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RD Chapter 8- Lines and Angles Ex-8.4 Interview Questions Answers

Question 1 : In figure, AB, CD and 1 and 2 are in the ratio 3 : 2.Determine all angles from 1 to 8.

Answer 1 :

Let 1 = 3xand 2 = 2x

From figure: 1 and 2 are linear pair of angles

Therefore, 1 + 2 = 180

3x + 2x = 180

5x = 180

x = 180 / 5

=> x = 36

So, 1 = 3x= 1080 and 2 = 2x= 720

As we know, vertically opposite angles are equal.

Pairs of vertically opposite angles are:

(1 = 3); (2 = 4) ; (5, 7) and (6 , 8)

1 = 3 = 108°

2 = 4 = 72°

5 = 7

6 = 8

We also know, if a transversal intersects any parallel lines,then the corresponding angles are equal

1 = 5 = 7 = 108°

2 = 6 = 8 = 72°

Answer: 1 =108°, 2 =72°, 3 =108°, 4 =72°, 5 =108°, 6 =72°, 7 =108° and 8 = 72°

Question 2 :

In figure, I, m andn are parallel lines intersected by transversal p at X, Y and Z respectively.Find 1, 2 and 3.

Answer 2 :

From figure:

Y =120° [Vertical opposite angles]

3 + Y = 180° [Linear pair angles theorem]

=> 3= 180– 120

=> 3= 60°

Line l is parallel to line m,

1 = 3 [ Corresponding angles]

1 = 60°

Also, line m is parallel to line n,

2 = Y [Alternate interior angles are equal]

2 =120°

Answer: 1 =60°, 2 =120° and 3 =60°.

Question 3 :

In figure, AB || CD|| EF and GH || KL. Find HKL.

Answer 3 :

Extend LK to meet line GF at point P.

From figure, CD || GF, so, alternate angles are equal.

CHG =HGP = 60°

HGP =KPF = 60° [Corresponding angles of parallellines are equal]

Hence, KPG=180 – 60 = 120°

=> GPK = AKL= 120° [Corresponding angles of parallellines are equal]

AKH = KHD = 25° [alternate angles of parallellines]

Therefore, HKL = AKH + AKL = 25 + 120 = 145°

Question 4 :

In figure, showthat AB || EF.

Answer 4 :

Produce EF to intersect AC at point N.

From figure, BAC =57° and

ACD =22°+35° = 57°

Alternative angles of parallel lines are equal

=> BA || EF …..(1)

Sum of Co-interior angles of parallel lines is 180°

EF || CD

DCE + CEF = 35 + 145 = 180° …(2)

From (1) and (2)

AB || EF

[Since, Lines parallel tothe same line are parallel to each other]

Hence Proved.

Question 5 :

In figure, if AB ||CD and CD || EF, find ACE.

Answer 5 :

Given: CD || EF

FEC + ECD = 180°

[Sum of co-interior anglesis supplementary to each other]

=> ECD =180° – 130° = 50°

Also, BA || CD

=> BAC = ACD = 70°

[Alternative angles ofparallel lines are equal]

But, ACE + ECD =70°

=> ACE =70° — 50° = 20°

Question 6 :

In figure, PQ || ABand PR || BC. If QPR = 102°, determine ABC. Give reasons.

Answer 6 :

Extend line AB to meet line PR at point G.

Given: PQ || AB,

QPR = BGR =102°

[Corresponding angles ofparallel lines are equal]

And PR || BC,

RGB+ CBG =180°

[Corresponding angles aresupplementary]

CBG =180° – 102° = 78°

Since, CBG = ABC

=>ABC =78°

Question 7 : In figure, state which lines are parallel and why?

Answer 7 :

We know, If a transversal intersects two lines such that a pairof alternate interior angles are equal, then the two lines are parallel

From figure:

=> EDC = DCA = 100°

Lines DE and AC are intersected by a transversal DC such thatthe pair of alternate angles are equal.

So, DE || AC

Question 8 :

In figure, if l||m,n || p and 1 = 85°, find 2.

Answer 8 :

Given: 1 = 85°

As we know, when a line cuts the parallel lines, the pair ofalternate interior angles are equal.

=> 1 = 3 = 85°

Again, co-interior angles are supplementary, so

2 + 3 = 180°

2 + 55°=180°

2 =180° – 85°

2 = 95°

Question 9 :

If two straightlines are perpendicular to the same line, prove that they are parallel to eachother.


Answer 9 :

Let lines l and m are perpendicular to n, then

1= 2=90°

Since, lines l and m cut by a transversal line n and thecorresponding angles are equal, which shows that, line l is parallel to line m.

Question 10 :

Prove that if thetwo arms of an angle are perpendicular to the two arms of another angle, thenthe angles are either equal or supplementary.

Answer 10 :

Let the angles be ACB and ABD

Let AC perpendicular to AB, and CD is perpendicular to BD.

To Prove : ACD = ABD OR ACD + ABD=180°

Proof :

In a quadrilateral,

A+ C+ D+ B =360°

[ Sum of angles ofquadrilateral is 360° ]

=> 180° + C + B = 360°

=> C + B = 360° –180°

Therefore, ACD + ABD = 180°

And ABD = ACD = 90°

Hence, angles are equal as well as supplementary.


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RD Chapter 8- Lines and Angles Ex-8.4 Contributors

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