 # 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 are available here. For students who feel stressed about searching for the most comprehensive and detailed NCERT Solutions for Class 7 Maths, we at BYJU’S have prepared step-by-step solutions with detailed explanations. We advise students who wish to score good marks in Maths, to go through these solutions and strengthen their knowledge.

Given below are the concepts covered in Chapter 6 – The Triangle and Its Properties of Class 7 Maths NCERT Solutions. The chapter contains 5 exercises covering problems related to these concepts.

• Medians of a Triangle
• Altitudes of a Triangle
• Exterior Angle of a Triangle and Its Property
• Angle Sum Property of a Triangle
• Two Special Triangles: Equilateral and Isosceles
• Sum of the Lengths of Two Sides of a Triangle
• Right-Angled Triangles and Pythagoras Property

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

### Access exercises of NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and Its Properties

Exercise 6.1 Solutions

Exercise 6.2 Solutions

Exercise 6.3 Solutions

Exercise 6.4 Solutions

Exercise 6.5 Solutions

### Access answers to NCERT Solutions Class 7 Maths Chapter 6 – The Triangle and Its Properties

Exercise 6.1 Page: 116

1. In Δ PQR, D is the mid-point of . (i) is ___.

Solution:-

Altitude

An altitude has one endpoint at a vertex of the triangle and another on the line containing the opposite side.

(ii) PD is ___.

Solution:-

Median

A median connects a vertex of a triangle to the mid-point of the opposite side.

(iii) Is QM = MR?

Solution:-

No, QM ≠ MR because D is the mid-point of QR.

2. Draw rough sketches for the following:

(a) In ΔABC, BE is a median.

Solution:-

A median connects a vertex of a triangle to the mid-point of the opposite side. (b) In ΔPQR, PQ and PR are altitudes of the triangle.

Solution:- An altitude has one endpoint at a vertex of the triangle and another on the line containing the opposite side.

(c) In ΔXYZ, YL is an altitude in the exterior of the triangle.

Solution:- In the figure, we may observe that for ΔXYZ, YL is an altitude drawn exteriorly to side XZ which is extended up to point L.

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

Solution:- Draw a line segment PS ⊥ BC. It is an altitude for this triangle. Here, we observe that the length of QS and SR is also the same. So PS is also a median of this triangle.

Exercise 6.2 Page: 118

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

(i) Solution:-

We know that,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 50o + 70o

= x = 120o

(ii) Solution:-

We know that,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 65o + 45o

= x = 110o

(iii) Solution:-

We know that,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 30o + 40o

= x = 70o

(iv) Solution:-

We know that,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 60o + 60o

= x = 120o

(v) Solution:-

We know that,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 50o + 50o

= x = 100o

(vi) Solution:-

We know that,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 30o + 60o

= x = 90o

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

(i) Solution:-

We know that,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x + 50o = 115o

By transposing 50o from LHS to RHS, it becomes – 50o

= x = 115o – 50o

= x = 65o

(ii) Solution:-

We know that,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= 70o + x = 100o

By transposing 70o from LHS to RHS, it becomes – 70o

= x = 100o – 70o

= x = 30o

(iii) Solution:-

We know that,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

The given triangle is a right-angled triangle. So, the angle opposite to the x is 90o.

= x + 90o = 125o

By transposing 90o from LHS to RHS, it becomes – 90o

= x = 125o – 90o

= x = 35o

(iv) Solution:-

We know that,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x + 60o = 120o

By transposing 60o from LHS to RHS, it becomes – 60o

= x = 120o – 60o

= x = 60o

(v) Solution:-

We know that,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

The given triangle is a right-angled triangle. So, the angle opposite to the x is 90o.

= x + 30o = 80o

By transposing 30o from LHS to RHS, it becomes – 30o

= x = 80o – 30o

= x = 50o

(vi) Solution:-

We know that,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

The given triangle is a right-angled triangle. So, the angle opposite to the x is 90o.

