R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 7 - Lines And Angles
Chapter 7 - Lines And Angles Exercise MCQ
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
Correct option: (d)
In a right triangle, one angle is 90° and the sum of acute angles of a right triangle is 90°.
An exterior angle of a triangle is 110° and its two interior opposite angles are equal. Each of these equal angles is
(a) 70°
(b) 55°
(c) 35°
(d)
Correct option: (b)
Let each interior opposite angle be x.
Then, x + x = 110° (Exterior angle property of a triangle)
⇒ 2x = 110°
⇒ x = 55°
The angles of a triangle are in the ratio 3:5:7 The triangle is
- Acute angled
- Obtuse angled
- Right angled
- an isosceles triangle
If one of the angles of triangle is 130° then the angle between the bisector of the other two angles can be
(a) 50°
(b) 65°
(c) 90°
(d) 155
Correct option: (d)
Let ∠A = 130°
In ΔABC, by angle sum property,
∠B + ∠C + ∠A = 180°
⇒ ∠B + ∠C + 130° = 180°
⇒ ∠B + ∠C = 50°
In the given figure, AOB is a straight line. The value of x is
(a) 12
(b) 15
(c) 20
(d) 25
Correct option: (b)
AOB is a straight line.
⇒ ∠AOB = 180°
⇒ 60° + 5x° + 3x° = 180°
⇒ 60° + 8x° = 180°
⇒ 8x° = 120°
⇒ x = 15°
The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle of the triangle is
(a) 120°
(b) 100°
(c) 80°
(d) 60°
Correct option: (c)
By angle sum property,
2x + 3x + 4x = 180°
⇒ 9x = 180°
⇒ x = 20°
Hence, largest angle = 4x = 4(20°) = 80°
In the given figure, ∠OAB = 110° and ∠BCD = 130° then ∠ABC is equal to
(a) 40°
(b) 50°
(c) 60°
(d) 70°
Correct option: (c)
Through B draw YBZ ∥ OA ∥ CD.
Now, OA ∥ YB and AB is the transversal.
⇒ ∠OAB + ∠YBA = 180° (interior angles are supplementary)
⇒ 110° + ∠YBA = 180°
⇒ ∠YBA = 70°
Also, CD ∥ BZ and BC is the transversal.
⇒ ∠DCB + ∠CBZ = 180° (interior angles are supplementary)
⇒ 130° + ∠CBZ = 180°
⇒ ∠CBZ = 50°
Now, ∠YBZ = 180° (straight angle)
⇒ ∠YBA + ∠ABC + ∠CBZ = 180°
⇒ 70° + x + 50° = 180°
⇒ x = 60°
⇒ ∠ABC = 60°
If two angles are complements of each other, then each angle is
- An acute angle
- An obtuse angle
- A right angle
- A reflex angle
Correct option: (a)
Two angles are said to be complementary, if the sum of their measures is 90°.
Clearly, the measures of each of the angles have to be less than 90°.
Hence, each angle is an acute angle.
An angle which measures more than 180° but less than 360°, is called
- An acute angle
- An obtuse angle
- A straight angle
- A reflex angle
Correct option: (d)
An angle which measures more than 180o but less than 360o is called a reflex angle.
The measure of an angle is five times its complement. The angle measures
- 25°
- 35°
- 65°
- 75°
Two complementary angles are such that twice the measure of the one is equal to three times the measure of the other. The larger of the two measures
- 72°o
- 54°
- 63°
- 36°
In the given figure, AOB is a straight line. If ∠AOC = 4x° and ∠BOC = 5x°, then ∠AOC =?
In the given figure, AOB is a straight line. If ∠AOC = (3x + 10) ° and ∠BOC = (4x - 26) °, then ∠BOC =?
- 96°
- 86°
- 76°
- 106°
In the given figure, AOB is a straight line. If ∠AOC = (3x - 10) °, ∠COD = 50° and ∠BOD = (x +20) °, then ∠AOC =?
- 40°
- 60°
- 80°
- 50°
Which of the following statements is false?
- Through a given point, only one straight line can be drawn
- Through two given points, it is possible to draw one and only one straight line.
