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

Triangles are one of the most fundamental shapes in geometry, and understanding their properties forms the basis for learning more advanced mathematical concepts. From identifying different types of triangles to learning the relationships between their angles and sides, this chapter helps students develop a strong knowledge of geometric properties. Knowing how to apply properties like the angle sum property, Pythagoras' theorem, and medians and altitudes is essential for solving the problems in this chapter. The NCERT Solutions help the students in practice, exam preparation, and strengthen conceptual clarity.

Latest: NCERT Class 7 Maths: Chapterwise Important Formulas with Examples and Points

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1. NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties - Important Points
2. NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties
3. NCERT Solutions for Class 7 Maths Chapter 6 Exercise
4. The Triangle and Its Properties Class 6 Maths Chapter 6 - Topics
5. NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties - Points to Remember
6. NCERT Solutions for Class 7 Maths Chapter Wise
7. NCERT Solutions for Class 7 Subject Wise

These Solutions for Class 7 Maths Chapter 6 are made by subject matter experts at Careers360 to help students learn easily. These solutions explain all the topics in a simple way so that students can understand them better and solve questions easily. To access expert solutions for other chapters as well, the students can visit the NCERT Solutions for Class 7 Maths.

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

Median of a triangle: A line that joins a vertex of a triangle to the midpoint of its opposite side.

A triangle has 3 medians

Exterior Angle: It is formed when a side of a triangle is produced.

Exterior Angle = Sum of the two interior opposite angles.

Angle sum property for a triangle: The Sum of all three interior angles = ${180}^{\circ }$

Types of triangles:

Equilateral Triangle:

• All sides are equal.
• All angles are equal (${60}^{\circ }$).

Isosceles Triangle:

• The two sides are equal.
• The angles opposite the equal sides are also equal.

Scalene Triangle:

• No sides are equal.
• No angles are equal.

Property of the length of the sides of a triangle:

The sum of any two sides > Third side.

Difference between any two sides < Third side.

Pythagoras property:

Square on the hypotenuse = the sum of the squares on its legs. (i.e) $c^2=a^2+b^2$

It is useful to decide whether a triangle is right-angled or not.

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

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NCERT Solutions for Class 7 Maths Chapter 6 Exercise

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

$\overline{PM}$ is _________________.

PD is _________________.

Is $QM=MR$ ?

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

PD is the median of the triangle.

No, $QM\ne MR$. As QD = DR.

2. Draw rough sketches for the following:

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

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

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

Answer: (a) In $\mathrm{\Delta }ABC$, BE is a median.

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

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

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

Answer: Yes, it is very much possible that the median and altitude of an isosceles triangle are 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:

Answer: 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) $x={30}^0+{60}^0={90}^0$

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

Answer: 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) $x+{30}^0={80}^0\Rightarrow x={80}^0-{30}^0={50}^0$

vi) $x+{35}^0={75}^0\Rightarrow {75}^0-{35}^0={40}^0$

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

Answer: 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

Answer: 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 the 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, $x={70}^{0}andy={60}^0$.

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

$y={80}^0$

Now, as we know, the sum of the 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, $x={50}^{0}andy={80}^0$.

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

$x={50}^0+{60}^0$

$x={110}^0$

Now, as we know, the sum of the 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, $x={110}^{0}andy={70}^0$.

iv) As we know, when two lines intersect, the opposite angles are equal. So

$x={60}^0$

Now, as we know, the sum of the 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, $x={60}^{0}andy={90}^0$

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

$y={90}^0$

Now, as we know, the sum of the 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, $x={45}^{0}andy={90}^0$

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

$y=x$

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

$x+x+x={180}^0$

$3x={180}^0$

$x={60}^0$

Hence, $x={60}^{0}andy={60}^0$.

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

Answer: As we know, according to the Triangle Inequality Law, the sum of the lengths of any two sides of a triangle is 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.

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

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

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

Yes, $OP+OQ>PQ$.

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

Yes, $OQ+OR>QR$

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

Yes, $OR+OP>RP$

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

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

So,

In $\mathrm{\Delta }ABM$:

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

In $\mathrm{\Delta }AMC$:

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

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

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

$AB+BC+CA>2AM$.

Hence, it is a true statement.

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

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

So,

In $\mathrm{\Delta }ABC$:

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

In $\mathrm{\Delta }CDA$ :

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

Adding (1) and (2), we get,

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

Hence, the given statement is True.

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

Answer: Let the intersection point of the two diagonals be O.

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

So,

In $\mathrm{\Delta }AOB$:

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

In $\mathrm{\Delta }BOC$ :

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

In $\mathrm{\Delta }COD$ :

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

In $\mathrm{\Delta }DOA$ :

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

Now, adding all four equations, we get

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

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

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

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

which can also be expressed as

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

Hence, this 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?

