NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 6 |

Chapter Name |
The Triangle and its Properties |

Exercise |
Ex 6.1, Ex 6.2, Ex 6.3, Ex 6.4, Ex 6.5. |

Number of Questions Solved |
21 |

Category |
NCERT Solutions |

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

### Chapter 6 The Triangle and its Properties Exercise 6.1

**Question 1.
**

**Solution:**

**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.

**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 same.

### Chapter 6 The Triangle and its Properties Exercise 6.2

**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,

- x = 50°+ 70° = 120°
- x = 65°+ 45° = 110°
- x = 30°+ 40°= 70°
- x = 60° + 60° = 120c
- x = 50° + 50° = 100c
- x = 30°+ 60° = 90°

**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,

### Chapter 6 The Triangle and its Properties Exercise 6.3

**Question 1.**

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

**Solution:
**

**Question 2.**

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

**Solution:
**

### Chapter 6 The Triangle and its Properties Exercise 6.4

**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 6

So, the given side lengths cannot form a triangle.

**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.

**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

**Question 4.**

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

**Solution:
**

**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)

**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.

### Chapter 6 The Triangle and its Properties Exercise 6.5

**Question 1.**

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

**Solution:
**

**Question 2.**

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

**Solution:
**

**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.

**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.

**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, ∆A

^{‘}BC is a right triangle, right-angled at B such that A’ B = 12 m, BC = 5 m. By Pythagoras theorem, we have

**Question 6.**

Angles Q and ii of a APQR are 25° and 65°. Write which of the following is true :

**(i)** PQ^{2} + QR^{2} = RP^{2}

**(ii)** PQ^{2} + RP^{2} = QR^{2}

**(iii)** RP^{2} + QR^{2} = PQ^{2
}

**Solution:
**

**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 BC^{2} = AC^{2} – AB^{2
}

**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.

We hope the NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties, drop a comment below and we will get back to you at the earliest.

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