AP State Syllabus AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.4

Question 1.

In the given triangles, find out ∠x, ∠y and ∠z

Solution:

In fig(i)

x° = 50° + 60°

(∵ exterior angle is equal to sum of the opposite interior angles)

∴ x= 110°

In fig (ii)

z° = 60° + 70°

(∵ exterior angle is equal to sum of the opposite interior angles)

∴ z = 130°

In fig (iii)

y° = 35° + 45° = 80°

(∵ exterior angle is equal to sum of the opposite interior angles)

Question 2.

In the given figure AS // BT; ∠4 = ∠5, \(\overline{\mathbf{S B}}\) bisects ∠AST. Find the measure of ∠1.

Solution:

Given AS // BT

∠4 = ∠5 and SB bisects ∠AST.

∴ By problem

∠2 = ∠3 …………..(1)

For the lines AS // BT

∠2 = ∠5 ( ∵alternate interior angles)

∴ In ΔBST

∠3 = ∠5 = ∠4

Hence ΔBST is equilateral triangle and each of its angle is equal to 60°.

∴∠3 = ∠2 = 60° [by eq. (1)]

Now ∠1 + ∠2 + ∠3 = 180°

∠1 + 60° + 60° = 180°

[ ∵ angles at a point on a line]

∴∠1 = 180° – 120° = 60°

Question 3.

In the given figure AB // CD; BC // DE then find the values of x and y.

Solution:

Given that AB // CD and BC // DE.

∴ 3x = 105° (∵ alternate interior angles for AB // CD)

x = \(\frac { 105° }{ 3 }\) = 35°

Also BC // DE

∴∠D = 105°

(∵ alternate interior angles)

Now in ΔCDE

24° + 105° + y = 180°

(∵ angle sum property)

∴ y = 180° – 129° = 51°

Question 4.

In the given figure BE ⊥ DA and CD ⊥ DA then prove that m∠1 = m∠3.

Solution:

Given that CD ⊥ DA and BE ⊥ DA.

⇒ Two lines CD and BE are perpendicular to the same line DA.

⇒ CD // BE (or)

∠D =∠E ⇒ CD // BE

(∵ corresponding angles for CD and BE and DA are transversal)

Now m∠1 = m∠3

(∵alternate interior angles for the lines CD // BE ; DB are transversal)

Hence proved.

Question 5.

Find the values of x, y for which lines AD and BC become parallel.

Solution:

For the lines AD and BC to be parallel x – y = 30° (corresponding angles) ……… (1)

2x = 5y ………….(2)

(∵ alternate interior angles)

Solving (1) & (2)

y = \(\frac{60}{3}\) = 20°

Substituting y = 20° in eq. (1)

x – 20° = 30°

⇒ x = 50°

∴ x = 50° and y = 20°

Question 6.

Find the values of x and y in the figure.

Solution:

From the figure y + 140° = 180°

(∵ linear pair of angles)

∴ y = 180° – 140° = 40°

And x° = 30° + y°

(∵ exterior angle = sum of the opposite interior angles)

x° = 30° + 40° = 70°

Question 7.

In the given figure segments shown by arrow heads are parallel. Find the values of x and y.

Solution:

From the figure

x° = 30° (∵ alternate interior angles)

y° = 45° + x° (∵ exterior angles of a triangle = sum of opp. interior angles)

y = 45° + 30° – 75°

Question 8.

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

Solution:

Given that ∠SPR = 135° and ∠PQT =110°

From the figure

∠SPR + ∠RPQ = 180°

∠PQT + ∠PQR = 180°

[∵ linear pair of angles]

⇒ ∠RPQ = 180° – ∠SPR

= 180° – 135° = 45°

⇒ ∠PQR = 180° – ∠PQT

= 180°-110° = 70°

Now in APQR

∠RPQ + ∠PQR + ∠PRQ = 180°

[∵ angle sum property]

∴ 45° + ’70° + ∠PRQ = 180°

∴ ∠PRQ = 180°-115° = 65°

Question 9.

In the given figure ∠X = 62° ; ∠XYZ = 54°. In ΔXYZ. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respec-tively find ∠OZY and ∠YOZ.

Solution:

Given that ∠X = 62° and ∠Y = 54°

YO arid ZO are bisectors of ∠Y and ∠Z.

In ΔXYZ

∠X + ∠XYZ + ∠XZY = 180° .

62° + 54° + ∠XZY = 180°

=> ∠XZY = 180°- 116° = 64°

Also in Δ𝜏OYZ

∠OYZ = 1/2 ∠XYZ = 1/2 x 54° = 27°

(∵ YO is bisector of ∠XYZ)

∠OZY = 1/2 ∠XZY = 1/2 x 64° = 32

(∵ OZ is bisector of ∠XYZ)

And ∠OYZ + ∠OZY + ∠YOZ = 180°

(∵ angle sum property, ΔOYZ)

⇒ 27 + 32° + ∠YOZ = 180°

⇒ ∠YOZ = 180° – 59° = 121°

Question 10.

In the given figure if AB // DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Solution:

Given that AB // DE, ∠CDE = 53°;

∠BAC = 35°

Now ∠E = 35°

( ∵ alternate interior angles)

Now in ∆CDE

∠C + ∠D + ∠E = 180°

(∵angle sum property, ACDE)

∴ ∠DCE + 53° + 35° = 180°

⇒ ∠DCE = 180° – 88° = 92°

Question 11.