= x + 35o = 75o

By transposing 35o from LHS to RHS, it becomes – 35o

= x = 75o – 35o

= x = 40o

Exercise 6.3 Page: 121

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

(i) Solution:-

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= ∠BAC + ∠ABC + ∠BCA = 180o

= x + 50o + 60o = 180o

= x + 110o = 180o

By transposing 110o from LHS to RHS, it becomes – 110o

= x = 180o – 110o

= x = 70o

(ii) Solution:-

We know that,

The sum of all the interior angles of a triangle is 180o.

The given triangle is a right-angled triangle. So, the ∠QPR is 90o.

Then,

= ∠QPR + ∠PQR + ∠PRQ = 180o

= 90o + 30o + x = 180o

= 120o + x = 180o

By transposing 110o from LHS to RHS, it becomes – 110o

= x = 180o – 120o

= x = 60o

(iii) Solution:-

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= ∠XYZ + ∠YXZ + ∠XZY = 180o

= 110o + 30o + x = 180o

= 140o + x = 180o

By transposing 140o from LHS to RHS, it becomes – 140o

= x = 180o – 140o

= x = 40o

(iv) Solution:-

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= 50o + x + x = 180o

= 50o + 2x = 180o

By transposing 50o from LHS to RHS, it becomes – 50o

= 2x = 180o – 50o

= 2x = 130o

= x = 130o/2

= x = 65o

(v) Solution:-

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= x + x + x = 180o

= 3x = 180o

= x = 180o/3

= x = 60o

∴ the given triangle is an equiangular triangle.

(vi) Solution:-

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= 90o + 2x + x = 180o

= 90o + 3x = 180o

By transposing 90o from LHS to RHS, it becomes – 90o

= 3x = 180o – 90o

= 3x = 90o

= x = 90o/3

= x = 30o

Then,

= 2x = 2 × 30o = 60o

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

(i) Solution:-

We know that,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

Then,

= 50o + x = 120o

By transposing 50o from LHS to RHS, it becomes – 50o

= x = 120o – 50o

= x = 70

We also know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= 50o + x + y = 180o

= 50o + 70+ y = 180o

= 120o + y = 180o

By transposing 120o from LHS to RHS, it becomes – 120o

= y = 180– 120

= y = 60o

(ii) Solution:-

From the rule of vertically opposite angles,

= y = 80o

Then,

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= 50o + 80o + x = 180o

= 130+ x = 180o

By transposing 130o from LHS to RHS, it becomes – 130o

= x = 180– 130

= x = 50o

(iii) Solution:-

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= 50o + 60o + y = 180o

= 110+ y = 180o

By transposing 110o from LHS to RHS, it becomes – 110o

= y = 180– 110

= y = 70o

Now,

From the rule of linear pair,

= x + y = 180o

= x + 70o = 180o

By transposing 70o from LHS to RHS, it becomes – 70o

= x = 180o – 70

= x = 110o

(iv) Solution:-

From the rule of vertically opposite angles,

= x = 60o

Then,

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= 30o + x + y = 180o

= 30o + 60o + y = 180o

= 90+ y = 180o

By transposing 90o from LHS to RHS, it becomes – 90o

= y = 180– 90

= y = 90o

(v) Solution:-

From the rule of vertically opposite angles,

= y = 90o

Then,

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= x + x + y = 180o

= 2x + 90o = 180o

By transposing 90o from LHS to RHS, it becomes – 90o

= 2x = 180– 90

= 2x = 90o

= x = 90o/2

= x = 45o

(vi) Solution:-

From the rule of vertically opposite angles,

= x = y

Then,

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= x + x + x = 180o

= 3x = 180o

= x = 180o/3

= x = 60o

Exercise 6.4 Page: 126

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

(i) 2 cm, 3 cm, 5 cm

Solution:-

Clearly, we have

(2 + 3) = 5

5 = 5

Thus, the sum of any two of these numbers is not greater than the third.

