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


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

Exercise 6.1

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

$\overline{PM}$ is __________

$PD$ is __________

Is $QM=MR$?

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Ans: Given: $QD=DR$

As we know, the Median divides the line segment into two equal parts. So, $PD$ is the median. And $PM$ makes $90{}^\circ $ at line segment on $QR$

So, $\overline{PM}$ is altitude.

It states $QM\ne MR$ as $D$ is the mid-point of $QR$.


2. Draw rough sketches for the following:

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

Ans:

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Here, $BE$ is a median in $\Delta ABC$

As we know, the Median divides the line segment into two equal parts.

Then, $BE$ is a median and $AE=EC$

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

Ans: Here, $PQ$ and $PR$ intersect at $90{}^\circ $ So $PQ$ and $PR$ are the altitudes of the $\Delta PQR$ and $RP\bot QP$.

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

Ans: To draw altitude in the exterior of $\Delta XYZ$,

First, we extend the line segment to point at $L$, such as $YL\bot XL$.

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

Ans: An isosceles triangle is a triangle that has two sides of equal lengths.

So, let us consider the $\Delta ABC$ which is an isosceles triangle having$AB=AC$.

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Now, let us draw a line from the vertex $B$ which is perpendicular to the side $AC$ of the triangle.

This line behaves as a median and altitude of the given triangle because point $L$ is the middle point between the line segment $AC$. 


Exercise 6.2

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

(i)

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Ans: In fig (i) we know, 

Exterior angle of a side in a triangle is sum of interior opposite angles, so $x=50{}^\circ +70{}^\circ $

$\Rightarrow x=120{}^\circ $.

(ii)

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Ans: As we know in fig (ii), 

Exterior angle of a side in a triangle is sum of interior opposite angles, so $x=65{}^\circ +45{}^\circ $

$\Rightarrow x=110{}^\circ $.

(iii)

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Ans: As we know in fig (iii), 

Exterior angle of a side in a triangle is sum of interior opposite angles, so $x=30{}^\circ +40{}^\circ $ $\Rightarrow x=70{}^\circ $.

(iv)

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Ans: As we know in fig (iv), 

Exterior angle of a side in a triangle is sum of interior opposite angles, so $x=60{}^\circ +60{}^\circ $$\Rightarrow x=120{}^\circ $.

(v) 

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Ans: As we know in fig (v),

Exterior angle of a side in a triangle is sum of interior opposite angles, so $x=50{}^\circ +50{}^\circ $ $\Rightarrow x=100{}^\circ $.

(vi)

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As we know,

Exterior angle of a side in a triangle is sum of interior opposite angles, so $x=30{}^\circ +60{}^\circ $ $\Rightarrow x=90{}^\circ $.


2. Find the value of the unknown interior angle $x$ in the following figures:

(i) 

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Ans: As we know in fig (i),

Exterior angle of a side in a triangle is sum of interior opposite angles, so $x+50{}^\circ =115{}^\circ $$\Rightarrow x=115{}^\circ -50{}^\circ $

$\Rightarrow x=65{}^\circ $.

(ii)

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Ans: As we know in fig (ii),

Exterior angle of a side in a triangle is sum of interior opposite angles, so $x+70{}^\circ =100{}^\circ $ $\Rightarrow x=100{}^\circ -70{}^\circ $

$\Rightarrow x=30{}^\circ $.

(iii)

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Ans: As we know in fig (iii),

Exterior angle of a side in a triangle is sum of interior opposite angles, so $x+90{}^\circ =125{}^\circ $ $\Rightarrow x=125{}^\circ -90{}^\circ $

$\Rightarrow x=35{}^\circ $.

(iv)

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Ans: As we know in fig (iv),

Exterior angle of a side in a triangle is sum of interior opposite angles, so $x+60{}^\circ =120{}^\circ $$\Rightarrow x=120{}^\circ -60{}^\circ $

$\Rightarrow x=60{}^\circ $.


(v) 

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Ans: As we know in fig (v),

Exterior angle of a side in a triangle is sum of interior opposite angles, so $x+30{}^\circ =80{}^\circ $$\Rightarrow x=80{}^\circ -30{}^\circ $

$\Rightarrow x=50{}^\circ $.


