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

Fill the following blanks.

$\overline{PM}$ is ___________ 

$PD$ is _______

Is $QM=MR$?

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

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 $△ABC$, $BE$ 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 $△ABC$, where $BE$ is a median, is drawn below.

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

(b) In $△PQR$, $PQ$ 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 $△PQR$, where $PQ$ and $PR$ are the altitudes of the triangle is drawn below.

(c) In $△XYZ$, $YL$ 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 $△XYZ$, $YL$ 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 $△ABC$ 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 $△ABC$. Hence verified, $AL$ is the median and altitude of the given triangle $△ABC$.

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 $△ABC$, $\mathrm{\angle }BAC+\mathrm{\angle }ACB+\mathrm{\angle }ABC={180}^{\circ }$.

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

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

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 $△PQR$, $\mathrm{\angle }RPQ+\mathrm{\angle }PQR+\mathrm{\angle }QRP={180}^{\circ }$.

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

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

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 $△XYZ$, $\mathrm{\angle }ZXY+\mathrm{\angle }XYZ+\mathrm{\angle }YZX={180}^{\circ }$.

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

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

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.

$$\Rightarrow {\displaystyle \frac{2x}{2}}={\displaystyle \frac{{130}^{\circ }}{2}}$$

 $$\Rightarrow x={65}^{\circ }$$

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.

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

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

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.

$$\Rightarrow {\displaystyle \frac{3x}{3}}={\displaystyle \frac{{90}^{\circ }}{3}}$$

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

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.

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

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

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.

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

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

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

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 $△ABC$, $AB+BC>AC$.

In $△ADC$, $AD+DC>AC$.

In $△DCB$, $DC+CB>DB$.

In $△ADB$, $AD+AB>DB$.

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.

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

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

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 $△AOB$, $AB<OA+OB$.

In $△BOC$, $BC<OB+OC$.

In $△COD$, $DC<OC+OD$.

In $△AOD$, $DA<OD+OA$.

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

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

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

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 $△PQR$, ${\left(\mathrm{P}R\right)}^2={\left(PQ\right)}^2+{\left(RQ\right)}^2$ 

$L.H.S.={\left(2.5\right)}^2=6.25cm$ 

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

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 $A^{\prime }CB$ represents the tree before it broken at the point $C$ and let the top $A^{\prime }$ touches the ground at $A$ after it broke. Then $△ABC$ is a right angled triangle, right angled at $B$.

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

Using Pythagoras theorem, In $△ABC$

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

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

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

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

$AC=13m$

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 $△PQR$ 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 $△PQR$,

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

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

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

Thus, $△PQR$ 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

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.

Other Study Material for CBSE Class 7 Maths Chapter 6

Chapter-Specific NCERT Solutions for Class 7 Maths

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