 9 CBSE NCERT solutions for class 7 maths chapter 9 Rational Numbers Chapter 10 NCERT solutions for class 7 maths chapter 10 Practical Geometry Chapter 11 Solutions of NCERT for class 7 maths chapter 11 Perimeter and Area Chapter 12 CBSE NCERT solutions for class 7 maths chapter 12 Algebraic Expressions Chapter 13 NCERT solutions for class 7 maths chapter 13 Exponents and Powers Chapter 14 Solutions of NCERT for class 7 maths chapter 14 Symmetry

## Important terms of NCERT class 7 maths chapter 6 the triangles and its properties -

The following terms are discussed in the chapter and questions based on these points are covered in the CBSE NCERT solutions for class 7 maths chapter 6 the triangle and its properties.

• Elements of a triangle- The six elements of a triangle are its three sides and the three angles.

• Median of a triangle- It is a line segment joining a vertex with the midpoint of its opposite side of the triangle. A triangle has three medians.

• The altitude of a triangle- It is a perpendicular line segment from a vertex to its opposite side of the triangle. A triangle has three altitudes.
• The angle sum property of a triangle- According to this the total measure of the three angles or the sum of all three angles of a triangle is
$\dpi{100} 180^o$.
• Equilateral triangle- It is a triangle with all sides are equal in length. In an equilateral triangle, all three angles have measure 60°.
• Isosceles triangle- It is a triangle is said to be an isosceles triangle if at least any two of its sides are equal in length.

Students can expect a similar type of questions discussed in the NCERT solutions for class 7 maths chapter 6 the triangle and its properties in exams.