# NCERT Solutions | Class 7 Maths Chapter 6 | The Triangles and its Properties

## CBSE Solutions | Maths Class 7

Check the below NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and its Properties Pdf free download. NCERT Solutions Class 7 Maths  were prepared based on the latest exam pattern. We have Provided The Triangles and its Properties Class 7 Maths NCERT Solutions to help students understand the concept very well.

### NCERT | Class 7 Maths

Book: National Council of Educational Research and Training (NCERT) Central Board of Secondary Education (CBSE) 7th Maths 6 The Triangles and its Properties English

#### The Triangles and its Properties | Class 7 Maths | NCERT Books Solutions

You can refer to MCQ Questions for Class 7 Maths Chapter 6 The Triangles and its Properties to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

## NCERT Solutions for Class 7 Maths Chapte 6 The Triangle and its Properties Exercise 6.1

### Ex 6.1 Class 7 Maths Question 2.

Draw rough sketches for the following :
(a) In ∆ABC, BE is a median.
(b) in ∆PQR, PQ and PR are altitudes of the triangle.
(c) In ∆XYZ, YL is an altitude In the exterior of the triangle.

#### Solution:

(a) Rough sketch of median BE of ∆ABC is as shown.

(b) Rough sketch of altitudes PQ and PR of ∆PQR is as shown.

(c) Rough sketch of an exterior altitude YL of ∆XYZ is as shown.

### Ex 6.1 Class 7 Maths Question 3.

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

#### Solution:

Draw a line segment BC. By paper folding, locate the perpendicular bisector of BC. The folded crease meets BC at D, its mid-point.
Take any point A on this perpendicular bisector. Join AB and AC. The triangle thus obtained is an isosceles ∆ABC in which AB = AC.
Since D is the mid-point of BC, so AD is its median. Also, AD is the perpendicular bisector of BC. So, AD is the altitude of ∆ABC.
Thus, it is verified that the median and altitude of an isosceles triangle are the same.

## NCERT Solutions for Class 7 Maths Chapte 6 The Triangle and its Properties Exercise 6.2

### Ex 6.2 Class 7 Maths Question 1.

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

#### Solution:

Since, in a triangle an exterior angle is equal to the sum of the two interior opposite angles, therefore,

1. x = 50°+ 70° = 120°
2. x = 65°+ 45° = 110°
3. x = 30°+ 40°= 70°
4. x = 60° + 60° = 120c
5. x = 50° + 50° = 100c
6. x = 30°+ 60° = 90°

### Ex 6.2 Class 7 Maths Question 2.

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

#### Solution:

We know that in a triangle, an exterior angle is equal to the sum of the two interior opposite angles. Therefore,

## NCERT Solutions for Class 7 Maths Chapte 6 The Triangle and its Properties Exercise 6.3

### Ex 6.3 Class 7 Maths Question 1.

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

### Ex 6.3 Class 7 Maths Question 2.

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

## NCERT Solutions for Class 7 Maths Chapte 6 The Triangle and its Properties Exercise 6.4

### Ex 6.4 Class 7 Maths Question 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

#### Solution:

(i) Since, 2 + 3 > 5
So the given side lengths cannot form a triangle.
(ii) We have, 3 + 6 > 7, 3 + 7 > 6 and 6 + 7 > 3
i. e., the sum of any two sides is greater than the third side.
So, these side lengths form a triangle.
(iii) We have, 6 + 3 > 2, 3 + 2 $$\ngtr$$ 6
So, the given side lengths cannot form a triangle.

### Ex 6.4 Class 7 Maths Question 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:

(i) Yes, OP + OQ > PQ because on joining OP and OQ, we get a ∆OPQ and in a triangle, sum of the lengths of any two sides is always greater than the third side.

(ii) Yes, OQ + OR > QR, because on joining OQ and
OR, we get a ∆OQR and in a triangle, sum of the length of any two sides is always greater than the third side.

(iii) Yes, OR + OP > RP, because on joining OR and OP, we get a ∆OPR and in a triangle, sum of the lengths of any two sides is always greater than the third side.

### Ex 6.4 Class 7 Maths Question 3.

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

#### Solution:

Using triangle inequality property in triangles ABM and AMC, we have
AB + BM > AM …(1) and, AC + MC > AM …(2)
Adding (1) and (2) on both sides, we get
AB + (BM + MC) + AC > AM + AM
⇒ AB + BC + AC > 2AM

### Ex 6.4 Class 7 Maths Question 4.

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

### Ex 6.4 Class 7 Maths Question 5.

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

#### Solution:

Let ABCD be a quadrilateral and its diagonals AC and BD intersect at O. Using triangle inequality property, we have
In ∆OAB
OA + OB > AB ……(1)

### Ex 6.4 Class 7 Maths Question 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:

Let x cm be the length of the third side.
Thus, 12 + 15 > x, x + 12 > 15 and x + 15 > 12
⇒ 27 > x, x > 3 and x > -3
The numbers between 3 and 27 satisfy these.
∴ The length of the third side could be any length between 3 cm and 27 cm.

## NCERT Solutions for Class 7 Maths Chapte 6 The Triangle and its Properties Exercise 6.5

### Ex 6.5 Class 7 Maths Question 1.

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

### Ex 6.5 Class 7 Maths Question 2.

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

### Ex 6.5 Class 7 Maths Question 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:

Hence, the distance of the foot of the ladder from the wall is 9 m.

### Ex 6.5 Class 7 Maths Question 4.

Which of the following can be the sides of a right triangle?
(i) 2.5 em, 6.5 cm, 6 cm.
(ii) 2 cm, 2 cm, 5 cm.
(iii) 1.5 cm, 2 cm, 2.5 cm.
In the case of right-angled triangles, identify the right angles.

#### Solution:

Thus, the given sides form a right-triangle and the right-angle is opposite to the side of length 2.5 cm.

### Ex 6.5 Class 7 Maths Question 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 ACB be the tree before it broke at the point C. and let its top A touch the ground at A after it broke. Then, ∆ABC is a right triangle, right-angled at B such that A’ B = 12 m, BC = 5 m. By Pythagoras theorem, we have

### Ex 6.5 Class 7 Maths Question 6.

Angles Q and ii of a APQR are 25° and 65°. Write which of the following is true :
(i) PQ2 + QR2 = RP2
(ii) PQ2 + RP2 = QR2
(iii) RP2 + QR2 = PQ2

### Ex 6.5 Class 7 Maths Question 7.

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

#### Solution:

Let ABCD be a rectangle such that AB = 40 m and AC = 41 m.
In right-angled ∆ABC, right-angled at B, by Pythagoras theorem, we have BC2 = AC2 – AB2

### Ex 6.5 Class 7 Maths Question 8.

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

#### Solution:

Let ABCD be the rhombus such that AC = 30 cm and BD = 16 cm.
We know that the diagonals of a rhombus bisect each other at right angles.