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NCERT Exemplar Class 9 Maths
Chapter 6- Lines and Angles
Chapter 6
Lines and Angles
Exercise 6.1 (8 Multiple Choice Questions and Answers)
Question: 1
In Fig. 6.1, if AB CD EF, PQ RS, RQD 25 and
CQP 60 , then QRS is equal to:
(a) 85
(b) 135
(c) 145
(d) 110
Chapter 6
Lines and Angles
Solution:
c
Given that,
PQ RS
PQC BRS 60
[Alternate exterior angles and]
and DQR QRA 25 [Alternate interior angles and
DQR 25 ]
QRS QRA ARS QRA 180 B P( RS Q) C 60
[Linear pair]
25 180 60
205 60
145
Hence, option (c) is correct.
Question: 2
If one angle of a triangle is equal to the sum of the other
two angles, then the triangle is:
(a) an isosceles triangle
Chapter 6
Lines and Angles
(b) an obtuse triangle
(c) an equilateral triangle
(d) a right triangle
Solution:
d
Question: 3
An exterior angle of a triangle is and its two interior
opposite angles are equal. Each of these equal angles is:
(a) 1
372
105
(b) 1
522
(c) 1
722
(d) 75
Solution:
b
Given,
Chapter 6
Lines and Angles
Exterior angle = 105
Let one of interior angle be x .
Sum of two opposite interior angles Exterior angle
x x 105
2x 105
x2
105
1
x 522
Hence, each equal angle of triangle is 1
522
Question 4
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
Chapter 6
Lines and Angles
Solution:
a
Given that,
The ratio of angles of a triangle is 5 : 3 : 7.
Let the angles of a triangle be A, B and C .
Let, A 5x B 3x , and C 7x
In ABC, A B C 180
( The sum of the angles of a triangle is180.)
5x 3x 7x 180
1 5x 180
x 12
A 5x 5 12 60
B 3x 3 12 36
and C 7x 7 12 84
Since all the angles are less than 90° ,the triangle is an
acute angled triangle.
Chapter 6
Lines and Angles
Question: 5
If one of the angles of a triangle is 130°, then the angle
formed when the bisectors of the other two angles meet
can be:
(a) 50
(b) 65
(c) 145
(d) 155
Solution:
d
Let the angles of ABC be A B , and C .
In ABC ,
A B C 180
Chapter 6
Lines and Angles
(The sum of the angles of a triangle is180.)
1 1 1 180 A B C 2 2 2 2
(Dividing both the sides by 2)
1 1 180 1 B C A2 2 2 2
1 1 1 1 1 1 B C 90 A B C 90 A2 2 2 2 2 2
Now, OB and OC are the angle bisectors of
ABC and ACB
Hence, 1
OBC B.2
And1
BCO C2
.
In OBC, OBC BCO COB 180
1 1 B + C 180 COB2 2
1 90 A 180 COB
2 (From equation i)
Chapter 6
Lines and Angles
1COB 90 A
2
130COB 90
2
COB 90 65
COB 155
Question: 6
In Fig. 6.2, POQ is a line. The value of x is:
(a) 20
(b) 25
(c) 30
(d) 35
Chapter 6
Lines and Angles
Solution:
a
Given,
POQ is a line segment.
Hence, POQ 180
POA AOB BOQ 180
Putting POA 40 , AOB 4 , x and BOQ 3 x
40 4 3 180 x x
x 7 140
x 7 140
x 20
Question 7
In Fig. 6.3, if OP RS O, PQ 110 and QRS 130 ,
then PQR is equal to:
Chapter 6
Lines and Angles
(a) 40
(b) 50
(c) 60
(d) 70
Solution:
c
producing OP such that it intersects RQ at point x.
Now, OP RS and RX is transversal.
Hence, RXP QRS
[The alternate interior angles are equal.]
RXP 130 … (i) [ QRS 130 ]
Now, RQ is a line segment.
Chapter 6
Lines and Angles
So, PXQ RXP 180
PXQ 180 RXP
PXQ 180 130 [From eqn (i)]
PXQ 50
In PQX, OPQ is an exterior angle.
Hence, OPQ PXQ PQX
[∴ Exterior angle = sum of two opposite interior angles.]
110 50 PQX
PQX 110 50
PQX 60
Question 8
Angles of a triangle are in the ratio 2 : 4 : 3.
The smallest angle of the triangle is
(a) 60
(b) 40
(c) 80
Chapter 6
Lines and Angles
(d) 20
Solution:
b
Given that,
The ratio of angles of a triangle is 2 : 4 : 3.