- Two straight lines can intersect only at one point
- A line segment can be produced to any desired length.
Correct option: (a)
Option (a) is false, since through a given point we can draw an infinite number of straight lines.
An angle is one-fifth of its supplement. The measure of the angle is
- 15°
- 30°
- 75°
- 150°
In the adjoining figure, AOB is straight line. If x:y:z = 4:5:6, then y = ?
- 60°
- 80°
- 48°
- 72°
In the given figure, straight lines AB and CD intersect at O. If ∠AOC =ϕ, ∠BOC = θ and θ = 3 ϕ, then ϕ =?
- 30°
- 40°
- 45°
- 60°
In the given figure, straight lines AB and CD intersect at O. If ∠AOC + ∠BOD = 130°, then ∠AOD =?
- 65°
- 115°
- 110°
- 125°
In the given figure AB is a mirror, PQ is the incident ray and and QR is the reflected ray. If ∠PQR = 108°, then ∠ AQP =?
- 72°
- 18°
- 36°
- 54°
In the given figure, AB ∥ CD, If ∠BAO = 60° and ∠OCD = 110° then ∠AOC = ?
(a) 70°
(b) 60°
(c) 50°
(d) 40°
Correct option: (c)
Let ∠AOC = x°
Draw YOZ ∥ CD ∥ AB.
Now, YO ∥ AB and OA is the transversal.
⇒ ∠YOA = ∠OAB = 60° (alternate angles)
Again, OZ ∥ CD and OC is the transversal.
⇒ ∠COZ + ∠OCD = 180° (interior angles)
⇒ ∠COZ + 110° = 180°
⇒ ∠COZ = 70°
Now, ∠YOZ = 180° (straight angle)
⇒ ∠YOA + ∠AOC + ∠COZ = 180°
⇒ 60° + x + 70° = 180°
⇒ x = 50°
⇒ ∠AOC = 50°
In the given figure, AB ‖ CD. If ∠AOC = 30° and ∠OAB = 100°, then ∠OCD =?
- 130°
- 150°
- 80°
- 100°
In the given figure, AB ‖ CD. If ‖CAB = 80o and ∠EFC= 25°, then ∠CEF =?
- 65°
- 55°
- 45°
- 75°
In the given figure, AB ‖ CD, CD ‖ EF and y:z = 3:7, then x = ?
- 108°
- 126°
- 162°
- 63°
In the given figure, AB ‖ CD. If ∠APQ = 70° and ∠RPD = 120°, then ∠QPR =?
- 50°
- 60°
- 40°
- 35°
In the given figure AB ‖ CD. If ∠EAB = 50° and ∠ECD=60°, then ∠AEB =?
- 50°
- 60°
- 70°
- 50°
In the given figure, ∠OAB = 75°, ∠OBA=55° and ∠OCD = 100°. Then ∠ODC=?
- 20°
- 25°
- 30°
- 35°
In the adjoining figure y =?
- 36°
- 54°
- 63°
- 72°
Chapter 7 - Lines And Angles Exercise Ex. 7A
Define the following terms:
(i) Angle (ii) Interior of an angle (iii) Obtuse angle (iv) Reflex angle (v) Complementary angles (vi) Supplementary angles
(i) Angle: Two rays having a common end point form an angle.
(ii) Interior of an angle: The interior of AOB is the set of all points in its plane, which lie on the same side of OA as B and also on same side of OB as A.
(iii) Obtuse angle: An angle whose measure is more than 90o but less than 180o, is called an obtuse angle.
(iv) Reflex angle: An angle whose measure is more than 180o but less than 360o is called a reflex angle.
(v) Complementary angles: Two angles are said to be complementary, if the sum of their measures is 90o.
(vi) Supplementary angles: Two angles are said to be supplementary, if the sum of their measures is 180o.
Find the complement of each of the following angles.
16o
Complement of 16o = 90 - 16o = 74o
Find the complement of each of the following angles.
46o 30
Complement of 46o 30' = 90o - 46o 30' = 43o 30'
Find the complement of each of the following angle:
55°
Complement of 55° = 90° - 55° = 35°
Find the complement of each of the following angle:
90°
Complement of 90° = 90° - 90° = 0°
Find the supplement of each of the following angles.
