Geometry: Lines and Angles

(1)

0 | P a g e

Windows User

9811291604 8860116040

Geometry:

Lines and Angles

(2)

1 | P a g e

Point: A point is a dot made by a sharp pen or pencil. It is represented by capital letter.

Line: A straight and endless path on both the directions is called a line.

Line segment: A line segment is a straight path between two points.

Ray: A ray is a straight path which goes forever in one direction.

Collinear points: If three or more than three points lie on the same line, then they are called collinear points.

Non-collinear points: If three or more than three points does not lie on the same line, then they are called non-collinear points.

Angle: The space between two straight lines that diverge from a common point or between two planes that extend from a common line.

Types of Angles

(3)

2 | P a g e

1. Acute angle: An angle between 0° and 90° is called acute angle.

2. Right angle: An angle which is equal to 90° is called right angle.

3. Obtuse angle: An angle which is more than 90° but less than 180° is called obtuse angle.

4. Straight angle: An angle whose measure is 180° is called straight angle.

5. Reflex angle: An angle whose measure is between 180° and 360° is called reflex angle.

(4)

3 | P a g e

6. Complete angle: An angle which is equal to 360° is called complete angle

Pairs of Angles

1.Complementary angles: Two angles are said to be complementary if the sum of their degree measure is 90°.

For example, pair of complementary angles are 35° and 55°.

2. Supplementary angles: Two angles are said to be supplementary if the sum of their degree measure is 180°.

∠AOC + ∠BOC = 180°

3. Bisector of angle: A ray which divides an angle into two equal parts is called bisector of the angle.

∠AOC = ∠BOC

4. Adjacent angles: Two angles are said to be adjacent angles if

(5)

4 | P a g e

 They have a common vertex (O)

 They have a common arm (OC)

 and their non-common arms are on either side of common arm (OA and OB).

∠AOB = ∠AOC +∠BOC

5. Linear pair: Two adjacent angles are said to be linear pair if their sum is equal to 180°.

∠AOC + ∠BOC = 180°

Axiom 6.1: If a ray stands on a line, then the sum of two adjacent angles so formed is 180°.

Axiom 6.2: If the sum of two adjacent angles is 180°, then the non-common arms of the angles form a line.

6. Vertically opposite angles: Vertically opposite angles are those angles which are opposite to each other (or not adjacent) when two lines cross each other.

Theorem 6.1: If two lines intersect each other, then the vertically opposite angles are equal.

To prove: If lines AB and CD mutually intersect at point O, then (a) ∠AOC = ∠BOD (Vertically opposite angles)

(b) ∠AOD = ∠BOC

Proof: Lines AB intersect CD at O.

∠1 + ∠2 = 180° (Linear pair)

(6)

5 | P a g e

∠2 + ∠3 = 180° (Linear pair)

From eqn. (1) and (2), ∠1 + ∠2 = ∠2 + ∠3

⇒ ∠1 = ∠3 ⇒ ∠AOD = ∠BOC Similarly, ∠AOC = ∠BOD Parallel Lines

If distance between two lines is the same at each and every point on two lines, then two lines are said to be parallel.

If lines l and m do not intersect each other at any point then l || m.

Transversal line: A line is said to be transversal which intersect two or more lines at distinct points.

1. Corresponding angles: Pair of angles having different vertex but lying on same side of the transversal are called corresponding angles. Note that in each pair one is interior and other is exterior angle.

 ∠1 and ∠2

 ∠3 and ∠4

 ∠5 and ∠6

 ∠1 and ∠8

(7)

6 | P a g e

These angles are pair of corresponding angles.

2. Alternate interior angles: Pair of angles having distinct vertices and lying can either side of the transversal are called alternate interior angles.

 ∠1 and ∠2

 ∠3 and ∠4

These angles are alternate interior angles

3. Consecutive interior angles: Pair of interior angles of same side of transversal line.

 ∠1 and ∠2

 ∠2 and ∠4

These angles are consecutive interior angles or co-interior angles

(8)

7 | P a g e

Axiom 6.3: If two parallel lines are intersected by a transversal then each pair of corresponding angles are equal.