Hence, it is not possible to draw a triangle whose sides are 2 cm, 3 cm and 5 cm.

(ii) 3 cm, 6 cm, 7 cm

Solution:-

Clearly, we have

(3 + 6) = 9 > 7

(6 + 7) = 13 > 3

(7 + 3) = 10 > 6

Thus, the sum of any two of these numbers is greater than the third.

Hence, it is possible to draw a triangle whose sides are 3 cm, 6 cm and 7 cm.

(iii) 6 cm, 3 cm, 2 cm

Solution:-

Clearly, we have

(3 + 2) = 5 < 6

Thus, the sum of any two of these numbers is less than the third.

Hence, it is not possible to draw a triangle whose sides are 6 cm, 3 cm and 2 cm.

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? Solution:-

If we take any point O in the interior of a triangle PQR and join OR, OP, OQ.

Then, we get three triangles ΔOPQ, ΔOQR and ΔORP are shown in the figure below. We know that,

The sum of the length of any two sides is always greater than the third side.

(i) Yes, ΔOPQ has sides OP, OQ and PQ.

So, OP + OQ > PQ

(ii) Yes, ΔOQR has sides OR, OQ and QR.

So, OQ + OR > QR

(iii) Yes, ΔORP has sides OR, OP and PR.

So, OR + OP > RP

3. AM is a median of a triangle ABC.

Is AB + BC + CA > 2 AM?

(Consider the sides of triangles ΔABM and ΔAMC.) Solution:-

We know that,

The sum of the length of any two sides is always greater than the third side.

Now consider the ΔABM,

Here, AB + BM > AM … [equation i]

Then, consider the ΔACM

Here, AC + CM > AM … [equation ii]

By adding equations [i] and [ii], we get,

AB + BM + AC + CM > AM + AM

From the figure we have, BC = BM + CM

AB + BC + AC > 2 AM

Hence, the given expression is true.

Is AB + BC + CD + DA > AC + BD? Solution:-

We know that,

The sum of the length of any two sides is always greater than the third side.

Now consider the ΔABC,

Here, AB + BC > CA … [equation i]

Then, consider the ΔBCD

Here, BC + CD > DB … [equation ii]

Consider the ΔCDA

Here, CD + DA > AC … [equation iii]

Consider the ΔDAB

Here, DA + AB > DB … [equation iv]

By adding equations [i], [ii], [iii] and [iv], we get,

AB + BC + BC + CD + CD + DA + DA + AB > CA + DB + AC + DB

2AB + 2BC + 2CD + 2DA > 2CA + 2DB

Take out 2 on both the side,

2(AB + BC + CA + DA) > 2(CA + DB)

AB + BC + CA + DA > CA + DB

Hence, the given expression is true.

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

Solution:-

Let us consider ABCD as a quadrilateral, and P is the point where the diagonals intersect. As shown in the figure below. We know that,

The sum of the length of any two sides is always greater than the third side.

Now consider the ΔPAB,

Here, PA + PB < AB … [equation i]

Then, consider the ΔPBC

Here, PB + PC < BC … [equation ii]

Consider the ΔPCD

Here, PC + PD < CD … [equation iii]

Consider the ΔPDA

Here, PD + PA < DA … [equation iv]

By adding equations [i], [ii], [iii] and [iv], we get,

PA + PB + PB + PC + PC + PD + PD + PA < AB + BC + CD + DA

2PA + 2PB + 2PC + 2PD < AB + BC + CD + DA

2PA + 2PC + 2PB + 2PD < AB + BC + CD + DA

2(PA + PC) + 2(PB + PD) < AB + BC + CD + DA

From the figure, we have, AC = PA + PC and BD = PB + PD

Then,

2AC + 2BD < AB + BC + CD + DA

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

Hence, the given expression 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?

Solution:-

We know that,

The sum of the length of any two sides is always greater than the third side.