Let angles of a triangle be A, B and C .
A 2x B 4x , and C 3x
In ABC, A B C 180
[The sum of the angles of a triangle is180]
2x 4x 3x 180
x 9 180
x 20
x A 2 2 20 40
B 4x 4 20 80
and C 3x 3 20 60
Hence, option (b) is correct.
Chapter 6
Lines and Angles
Exercise 6.2
Question 1
For what value of x y in Fig. 6.4 will ABC be a line?
Justify your answer.
Solution:
For ABC to be a line, the sum of the two adjacent angles
must be 180 or ABD and DBC must form a linear pair.
i.e., x y 180 .
Question 2
Can a triangle have all the angles less than60? Give
reason for your answer.
Chapter 6
Lines and Angles
Solution:
No, a triangle cannot have all angles less than60 ,
because if all angles are less than60 , then their sum will
be less than 180 (not equal to180).
Hence, it will not be a triangle.
Question: 3
Can a triangle have two obtuse angles? Give reason for
your answer.
Solution:
No, because if a triangle has two obtuse angles i.e., more
than 90 angle, then the sum of all three angles of a
triangle will be greater than 180
(not equal to180).
Hence, it will not be a triangle.
Chapter 6
Lines and Angles
Question: 4
How many triangles can be drawn with angles
measuring as45, 64and 72? Give reason for your
answer.
Solution:
None.
The sum of given angles 45 64 72 181 180 .
Hence, we see that sum of all three angles is not equal to
180. So, we cannot draw a triangle with the given angles.
Question: 5
How many triangles can be drawn with angles as 53,
64and63? Give reason for your answer.
Solution:
The sum of the given angles 53 64 63 180 .
Since, the sum of all interior angles of the triangle is 180,
infinitely many triangles can be drawn.
Chapter 6
Lines and Angles
Question: 6
In Fig. 6.5, find the value of x for which the lines l and m
are parallel.
Solution:
In the given figure,l m .
Using the properties of parallel line (if a transversal
intersects two parallel lines, then the sum of interior
angles on the same side of a transversal is
supplementary), we have:
x 44 180
x 180 44
x 136
Chapter 6
Lines and Angles
Question: 7
Two adjacent angles are equal. Is it necessary that each
of these angles will be a right angle? Justify your
answer.
Solution:
No, because each of these will be a right angle only when
they form a linear pair.
Question: 8
If one of the angles formed by two intersecting lines is a
right angle, what can you say about the other three
angles? Give reason for your answer.
Solution:
Let two intersecting lines be l and m. As one of the angles
is a right angle, then it means that lines l and m are
perpendicular to each other. By using linear pair axiom,
other three angles will be right angles.
Chapter 6
Lines and Angles
Question: 9
In Fig. 6.6, which of the two lines are parallel and why?
Solution:
In case (i),
The sum of two interior angles132 48 180 .
Here, the sum of two interior angles on the same side of
line n is180. Hence, l and m are the parallel lines.
In case (ii)
The sum of two interior angles73 106 179 180 .
Chapter 6
Lines and Angles
Here, the sum of two interior angles on the same side of
the transversal is not equal to180. Hence, lines p and q
are not the parallel lines.
Question: 10
Two lines l and m are perpendicular to the same line n.
Are l and m perpendicular to each other? Give reason for
your answer.
Solution:
No.
Let lines l and m be two lines which are perpendicular to
the line n.
1 2 90 90 180
[ l n andm n ]
Chapter 6
Lines and Angles
Exercise: 6.3
Question: 1
In Fig. 6.9, OD is the bisector of AOC , OE is the
bisector of BOC andOD OE . Show that the points A,
O and B are collinear.
Solution:
Given:
In figure, OD OE and OD is the bisector of AOC and OE
is the bisector of BOC .
To prove:
Points A, O and B are collinear i.e., AOB is a straight line.
Proof:
OD and OE bisect angles AOC and BOC respectively.
Chapter 6
Lines and Angles
AOC = 2 DOC …….. (i) and
BOC = 2 COE …….. (ii)
On adding equations (i) and (ii), we get
AOC + COB = 2 DOC + 2 COE
AOC + COB = 2( DOC + COE)
AOC + COB = 2 DOE
AOC + COB = 2 90 OD OE][
AOC + COB =180
AOB =180
So, AOC and COB are linear pairs or AOB is a straight
line.
Hence, points A, O and B are collinear.
