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NCERT exemplar Class 9 Maths solutions chapter 6 provide students with detailed answers for chapter 6- Lines and Angles. These problems and their solutions are devised to understand the chapter better and extremely useful in understanding Lines and Angles concepts. The NCERT exemplar Class 9 Maths chapter 6 solutions are intricate and helpful in pinpointing the critical understanding of concepts and to solve questions in NCERT Class 9 Maths Book. The exemplar questions and answers are based on the recommended CBSE Class 9 Syllabus. These Class 9 Maths NCERT exemplar solutions chapter 6 Lines and Angles solutions are a rich source of reference material equipped with numerous practice problems of lines and angles.
Question:1
Write the correct answer in each of the following:
Answer:
Here in this question, it is given,
Hence, option C is the right answer.
Question:2
Write the correct answer in each of the following:
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
(A) an isosceles triangle
(B) an obtuse triangle
(C) an equilateral triangle
(D) a right triangle
Answer: [D] a right triangle
In a
Thus triangle ABC is a right-angled triangle.
Hence Option (D) is the correct answer.
Question:3
Answer:
Given ® Exterior Angle =Let the two equal interior angle be .
Sum of opposite interior angles of a triangle = Exterior angle of the triangle.
So, option (B) is the correct Answer
Question:4
Write the correct answer in each of the following: The angles of a triangle are in the ratio 5 : 3 : 7. The triangle is
(A) an acute angled triangle
(B) an obtuse angled triangle
(C) a right triangle
(D) an isosceles triangle
Given: The ratio of angles of triangles is 5 : 3 : 7
Let angles of the triangle be
Then,
In
All angles are less than 90° hence acute.
Question:5
(A)
(B)
(C)
(D)
Answer: (D)In
[Sum of all interiors angles of triangle is ]
,
Put value of equation (1) in equation (2)
Question:6
Write the correct answer in each of the following:
In Fig. 6.2, POQ is a line. The value of x is
Answer:
Given POQ is a line(angle on a straight line)
Question:7
Write the correct answer in each of the following:
,
Answer:
Also, for line OP
Now, we know that the sum of angles of a triangle is 180^{0}
,
Hence option C is the correct answer
Question:8
Answer:
in the ratio 2:4:3Hence option (B) is correct
Question:1
For what value of x+y in Fig. 6.4 will ABC be a line? Justify your answer.
Answer:
Here x & y are two adjacent angles and for straight-line x+y=180^{0}Justification: As we know that if the sum of two adjacent angles is 180^{0} it represents the straight line.
Question:2
Can a triangle have all angles less than 600? Give reason for your answer.
Answer:
In a triangle, the sum of interior angles is always equal to 180^{0}So it is not possible to have all angles less than 60^{0 }because their sum will not be equal to 180^{0}
Hence, the answer is No
Question:3
Can a triangle have two obtuse angles? Give reason for your answer
Answer: NoNo, a triangle cannot have two obtuse angles
Reason: the obtuse angle is greater than 90^{0}
Let us consider the smallest obtuse angle, i.e., 91^{0}
Let the two obtuse angles be equal to 91^{o}, so the sum of these two obtuse angles is equal to 182^{0}
But the sum of interior angles of a triangle is always equal to 180^{0}
So the given condition is not possible.
Hence, we cannot create a triangle with two obtuse angles
Question:4
How many triangles can be drawn having its angles as 450, 640 and 720? Give reason for your answer.
Answer: ZeroThe three angles given are 45^{0}, 64^{0} and 72^{0}
Now, the sum of these angles = 45^{0}+64^{0}+72^{0}
= 181^{o}
Sum of angles of a triangle cannot be more than 180^{0}.
Hence, the triangle is not possible
So the answer is zero.
Question:5
How many triangles can be drawn having its angles as 530, 640 and 630? Give reason for your answer.
Answer: infinitely many
Infinitely many triangles can be drawn having its angles as 53^{0}, 64^{0} and 63^{0}
The sum of given angles = 53^{0} + 64^{0+} 63^{0} = 180^{0}
Here, we can see that sum of all interior angles of triangles is 180^{0}, so infinitely many triangles can be drawn depending on the lengths of its sides.
So, infinitely many triangles can be drawn from the given angles.
Question:6
In Fig. 6.5, find the value of x for which the lines l and m are parallel.
Answer:
In the given figure, the lines l and m are parallel.
Line n is a transversal line.
So, and 44^{o} are co-interior angles.
We know that the sum of co-interior angles is 180^{o}.
Hence
Question:7
Answer: NoGiven that two adjacent angles are equal.
No it is not necessary that each of these angles will be a right angle.
Let us see why.
Adjacent angles: Two angles are said to be adjacent only when they have common vertex, a common side but they do not overlap.