(vi)

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Ans: As we know in fig (vi),

Exterior angle of a side in a triangle is sum of interior opposite angles, so $x+35{}^\circ =75{}^\circ $$\Rightarrow x=75{}^\circ -35{}^\circ $

$\Rightarrow x=40{}^\circ $.


Exercise 6.3

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

(i)  

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Ans: In a triangle, all interior angles are $180{}^\circ $, this property is known as angle sum property of triangle.

In $\Delta ABC$, 

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

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

$\Rightarrow x=180{}^\circ -110{}^\circ $

$\Rightarrow x=70{}^\circ $.

(ii) 

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Ans: In a triangle, all interior angles are $180{}^\circ $, this property is known as angle sum property of triangle.

In $\Delta PQR$, 

$\angle P+\angle Q+\angle R=180{}^\circ $

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

$\Rightarrow x=180{}^\circ -120{}^\circ $

$\Rightarrow x=60{}^\circ $.

(iii)

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Ans: In a triangle, all interior angles are $180{}^\circ $, this property is known as angle sum property of triangle.

In $\Delta ABC$, 

$\angle X+\angle Y+\angle Z=180{}^\circ $

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

$\Rightarrow x=180{}^\circ -140{}^\circ $

$\Rightarrow x=40{}^\circ $.

(iv)

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Ans: In a triangle, all interior angles are $180{}^\circ $, this property is known as angle sum property of triangle.

In given triangle, 

$x+x+50{}^\circ =180{}^\circ $

$\Rightarrow 2x=180{}^\circ -50{}^\circ $

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

$\Rightarrow x=65{}^\circ $.

(v) 

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Ans: In a triangle, all interior angles are $180{}^\circ $, this property is known as angle sum property of triangle.

In given triangle, 

$x+x+x=180{}^\circ $

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

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

$\Rightarrow x=60{}^\circ $.

(vi)

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Ans: In a triangle, all interior angles are $180{}^\circ $, this property is known as angle sum property of triangle.

In given triangle, 

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

$\Rightarrow 3x=180{}^\circ -90{}^\circ $

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

$\Rightarrow x=30{}^\circ $.


2. Find the value of the unknowns $x$ and $y$ in the following figures:

(i)  

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Ans: In a triangle, all interior angles are $180{}^\circ $, this property is known as angle sum property of triangle.

In given triangle, 

$y+120{}^\circ =180{}^\circ $ 

$\Rightarrow y=180{}^\circ -120{}^\circ $

$\Rightarrow y=60{}^\circ $ and 

$60{}^\circ +x+50{}^\circ =180{}^\circ $

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

$\Rightarrow x=70{}^\circ $

Value of $x,y$ is $70{}^\circ ,60{}^\circ $.

(ii) 

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Ans: In a triangle, all interior angles are $180{}^\circ $, this property is known as angle sum property of triangle.

In given triangle, 

$y=80{}^\circ $  (as vertically Opposite angles are equal)

and $50{}^\circ +x+y=180{}^\circ $

$\Rightarrow 50{}^\circ +80{}^\circ +x=180{}^\circ $

$\Rightarrow y=180{}^\circ -130{}^\circ $

$\Rightarrow x=50{}^\circ $

Value of $x,y$ is $50{}^\circ ,80{}^\circ $ .

(iii)

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Ans: In a triangle, all interior angles are $180{}^\circ $, this property is known as angle sum property of triangle.

In given triangle,

$y+60{}^\circ +50{}^\circ =180{}^\circ $ 

$\Rightarrow y=180{}^\circ -110{}^\circ $

$\Rightarrow y=70{}^\circ $ and 

$x+y=180{}^\circ $

$\Rightarrow x=180{}^\circ -70{}^\circ $

$\Rightarrow x=110{}^\circ$

Value of $x,y$ is $110{}^\circ ,70{}^\circ $.

(iv)

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Ans: In a triangle, all interior angles are $180{}^\circ $, this property is known as angle sum property of triangle.

In given triangle, 

$x=60{}^\circ $ (as vertically Opposite angles are equal)

and $30{}^\circ +60{}^\circ +y=180{}^\circ $

$\Rightarrow y=180{}^\circ -90{}^\circ $

$\Rightarrow y=90{}^\circ $

Value of $x,y$ is $60{}^\circ ,90{}^\circ $.