In the given figure if line segments PQ and RS intersect at point T, such that∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Solution:

Given that ∠PRT = 40°; ∠RPT = 95°;

∠TSQ = 75°

In ∆PRT ∠P + ∠R + ∠PTR = 180°

(∵angle sum property)

95° + 40° + ∠PTR = 180°

⇒ ∠PTR = 180° – 135° = 45°

Now ∠PTR = ∠STQ

(∵ vertically opposite angles)

In ΔSTQ ∠S + ∠Q + ∠STQ = 180°

(∵ angle sum property)

75° + ∠SQT + 45° = 180°

∴ ∠SQT = 180° – 120° = 60°

Question 12.

In the given figure, ABC is a triangle in which ∠B = 50° and ∠C = 70°. Sides AB and AC are produced. If ∠ is the measure of angle between the bisec¬tors of the exterior angles so formed, then find ‘z’.

Solution:

Given that ∠B = 50°; ∠C = 70°

Angle between bisectors of exterior angles B and C is ∠.

From the figure

50° + 2x = 180°

70° + 2y = 180°

(∵ linear pair of angles)

∴ 2x= 180°-50°

2x= 130°

x = \(\frac{130}{2}\)

= 65°

2y= 180°-70°

2y= 110°

x = \(\frac{110°}{2}\)

= 55°

Now in ΔBOC

x + y + ∠ = 180° (∵ angle sum property)

65° + 55° + ∠ = 180°

z = 180° -120° = 60°

Question 13.

In the given figure if PQ ⊥ PS; PQ // SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Solution:

Given that PQ ⊥ PS ; PQ // SR

∠SQR = 28°, ∠QRT = 65°

From the figure

∠QSR = x° (∵ alt. int. angles for the lines PQ // SR)

Also 65° = x + 28° (∵ ext. angles = sum of the opp. interior angles)

∴ x° = 65° – 28° = 37°

And x° + y° = 90°

[ ∵ PQ ⊥ PS and PQ // SR. ⇒ ∠P = ∠S]

37° + y = 90°

∴ y = 90° – 37° = 53°

Question 14.

In the given figure ΔABC side AC has been produced to D. ∠BCD = 125° and ∠A: ∠B = 2:3, find the measure of ∠A and ∠B

Solution:

Given that ∠BCD = 125°

∠A : ∠B = 2 : 3

Sum of the terms of the ratio

∠A : ∠B = 2 + 3 = 5

We know that ∠A + ∠B = ∠BCD

(∵ exterior angles of triangle is equal to sum of its opp. interior angles)

∴ ∠A = \(\frac{2}{5}\) x 125° = 50°

∠B = \(\frac{3}{5}\) x 125° = 75°

Question 15.

In the given figure, it is given that, BC // DE, ∠BAC = 35° and ∠BCE = 102°. Find the measure of 0 ∠BCA i0 ∠ADE and iii) ∠CED.

Solution:

Given that BC // DE ; ∠BAC = 35°;

∠BCE = 102°

i) From the figure

102° + ∠BCA = 180°

(∵ linear pair of angles)

∴ ∠BCA = 180° – 102° = 78°

ii) ∠ADE + ∠CBD = 180°

(∵ interior angles on the same side of the transversal)

∠ADE + (78° + 35°) = 180°

(∵ ∠CBD = ∠BAC + ∠BCA)

∴ ∠ADE = 180° – 113° = 67°

iii) From the figure .

∠CED = ∠BCA = 78°

(∵ corresponding angles)

Question 16.

In the given figure, it is given that AB = AC; ∠BAC = 36°; ∠ADB = 45° and ∠AEC = 40°. Find i) ∠ABC

i) ∠ACB iii) ∠DAB iv) ∠EAC.

Solution:

Given that AB = AC; ∠BAC = 36°,

∠ADB = 45°, ∠AEC = 40°

(i) & (ii)

In ∆ABC ; AB = AC

⇒ ∠ABC = ∠ ACB

And 36° + ∠ABC + ∠ACB = 180°

(∵ angle sum property)

∴ ∠ABC = \(\frac{180^{\circ}-36^{\circ}}{2}=\frac{144^{\circ}}{2}=72^{\circ}\)

∠ACB = 72°

iii) From the figure

∠ABD + ∠ABC = 180°

∠ABD = 180° – 72° = 1086

In ΔABD

∠DAB + ∠ABD + ∠D = 180°

∠DAB + 108° + 45° = 180°

∠DAB = 180° – 153° = 27°

iv) In ΔADE

∠D + ∠A + ∠E = 180°

45° + ∠A + 40° = 180°

⇒ ∠A = 180° -85° = 95°

But ∠A = ∠DAB + 36° + ∠EAC

95° = 27°, + 36° + ∠EAC

∴ ∠EAC = 95° – 63° = 32°

Question 17.

Using information given in the figure, calculate the values of x and y.

Solution:

From the figure In ∆ACB

34° + 62° + ∠ACB = 180°

(∵ angle sum property)

∴ ∠ACB = 180° – 96° = 84° .

And x + ∠ACB = 180°

(∵ linear pair of angles) .

∴ x + 84° = 180°

x = 180°-84° = 96°

(OR)

x = 34° + 62° = 96°

( ∵ x is exterior angle, ∆ABC)

y = 24° + x°

= 24° + 96° = 120°

(∵ y is exterior angle, ∆DCE)