The following figure shows adjacent angles ( and ), (<DBC,)
But for the case that both adjacent angles are right angles, they should form a linear pair or we can say that they should lie on the same line, but such condition is not given to us.
Question:8
Answer: All the other angles should be at right angles.If one of the angle is formed by two intersecting line is a right angle,
Hence all the other angles should be at right angles.
Question:9
In Fig.6.6, which of the two lines are parallel and why?
Answer: l and m are parallel
Question:10
Answer: No.The two lines l and m are perpendicular to the same line n.
We can have 3 possible scenarios:
1. The lines l and m may lie on the same side of n
As we can see corresponding angles are equal, so l is parallel to m
l and m are not perpendicular to each other
2. The lines l and m may lie on the opposite side of n
Extend lA to point B and mC to point D
As we can see that interior opposite angles are equal, so l is parallel to m
l and m are not perpendicular to each other
3. The lines l and m are the same and overlap each other.
Again, l and m are not perpendicular to each other
Hence, the lines l and m are not perpendicular.
Question:1
Answer:
Given in the figure , OD and OE are bisectors of and respectively
To show: Points A, O & B are collinear, i.e., AOB is a straight line
Proof: Since OD and OE bisect angles and respectively
… (i)
And … (ii)
On adding equation (i) and (ii) we get
(Given )
So, & are forming a linear pair.
AOB is a straight line.
Therefore, points A, O and B are collinear.
Hence proved
Question:2
. Show that the lines m and n are parallel.
Answer:
Given: The figure and
To show :
Proof: and
Here, and
Here (Vertically opposite angles)
Now,
We know if the sum of two interior angles on the same side is then the lines are parallel
Hence proved, .
Question:3
Answer:
Given: In the figure, , AP and BQ are the Bisectors of and
To prove:
Proof: Since and t is a transversal therefore
(alternate interior angles)
(Divide both sides by 2)
(AP & BQ are the bisectors of & )
Now consider, two lines AP and BQ with transversal AB
and are alternate interior angles and these are equal.
Hence,
Hence proved
Question:4
If in Figure bisectors AP and BQ of the alternate interior angles are parallel, then show that .
Answer:
In the figure , AP and BQ are the bisectors of the alternate interior angles and .
To show :
Proof: Since and t is transversal therefore
(Alternate Interior Angles)
(Multiplying both sides by 2)
Now, AP and BQ are the bisectors of alternate interior angle and
So,
Now consider lines l and m
(alternate interior angles are equal)
Hence
Hence proved
Question:5
In Fig. 6.12, . Show that [Hint: Produce DE to intersect BC at P (say)].
Answer:
Produce DE to intersect BC at P
Now, and DP is the transversal
(corresponding angles) … (i)
Now and BC is transversal,
(corresponding angles) … (ii)
From (i) and (ii) we get
Hence, Proved
Question:6
Answer:
Given
and
To show
Construction:
Extend EF to point P on AB
Proof: In figure, , so
(Sum of co-interior angles is ) …(i)
Now, and PE is the transversal line.
(Corresponding angle) …(ii)
From Equations (i) and (ii)
Question:7
In Fig. 6.14, and AP and BP are bisectors of and , respectively. Find
Answer:
and AP and PB are the bisectors of and ,
We know that the interior angles on the same side of the transversal are supplementary
So,
…(i)
AP and BP are the bisectors of and respectively.
…(ii)
…(iii)
On adding equation (ii) and (iii) we get
From equation (i)
In ,
Question:8
Answer:
Angles of the triangle are in the ratio- 2 : 3 : 4
Let the angles are 2x, 3x, 4x then:
(angle sum property)
Then the angles of the triangle are:
This triangle is a scalene triangle as all the angles are of different measure.
Question:9
A triangle ABC is right angled at A. L is a point on BC such that . Prove that .
Answer:
Proof: In and ,
(each ) (i)
And (Common angle) (ii)
In ,
(angle sum property)
(from i)
In ,
(angle sum property)
(from i)
(from ii)
Hence,
Hence proved
Question:10
Answer:
Given: Two lines m and n are parallel and another two lines p and q are respectively perpendicular to m and n, i.e., .
To prove
Proof:
Since m||n and p is perpendicular to m and n.
Similarly, q is perpendicular to m and n.
Now for lines p and q, m is the transversal
So we can see that all the conditions are fulfilled for the lines to be parallel, i.e., Corresponding angles are equal, sum of cointerior angles is 180^{o}, alternate angles are equal.
Hence,
Hence proved
Question:1
If two lines intersect, prove that the vertically opposite angles are equal.
Answer:
It is given that if two lines intersect, the vertically opposite angles are equal.
Proof:
Now let AB and CD be two lines intersecting at point O.