(v) 

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Ans: In a triangle, all interior angles are $180{}^\circ $, this property is known as angle sum property of triangle.

In given triangle, 

$y=90{}^\circ $  (as vertically Opposite angles are equal)

and $x+x+90{}^\circ =180{}^\circ $

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

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

$\Rightarrow x=45{}^\circ $

Value of $x,y$ is $45{}^\circ ,90{}^\circ $.

(vi)

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Ans: In a triangle, all interior angles are $180{}^\circ $, this property is known as angle sum property of triangle.

In given triangle, 

$y=x$ (as vertically Opposite angles are equal)

and $x+x+x=180{}^\circ $

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

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

$\Rightarrow x=60{}^\circ $

Value of $x,y$ is $60{}^\circ ,60{}^\circ $.


Exercise 6.4

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

(i). $2\ \text{cm}$, $3\ \text{cm}$, $5\ \text{cm}$

Ans: As we know,

In a triangle, the sum of the length of any two sides should be greater than the length of the third side. i.e., $2+3>5$ is not possible

$5+3>2$ is possible

$2+5>3$ is possible

Hence, the triangle is not possible.

(ii). $3\ \text{cm}$, $6\ \text{cm}$, $7\ \text{cm}$

Ans: As we know,

In a triangle, the sum of the length of any two sides should be greater than the length of the third side.$6+3>7$ is possible

$7+3>6$ is possible

\[6+7>3\] is possible

Hence, the triangle is possible.

(iii). $6\ \text{cm}$, $3\ \text{cm}$, $2\ \text{cm}$

Ans: As we know,

In a triangle, the sum of the length of any two sides should be greater than the length of the third side. $6+3>2$ is possible

$6+2>3$ is possible

$2+3>6$ is not possible

Hence, the triangle is not possible.


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

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(i). Is $OP+OQ>PQ$?

Ans: YES

As we know, in a triangle, the sum of the length of any two sides should be greater than the length of the third side.

So, In $\Delta POQ$,

$\Rightarrow OP+OQ>PQ$.

(ii). Is $OQ+OR>QR$?

Ans: Yes

As we know, in a triangle, the sum of the length of any two sides should be greater than the length of the third side.

So, In $\Delta ROQ$,

$\Rightarrow OR+OQ>QR$.

(iii). Is $OR+OP>RP$?

Ans: Yes

As we know, in a triangle, the sum of the length of any two sides should be greater than the length of the third side.

So, In $\Delta POR$,

$\Rightarrow OP+OR>PR$


3. $AM$ is a median of a triangle $ABC$. Is $AB+BC+CA>2AM$? (Consider the sides of triangles $\Delta ABM$ and $\Delta AMC$).

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Ans: As we know that,

The sum of any two sides in a triangle should be greater than the length of the third side.

So, In $\Delta ABM$,

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

In $\Delta AMC$,

$AC+MC>AM\ \ ......\text{(2)}$

Adding both equations, we get –

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

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

So, the statement $AB+AC+BC>2AM$ is TRUE.


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

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Ans: As we know,

In a triangle, sum of the length of any two sides should be greater than length of third side. So, In $\Delta ABC$, $AB+BC>AC\ \ ......\text{(1)}$

In $\Delta ADC$, $AD+DC>AC\ \ ......\text{(2)}$

In $\Delta DCB$, $DC+CB>BD\ \ ......\text{(3)}$

In $\Delta ADB$, $AB+AD>BD\ \ ......\text{(4)}$

Adding all four equations, we get $AB+BC+AD+DC+DC+CB+AD+AB>AC+AC+BD+BD$

$\Rightarrow 2AB+2BC+2AD+2DC>2AC+2BD$

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

Hence the given statement $AB+BC+AD+DC>AC+BD$ is TRUE.


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

Ans: As we know,

In a triangle, sum of the length of any two sides should be greater than length of third side.