If AB || CD, then

 ∠PEB = ∠EFD

 ∠PEA = ∠EFC

 ∠BEF = ∠DFQ

 ∠AEF = ∠CFQ

Theorem 6.2: If two parallel lines are intersected by a transversal then pair of alternate interior angles are equal.

If AB || CD, then ?

 ∠AEF = ∠EFD

 ∠BEF = ∠CFE

(9)

8 | P a g e

Theorem 6.3: If two parallel lines are intersected by a transversal then the ! sum of consecutive interior angles of same side of transversal is equal to 180°. If AB || CD then (i) ∠BEF + ∠DFE = 180°

(ii) ∠AEF + ∠CFE = 180°

Axiom 6.4: If two lines are intersected by a transversal and a pair of corresponding angles are equal, then two lines are parallel.

(i) If ∠PEB = ∠EFD (corresponding angles), then AB || CD

Theorem 6.4: If two lines intersected by a transversal and a pair of alternate interior angles are equal, then two lines are parallel. If ∠AEF = ∠EFD (alternate interior angles),

(10)

9 | P a g e then AB || CD.

Theorem 6.5: If two lines are intersected by a transversal and the sum of consecutive interior angles of same side of transversal is equal to 180°, the lines are parallel. If

∠AEF + ∠CFE = 180°, then AB || CD.

Theorem 6.6: Lines which are parallel to the same line are parallel to each other.

If AB || EF and CD || EF then AB || CD

Theorem 6.7: The sum of the angles of a triangle is equal to 180°.

Given: ΔABC

To prove: ∠A + ∠B + ∠C = 180°

Construction: Draw DE || BC Proof: DE || BC

then ∠1 = ∠4 …(1) (alternate interior angles)

∠2 = ∠5 …(2) (alternate interior angles) Adding equations (1) and (2),

∠1 + ∠2 = ∠4 +∠5

Adding ∠3 on both sides,

(11)

10 | P a g e

∠1 +∠2 + ∠3 = ∠3 + ∠4 + ∠5

⇒ ∠A + ∠B + ∠C = 180° (Sum of angles at a point on same side of a line is 180°) Theorem 6.8: If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

Given: AABC in which, side BC is produced to D.

To Prove: ∠ACD = ∠BAC + ∠ABC

Proof: ∠ACD + ∠ACB = 180° …(1) (Linear pair)

∠ABC + ∠ACB + ∠BAC = 180° …(2) From eqn. (1) and (2), ∠ACD + ∠ACB

= ∠ABC + ∠ACB + ∠BAC

= ∠ACD = ∠ABC + ∠BAC

NCERT Solutions for Class 9 Maths Chapter 6 Ex 6.1

Ex 6.1 Class 9 Maths Question 1

In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Solution:

Since AB is a straight line,

∴ ∠AOC + ∠COE + ∠EOB = 180°

or (∠AOC + ∠BOE) + ∠COE = 180° or 70° + ∠COE = 180° [ ∵∠AOC + ∠BOE = 70° (Given)]

or ∠COE = 180° – 70° = 110°

∴ Reflex ∠COE = 360° – 110° = 250°

(12)

11 | P a g e

Also, AB and CD intersect at O.

∴∠COA = ∠BOD [Vertically opposite angles]

But ∠BOD = 40° [Given]

∴ ∠COA = 40°

Also, ∠AOC + ∠BOE = 70°

∴ 40° + ∠BOE = 70° or ∠BOE = 70° -40° = 30°

Thus, ∠BOE = 30° and reflex ∠COE = 250°.

Ex 6.1 Class 9 Maths Question 2.

In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c.

Solution:

Since XOY is a straight line.

∴ b+a+∠POY= 180°

But ∠POY = 90° [Given]

∴ b + a = 180° – 90° = 90° …(i) Also a : b = 2 : 3 ⇒ b = 3a2 …(ii) Now from (i) and (ii), we get

3a2 + A = 90°

⇒ 5a2 = 90°

⇒ a = 90∘5

×2=36

^∘^{= 36°}

From (ii), we get b = 32 x 36° = 54°

Since XY and MN interstect at O,

∴ c = [a + ∠POY] [Vertically opposite angles]

or c = 36° + 90° = 126°

Thus, the required measure of c = 126°.

