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 want 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

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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 end point at a vertex of the triangle and other 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 end point at a vertex of the triangle and other 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 same.

Solution:-

Draw a Line segment PS ⊥ BC. It is an altitude for this triangle. Here we observe that length of QS and SR is also 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 = 50^o + 70^o

= x = 120^o

(ii)

Solution:-

We Know That,

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

= x = 65^o + 45^o

= x = 110^o

(iii)

Solution:-

We Know That,

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

= x = 30^o + 40^o

= x = 70^o

(iv)

Solution:-

We Know That,

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

= x = 60^o + 60^o

= x = 120^o

(v)

Solution:-

We Know That,

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

= x = 50^o + 50^o

= x = 100^o

(vi)

Solution:-

We Know That,

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

= x = 30^o + 60^o

= x = 90^o

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 + 50^o = 115^o

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

= x = 115^o – 50^o

= x = 65^o

(ii)

Solution:-

We Know That,

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

= 70^o + x = 100^o

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

= x = 100^o – 70^o

= x = 30^o

(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 90^o.

= x + 90^o = 125^o

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

= x = 125^o – 90^o

= x = 35^o

(iv)

Solution:-

We Know That,

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

= x + 60^o = 120^o

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

= x = 120^o – 60^o

= x = 60^o

(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 90^o.

= x + 30^o = 80^o

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

= x = 80^o – 30^o

= x = 50^o

(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 90^o.

= x + 35^o = 75^o

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

= x = 75^o – 35^o

= x = 40^o

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 180^o.

Then,

= ∠BAC + ∠ABC + ∠BCA = 180^o

= x + 50^o + 60^o = 180^o

= x + 110^o = 180^o

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

= x = 180^o – 110^o

= x = 70^o

(ii)

Solution:-

We know that,

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

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

Then,

= ∠QPR + ∠PQR + ∠PRQ = 180^o

= 90^o + 30^o + x = 180^o

= 120^o + x = 180^o

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

= x = 180^o – 120^o

= x = 60^o

(iii)

Solution:-

We know that,

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

Then,

= ∠XYZ + ∠YXZ + ∠XZY = 180^o

= 110^o + 30^o + x = 180^o

= 140^o + x = 180^o

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

= x = 180^o – 140^o

= x = 40^o

(iv)

Solution:-

We know that,

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

Then,

= 50^o + x + x = 180^o

= 50^o + 2x = 180^o

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

= 2x = 180^o – 50^o

= 2x = 130^o

= x = 130^o/2

= x = 65^o

(v)

Solution:-

We know that,

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

Then,

= x + x + x = 180^o

= 3x = 180^o

= x = 180^o/3

= x = 60^o

∴The given triangle is an equiangular triangle.

(vi)

Solution:-

We know that,

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

Then,

= 90^o + 2x + x = 180^o

= 90^o + 3x = 180^o

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

= 3x = 180^o – 90^o

= 3x = 90^o

= x = 90^o/3

= x = 30^o

Then,

= 2x = 2 × 30^o = 60^o

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,

= 50^o + x = 120^o

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

= x = 120^o – 50^o

= x = 70^o 

We also know that,

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

Then,

= 50^o + x + y = 180^o

= 50^o + 70^o + y = 180^o

= 120^o + y = 180^o

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

= y = 180^o – 120^o 

= y = 60^o

(ii)

Solution:-

From the rule of vertically opposite angles,

= y = 80^o

Then,

We know that,

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

Then,

= 50^o + 80^o + x = 180^o

= 130^o + x = 180^o

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

= x = 180^o – 130^o 

= x = 50^o

(iii)

Solution:-

We know that,

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

Then,

= 50^o + 60^o + y = 180^o

= 110^o + y = 180^o

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

= y = 180^o – 110^o 

= y = 70^o

Now,

From the rule of linear pair,

= x + y = 180^o

= x + 70^o = 180^o

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

= x = 180^o – 70

= x = 110^o

(iv)

Solution:-

From the rule of vertically opposite angles,

= x = 60^o

Then,

We know that,

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

Then,

= 30^o + x + y = 180^o

= 30^o + 60^o + y = 180^o

= 90^o + y = 180^o

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

= y = 180^o – 90^o 

= y = 90^o

(v)

Solution:-

From the rule of vertically opposite angles,

= y = 90^o

Then,

We know that,

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

Then,

= x + x + y = 180^o

= 2x + 90^o = 180^o

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

= 2x = 180^o – 90^o 

= 2x = 90^o

= x = 90^o/2

= x = 45^o

(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 180^o.

Then,

= x + x + x = 180^o

= 3x = 180^o

= x = 180^o/3

= x = 60^o

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 is 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 equation [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.

4. ABCD is a quadrilateral.

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 equation [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 is quadrilateral and P is the point where the diagonals are 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 equation [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 triangle are 12 cm and 15 cm.

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

12 + 15 = 27 cm.

Then, it is given that the third side is cannot 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 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 square 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 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 square 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 square 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 square of two side of triangle is equal to the square of third side,

∴The given triangle is right-angled triangle.

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 square of two side of triangle is not equal to the square of third side,

∴The given triangle is not 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 square of two side of triangle is equal to the square of third side,

∴The given triangle is right-angled triangle.

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 is the point where tree is broken at the height 5 m from the ground.

Tree top 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 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 25^o and 65^o.

Write which of the following is true:

(i) PQ² + QR² = RP²

(ii) PQ² + RP² = QR²

(iii) RP² + QR² = PQ²

Solution:-

Given that ∠Q = 25^o, ∠R = 65^o

Then, ∠P =?

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

∠PQR + ∠QRP + ∠RPQ = 180^o

25^o + 65^o + ∠RPQ = 180^o

90^o + ∠RPQ = 180^o

∠RPQ = 180 – 90

∠RPQ = 90^o

Also, we know that 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 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 rhombus has equal length and its diagonal PR and SQ are intersecting each other at a point O. Diagonals in rhombus bisect each other at 90^o.

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 side of rhombus is 17 cm

Now,

Perimeter of rhombus = 4 × side of the rhombus

= 4 × 17

= 68 cm

∴ Perimeter of rhombus is 68 cm.

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

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, second exercise has questions of finding the altitudes of triangles, third exercise deals with finding the exterior angle of a triangle and Its property, fourth exercises 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 its properties.

Is BYJU’S website providing answers for NCERT Solutions for Class 7 Maths Chapter 6 Quadratic Equations?

Yes, you can avail the PDFs of NCERT Solutions for Class 7 Maths Quadratic Equations. These solutions are formulated by expert faculty at BYJU’S in an 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.

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.