So, In $\Delta AOB$, $AB<AO+OB\ \ ......\text{(1)}$

In $\Delta BOC$, $BC<OB+OC\ \ ......\text{(2)}$

In $\Delta COD$, $CD<OC+OD\ \ ......\text{(3)}$

In $\Delta AOD$, $AD<OD+OA\ \ ......\text{(4)}$

Adding all four equations, we get –$AB+BC+AD+DC<OA+OB+OB+OC+OC+OD+OD+OA$

$\Rightarrow AB+BC+AD+DC<2\left( OA+OC \right)+2\left( OB+OD \right)$

$\Rightarrow AB+BC+AD+DC<2\left( AC+BD \right)$

Hence the given statement $AB+BC+AD+DC<2\left( AC+BD \right)$ is TRUE.


6. The lengths of two sides of a triangle are $12\ \text{cm}$ and $15\ \text{cm}$. Between what two measures should the length of the third shell fall?

Ans: As we know, 

In a triangle, the sum of the length of any two sides should be greater than the length of the third side.

Given: Two sides of triangle $12$ cm and $15$ cm.

Let third sides be $A$ cm

According to the property,

$12+15>A$ 

$27>A\ \ ......\text{(1)}$ and $12+A>15$

$A>3\ \ ......\text{(2)}$

From both equations, we have

$3<A<27$

The length of the third should be greater than $3$ cm and less than $27$ cm. 


Exercise 6.5

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

Ans:

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Given: $PQ=10$ cm and $PR=24$ cm

Let $QR=x$

According to the Pythagoras theorem

${{\left( Hypotenuse \right)}^{2}}={{\left( Base \right)}^{2}}+{{\left( Perpendicular \right)}^{2}}$

In $\Delta PQR$ ,

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

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

$\Rightarrow {{x}^{2}}=100+576$

$\Rightarrow {{x}^{2}}=676$

$\Rightarrow x=\sqrt{676}$

$\Rightarrow x=26$ cm

The length of $QR$ is $26$ cm.


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

Ans:

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Given: $AB=25$ cm and $AC=7$ cm

Let $BC=x$

According to the Pythagoras theorem ${{\left( Hypotenuse \right)}^{2}}={{\left( Base \right)}^{2}}+{{\left( Perpendicular \right)}^{2}}$

In $\Delta ABC$,

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

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

$\Rightarrow 625=49+{{x}^{2}}$

$\Rightarrow {{x}^{2}}=625-49$

$\Rightarrow {{x}^{2}}=576$

$\Rightarrow x=\sqrt{576}$

$\Rightarrow x=24$ cm

The length of $BC$ is $24$ cm.


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.

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Ans: Given: $Hypotenuse=15$ cm and $Perpendicular=7$ cm

Let $Base=x$

According to the Pythagoras theorem ${{\left( Hypotenuse \right)}^{2}}={{\left( Base \right)}^{2}}+{{\left( Perpendicular \right)}^{2}}$$\Rightarrow {{15}^{2}}={{\left( 12 \right)}^{2}}+{{x}^{2}}$

$\Rightarrow 225=144+{{x}^{2}}$

$\Rightarrow {{x}^{2}}=225-144$

$\Rightarrow {{x}^{2}}=81$

$\Rightarrow x=9$ cm

The distance of the foot from ladder is $9$ cm.


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:

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According to the Pythagoras theorem ${{\left( Hypotenuse \right)}^{2}}={{\left( Base \right)}^{2}}+{{\left( Perpendicular \right)}^{2}}$$\Rightarrow {{\left( 6.5 \right)}^{2}}={{\left( 2.5 \right)}^{2}}+{{\left( 6 \right)}^{2}}$

$\Rightarrow 42.25=6.25+36$

LHS = RHS

Hence, the sides are of the right-angled triangle.

Hypotenuse’s length is $6.5$ cm

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

Ans: According to the Pythagoras theorem ${{\left( Hypotenuse \right)}^{2}}={{\left( Base \right)}^{2}}+{{\left( Perpendicular \right)}^{2}}$$\Rightarrow {{\left( 5 \right)}^{2}}={{\left( 2 \right)}^{2}}+{{\left( 2 \right)}^{2}}$

$\Rightarrow 25=8$

LHS $\ne $ RHS

The sides are not of the right-angled triangle.