In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Solution:

(13)

12 | P a g e

ST is a straight line.

∴ ∠PQR + ∠PQS = 180° …(1) [Linear pair]

Similarly, ∠PRT + ∠PRQ = 180° …(2) [Linear Pair]

From (1) and (2), we have

∠PQS + ∠PQR = ∠PRT + ∠PRQ But ∠PQR = ∠PRQ [Given]

∴ ∠PQS = ∠PRT

In figure, if x + y = w + ⇒, then prove that AOB is a line.

Solution:

Sum of all the angles at a point = 360°

∴ x + y + ⇒ + w = 360° or, (x + y) + (⇒ + w) = 360°

But (x + y) = (⇒ + w) [Given]

∴ (x + y) + (x + y) = 360° or, 2(x + y) = 360°

or, (x + y) = 360∘2 = 180°

∴ AOB is a straight line.

In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

Solution:

rara POQ is a straight line. [Given]

∴ ∠POS + ∠ROS + ∠ROQ = 180°

But OR ⊥ PQ

∴ ∠ROQ = 90°

⇒ ∠POS + ∠ROS + 90° = 180°

(14)

13 | P a g e

⇒ ∠POS + ∠ROS = 90°

⇒ ∠ROS = 90° – ∠POS … (1)

Now, we have ∠ROS + ∠ROQ = ∠QOS

⇒ ∠ROS + 90° = ∠QOS

⇒ ∠ROS = ∠QOS – 90° ……(2) Adding (1) and (2), we have 2 ∠ROS = (∠QOS – ∠POS)

∴ ∠ROS = 12

(∠QOS−∠POS)

It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.

Solution:

XYP is a straight line.

∴ ∠XYZ + ∠ZYQ + ∠QYP = 180°

⇒ 64° + ∠ZYQ + ∠QYP = 180°

[∵ ∠XYZ = 64° (given)]

⇒ 64° + 2∠QYP = 180°

[YQ bisects ∠ZYP so, ∠QYP = ∠ZYQ]

⇒ 2∠QYP = 180° – 64° = 116°

⇒ ∠QYP = 116∘2 = 58°

∴ Reflex ∠QYP = 360° – 58° = 302°

Since ∠XYQ = ∠XYZ + ∠ZYQ

⇒ ∠XYQ = 64° + ∠QYP [∵∠XYZ = 64°(Given) and ∠ZYQ = ∠QYP]

⇒ ∠XYQ = 64° + 58° = 122° [∠QYP = 58°]

Thus, ∠XYQ = 122° and reflex ∠QYP = 302°.

Ex 6.2

In figure, find the values of x and y and then show that AB || CD.

(15)

14 | P a g e Solution:

In the figure, we have CD and PQ intersect at F.

∴ y = 130° …(1)

[Vertically opposite angles]

Again, PQ is a straight line and EA stands on it.

∠AEP + ∠AEQ = 180° [Linear pair]

or 50° + x = 180°

⇒ x = 180° – 50° = 130° …(2) From (1) and (2), x = y

As they are pair of alternate interior angles.

∴ AB || CD

In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Solution:

(16)

15 | P a g e

AB || CD, and CD || EF [Given]

∴ AB || EF

∴ x = z [Alternate interior angles] ….(1) Again, AB || CD

⇒ x + y = 180° [Co-interior angles]

⇒ z + y = 180° … (2) [By (1)]

But y : z = 3 : 7

z = 73 y = 73(180°- z) [By (2)]

⇒ 10z = 7 x 180°

⇒ z = 7 x 180° /10 = 126°

From (1) and (3), we have x = 126°.

In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Solution:

AB || CD and GE is a transversal.

