**Telangana SCERT Solution Class IX (9) Math Chapter 4 Lines and Angles Exercise 4.3**

**(1)**

(i) l ∥ m, the question itself suggest that l ∥ m.

(ii) ∠1 = ∠5 [∵Corresponding angles]

(iii) ∠5 + ∠8 = 180 [∵supplementary angle]

(iv) ∠1 + ∠8 = 180^{o}

∠1 = ∠5 [∵Corresponding angle]

∠5 + ∠8 = 180^{o} [∵supplementary angle]

∴ ∠1 + ∠8 = 180^{o}

v) <1 is supplement of <8 because sum of <1 and <8 as shown above produces 180^{o}

**(2) In the adjacent figure AB || CD; CD || EF and y : z = 3 : 7, find x.**

Given, AB ∥ CD; CD ∥ EF

AB ∥ EF

Let, y = 3a

Z = 7a

Y : Z = 3 : 7

∠K = 180^{o} – ∠7 [∵supplementary angle]

= 180^{o} – 7a

Now, ∠K = ∠Y [∵Corresponding angle]

180^{o} – 7a = 3a [putting the value of ∠K and ∠Y]

Or, 180^{o} = 10a

Or, a = 180^{0}/10

= 18^{o}

Y = 3a

= 3 x 18

= 54^{o}

Now,

∠X + ∠Y = 180^{o} [∵co interior angles]

∴ ∠x = 180^{o} – Y

= 180^{o} – 54

= 126^{o }

**(3) In the adjacent figure AB || CD, EF ⊥ CD and ∠DEG = 126°, find ∠AGE, ∠FEG and ∠EGF**

Given,

AB ∥ CD, EF ⊥ CD

∠DEG = 126^{o}

∠FEG = ∠DEG – 90^{o }[EF ⊥ CD]

= 126^{o} – 90^{o}

= 36^{o}

∠AGE = ∠DEG [∵ alternate angle]

Or, ∠AGE = 126^{o}

<EGF = <CEG [∵ alternate angle]

∠CEG + ∠DEG = 180^{o} [∵supplementary angle]

Or, ∠CEG = 180^{o} – 126^{o}

= 54^{o}

∠EGF = ∠CEG = 54^{o}

**(4) In the adjacent figure PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠SRQ.**

Given, PQ ∥ ST

∠PQR = 110^{o}, ∠RST = 130^{o}

Draw a line parallel to ST through point R

MN ∥ ST

∠MRQ + ∠PQR = 180^{o} [Co interior angle]

Or, ∠MRQ + 110^{o} = 180^{o}

Or, ∠MRQ = 180^{o} – 110^{o}

= 70^{o}

∠NRS + ∠RST = 180^{o} [Co – interior angle]

Or, ∠NRS + 130^{o} = 130^{o}

Or, ∠NRS = 180^{o} – 130^{o}

= 50^{o}

∠SRQ + ∠NRS + ∠MRQ = 180^{o}

Or, ∠SRQ + 50^{o} + 70^{o} = 180^{o}

Or, ∠SRQ = 180^{o} – 120^{o}

∠SRQ = 60^{o}

**(5) In the adjacent figure m || n. A, B are any two points on m and n respectively. Let ‘C’ be an interior, point between the lines m and n. Find ∠ACB.**

Given,

M ∥ N

Draw a line K parallel to m and n passing through point c.