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

Ans:

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According to the Pythagoras theorem ${{\left( Hypotenuse \right)}^{2}}={{\left( Base \right)}^{2}}+{{\left( Perpendicular \right)}^{2}}$

$\Rightarrow {{\left( 2.5 \right)}^{2}}={{\left( 1.5 \right)}^{2}}+{{\left( 2 \right)}^{2}}$

$\Rightarrow 6.25=2.25+4$

$\Rightarrow 6.25=6.25$

LHS = RHS

Hence, the sides are of the right-angled triangle.

Hypotenuse’s length is $2.5$ cm


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:

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

Let $AC=x$

According to the Pythagoras theorem ${{\left( Hypotenuse \right)}^{2}}={{\left( Base \right)}^{2}}+{{\left( Perpendicular \right)}^{2}}$

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

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

$\Rightarrow {{x}^{2}}=144+25$

$\Rightarrow {{x}^{2}}=169$

$\Rightarrow x=\sqrt{169}$

$\Rightarrow x=13$ m

The length of $AC$ is $13$ m.

Now, we will calculate the total length of the tree –

As, $A'B=AC+BC$.

$\Rightarrow A'B=13+5$

$\Rightarrow A'B=18\ \text{m}$


6. Angles $Q$ and $R$ of a $\Delta PQR$ 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). $R{{P}^{2}}+Q{{R}^{2}}=P{{Q}^{2}}$

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Ans: Given: $\angle Q=25{}^\circ $ and $\angle R=65{}^\circ $

As we know,

In a triangle, all interior angles are $180{}^\circ $, this property is known as angle sum property of triangle. And

In $\Delta PQR$, 

$\angle P+\angle Q+\angle R=180{}^\circ $

$\Rightarrow \angle P+90{}^\circ =180{}^\circ $

$\Rightarrow \angle P=90{}^\circ $

Side opposite to $\angle P$ is $QR$,

According to Pythagoras theorem,${{\left( QR \right)}^{2}}={{\left( PR \right)}^{2}}+{{\left( QP \right)}^{2}}$

Hence, the correct option is Option (2).


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

Ans:

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Given: $AC=40$ cm and $AD=41$ cm 

Let $DC=x$

According to the Pythagoras theorem ${{\left( Hypotenuse \right)}^{2}}={{\left( Base \right)}^{2}}+{{\left( Perpendicular \right)}^{2}}$

In $\Delta ACD$, 

${{\left( AD \right)}^{2}}={{\left( AC \right)}^{2}}+{{\left( DC \right)}^{2}}$

$\Rightarrow {{41}^{2}}={{40}^{2}}+{{x}^{2}}$

$\Rightarrow 1681=1600+{{x}^{2}}$

$\Rightarrow {{x}^{2}}=81$

$\Rightarrow x=\sqrt{81}$

\[\Rightarrow x=9\] cm

The breadth of rectangle is $9$ cm

Perimeter of rectangle is $2\left( l+b \right)$

$\Rightarrow 2\left( 40+9 \right)$

$\Rightarrow 2\left( 49 \right)$

$\Rightarrow 98$ cm

The perimeter of rectangle is $98$ cm.


8. The diagonals of a rhombus measures $16$ cm and $30$ cm. find its perimeter.

Ans:

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

We know that, intersection angle of diagonals of rhombus is $90{}^\circ $ 

So, $OD=\dfrac{DB}{2}=8$ cm and 

$OC=\dfrac{AC}{2}=15$ cm

According to the Pythagoras theorem ${{\left( Hypotenuse \right)}^{2}}={{\left( Base \right)}^{2}}+{{\left( Perpendicular \right)}^{2}}$

In $\Delta OCD$,

${{\left( CD \right)}^{2}}={{\left( OC \right)}^{2}}+{{\left( OD \right)}^{2}}$

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

$\Rightarrow {{\left( CD \right)}^{2}}=64+225$

$\Rightarrow {{\left( CD \right)}^{2}}=289$

$\Rightarrow CD=\sqrt{289}$

$\Rightarrow CD=17$ cm

Perimeter of rhombus is $4\times side$,

$\Rightarrow 4\times 17$

$\Rightarrow 68$ cm

Perimeter of Rhombus is $68$ cm.


NCERT Solutions for Class 7 Maths Chapter 6 Free PDF download

In the official website of CoolGyan, the NCERT Solutions for Class 7 Maths Chapter 6 - The Triangle and its Properties' free PDF book is available for students so that they can practice with no interruptions.