75o 36'
Supplement of 75o 36' = 180o - 75o 36' = 104o 24'
Find the supplement of each of the following angle:
42°
Supplement of 42° = 180° - 42° = 138°
Find the supplement of each of the following angle:
90°
Supplement of 90° = 180° - 90° = 90°
Find the supplement of each of the following angle:
124°
Supplement of 124° = 180° - 124° = 56°
Find the measure of an angle which is
(i) equal to its complement, (ii) equal to its supplement.
(i) Let the required angle be xo
Then, its complement = 90o - xo
The measure of an angle which is equal to
its complement is 45o.
(ii) Let the required angle be xo
Then, its supplement = 180o - xo
The measure of an angle which is equal to
its supplement is 90o.
Find the measure of an angle which is 36o more than its complement.
Let the required angle be xo
Then its complement is 90o - xo
The measure of an angle which is 36o
more than its complement is 63o.
Find the measure of an angle which is 30° less than its supplement.
Let the measure of the required angle = x°
Then, measure of its supplement = (180 - x)°
It is given that
x° = (180 - x)° - 30°
⇒ x° = 180° - x° - 30°
⇒ 2x° = 150°
⇒ x° = 75°
Hence, the measure of the required angle is 75°.
Find the angle which is four times its complement.
Let the required angle be xo
Then, its complement = 90o - xo
The required angle is 72o.
Find the angle which is five times its supplement.
Let the required angle be xo
Then, its supplement is 180o - xo
The required angle is 150o.
Find the angle whose supplement is four times its complement.
Let the required angle be xo
Then, its complement is 90o - xo and its supplement is 180o - xo
That is we have, The required angle is 60o.
Find the angle whose complement is one-third of its supplement.
Let the required angle be xo
Then, its complement is 90o - xo and its supplement is 180o - xo
The required angle is 45o.
Two complementary angles are in the ratio 4: 5. Find the angles.
Let the two required angles be xo and 90o - xo.
Then
5x = 4(90 - x)
5x = 360 - 4x
5x + 4x = 360
9x = 360
Thus, the required angles are 40o and 90o - xo = 90 o - 40o = 50o.
Find the value of x for which the angles (2x - 5)° and (x - 10)° are the complementary angles.
(2x - 5)° and (x - 10)° are complementary angles.
∴ (2x - 5)° + (x - 10)° = 90°
⇒ 2x - 5° + x - 10° = 90°
⇒ 3x - 15° = 90°
⇒ 3x = 105°
⇒ x = 35°
Chapter 7 - Lines And Angles Exercise Ex. 7B
In the given figure, AOB is a straight line. Find the value of x.
Since BOC and
COA form a linear pair of angles, we have
BOC +
COA = 180o
xo + 62o = 180o
x = 180 - 62
x = 118o
In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find ∠AOC and ∠BOD.
∠AOB is a straight angle.
⇒ ∠AOB = 180°
⇒ ∠AOC + ∠COD + ∠BOD = 180°
⇒ (3x - 7)° + 55° + (x + 20)° = 180°
⇒ 4x + 68° = 180°
⇒ 4x = 112°
⇒ x = 28°
Thus, ∠AOC = (3x - 7)° = 3(28°) - 7° = 84° - 7° = 77°
And, ∠BOD = (x + 20)° = 28° + 20° = 48°
In the given figure, AOB is a straight line. Find the value of x. Hence, find AOC,
COD and
BOD.
Since BOD and
DOA from a linear pair of angles.
BOD +
DOA = 180o
BOD +
DOC +
COA = 180o
xo + (2x - 19)o + (3x + 7)o = 180o
6x - 12 = 180
6x = 180 + 12 = 192
x = 32
AOC = (3x + 7)o = (3
32 + 7)o = 103o
COD = (2x - 19)o = (2
32 - 19)o = 45o
and BOD = xo = 32o
In the given figure, x: y: z = 5: 4: 6. If XOY is a straight line, find the values of x, y and z.
x: y: z = 5: 4: 6
The sum of their ratios = 5 + 4 + 6 = 15
But x + y + z = 180o
[Since, XOY is a straight line]
So, if the total sum of the measures is 15, then the measure of x is 5.