From the question, it is given that two sides of the triangle are 12 cm and 15 cm.

So, the third side length should be less than the sum of the other two sides,

12 + 15 = 27 cm

Then, it is given that the third side can not be less than the difference of the two sides, 15 – 12 = 3 cm

So, the length of the third side falls between 3 cm and 27 cm.

Exercise 6.5 Page: 130

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

Solution:-

Let us draw a rough sketch of a right-angled triangle. By the rule of Pythagoras’ Theorem,

Pythagoras’ theorem states that for any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of squares on the legs.

In the above figure, RQ is the hypotenuse,

QR2 = PQ2 + PR2

QR2 = 102 + 242

QR2 = 100 + 576

QR2 = 676

QR = √676

QR = 26 cm

Hence, the length of the hypotenuse QR = 26 cm

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

Solution:-

Let us draw a rough sketch of the right-angled triangle. By the rule of Pythagoras’ Theorem,

Pythagoras’ theorem states that for any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of squares on the legs.

In the above figure, RQ is the hypotenuse,

AB2 = AC2 + BC2

252 = 72 + BC2

625 = 49 + BC2

By transposing 49 from RHS to LHS, it becomes – 49

BC2 = 625 – 49

BC2 = 576

BC = √576

BC = 24 cm

Hence, the length of the BC = 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. Solution:-

By the rule of Pythagoras’ Theorem,

Pythagoras’ theorem states that for any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of squares on the legs.

In the above figure, RQ is the hypotenuse,

152 = 122 + a2

225 = 144 + a2

By transposing 144 from RHS to LHS, it becomes – 144

a2 = 225 – 144

a2 = 81

a = √81

a = 9 m

Hence, the length of a = 9 m

4. Which of the following can be the sides of a right triangle?

(i) 2.5 cm, 6.5 cm, 6 cm

(ii) 2 cm, 2 cm, 5 cm

(iii) 1.5 cm, 2cm, 2.5 cm

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

Solution:-

(i) Let a = 2.5 cm, b = 6.5 cm, c = 6 cm

Let us assume the largest value is the hypotenuse side, i.e., b = 6.5 cm.

Then, by Pythagoras’ theorem,

b2 = a2 + c2

6.52 = 2.52 + 62

42.25 = 6.25 + 36

42.25 = 42.25

The sum of squares of two sides of the triangle is equal to the square of the third side,

∴ the given triangle is a right-angled triangle.

The right angle lies on the opposite of the greater side, 6.5 cm.

(ii) Let a = 2 cm, b = 2 cm, c = 5 cm

Let us assume the largest value is the hypotenuse side, i.e. c = 5 cm.

Then, by Pythagoras’ theorem,

c2 = a2 + b2

52 = 22 + 22

25 = 4 + 4

25 ≠ 8

The sum of squares of two sides of the triangle is not equal to the square of the third side,

∴ the given triangle is not a right-angled triangle.

(iii) Let a = 1.5 cm, b = 2 cm, c = 2.5 cm

Let us assume the largest value is the hypotenuse side, i.e., b = 2.5 cm.

Then, by Pythagoras’ theorem,

b2 = a2 + c2

2.52 = 1.52 + 22

6.25 = 2.25 + 4

6.25 = 6.25

The sum of squares of two sides of the triangle is equal to the square of the third side,

∴ the given triangle is a right-angled triangle.

The right angle lies on the opposite of the greater side 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.

Solution:-

Let ABC is the triangle and B be the point where the tree is broken at the height of 5 m from the ground.

Treetop touches the ground at a distance of AC = 12 m from the base of the tree, By observing the figure, we came to conclude that a right-angle triangle is formed at A.

