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 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,

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°

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) PQ² + QR² = RP²
(ii) PQ² + RP² = QR²
(iii) RP² + QR² = PQ²

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 BC² = AC² – AB²

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.

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