If the sum of angles is 180o, then, measure of
And, if the total sum of the measures is 15, then the measure of y is 4.
If the sum of the angles is 180o, then, measure of
And z = 180o -
x -
y
= 180o - 60o - 48o
= 180o - 108o = 72o
x = 60, y = 48 and z = 72.
In the given figure, what value of x will make AOB, a straight line?
AOB will be a straight line, if two adjacent angles form a linear pair.
BOC
+
AOC = 180o
(4x - 36)o + (3x + 20)o
= 180o
4x - 36 + 3x + 20 = 180
7x - 16 = 180o
7x = 180 + 16 = 196
The value of x = 28.
Two
lines AB and CD intersect at O. If AOC = 50o, find
AOD,
BOD and
BOC.
Since
AOC and
AOD form a linear pair.
AOC
+
AOD = 180o
50o +
AOD = 180o
AOD = 180o - 50o = 130o
AOD
and
BOC are vertically opposite angles.
AOD =
BOC
BOC = 130o
BOD
and
AOC are vertically opposite angles.
BOD
=
AOC
BOD = 50o
In the given figure, three coplanar lines AB,CD and EF intersect at a point O, forming angles as shown. Find the values of x,y,z and t.
Since COE and
DOF are vertically opposite angles, we have,
COE =
DOF
z = 50o
Also BOD and
COA are vertically opposite angles.
So, BOD =
COA
t = 90o
As COA and
AOD form a linear pair,
COA +
AOD = 180o
COA +
AOF +
FOD = 180o [
t = 90o]
t + x + 50o = 180o
90o + xo + 50o = 180o
x + 140 = 180
x = 180 - 140 = 40
Since EOB and
AOF are vertically opposite angles
So, EOB =
AOF
y = x = 40
Thus, x = 40 = y = 40, z = 50 and t = 90
In
the given figure, three coplanar lines AB,CD and EF intersect at a point O.
Find the value of x. Hence, find AOD,
COE and
AOE.
Since
COE and
EOD form a linear pair of angles.
COE +
EOD = 180o
COE +
EOA +
AOD = 180o
5x +
EOA + 2x = 180
5x +
BOF + 2x = 180
[EOA and
BOF are vertically opposite angles so,
EOA =
BOF]
5x + 3x + 2x = 180
10x = 180
x = 18
Now AOD = 2xo = 2
18o
= 36o
COE = 5xo = 5
18o
= 90o
and, EOA =
BOF = 3xo = 3
18o
= 54o
Two adjacent angles on a straight line are in the ratio 5 : 4. Find the measure of each one of these angles.
Let the two adjacent angles be 5x and 4x.
Now, since these angles form a linear pair.
So, 5x + 4x = 180o
9x = 180o
The required angles are 5x = 5x = 5
20o
= 100o
and 4x = 4 20o
= 80o
If two straight lines intersect each other in such a way that one of the angles formed measures 90o, show that each of the remaining angles measures 90o.
Let
two straight lines AB and CD intersect at O and let AOC = 90o.
Now, AOC =
BOD
[Vertically opposite angles]
BOD = 90o
Also,
as AOC and
AOD form a linear pair.
90o +
AOD = 180o
AOD = 180o - 90o = 90o
Since, BOC =
AOD
[Verticallty opposite angles]
BOC = 90o
Thus, each of the remaining angles is 90o.
Two
lines AB and CD intersect at a point O such that BOC +
AOD = 280o, as shown in the figure. Find
all the four angles.
Since,
AOD and
BOC are vertically opposite angles.
AOD
=
BOC
Now,
AOD +
BOC = 280o [Given]
AOD +
AOD = 280o
2
AOD = 280o
AOD =
BOC =
AOD = 140o
As,
AOC and
AOD form a linear pair.
So, AOC +
AOD = 180o
AOC + 140o = 180o
AOC = 180o - 140o = 40o
Since,
AOC and
BOD are vertically opposite angles.
AOC
=
BOD
BOD = 40o
BOC = 140o,
AOC = 40o ,
AOD = 140o and
BOD = 40o.