From the rule of Pythagoras’ theorem,

BC2 = AB2 + AC2

BC2 = 52 + 122

BC2 = 25 + 144

BC2 = 169

BC = √169

BC = 13 m

Then, the original height of the tree = AB + BC

= 5 + 13

= 18 m

6. Angles Q and R of a ΔPQR are 25o and 65o.

Write which of the following is true:

(i) PQ2 + QR2 = RP2

(ii) PQ2 + RP2 = QR2

(iii) RP2 + QR2 = PQ2 Solution:-

Given that ∠Q = 25o, ∠R = 65o

Then, ∠P =?

We know that sum of the three interior angles of a triangle is equal to 180o.

∠PQR + ∠QRP + ∠RPQ = 180o

25o + 65o + ∠RPQ = 180o

90o + ∠RPQ = 180o

∠RPQ = 180 – 90

∠RPQ = 90o

Also, we know that the side opposite to the right angle is the hypotenuse.

∴ QR2 = PQ2 + PR2

Hence, (ii) is true.

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

Solution:- Let ABCD be the rectangular plot.

Then, AB = 40 cm and AC = 41 cm

BC =?

According to Pythagoras’ theorem,

From the right angle triangle ABC, we have

= AC2 = AB2 + BC2

= 412 = 402 + BC2

= BC2 = 412 – 402

= BC2 = 1681 – 1600

= BC2 = 81

= BC = √81

= BC = 9 cm

Hence, the perimeter of the rectangle plot = 2 (length + breadth)

Where, length = 40 cm, breadth = 9 cm

Then,

= 2(40 + 9)

= 2 × 49

= 98 cm

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

Solution:- Let PQRS be a rhombus, all sides of the rhombus have equal length, and its diagonal PR and SQ are intersecting each other at point O. Diagonals in the rhombus bisect each other at 90o.

So, PO = (PR/2)

= 16/2

= 8 cm

And, SO = (SQ/2)

= 30/2

= 15 cm

Then, consider the triangle POS and apply the Pythagoras theorem,

PS2 = PO2 + SO2

PS2 = 82 + 152

PS2 = 64 + 225

PS2 = 289

PS = √289

PS = 17 cm

Hence, the length of the side of the rhombus is 17 cm

Now,

The perimeter of the rhombus = 4 × side of the rhombus

= 4 × 17

= 68 cm

∴ the perimeter of the rhombus is 68 cm.

## Frequently Asked Questions on NCERT Solutions for Class 7 Maths Chapter 6

Q1

### How many exercises are there in NCERT Solutions for Class 7 Maths Chapter 6?

The fourth Chapter of NCERT Solutions for Class 7 Maths has 5 exercises. The first exercise deals with the topic of determining the medians of triangles, the second exercise has questions about finding the altitudes of triangles, the third exercise deals with finding the exterior angle of a triangle and its property, the fourth exercise has questions based on two special triangles: equilateral and isosceles and last exercise has the question based on the right-angled triangles and Pythagoras property. By solving these exercises, students are able to answer all the questions based on triangles and their properties.
Q2

### Is BYJU’S website providing answers for NCERT Solutions for Class 7 Maths Chapter 6 Triangles and Its Properties?

Yes, you can avail of the PDFs of NCERT Solutions for Class 7 Maths Quadratic Equations. These solutions are formulated by expert faculty at BYJU’S in a unique way. Also, they provide solutions for Class 1 to 12 NCERT Textbooks in free PDFs. Who wish to score high in exams are advised to solve the NCERT Textbook.
Q3

### Mention the important concepts that you learn in NCERT Solutions for Class 7 Maths Chapter 6 Triangles and Its Properties.

The concepts presented in NCERT Solutions for Class 7 Maths Chapter 6 Triangles and Its Properties are
1. Medians of a Triangle
2. Altitudes of a Triangle
3. Exterior Angle of a Triangle and Its Property
4. Angle Sum Property of a Triangle
5. Two Special Triangles: Equilateral and Isosceles
6. Sum of the Lengths of Two Sides of a Triangle
7. Right-Angled Triangles and Pythagoras Property