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

As we know that the sum of the two sides of ANY triangle is always greater than the third side(Triangle 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 ofthe third side of the triangle must be between 3cm to 27cm.

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

Answer: 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,

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

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

$(QR{)}^2=100+576$

$(QR{)}^2=676$

$QR=26cm$

Hence, the Length of QR is 26 cm.

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

Answer: 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, the length of BC is 24 cm.

3. A 15m long ladder reached a window 12m high from the ground on placing it against a wall at a distance of a.. Find the distance of the foot of the ladder from the wall.

Answer: Here, as we can see, the ladder with the wall forms a right-angled triangle with the vertical height of the wall = perpendicular = 12m

length of ladder = Hypotenuse = 15m

Now, as we know

In a Right-angled Triangle: By Pythagoras' Theorem,

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

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

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

$(Base{)}^2=225-144$

$(Base{)}^2=81$

$Base=9m$

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?

(i) $2.5cm$ , $6.5cm$ , 6 cm.

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

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

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

Answer: As we know,

In a Right-angled Triangle: By Pythagoras Theorem,

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

(i) $2.5cm$, $6.5cm$, 6cm.

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

Hypotenuse = 6.5cm

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 side of the longest side (hypotenuse). So the right angle is at the place where the 2.5cm side and the 6cm side meet.

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

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

Hypotenuse = 5cm

Verifying the Pythagoras theorem,

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

$25=4+4$

$25\ne 8$

Hence it is Not a right-angled triangle.

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

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

Hypotenuse = 2.5cm

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.

5. A tree is broken at a height of 5m from the ground and its top touches the ground at a distance of 12m from the base of the tree. Find the original height of the tree.

Answer: As we can see the tree makes a right angle with

Perpendicular = 5m

Base = 12m

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

= 5m + 13m

= 18m.

Hence the original height of the tree was 18m.

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

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

$\mathrm{\angle }P+\mathrm{\angle }Q+\mathrm{\angle }R={180}^0$

$\mathrm{\angle }P+{25}^0+{65}^0={180}^0$

$\mathrm{\angle }P={180}^0-{25}^0-{65}^0$

$\mathrm{\angle }P={90}^0$

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

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

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

$P{Q}^2+R{P}^2=Q{R}^2$

Hence option (ii) is correct.

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

Answer: As we can see in the rectangle,

By Pythagoras theorem,

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

Now as given in the question,

Diagonal = 41cm.

Length = 40cm.

So, putting these values 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 9cm.

So, the perimeter of the rectangle = 2 ( Length + Width )

= 2 ( 40cm + 9cm )

= 2 x 49cm

= 98cm

Hence the perimeter of the rectangle is 98cm.

8. The diagonals of a rhombus measure 16cm and 30cm. Find its perimeter.

Answer: 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 diagonals.

So, by Pythagoras' Theorem, we can say that

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

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

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

$(Side{)}^2=64+225$

$(Side{)}^2=289$

$Side=17cm$

Hence side of the rhombus is 17cm.

So, the Perimeter of the rhombus = 4 x 17cm

= 68cm.

Hence, the perimeter of the rhombus is 68cm.

The Triangle and Its Properties Class 6 Maths Chapter 6 - Topics

• Medians Of A Triangle
• Altitudes Of Triangle
• Exterior Angle Of A Triangle and Its Properties
• Angle Sum Property Of A Triangle
• Two Special Triangles: Equilateral And Isosceles
• Sum Of The Lengths Of Two Sides Of A Triangle
• Right-Angle Triangles And Pythagoras' Property

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

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 ${180}^{\circ }$.

Equilateral triangle: It is a triangle with all sides equal in length. In an equilateral triangle, all three angles have a measure of ${60}^{\circ }$.

Isosceles triangle: A triangle is said to be an isosceles triangle if at least two of its sides are equal in length.

Exterior angle of the triangle: The sum of the two interior opposite angles.

Angle sum property for a triangle: The sum of all three interior angles = ${180}^{\circ }$

Relationship between the sides of the triangle:
The sum of any two sides of the triangle > Third side.
Difference between any two sides < Third side.

Pythagoras' property: $a^2+b^2=c^2$, where c is the hypotenuse and a,b are the other two sides of the triangle.

NCERT Solutions for Class 7 Maths Chapter Wise

NCERT Solutions for Class 7 Subject Wise

Students can refer to the link below to access the subject-wise solutions for Class 7.

• NCERT Solutions for Class 7 Maths
• NCERT Solutions for Class 7 Science

Students can also check the NCERT Books and the NCERT Syllabus here:

• NCERT Syllabus Class 7 Maths
• NCERT Books Class 7
• NCERT Syllabus Class 7