Two lines AB and CD intersect each other at a point O such that ∠AOC : ∠AOD = 5 : 7. Find all the angles.
Let ∠AOC = 5x and ∠AOD = 7x
Now, ∠AOC + ∠AOD = 180° (linear pair of angles)
⇒ 5x + 7x = 180°
⇒ 12x = 180°
⇒ x = 15°
⇒ ∠AOC = 5x = 5(15°) = 75° and ∠AOD = 7x = 7(15°) = 105°
Now, ∠AOC = ∠BOD (vertically opposite angles)
⇒ ∠BOD = 75°
Also, ∠AOD = ∠BOC (vertically opposite angles)
⇒ ∠BOC = 105°
In the given figure, three lines AB, CD and EF intersect at a point O such that ∠AOE = 35° and ∠BOD = 40°. Find the measure of ∠AOC, ∠BOF, ∠COF and ∠DOE.
∠BOD = 40°
⇒ AOC = ∠BOD = 40° (vertically opposite angles)
∠AOE = 35°
⇒ ∠BOF = ∠AOE = 35° (vertically opposite angles)
∠AOB is a straight angle.
⇒ ∠AOB = 180°
⇒ ∠AOE + ∠EOD + ∠BOD = 180°
⇒ 35° + ∠EOD + 40° = 180°
⇒ ∠EOD + 75° = 180°
⇒ ∠EOD = 105°
Now, ∠COF = ∠EOD = 105° (vertically opposite angles)
In the given figure, the two lines AB and CD intersect at a point O such that ∠BOC = 125°. Find the values of x, y and z.
∠AOC + ∠BOC = 180° (linear pair of angles)
⇒ x + 125 = 180°
⇒ x = 55°
Now, ∠AOD = ∠BOC (vertically opposite angles)
⇒ y = 125°
Also, ∠BOD = ∠AOC (vertically opposite angles)
⇒ z = 55°
If two straight lines intersect each other then prove that the ray opposite to the bisector of one of the angles so formed bisects the vertically opposite angle.
Given : AB and CD are two lines which are
intersecting at O. OE is a ray bisecting the BOD. OF is a ray opposite to ray OE.
To Prove: AOF
=
COF
Proof : Since are two
opposite rays,
is a straight
line passing through O.
AOF =
BOE
and COF =
DOE
[Vertically opposite angles]
But BOE =
DOE (Given)
AOF =
COF
Hence, proved.
Prove that the bisectors of two adjacent supplementary angles include a right angle.
Given: is
the bisector of
BCD and
is the bisector of
ACD.
To Prove: ECF
= 90o
Proof: Since ACD and
BCD forms a linear pair.
ACD +
BCD = 180o
ACE +
ECD +
DCF +
FCB = 180o
ECD +
ECD +
DCF +
DCF = 180o
because
ACE =
ECD
and DCF =
FCB
2(
ECD) + 2 (
CDF) = 180o
2(
ECD +
DCF) = 180o
ECD +
DCF =
ECF = 90o (Proved)
Chapter 7 - Lines And Angles Exercise Ex. 7C
In the given figure, l ∥ m and a transversal t cuts them. If ∠1 = 120°, find the measure of each of the remaining marked angles.
Given, ∠1 = 120°
Now, ∠1 + ∠2 = 180° (linear pair)
⇒ 120° + ∠2 = 180°
⇒ ∠2 = 60°
∠1 = ∠3 (vertically opposite angles)
⇒ ∠3 = 120°
Also, ∠2 = ∠4 (vertically opposite angles)
⇒ ∠4 = 60°
Line l ∥ line m and line t is a transversal.
⇒ ∠5 = ∠1 = 120° (corresponding angles)
∠6 = ∠2 = 60° (corresponding angles)
∠7 = ∠3 = 120° (corresponding angles)
∠8 = ∠4 = 60° (corresponding angles)
In the given figure, l ∥ m and a transversal t cuts them. If ∠7 = 80°, find the measure of each of the remaining marked angles.