From the figure, we have two pairs of vertically opposite angles namely:
(i) and
(ii) and
Now we have to prove that
And
Now ray OA stands on line CD
… (i) (linear pair angles)
Similarly, can we write
… (ii) (linear pair angles)
From equation (i) and (ii) comparing
Similarly, we can prove that
Hence Proved.
Question:2
Bisectors of interior and exterior of a intersect at the point T. Prove that .
Answer:
According to the question,
Bisectors of interior and exterior of a intersect at the point T
… (1)
And … (2)
Now from we have
… (3) (exterior angle is equal to the sum of interior opposite angles)
And from we have
(exterior angle is equal to the sum of interior opposite angles)
Or using (1 and 2)
Or
Using (3),
So,
Question:3
Answer:
Given: A transversal EF cuts two parallel line AB and CD at point G & H. GL and HM are bisectors of angles
To prove:
Proof: (Corresponding angles)
These are the corresponding angle formed by the line GL and HM, where EF is the transversal.
Hence proved
Question:4
Answer:
Given: Consider a line R and a point P
Construction:
Draw two lines (m and n) passing through P which are perpendicular to line R.
To prove: Only one perpendicular line can be drawn through a point P
Proof: In
{angle sum property}
So lines n and m will coincide
Therefore we can draw only one perpendicular to a given line.
Question:5
Answer:
Given: Let line x and y are two intersecting lines. Let n and p be another two lines which are perpendicular to x and y
To prove: n and p intersect at a point
Proof: Let lines n and p are not intersecting then … (1)
Since n and p are parallel and n is perpendicular to x and p is perpendicular to y respectively
So,
But, it is a contradiction as it is given that x and y are two intersecting lines
Thus our assumption is wrong.
n and p intersect at a point
Question:6
Prove that a triangle must have at least two acute angles.
Answer:
It is given that a triangle must have at least two acute angles.
An acute angle is less than 90 degrees
Let us assume that a triangle does not have two acute angles.
So, it has two angles that are either right angles (=90 degrees) or obtuse angles (greater than 90 degrees)
So let two right angles are present,
So using angle sum property of a triangle, third angle must be zero which is not possible.
Also let one angle be right and one be obtuse. We can take the smallest obtuse angle, i.e.,
So using angle sum property of a triangle, third angle must be negative which is not possible.
Again, if both the angles are obtuse the third angle must be negative which is not possible.
So it is necessary for a triangle to have at least two acute angles.
Hence proved
Question:7
In Fig. 6.17, , PA is the bisector of and . Prove that .
Answer:
Given : In ,
PA is the bisector of and .
To prove :
Proof: Since PA is the bisector of , we have
… (i)
In right angled we have
(Angle sum property)
… (ii)
Now
using (i) & (ii)
Since
Hence proved
The major topics covered through the NCERT exemplar Class 9 Maths solutions chapter 6 are as follows:
◊ Lines are discussed in detail with the understanding of line segments.
◊ Concepts of parallel lines.
◊ Identify if two lines are intersecting or not.
◊ Concepts of collinearity of three or more than three points.
◊ NCERT exemplar Class 9 Maths chapter 6 solutions discusses the angles in detail.
◊ It includes the understanding of right-angle, acute angle, obtuse angle, complementary angles, et cetera.
◊ The chapter includes knowledge of adjacent angles, vertically opposite angles, and their relations.
◊ This chapter talks about the angle sum property of a triangle which says that all three angles sum to 180°.
These Class 9 Maths NCERT exemplar chapter 6 solutions provide a student with basic concepts of Lines and Angles to progress with geometry learning. In this chapter, Lines and Angles are explained separately. Students can easily attempt and solve other books such as RS Aggarwal Class 9 Maths, NCERT Class 9 Maths, RD Sharma Class 9 Maths, etcetera.
NCERT exemplar Class 9 Maths solutions chapter 6 pdf download feature is available for the students to provide them with coherent learning experience while studying NCERT exemplar Class 9 Maths chapter 6.
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
No, parallel lines can never intersect each other in other words we can say that Parallel line meets or intersect at infinity.
No, the sum of all three interior angles has to be 180°, and all the angles will be non-zero angles. If two angles are right-angle, then these two corresponding lines will be parallel to each other and never intersect to form a triangle.
Yes, the concepts of lines and angles are beneficial in understanding problems of physics and mathematics. Mathematics is not the part of the syllabus of NEET; however, knowledge of lines and triangle is very much required in physics
Generally, a total of 3-4 questions appear yearly on the final examination, which includes the MCQs, very short answers, and occasionally a long-answer type question. These NCERT exemplar Class 9 Maths solutions chapter 6 are sufficient to grasp the concepts and practice required to attempt the paper successfully.
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