# AP 9th Class Maths Important Questions Chapter 4 Lines and Angles

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°