Given, ∠7 = 80°
Now, ∠7 + ∠8 = 180° (linear pair)
⇒ 80° + ∠8 = 180°
⇒ ∠8 = 100°
∠7 = ∠5 (vertically opposite angles)
⇒ ∠5 = 80°
Also, ∠6 = ∠8 (vertically opposite angles)
⇒ ∠6 = 100°
Line l ∥ line m and line t is a transversal.
⇒ ∠1 = ∠5 = 80° (corresponding angles)
∠2 = ∠6 = 100° (corresponding angles)
∠3 = ∠7 = 80° (corresponding angles)
∠4 = ∠8 = 100° (corresponding angles)
In the given figure, l ∥ m and a transversal t cuts them. If ∠1 : ∠2 = 2 : 3, find the measure of each of the marked angles.
Given, ∠1 : ∠2 = 2 : 3
Now, ∠1 + ∠2 = 180° (linear pair)
⇒ 2x + 3x = 180°
⇒ 5x = 180°
⇒ x = 36°
⇒ ∠1 = 2x = 72° and ∠2 = 3x = 108°
∠1 = ∠3 (vertically opposite angles)
⇒ ∠3 = 72°
Also, ∠2 = ∠4 (vertically opposite angles)
⇒ ∠4 = 108°
Line l ∥ line m and line t is a transversal.
⇒ ∠5 = ∠1 = 72° (corresponding angles)
∠6 = ∠2 = 108° (corresponding angles)
∠7 = ∠3 = 72° (corresponding angles)
∠8 = ∠4 = 108° (corresponding angles)
For what value of x will the lines l and m be parallel to each other?
Lines l and m will be parallel if 3x - 20 = 2x + 10
[Since, if corresponding angles are equal, lines are parallel]
3x - 2x = 10 + 20
x = 30
For what value of x will the lines l and m be parallel to each other?
*Question modified, back answer incorrect.
For lines l and m to be parallel to each other, the corresponding angles (3x + 5)° and (4x)° should be equal.
⇒ (3x + 5)° = 4x°
⇒ x = 5°
In the given figure, AB || CD and BC || ED. Find the value of x.
Since AB || CD and BC is a transversal.
So, BCD =
ABC = xo [Alternate angles]
As BC || ED and CD is a transversal.
BCD +
EDC = 180o
BCD + 75o =180o
BCD = 180o - 75o = 105o
ABC = 105o [since
BCD =
ABC]
xo =
ABC = 105o
Hence, x = 105.
In the given figure, AB || CD || EF. Find the value of x.
Since AB || CD and BC is a transversal.
So, ABC =
BCD [atternate interior angles]
70o = xo +
ECD(i)
Now, CD || EF and CE is transversal.
So,ECD +
CEF = 180o [sum of consecutive interior angles is 180o]
ECD + 130o = 180o
ECD = 180o - 130o = 50o
Putting ECD = 50o in (i) we get,
70o = xo + 50o
x = 70 - 50 = 20
In the give figure, AB ∥ CD. Find the values of x, y and z.
AB ∥ CD and EF is transversal.
⇒ ∠AEF = ∠EFG (alternate angles)
Given, ∠AEF = 75°
⇒ ∠EFG = y = 75°
Now, ∠EFC + ∠EFG = 180° (linear pair)
⇒ x + y = 180°
⇒ x + 75° = 180°
⇒ x = 105°
∠EGD = ∠EFG + ∠FEG (Exterior angle property)
⇒ 125° = y + z
⇒ 125° = 75° + z
⇒ z = 50°
Thus, x = 105°, y = 75° and z = 50°
Ineach of the figures given below, AB || CD. Find the value of x in each case.
Through E draw EG || CD. Now since EG||CD and ED is a transversal.
So,GED =
EDC = 65o[Alternate interior angles]
Since EG || CD and AB || CD,
EG||AB and EB is transversal.
So,BEG =
ABE = 35o[Alternate interior angles]
So,DEB = xo
BEG +
GED = 35o + 65o = 100o.
Hence, x = 100.
Ineach of the figures given below, AB || CD. Find the value of x in each case.
Through O draw OF||CD.
Now since OF || CD and OD is transversal.
CDO +
FOD = 180o
[sum of consecutive interior angles is 180o]
25o +
FOD = 180o
FOD = 180o - 25o = 155o
As OF || CD and AB || CD [Given]
Thus, OF || AB and OB is a transversal.