∴ ∠AGE = ∠GED [Alternate interior angles]

But ∠GED = 126° [Given]

∴∠AGE = 126°

Also, ∠GEF + ∠FED = ∠GED

or ∠GEF + 90° = 126° [∵ EF ⊥ CD (given)]

x = z [Alternate interior angles]… (1) Again, AB || CD

⇒ x + y = 180° [Co-interior angles]

∠GEF = 126° -90° = 36°

Now, AB || CD and GE is a transversal.

∴ ∠FGE + ∠GED = 180° [Co-interior angles]

or ∠FGE + 126° = 180°

or ∠FGE = 180° – 126° = 54°

Thus, ∠AGE = 126°, ∠GEF=36° and ∠FGE = 54°.

In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS.

Solution:

(17)

16 | P a g e

Draw a line EF parallel to ST through R.

Since PQ || ST [Given]

and EF || ST [Construction]

∴ PQ || EF and QR is a transversal

⇒ ∠PQR = ∠QRF [Alternate interior angles] But ∠PQR = 110° [Given]

∴∠QRF = ∠QRS + ∠SRF = 110° …(1) Again ST || EF and RS is a transversal

∴ ∠RST + ∠SRF = 180° [Co-interior angles] or 130° + ∠SRF = 180°

⇒ ∠SRF = 180° – 130° = 50°

Now, from (1), we have ∠QRS + 50° = 110°

⇒ ∠QRS = 110° – 50° = 60°

Thus, ∠QRS = 60°.

In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Solution:

We have AB || CD and PQ is a transversal.

∴ ∠APQ = ∠PQR

[Alternate interior angles]

⇒ 50° = x [ ∵ ∠APQ = 50° (given)]

Again, AB || CD and PR is a transversal.

∴ ∠APR = ∠PRD [Alternate interior angles]

⇒ ∠APR = 127° [ ∵ ∠PRD = 127° (given)]

⇒ ∠APQ + ∠QPR = 127°

⇒ 50° + y = 127° [ ∵ ∠APQ = 50° (given)]

⇒ y = 127°- 50° = 77°

Thus, x = 50° and y = 77°.

In figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes

(18)

17 | P a g e

the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

Solution:

Draw ray BL ⊥PQ and CM ⊥ RS

∵ PQ || RS ⇒ BL || CM [∵ BL || PQ and CM || RS]

Now, BL || CM and BC is a transversal.

∴ ∠LBC = ∠MCB …(1) [Alternate interior angles]

Since, angle of incidence = Angle of reflection

∠ABL = ∠LBC and ∠MCB = ∠MCD

⇒ ∠ABL = ∠MCD …(2) [By (1)]

Adding (1) and (2), we get

∠LBC + ∠ABL = ∠MCB + ∠MCD

⇒ ∠ABC = ∠BCD

i. e., a pair of alternate interior angles are equal.

∴ AB || CD.

Ex 6.3

In figure, sides QP and RQ of ∆PQR are produced to points S and T, respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

(19)

We have, ∠TQP + ∠PQR = 180°

[Linear pair]

⇒ 110° + ∠PQR = 180°

⇒ ∠PQR = 180° – 110° = 70°

Since, the side QP of ∆PQR is produced to S.

⇒ ∠PQR + ∠PRQ = 135°

[Exterior angle property of a triangle]

⇒ 70° + ∠PRQ = 135° [∠PQR = 70°]

⇒ ∠PRQ = 135° – 70° ⇒ ∠PRQ = 65°

In figure, ∠X = 62°, ∠XYZ = 54°, if YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ.

Solution:

In ∆XYZ, we have ∠XYZ + ∠YZX + ∠ZXY = 180°

[Angle sum property of a triangle]

But ∠XYZ = 54° and ∠ZXY = 62°

∴ 54° + ∠YZX + 62° = 180°

⇒ ∠YZX = 180° – 54° – 62° = 64°

YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively.

∴ ∠OYZ = 12

∠XYZ

⁼¹²(54°) = 27°

and ∠OZY = 12

∠YZX

= 12(64°) = 32°

Now, in ∆OYZ, we have

∠YOZ + ∠OYZ + ∠OZY = 180°

⇒ ∠YOZ + 27° + 32° = 180°

(20)

19 | P a g e

⇒ ∠YOZ = 180° -27° – 32° = 121°

Thus, ∠OZY = 32° and ∠YOZ = 121°

In figure, if AB || DE, ∠BAC = 35° and ∠CDE = 539 , find ∠DCE.