Question: 2
In Fig. 6.10, 1 = 60 and 6 = 120 .Show that the lines m
and n are parallel.
Chapter 6
Lines and Angles
Solution:
Given:
In figure, 1 = 60 and 6 = 120
To prove:
m n .
Proof:
Since, 1 = 60 and 6 = 120
Here, 1 = 3 [Vertically opposite angles]
3 = 1 = 60
Now, 3 + 6 = 60 120
3 + 6 = 180
Chapter 6
Lines and Angles
The sum of two interior angles on the same side of the
transversal is180.
Hence, the lines are parallel orm n .
Question: 3
AP and BQ are the bisectors of the two alternate interior
angles formed by the intersection of a transversal t with
parallel lines l and m (Fig. 6.11). Show thatAP BQ .
Chapter 6
Lines and Angles
Solution:
Given:
In figure, l m , AP and BQ are the bisectors of
EAB and ABH respectively.
To prove:
AP BQ
Proof:
Since, l mand t is transversal,
Chapter 6
Lines and Angles
therefore, EAB = ABH [Alternate interior angles]
EAB = ABH1 1
2 2 [Dividing both sides by 2]
PAB = ABQ
[AP and BQ are the bisectors of EAB and ABH ]
Since, PAB and ABQ are alternate interior angles
formed when transversal AB intersects lines AP and BQ,
hence,AP BQ .
Question: 4
If in Fig. 6.11, bisectors AP and BQ of the alternate
interior angles are parallel, then show that l m .
Solution:
Given:
Chapter 6
Lines and Angles
In figure, AP BQ , AP and BQ are the bisectors of
alternate interior angles EAB and ABH .
To prove:
l m
Proof:
Since, AP BQand t is the transversal, therefore
PAB = ABQ [Alternate interior angles].
2 PAB = 2 ABQ [Multiplying both sides by 2]
EAB = ABH
As the alternate interior angles EAB and ABH are
equal, the lines l and m are parallel. [If two alternate
interior angles are equal, then the lines are parallel.]
Chapter 6
Lines and Angles
Question: 5
In Fig. 6.12, BA ED andBC EF . Show that
ABC = DEF
[Hint: Produce DE to intersect BC at P (say)].
Solution:
Given:
BA ED , andBC EF .
To prove: ABC = DEF
Construction:
Produce DE which intersects BC at P.
Chapter 6
Lines and Angles
Proof:
In figure,
BA ED BA DP
ABP = EPC [Corresponding angles]
ABC = EPC … (i)
Again, BC EF PC EF
EPC = DEF … (ii)
[Corresponding angles]
From equations (i) and (ii), we get
ABC= DEF
Hence, it is proved.
Chapter 6
Lines and Angles
Question: 6
In Fig. 6.13, BA ED and BC EF .
Show that ABC+ DEF = 180 .
Solution:
Given:
In the figure,BA ED andBC EF .
To prove:
ABC+ DEF = 180 .
Construction:
Draw a ray PE opposite to ray EF.
Chapter 6
Lines and Angles
Proof:
In figure, BC EFor BC PF
EPB+ PBC = 180 ... (i)
[The sum of the interior angles on the same side of the
transversal is 180°.]
Now, AB ED and PE is a transversal line.
EPB= DEF ... (ii) [Corresponding angles]
From equations (i) and (ii), we get
DEF+ PBC = 180
ABC+ DEF = 180 [ PBC= ABC ]
Hence, it is proved.
Chapter 6
Lines and Angles
Question: 7
In Fig. 6.14, DE QR and AP and BP are the bisectors of
EAB and RBA , respectively. Find APB .
Solution:
Given:
DE QR , and AP and PB are the bisectors of EAB and
RBA , respectively.
To find:
APB
Proof:
As DE QR and AB is a transversal, the sum of the interior
angles on the same side of the transversal is
supplementary.
Chapter 6
Lines and Angles
EAB+ RBA = 180
EAB + RBA =1 1 180
2 2 2
[Dividing both sides by 2]
EAB + RBA = 901 1
2 2
… (i)
Since AP and BP are the bisectors of EAB and RBA
respectively,
BAP = EAB1
2 … (ii)
and ABP = RBA1
2 … (iii)
On adding equations (ii) and (iii), we get
BAP + ABP= EAB + RBA1 1
2 2
Using equation (i), we get
BAP + ABP = 90 … (iv)
Chapter 6
Lines and Angles
In ABP ,
BAP + ABP + APB = 180
[The sum of the interior angles in a triangle is180.]