So,ABO +
FOB = 180o [sum of consecutive interior angles is 180o]
55o +
FOB = 180o
FOB = 180o - 55o = 125o
Now, xo = FOB +
FOD = 125o + 155o = 280o.
Hence, x = 280.
Ineach of the figures given below, AB || CD. Find the value of x in each case.
Through E, draw EF || CD.
Now since EF || CD and EC is transversal.
FEC +
ECD = 180o
[sum of consecutive interior angles is 180o]
FEC + 124o = 180o
FEC = 180o - 124o = 56o
Since EF || CD and AB ||CD
So, EF || AB and AE is a trasveral.
So,BAE +
FEA = 180o
[sum of consecutive interior angles is 180o]
116o +
FEA = 180o
FEA = 180o - 116o = 64o
Thus,xo = FEA +
FEC
= 64o + 56o = 120o.
Hence, x = 120.
In the given figure, AB || CD. Find the value of x.
Through C draw FG || AE
Now, since CG || BE and CE is a transversal.
So, GCE =
CEA = 20o [Alternate angles]
DCG = 130o -
GCE
= 130o - 20o = 110o
Also, we have AB || CD and FG is a transversal.
So, BFC =
DCG = 110o [Corresponding angles]
As, FG || AE, AF is a transversal.
BFG =
FAE [Corresponding angles]
xo =
FAE = 110o.
Hence, x = 110
In the given figure, AB || PQ. Find the values of x and y.
Since AB || PQ and EF is a transversal.
So, CEB =
EFQ [Corresponding angles]
EFQ = 75o
EFG +
GFQ = 75o
25o + yo = 75o
y = 75 - 25 = 50
Also, BEF +
EFQ = 180o [sum of consecutive interior angles is 180o]


= 180o - 75o
BEF = 105o
FEG +
GEB =
BEF = 105o
FEG = 105o -
GEB = 105o - 20o = 85o
In EFG we have,
xo + 25o + FEG = 180o
Hence, x = 70.
In the given figure, AB || CD. Find the value of x.
Since AB || CD and AC is a transversal.
So, BAC +
ACD = 180o [sum of consecutive interior angles is 180o]
ACD = 180o -
BAC
= 180o - 75o = 105o
ECF =
ACD [Vertically opposite angles]
ECF = 105o
Now in CEF,
ECF +
CEF +
EFC =180o

x = 180 - 30 - 105 = 45
Hence, x = 45.
In the given figure, AB || CD. Find the value of x.
Since AB || CD and PQ a transversal.
So, PEF =
EGH [Corresponding angles]
EGH = 85o
EGH and
QGH form a linear pair.
So, EGH +
QGH = 180o
QGH = 180o - 85o = 95o
Similarly, GHQ + 115o = 180o
GHQ = 180o - 115o = 65o
In GHQ, we have,
xo + 65o + 95o = 180o
x = 180 - 65 - 95 = 180 - 160
x = 20
In the given figure, AB || CD. Find the values of x, y and z.
Since AB || CD and BC is a transversal.
So, ABC =
BCD
x = 35
Also, AB || CD and AD is a transversal.
So, BAD =
ADC
z = 75
In ABO, we have,
xo + 75o + yo = 180o
35 + 75 + y = 180
y = 180 - 110 = 70
x = 35, y = 70 and z = 75.
In the given figure, AB || CD. Prove that p + q - r = 180.
Through F, draw KH || AB || CD
Now, KF || CD and FG is a transversal.
KFG =
FGD = ro (i)
[alternate angles]
Again AE || KF, and EF is a transversal.
So,AEF +
KFE = 180o
KFE = 180o - po (ii)
Adding (i) and (ii) we get,
KFG +
KFE = 180 - p + r
EFG = 180 - p + r
q = 180 - p + r
i.e.,p + q - r = 180
In the given figure, AB || CD and EF || GH. Find the values of x, y, z and t.
PRQ = xo = 60o [vertically opposite angles]
Since EF || GH, and RQ is a transversal.
So, x =
y [Alternate angles]
y = 60
AB || CD and PR is a transversal.