Solution:

AB || DE and AE is a transversal.

So, ∠BAC = ∠AED [Alternate interior angles]

and ∠BAC = 35° [Given]

∴ ∠AED = 35°

Now, in ∆CDE, we have ∠CDE + ∠DEC + ∠DCE = 180°

{Angle sum property of a triangle]

∴ 53° + 35° + ∠DCE =180°

[∵ ∠DEC = ∠AED = 35° and∠CDE = 53° (Given)]

⇒ ∠DCE = 180° – 53° – 35° = 92°

Thus, ∠DCE = 92°

In figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠TSQ

= 75°, find ∠ SQT.

Solution:

In ∆PRT, we have ∠P + ∠R + ∠PTR = 180°

⇒ 95° + 40° + ∠PTR =180°

[ ∵ ∠P = 95°, ∠R = 40° (given)]

⇒ ∠PTR = 180° – 95° – 40° = 45°

But PQ and RS intersect at T.

∴ ∠PTR = ∠QTS

(21)

20 | P a g e

[Vertically opposite angles]

∴ ∠QTS = 45° [ ∵ ∠PTR = 45°]

Now, in ∆ TQS, we have ∠TSQ + ∠STQ + ∠SQT = 180°

∴ 75° + 45° + ∠SQT = 180° [ ∵ ∠TSQ = 75° and ∠STQ = 45°]

⇒ ∠SQT = 180° – 75° – 45° = 60°

Thus, ∠SQT = 60°

In figure, if PQ ⊥ PS, PQ||SR, ∠SQR = 2S° and ∠QRT = 65°, then find the values of x and y.

Solution:

In ∆ QRS, the side SR is produced to T.

∴ ∠QRT = ∠RQS + ∠RSQ

But ∠RQS = 28° and ∠QRT = 65°

So, 28° + ∠RSQ = 65°

⇒ ∠RSQ = 65° – 28° = 37°

Since, PQ || SR and QS is a transversal.

∴ ∠PQS = ∠RSQ = 37°

[Alternate interior angles]

⇒ x = 37°

Again, PQ ⊥ PS ⇒ AP = 90°

Now, in ∆PQS,

we have ∠P + ∠PQS + ∠PSQ = 180°

⇒ 90° + 37° + y = 180°

⇒ y = 180° – 90° – 37° = 53°

Thus, x = 37° and y = 53°

In figure, the side QR of ∆PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that

(22)

In ∆PQR, side QR is produced to S, so by exterior angle property,

∠PRS = ∠P + ∠PQR

⇒ 12∠PRS = 12∠P + 12∠PQR

⇒ ∠TRS = 12∠P + ∠TQR …(1)

[∵ QT and RT are bisectors of ∠PQR and ∠PRS respectively.]

Now, in ∆QRT, we have

∠TRS = ∠TQR + ∠T …(2)

From (1) and (2),

we have ∠TQR + 12∠P = ∠TQR + ∠T

⇒ 12∠P = ∠T

⇒ 12∠QPR = ∠QTR or ∠QTR = 12∠QPR

NCERT Exemplar

Exercise 6.1: Multiple Choice Questions (MCQs)

Question 1:

In figure, if AB || CD || EE, PQ || RS, ∠RQD = 25° and ∠CQP = 60°, then ∠QRS is equal to

(a) 85° (b)135° (c)145° (d) 110°

Solution:

(c) Given, PQ || RS

∠PQC = ∠BRS = 60° [alternate exterior angles and ∠PQC = 60° (given)] and ∠DQR =

(23)

22 | P a g e

∠QRA = 25° [alternate interior angles]

[∠DQR = 25°, given]

∠QRS = ∠QRA + ∠ARS

= ∠QRA + (180° – ∠BRS) [linear pair axiom]

= 25° + 180° – 60°= 205° – 60°= 145°

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 (b) an obtuse triangle

(c) an equilateral triangle (d) a right triangle Solution:

(d) Let the angles of a AABC be ∠A, ∠B and ∠C.