90 + APB = 180
[From equation (iv)]
APB 180 90 = 90
Hence, APB 90
Question: 8
The angles of a triangle are in the ratio 2 : 3 : 4. Find the
angles of the triangle.
Solution:
Given,
The ratio of the angles of a triangle is 2 : 3 : 4.
Let the angles of a triangle be A, B and C .
A 2x B 3x , and C 4x
In ABC, A B C 180
Chapter 6
Lines and Angles
[The sum of the angles of a triangle is180 ]
2x 3x 4x = 180
9x 180
x 20
A 2x 2 20 40
B 3x 3 20 60
and C 4x 4 20 80
Question: 9
A triangle ABC is right angled at A. L is a point on BC
such thatAL BC . Prove that BAL ACB .
Solution:
Given:
In ABC, A 90 and AL BC
To prove:
BAL ACB
Chapter 6
Lines and Angles
Proof:
In ABC and LAC ,
BAC ALC … (i)
and ABC ABL ..… (ii) [Common angle]
Adding equations (i) and (ii), we get
BAC ABC ALC + ABL ..… (iii)
In ABC, BAC ACB ABC 180
[The sum of the angles of a triangle is180 .]
BAC ABC 180 ACB ..… (iv)
In ABL, ABL ALB BAL 180
[The sum of the angles of a triangle is180 ]
ABL ALC 180 BAL ..… (v)
Chapter 6
Lines and Angles
[ ALC ALB 90 ]
Substituting the value from equation (iv) and (v) in
equation (iii), we get
180 ACB 180 BAL
ACB BAL
Hence, it is proved.
Question: 10
Two lines are respectively perpendicular to two parallel
lines. Show that they are parallel to each other.
Solution:
Given:
Two lines m and n are parallel and another two lines p
and q are perpendicular to m and n respectively.
Or p m and p n, q m and q n
To prove:
Chapter 6
Lines and Angles
p q
Proof:
Since,m n and p is perpendicular to m and n.
So,
p is perpendicular to m …(i)
p is perpendicular to n …(ii)
Since, m ∥ n and q is perpendicular to m and n.
So,
q is perpendicular to m …(iii)
q is perpendicular to n …(iv)
From the equations (i) and (iii) [Or from (ii) and (iv)], we
get,
p q .
Chapter 6
Lines and Angles
[If two lines are perpendicular to the same line, the lines
are parallel to each other.]
Hence, it isp q .
Chapter 6
Lines and Angles
Exercise: 6.4
Question: 1
If two lines intersect, prove that the vertically opposite
angles are equal.
Solution:
Given:
Two lines AB and CD intersect at point O.
To prove:
(i) AOC BOD
(ii) AOD BOC
Proof:
(i) Since, ray OA stands on line CD,
Chapter 6
Lines and Angles
AOC AOD 180 … (i)
[Linear pair axiom]
Similarly, ray OD stands on line AB.
AOD BOD 180 … (ii)
From equations (i) and (ii), we get
AOC AOD AOD BOD
AOC BOD
(ii) Since, ray OD stands on line AB,
AOD BOD 180 … (iii)
[Linear pair axiom]
Similarly, ray OB stands on line CD.
DOB BOC 180 … (iv)
From equations (iii) and (iv), we get
AOD BOD DOB BOC
AOD BOC
Chapter 6
Lines and Angles
Question: 2
Bisectors of interior B and exterior ACD of
ABC intersect at the point T.
Prove that:
1BTC BAC
2 .
Solution:
Given:
In ABC , the bisectors of ABC and ACD meet at point
T.
Construction:
Produce BC to D.
To Prove:
1BTC BAC
2
Chapter 6
Lines and Angles
Proof:
In ABC, ACD is an exterior angle.
ACD = ABC + CAB
[The exterior angle of a triangle is equal to the sum of two
opposite interior angles.]
1 1 1ACD CAB + ABC
2 2 2
[Dividing both sides by 2]
1 1TCD CAB + ABC
2 2 … (i)
[∵ CT is a bisector of1
ACD ACD TCD2
]
In BTC ,
TCD BTC + CBT
[The exterior angle of a triangle is equal to the sum of two
interior opposite angles.]
1TCD BTC + ABC
2 … (ii)
[∵ BT bisector of 1
ABC CBT ABC2
]
Chapter 6
Lines and Angles
From equations (i) and (ii), we get
1 1 1CAB + ABC BTC ABC
2 2 2
1CAB BTC
2
or 1
BAC BTC2
Hence, it is proved.