So, [Alternate angles]
[since
]
x +
QRD = 110o
QRD = 110o - 60o = 50o
In QRS, we have,
QRD + to + yo = 180o
50 + t + 60 = 180
t = 180 - 110 = 70
Since, AB || CD and GH is a transversal
So, zo = to = 70o [Alternate angles]
x = 60 , y = 60, z = 70 and t = 70
In the given figure, AB ∥ CD and a transversal t cuts them at E and F respectively. If EG and FG are the bisectors of ∠BEF and ∠EFD respectively, prove that ∠EGF = 90°.
AB ∥ CD and a transversal t cuts them at E and F respectively.
⇒ ∠BEF + ∠DFE = 180° (interior angles)
⇒ ∠GEF + ∠GFE = 90° ….(i)
Now, in ΔGEF, by angle sum property
∠GEF + ∠GFE + ∠EGF = 180°
⇒ 90° + ∠EGF = 180° ….[From (i)]
⇒ ∠EGF = 90°
In the given figure, AB ∥ CD and a transversal t cuts them at E and F respectively. If EP and FQ are the bisectors of ∠AEF and ∠EFD respectively, prove that EP ∥ FQ.
Since AB ∥ CD and t is a transversal, we have
∠AEF = ∠EFD (alternate angles)
⇒ ∠PEF = ∠EFQ
But, these are alternate interior angles formed when the transversal EF cuts EP and FQ.
∴ EP ∥ FQ
In the given figure, BA ∥ ED and BC ∥ EF. Show that ∠ABC = ∠DEF.
Construction: Produce DE to meet BC at Z.
Now, AB ∥ DZ and BC is the transversal.
⇒ ∠ABC = ∠DZC (corresponding angles) ….(i)
Also, EF ∥ BC and DZ is the transversal.
⇒ ∠DZC = ∠DEF (corresponding angles) ….(ii)
From (i) and (ii), we have
∠ABC = ∠DEF
In the given figure, BA ∥ ED and BC ∥ EF. Show that ∠ABC + ∠DEF = 180°.
Construction: Produce ED to meet BC at Z.
Now, AB ∥ EZ and BC is the transversal.
⇒ ∠ABZ + ∠EZB = 180° (interior angles)
⇒ ∠ABC + ∠EZB = 180° ….(i)
Also, EF ∥ BC and EZ is the transversal.
⇒ ∠BZE = ∠ZEF (alternate angles)
⇒ ∠BZE = ∠DEF ….(ii)
From (i) and (ii), we have
∠ABC + ∠DEF = 180°
In the given figure, m and n are two plane mirrors perpendicular to each other. Show that the incident ray CA is parallel to the reflected ray BD.
Let the normal to mirrors m and n intersect at P.
Now, OB ⊥ m, OC ⊥ n and m ⊥ n.
⇒ OB ⊥ OC
⇒ ∠APB = 90°
⇒ ∠2 + ∠3 = 90° (sum of acute angles of a right triangle is 90°)
By the laws of reflection, we have
∠1 = ∠2 and ∠4 = ∠3 (angle of incidence = angle of reflection)
⇒ ∠1 + ∠4 = ∠2 + ∠3 = 90°
⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ ∠CAB + ∠ABD = 180°
But, ∠CAB and ∠ABD are consecutive interior angles formed, when the transversal AB cuts CA and BD.
∴ CA ∥ BD
In the figure given below, state which lines are parallel and why?
In the given figure,
∠BAC = ∠ACD = 110°
But, these are alternate angles when transversal AC cuts AB and CD.
Hence, AB ∥ CD.
Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.
Let the two parallel lines be m and n.
Let p ⊥ m.
⇒ ∠1 = 90°
Let q ⊥ n.
⇒ ∠2 = 90°
Now, m ∥ n and p is a transversal.
⇒ ∠1 = ∠3 (corresponding angles)
⇒ ∠3 = 90°
⇒ ∠3 = ∠2 (each 90°)
But, these are corresponding angles, when transversal n cuts lines p and q.
∴ p ∥ q.
Hence, two lines which are perpendicular to two parallel lines, are parallel to each other.
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