Given, ∠A = ∠B+∠C …(i)

InMBC, ∠A+ ∠B+ ∠C-180° [sum of all angles of atriangle is 180°]…(ii) From Eqs. (i) and (ii),

∠A+∠A = 180° => 2 ∠A = 180°

=> 180° /2

∠A = 90°

Hence, the triangle is a right triangle.

Question 3:

An exterior angle of a triangle is 105° and its two interior opposite angles are equal.

Each of these equal angles is

(a) 37 ½° (b) 52 ½° (c)72 ½° (d) 75°

Solution:

Let one of interior angle be x°.

∴ Sum of two opposite interior angles = Exterior angle

∴ x° + x° = 105°

2x° = 105°

x° = 105°/2 x°=52 ½°

Hence, each angle of a triangle is 52 ½°.

Question 4:

If the angles of a triangle are in the ratio 5:3:7, then the triangle is (a) an acute angled triangle

(b) an obtuse angled triangle (c) a right angled triangle (d) an isosceles triangle Solution:

(a) Given, the ratio of angles of a triangle is 5 : 3 : 7.

Let angles of a triangle be ∠A,∠B and ∠C.

Then, ∠A = 5x, ∠B = 3x and ∠C = 7x

In ΔABC, ∠A + ∠B + ∠C = 180° [since, sum of all angles of a triangle is 180°]

5x + 3x + 7x = 180°

(24)

23 | P a g e

=> 15x = 180°

x = 180°/15= 12°

∠A = 5x = 5 x 12° = 60°

∠B = 3x= 3 x 12°= 36°

and ∠C =7x = 7 x 12° = 84°

Since, all angles are less than 90°, hence the triangle is an acute angled triangle.

Question 5:

If one of the angles of a triangle is 130°, then the angle between the bisectors of the other two angles can be

(a) 50° (b) 65° (c) 145° (d) 155°

Solution:

(d) Let angles of a triangle be ∠A, ∠B and ∠C.

Question 6:

In the figure, POQ is a line. The value of x is

(25)

24 | P a g e

(a)20° (b)25° (c)30° (d) 35°

Thinking Process

When two or more rays are initiated from a same point of a line, then the sum of all angles made between the rays and line at the same point is 180°.

Solution:

Question 7:

In the figure, if OP || RS, ∠OPQ = 110° and ∠QRS = 130°, then ∠PQR is equal to

(a) 40° (b) 50° (c) 60° (d) 70°

(26)

25 | P a g e Solution:

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° (d) 20°

Thinking Process

Use the concept, the sum of all angles in a triangle is 180°. Further, simplify it and get the smallest angle.

Solution:

(b) Given, the ratio of angles of a triangle is 2 : 4 : 3.

Let the angles of a triangle be ∠A, ∠B and ∠C.

∠A = 2x, ∠B = 4x

∠C = 3x , ∠A+∠B+ ∠C= 180°

[sum of all the angles of a triangle is 180°]

2x + 4x + 3x = 180°

9x = 180°

x=180°/9 =20°

∠A=2x=2 x 20° = 40°

∠B = 4x = 4 x 20° = 80°

∠C = 3x = 3 x 20° = 60°

Hence, the smallest angle of a triangle is 40°.

Exercise 6.2: Very Short Answer Type Questions

(27)

26 | P a g e Question 1:

For what value of x + y in figure 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° i.e.,x + y = 180°.

Question 2:

Can a triangle have all angles less than 60°? Give reason for your answer.

Solution:

No, a triangle cannot have all angles less than 60°, because if all angles will be less than 60°, then their sum will not be equal to 180°. 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 the triangle have two obtuse angles i.e., more than 90° angle, then the sum of all three angles of a triangle will not be equal to 180°.

Question 4:

How many triangles can be drawn having its angles as 45°, 64° and 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, no triangle can be drawn with the given angles.

Question 5:

How many triangles can be drawn having its angles as 53°, 64° and 63°? Give reason for your answer.

