NCERT Solutions for Class 7 Maths Chapter 6 Triangle and its Properties
Neha Tanna29 Feb, 2024
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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| CBSE Syllabus Class 7 | |
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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 oCBSE Board Exam Centre List 2024
(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: 1211. 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 oRelated Links -
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 oNCERT Solutions for Class 7 Maths Chapter 6 FAQs
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