Question: 3
A transversal intersects two parallel lines. Prove that
the bisectors of any pair of corresponding angles so
formed are parallel.
Solution:
Given:
Two lines DE and QR are parallel. A transversal intersects
the lines DE and QR at A and B respectively. Also, BP and
AF are the bisectors of angles ABR and CAE
respectively.
To prove:
Chapter 6
Lines and Angles
BP AF
Proof:
Given, DE QR
CAE ABR
[Corresponding angles]
1 1CAE ABR
2 2
[Dividing both sides by 2]
CAF ABP
[∵ BP and AF are the bisectors of angles ABR and
CAE respectively]
This implies that AF is parallel to BP as the corresponding
angles CAF and ABP are equal.
Hence, AF BP .
Chapter 6
Lines and Angles
Question: 4
Prove that through a given point, we can draw only one
perpendicular on a given line.
[Hint: Use proof by contradiction].
Solution:
Given:
Consider a line l and a point P.
To prove:
Only one perpendicular can be drawn from P to l.
Construction:
Suppose that two lines PA and PB are passing through
the point P and they are perpendicular to l.
Proof:
Chapter 6
Lines and Angles
In APB ,
PAB P PBA 180
[The angle sum property of a triangle]
90 P 90 180 (since PA is perpendicular to
l and PB is perpendicular to l)
P 180 180
P 0
(Which is possible only when the lines PA and PB
coincide)
Hence, only one perpendicular line can be drawn through
a given point.
Question: 5
Prove that two lines that are respectively perpendicular
to two intersecting lines intersect each other.
[Hint: Use proof by contradiction].
Solution:
Given:
Chapter 6
Lines and Angles
Let lines l and m be two intersecting lines. Again, let n
and p be another two lines which are perpendicular to the
intersecting lines.
To prove:
Two lines n and p intersect at a point.
Proof:
Let us consider lines n and p are not intersecting,
and then it means they are parallel to each other
i.e., n p … (i)
Since, lines n and p are perpendicular to l and m
respectively, but from equation (i) n p , it is implied that
l m , it is a contradiction.
Thus, our assumption is wrong.
Hence, lines n and p intersect at a point.
Question: 6
Prove that a triangle must have at least two acute
angles.
Chapter 6
Lines and Angles
Solution:
Given:
ABC is a triangle
To prove:
ABC must have two acute angles.
Proof:
Let us consider the following cases:
Case I:
When two angles are90 :
Suppose two angles are B 90 and C 90 .
The sum of all three angles of a triangle is180 .
A B C 180
A 90 90 180
A 180 180 0 , which is not possible.
Hence, this case is rejected.
Case II:
When two angle are obtuse:
Chapter 6
Lines and Angles
Suppose two angles B and C are more than90 ,
the sum of all three angles of a triangle is180 .
A B C 180
A 180 ( B C) 180 − (an angle greater than180 )
A = negative angle, which is not possible.
Hence, this case is also rejected.
Case III:
When one angle is 90 and the other is obtuse:
Suppose B 90 and C is obtuse.
The sum of all three angles of a triangle is180 .
A B C 180
A 180 (90 C)
90 C
= negative angle, which is not possible.
Hence, this case is also rejected.
Case IV:
Chapter 6
Lines and Angles
When two angles are acute, then the sum of two angles is
less than 180 so that the third angle is also acute.
Hence, a triangle must have at least two acute angles.
Question: 7
In Fig. 6.17, Q R, PA is the bisector of QPR and
PM QR . Prove that
1
APM Q R2
.
Solution:
Given:
Q R, PA is the bisector of QPR and PM QR .
To prove:
1
APM Q R2
.
Chapter 6
Lines and Angles
Proof:
Since, PA is the bisector of QPR
QPA APR … (i)
In PQM, Q PMQ QPM 180
[𝐴ngle sum property of triangles]
Q 90 QPM 180
[ PMQ 90 ]
Q 90 QPM … (ii)
In PMR, PMR R RPM 180
[Angle sum property of triangles]
90 R RPM 180
[ PMQ 90 ]
R 180 90 RPM
R 90 RPM … (iii)
Subtracting equations (iii) from (ii), we get
Q R (90 QPM) (90 RPM)
Q R RPM QPM
Chapter 6
Lines and Angles
Q R ( RPA APM) ( QPA APM) … (iv)
Q R QPA APM QPA APM
[𝑈sing equation (i)]
Q R 2 APM
1APM ( Q R)
2
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