Solution:

Infinitely many triangles,

The sum of given angles = 53° + 64° + 63° = 180°

Here, we see that sum of all interior angles of triangle is 180°, so infinitely many triangles can be drawn.

Question 6:

In the figure, find the value of x for which the lines l and m are parallel.

(28)

In the given figure, l || m and we know that, if a transversal intersects two parallel lines, then sum of interior angles on the same side of a transversal is supplementary. x + 44°

= 180°

x = 180°-44°

=> x = 136° . 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 l and m makes a one right angle, then it means that lines I and m are perpendicular each other. By using linear pair axiom aniom, other three angles will be a right angle.

Question 9:

In the figure, which of the two lines are parallel and why?

(29)

In Fig. (i) sum of two interior angles 132° + 48° = 180° [∴ equal to 180°]

Here, we see that the sum of two interior angles on the same side of n is 180°, then they are the parallel lines.

In Fig. (ii), the sum of two interior angles 73° + 106° = 179° ≠ 180°. Here, we see that the sum of two interior angles on same side of r is not equal to 180°, then they 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, since, lines l and m are perpendicular to the line n.

∠1 = ∠2 = 90° [∴ l ⊥ n and min]

It implies that these are corresponding angles.

Hence, l|| m.

Exercise 6.3: Short Answer Type Questions

Question 1:

In the figure, OD is the bisector of ∠AOC, OE is the bisector of ∠BOC and OD ⊥ OE.

Show that the points A, 0 and B are collinear.

Thinking Process

For showing collinearity of A, O and B, we have to show that ∠AOB =180°.

(30)

Given In the figure, OD ⊥ OE, OD and OE are the bisectors of ∠AOC and ∠BOC.

To show Points A, O and B are collinear i.e., AOB is a straight line.

Proof Since, OD and OE bisect angles ∠AOC and ∠BOC, respectively.

∠AOC =2 ∠DOC …(i) and ∠COB = 2 ∠COE …(ii)

On adding Eqs. (i) and (ii), we get

∠AOC + ∠COB = 2 ∠DOC +2 ∠COE => ∠AOC +∠COB = 2(∠DOC +∠COE)

=> ∠AOC + ∠COB= 2 ∠DOE

=> ∠AOC+ ∠COB = 2 x 90° [∴ OD ⊥ OE]

=> ∠AOC + ∠COB = 180°

∴ ∠AOB = 180°

So, ∠AOC and ∠COB are forming linear pair.

Also, AOB is a straight line.

Hence, points A, O and B are collinear.

Question 2:

In the figure, ∠1 = 60° and ∠6 = 120°. Show that the lines m and n are parallel.

Solution:

Given In the figure ∠1 = 60° and ∠6 = 120°

To show 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°

We know that, if the sum of two interior angles on same side of l is 180°, then lines are parallel.

Hence, m || 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 (in the given figure). Show that AP || BQ.

(31)

Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively.

To prove AP|| BQ

Proof Since, l || m and t is transversal.

Therefore, ∠EAB = ∠ABH [alternate interior angles]

½ ∠EAB =½ ∠ABH [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 with two lines AP and BQ and transversal AB. Hence, AP || BQ.

Question 4:

In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l ||m.

Solution:

Given, In the figure AP|| BQ, AP and BQ are the bisectors of alternate interior angles

∠CAB and ∠ABF.

To show l || m

(32)

31 | P a g e

Proof Since, AP|| BQ and t is transversal, therefore ∠PAB = ∠ABQ [alternate interior angles]

=> 2 ∠PAB = 2 ∠ABQ [multiplying both sides by 2]

So, alternate interior angles are equal.

We know that, if two alternate interior angles are equal, then lines are parallel. Hence, l

|| m.

Question 5:

In the figure, BA || ED and BC || EF. Show that ∠ABC = ∠DEF.

Solution:

Given BA || ED and BC || EF.

To show ∠ABC = ∠DEF.

Construction Draw a ray EP opposite to ray ED.

(33)

32 | P a g e Question 6:

In the figure, BA || ED and BC || EF. Show that ∠ABC + ∠DEF = 180°.