∠X = ∠BCK [alternate angles]

∠Y = ∠ACK [alternate angles]

Now, ∠BCK + ∠ACK = Z

∴ Z = X + Y [putting the value of ∠BCK and ∠ACK]

**(6) Find the value of a and b, given that p || q and r || s.**

Given,

p ∥ q, r ∥ s

2a = 80 [Corresponding angles]

Or, a = 80^{o}/2

= 40^{o}

b + 80^{o} = 180 [Co-interior angles]

Or, b = 180^{o} – 80^{o}

= 100^{o}

**(7) If in the figure a || b and c || d, then name the angles that are congruent to (i) ∠1 (ii) ∠2.**

Given,

a ∥ b, c ∥ d

(i) ∠1 = ∠3 [∵vertically opposite angles]

∠1 = ∠5 [∵corresponding angles]

∠1 = ∠5 = ∠7 [∵vertically opposite angles]

∠1 = ∠9 [∵corresponding angles]

∠1 = ∠11 [∵alternate angles]

∠5 = ∠15 [∵alternate angles]

∠5 = ∠13 [∵corresponding angles]

Therefore, ∠1 = ∠3, ∠5, ∠7, ∠9, ∠11, ∠13, ∠15

These are all congruent angles of angle 1

(ii) ∠2 = ∠4 [vertically opposite angles]

∠2 = ∠6 [∵corresponding angles]

∠6 = ∠8 [∵vertically opposite angles]

∠2 = ∠10 [∵corresponding angles]

∠6 = ∠14 [∵corresponding angles]

∠10 = ∠12 [∵vertically opposite angles]

∠14 = ∠16 [∵vertically opposite angles]

∴ ∠2 = ∠4, ∠6, ∠8, ∠10, ∠12, ∠14, ∠16

There are all congruent angles of ∠2

**(8) In the figure the arrow head segments are parallel. find the value of x and y.**

Given, The arrow heats segments are parallel

∴ Y^{o} = 59^{o} [alternate angles]

X^{o} = 60^{o} [corresponding angles]

**(9) In the figure the arrow head segments are parallel then find the value of x and y**

Given, segments with arrow heeds are parallel

∴ Name the points on the figure.

∴ ∠DOE = ∠COF = 105^{0} [vertically opposite angles]

∴ y^{o} + ∠DOE + 35^{o} = 180^{o} [∵supplementary angles]

y^{o} = 180^{o} – 35^{o} – 105^{o}

= 40

Now, y^{o} = x^{o} [∵corresponding angles]

∴ x = 40^{o}

**(10) Find the value of x and y from the figure.**

From the figure considering the segments with arrow heads are parallel to each other.

Naming ray OA and segment OB

∴ <AOB = 180^{o} – 120^{o} [∵supplementary angles]

= 60^{o}

∴ X^{o} = AOB = 60^{o} [corresponding angles]

X = 3y + 6 = 60^{o} [corresponding angle]

Or, 3y = 60 – 6

Or, y = 54/3

Or, y = 18^{o}

**(11) From the figure find x and y.**

Considering

The arrow head segments are parallel

X + 65^{o} + 52^{o} = 180^{o} [∵co interior angles]

Or, x + 117 = 180^{o}

Or, x = 180 – 117

Or, x = 63^{o}

3y + 5 + 52^{o} + 90^{o} = 180 [∴ sum of interior angles of a triangle is 180^{o}]

Or, 3y + 147^{o} = 180^{o}

Or, 3y = 180^{o} – 147^{o}

Or, y = 33/3

= 11^{o}

**(12) Draw figures for the following statement. “If the two arms of one angle are respectively perpendicular to the two arms of another angle then the two angles are either equal or supplementary”.**

**(13) In the given figure, if AB || CD, ∠ APQ = 50° and ∠PRD = 127°, find x and y.**

**(15) In the figures given below AB || CD. EF is the transversal intersecting AB and CD at G and H**

**respectively. Find the values of x and y. Give reasons.**

**(16) In the adjacent figure, AB || CD, ‘t’ is a transversal intersecting E and F respectively. If ∠2 : ∠1 = 5 : 4, find the measure of each marked angles.**

**(17) In the adjacent figure AB || CD. Find the value of angles x°, y° and z°.**

**(18) In the adjacent figure AB || CD. Find the values of x, y and z**

**(19) In each of the following figures AB || CD. Find the value of x in each case.**

Thank u so much

Thank you for clearing our douts

Thank you for clearing our douts

Thank you for clearing our doubts

It’s good