Chapter 6 The Triangle and its Properties

 

NCERT Solutions for Class 7 Maths Chapter 6

6.1 Introduction

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

Exercise 6.1 with Three Problems (1 is a Long Answer, and 2 are Short Answer Type Questions).

 

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, along with a detailed explanation. So, students need to have a glance at NCERT Solutions for Class 7 Maths Chapter 6 PDF books.

Exercise 6.2 with Two Sums (Both are Short Answers).

 

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 some problems based on these to understand more clearly.

Exercise 6.3 with 2 Sums (2 Sums are Short Answers).

 

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 for various reasons. It is beneficial to check out the PDF once or twice for a better understanding of the topic. 

Exercise 6.4 with Six Problems (1 is Long Answer Type, and 5 are Short Answer Type Questions).

 

6.5 Angle Sum Property of a Triangle

In this section of ncert class 7 maths chapter 6 solution, 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. 

Exercise 6.5 with 8 Sums (6 are Short Answers, and 2 are Long Answers).

 

6.6 Two Special Triangles - Equilateral and Isosceles

In the introduction part, as students get aware of different triangles, this section will give a detailed description of the types of triangles based on the sides. They are as follows:

  • An equilateral triangle is a triangle in which all three sides are equal. Here the angle is 60°. 

  • Isosceles Triangle is a triangle where the two sides are equal. Here the angle might be 45°.

  • Scalene  Triangle is a triangle where all the sides are different from one another.

 

6.7 Sum of the Lengths of Two Sides of Triangle

Here students need to learn how to find the length of sides and add them. It is vital to sum the lengths of two sides which are higher than the length of the third side. It is proven and solved by many examples and experiments. So, to avoid further confusion, students can read the PDF through our official website. NCERT Solutions for Class 7 Maths Chapter 6 PDF is explained with diagrams in detail.

 

6.8  Right Angles Triangle and Pythagoras Property

In this last section, students are going to learn about another classification of triangles based on angles. They are:

  • Acute Angles Triangle is a triangle whose angle is less than 45°.

  • Obtuse Angle Triangle is a triangle whose angle is 60°.

  • Right Angle Triangle is the triangle whose angle is 90°.

These are the classification. NCERT Solutions for Class 7 Maths Chapter 6 is helpful to read and understand about Pythagoras Theorem.

The slanting line, which is opposite to the right angle is known as hypotenuse. The square of the hypotenuse is equal to the squares of the sum of the other two sides. This property is known as Pythagoras Theorem.

 

Key Features of NCERT Solutions for Class 7 Maths Chapter 6

Students can have several benefits by using NCERT Solutions for Class 7th Maths Chapter 6 PDF with solutions from the official website of CoolGyan. All the questions have been answered to make students understand all the concepts with ease. Practicing these chapter 6 maths class 11 ncert solutions will definitely help in preparing students to give exams and ace the topic. Some more key points are:

  • Our NCERT Solutions provides you with a better and simple understanding of complicated topics.

  • Our NCERT Solutions are useful to enrich the young mathematicians.

  • Our Solutions explain everything in a simple language to reach out to maximum students

FAQs (Frequently Asked Questions)

Q1: What will I Learn in Class 7 Maths Chapter 6 Chapter?

Ans: 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 - Alternate, corresponding, vertically opposite, adjacent.

  • Sum of two angles lying on the same plane.

  • Types of angles based on degrees – acute, obtuse and right-angled.

  • Sign conventions.

  • 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, for example, quadrilaterals like a rhombus.

  • The various terminologies

  • Angle sum property of triangles.

  • Isosceles and equilateral triangles.

  • Right-angles triangles.

  • Pythagoras property.

Q2: Explain the Basic Properties of the Triangles.

Ans: While solving the figure based problems, you need to know the below given point: 

  • 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 deemed to be 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).

Q3: How Many Exercises and Questions are there in Chapter 6 of Class 7 Maths?

Ans: 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 Sum of all interior angles of a triangle.

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

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

  • Exercise 6.5 consists of 8 questions. This particular exercise questions are mainly focused on Pythagoras property, a triangle and a quadrilateral.

Q 4: How to Lower the Fear of Maths with NCERT Solutions by CoolGyan?

Ans: 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.