Solution:

Given BA || ED and BC || EF To show, ∠ABC + ∠DEF = 180°

Construction Draw a ray PE opposite to ray EF.

Question 7:

In the figure, DE || QR and AP and BP are bisectors of ∠EAB and ∠RBA, respectively.

Find ∠APB.

(34)

Question 8:

A ΔABC is right angled at A. L is a point on BC such that AL ⊥ BC. Prove that ∠BAL =

∠ACB.

Solution:

Given In ΔABC, ∠A = 90° and AL ⊥ BC To prove ∠BAL = ∠ACB

Proof In ΔABC and ΔLAC, ∠BAC = ∠ALC [each 90°] …(i)

and ∠ABC = ∠ABL [common angle] …(ii)

On adding Eqs. (i) and (ii), we get

∠BAC + ∠ABC = ∠ALC + ∠ABL …(iii) Again, in ΔABC,

∠BAC + ∠ACB + ∠ABC = 180°

[sum of all angles of a triangle is 180°] =>∠BAC+∠ABC = 1 80°-∠ACB …(iv) In ΔABL,

∠ABL + ∠ALB + ∠BAL = 180°

(35)

34 | P a g e

[sum of all angles of a triangle is 180°] => ∠ABL+ ∠ALC = 180° – ∠BAL [∴ ∠ALC =

∠ALB= 90°] …(v)

On substituting the value from Eqs. (iv) and (v) in Eq. (iii), we get 180° – ∠ACS = 180° –

∠SAL

=> ∠ACB = ∠BAL Hence proved.

Question 9:

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 respectively perpendicular to m and n.

i.e., p ⊥ m, p ⊥ n, q ⊥ m, q ⊥ n To prove p||g

Proof Since, m || n and p is perpendicular to m and n.

So, sum of two interior angles is supplementary.

We know that, if a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel.

Hence, p||g.

Exercise 6.4: Long Answer Type Questions

Question 1:

If two lines intersect prove that the vertically opposite angles are equal Solution:

(36)

35 | P a g e

Given Two lines AB and CD intersect at point O.

Question 2:

Bisectors of interior ∠B and exterior ∠ACD of a ΔABC intersect at the point T. Prove that ∠BTC = ½ ∠BAC.

Thinking Process

For obtaining the interior required result use the property that the exterior angle of a triangle is equal to the sum of the two opposite angles of triangle.

Solution:

Given In AABC, produce SC to D and the bisectors of ∠ABC and ∠ACD meet at point

(37)

36 | P a g e

T. To prove ∠BTC = ½ ∠BAC

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 AB and CD are parallel and intersected by transversal t at P and 0, respectively. Also, EP and FQ are the bisectors of angles ∠APG and ∠CQP,

(38)

37 | P a g e respectively.

Question 4:

Prove that through a given point, we can draw only one perpendicular to a given line.

Solution:

Given Consider a line l and a point P.

(39)

38 | P a g e Question 5:

Prove that two lines that are respectively perpendicular to two intersecting lines intersect each other.

Solution:

Given Let lines l and m are two intersecting lines. Again, let n and p be another two lines which are perpendicular to the intersecting lines meet at point D.

To prove Two lines n and p intersecting at a point.

Proof Suppose we consider lines n and p are not intersecting, then it means they are parallel to each other i.e., n || p …(i)

Since, lines n and pare perpendicular to m and l, respectively.

But from Eq. (i) n || p it implies that l || m.

Hence, it is a contradiction.

Thus, our assumption is wrong.

Therefore, lines n and p intersect at a point.

Question 6:

Prove that a triangle must have atleast two acute angles.

Solution:

Given ΔABC is a triangle.

To prove ΔABC must have two acute angles Proof Let us consider the following cases Case I When two angles are 90°.

(40)

39 | P a g e

Suppose two angles are ∠B = 90° and ∠C = 90°

Question 7:

In the given figure, ∠Q > ∠R, PA is the bisector of ∠QPR and PM ⊥ QR.

Prove that ∠APM = ½(∠Q – ∠R).

(41)