RD Chapter 8- Lines and Angles Ex-8.1 |
RD Chapter 8- Lines and Angles Ex-8.2 |
RD Chapter 8- Lines and Angles Ex-8.3 |
RD Chapter 8- Lines and Angles Ex-VSAQS |

**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= 108^{0} and ∠2 = 2x= 72^{0}

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°

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°.

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°

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.

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°

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°

**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

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°

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.

**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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