NCERT Solutions for Class 7 Maths Chapter 6
NCERT Solutions for Class 7 Maths Chapter 6:
You may find The Triangle and Its Properties here. We have developed step-by-step solutions with thorough explanations for students who are anxious about finding the most thorough and detailed NCERT Solutions for Class 7 Maths Chapter 6. We suggest that students who want to do well in math review these NCERT Solutions for Class 7 Maths Chapter 6 and improve their skills.
NCERT Solutions for Class 7 Maths Chapter 6 PDF
Here are the links to download the NCERT Solutions for Class 7 Maths Chapter 6 in PDF format. The major objective is to help students understand and resolve these problems. We have prepared the NCERT Solutions for Class 7 Maths Chapter 6 and have solved the problems step-by-step with detailed explanations.
NCERT Solutions for Class 7 Maths Chapter 6 PDF
NCERT Solutions for Class 7 Maths Chapter 6 Triangle and its Properties
Below we have provided
NCERT Solutions
for Class 7 Maths Chapter 6 for students to help them understand the chapter better and to score good marks in their examination.
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 = 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
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(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
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(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
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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
