These AP 9th Class Maths Important Questions 4th Lesson Lines and Angles will help students prepare well for the exams.

## AP State Syllabus 9th Class Maths 4th Lesson Important Questions and Answers Lines and Angles

Question 1.

∠POR and ∠QOR is a linear pair. If ∠POR = 3x° and ∠QOR = (2x + 10)° then find the value of x.

Solution:

∠POR and ∠QOR is a linear pair.

So ∠POR + ∠QOR = 180°

3x + 2x + 10° = 180°

5x + 10° = 180°

5x = 180 – 10

5x = 170°

x = \(\frac{170}{5}\) = 34°

∴ x = 34°.

Question 2.

Draw an equilateral triangle whose sides are 6 cm each.

Solution:

Question 3.

Write all pairs of vertically opposite angles from the diagram.

Solution:

From figure Vertically opposite angles are respectively

∠AOB = ∠COD and ∠AOD = ∠BOC

Question 4.

An exterior angle of a triangle is 110° and one of the interior opposite angle is 30°. Find the other two angles of the triangle.

Solution:

∠A = 30°, ∠ACD = 110°, ∠B = x, ∠C = y 30 + x = 110°

(By exterior angle property)

x = 80°

y + 110° = 180° (∵ linear pair)

y = 70°

∴ Remaining angles are x = 80° and y = 70°.

Question 5.

Find the value of ‘x’ in the figure.

From Figure

(5x + 3)° + 97° = 180° {∵ Linear pair}

5x + 100° = 180°

5x = 180° – 100° ⇒ 5x = 80°

x = \(\frac{80}{5}\) = 16°

∴ x = 16°.

Question 6.

If ∠1 + ∠2 < 180° what can you say about lines l and m?

Solution:

∠1 & ∠2 are two interior angles that lie on same side of transversal ‘n’.

As their sum is less than 180°, the two lines intersect at that side.

∴ The two lines ‘l’ and ‘m’ are two intersecting lines at the side of the angles ∠1 & ∠2.

Question 7.

In the following figure \(\overline{\mathbf{A B}}\) is a straight line. OP and OQ are two rays. Find the value of x and also find ∠AOP and ∠AOQ.

Solution:

\(\overline{\mathbf{A B}}\) is a straight line

Sum of the angles formed at ‘O’ is 180°

∴ 3x – 17 + 2x + 5 + x = 180°

6x – 12 = 180°

6x = 192° ⇒ x = 32°

∴ ∠AOP = 3x -17 = 96 – 17 = 79°

∠POQ = 2x + 5 = 64 + 5 = 69°

∠AOQ = 79° + 69 = 148°.

Question 8.

l // m and ‘n’ is transversal.

If ∠b = (3x – 10)° and ∠h = (5x – 30)° then determine all the angles.

Solution:

In figure l // m, so ∠b = ∠h

3x – 10 = 5x – 30

-2x = -20 ⇒ x = 10°

∠b = 3x – 10 = 3 × 10 – 10 = 20°

∠a = 180 – ∠b = 180 – 20 = 160°

(corresponding angles)

∴ ∠b = ∠d = ∠f = ∠h = 20°

∠a = ∠c = ∠e = ∠g = 160°.

Question 9.

In the given figure, the lines, p, q and r are parallel to one another. Find the values of a, b, c, x, y and z.

Solution:

From given figure .

y = 115° {∵ Corresponding angles}

⇒ x + y = 180° {linear pair}

⇒ x = 180°- y = 180° – 115° = 65°

⇒ z = x°= 65° { ∵ Corresponding angles}

c° – 70°

a° + c° = 180° {linear pair}

⇒ a° + 70° = 180°

⇒ a° = 180° – 70° = 110°

b° = c° = 70°

{Vertically opposite angles}

So, a = 110°, b = 70°, c = 70°, x = 65°,

y = 115°, z = 65°

Question 10.

Calculate the values of ‘x’ and ‘y’ using the information given in the figure.

Solution:

From ΔABC

∠A + ∠B + ∠C = 180°

⇒ 40° + 60° + x° = 180°

⇒ x = 180° – 100° ⇒ x = 80°.

∠DCA = 180° – ∠x

[ ∵ ∠BCA, ∠DCA forms a linear pair]

∠DCA = 180° – 80° = 100°

∴ ∠y = ∠DCA + 20°[ ∵ exterior angle of ’ a triangle is equal to the = 100° + 20°

sum of its two opposite interior angles

∠y = 120°.

Question 11.

In the adjacent figure AB // CD. Find ‘ the values of x, y and z.

Solution:

Given that AB//CD

From the adj. fig. ∠AEP + ∠PED + ∠BEC

= 180° (∵ straight angle)

⇒ x° + 64° + x° = 180°

⇒ 2x = 180° – 64° = 116°

x = \(\frac{116^{\circ}}{2}\) = 58°

Now ∠z + ∠x = 180° [ ∵ AB//CD Zx, Zz are interior angles which are formed same side of the transversal line]

⇒∠z + 58° = 180° ⇒ ∠z = 180° – 58°

∴ ∠z = 122°.

from ΔAPE

∠A + ∠P + ∠E = 180°

⇒ 90° + ∠y + ∠x = 180°

⇒ 90° + ∠y + 58° = 180°

⇒ ∠y = 180° – 148° ⇒ ∠y = 32°

⇒ ∠x = 58°, ∠y = 32°, ∠z = 122°.

Question 12.

In the given figure \(\overline{\mathrm{AB}}\) || \(\overline{\mathrm{CD}}\). Find the values of x and y.

Solution:

Given that AB // CD.

From the figure x° + 60° + x° = 180°

( ∵ The angles at a point on the line)

2x = 180° – 60° ⇒ x = \(\frac{120^{\circ}}{2}\) = 60°

We know x° + x° + 2y° = 180°

[∵ interior angles on the same side of transversal]

2x°+ 2y° = 180°

2y° = 180° – 2x°

2y° = 180° – 2 (60°) = 60°

y = 30°