These AP 9th Class Maths Important Questions 6th Lesson Lines and Angles will help students prepare well for the exams.

## AP Board Class 9 Maths 6th Lesson Lines and Angles Important Questions

### 9th Class Maths Lines and Angles 2 Marks Important Questions

Question 1.

Find the measure of an angle supplement is equal to the angle itself.

Solution:

Let the measure of an angle be x°.

Then, the measure of its supplementary angle = 180° – x°

According to the question, x° = 180° -x°

⇒ x° + x° = 180°

⇒ 2x° = 180°

⇒ x° = 90°

Hence, the measure of the angle is 90°.

Question 2.

Two supplementary angles differ by 48°. Find the angles.

Solution:

Let the two supplementary angles be x° and y° (x° > y°). Then,

x° + y° = 180° …………. (1)

Again, according to the question,

x° + y° = 48° …………… (2)

Adding equation (1) and equation (2), we get

2x° = 180° + 48°

⇒ 2x° = 228°

⇒ x° = \(\frac{228^{\circ}}{2}\) = 144°

Subtracting equation (2) from equation (1), we get

2y° = 180° – 48°

⇒ 2y° – 132°

⇒ y° = \(\frac{132^{\circ}}{2}\) = 66°

Hence, the two supplementary angles are 114° and 66°.

Question 3.

Two complementary angles are in the ratio 4 : 5. Then find angles.

Solution:

Let the two complementary angles be x° and y°. Then,

x° + y° = 90° …………… (1)

According to the question,

x° : y° = 4 : 5 ⇒ \(\frac{x^{\circ}}{y^{\circ}}\) = \(\frac{4}{5}\)

⇒ 5x° = 4y° ⇒ 5x° – 4y° = 0° ……………… (2)

Multiplying equation (1) by 4, we get

4x° + 4y° = 360° ………….. (3)

Adding equation (2) and equation (3), we get

9x° = 360° ⇒ x° = \(\frac{360^{\circ}}{9}\) = 40° …………. (4)

Putting x° = 40° in equation (1), we get

40° + y° = 90°

y° = 90° – 40° = 50° …………. (5)

Hence the two complementary angles are 40° and 50°.

Question 4.

Difference between an angle and its supplement is 30°. Find the angles.

Solution:

Let the angle be x°.

Then, its supplement = 180° – x°

According to the question,

x° – (180° – x° ) = 30°

⇒ 2x° – 180° = 30°

⇒ 2x° = 210° ⇒ x° = \(\frac{210^{\circ}}{2^{\circ}}\) = 105°

(or)

(180° – x°) – x° = 30°

⇒ 180° – 2x° = 30°

⇒ -2x° = -150°

⇒ x° = \(\frac{-150^{\circ}}{-2^{\circ}}\) = 75°

Hence, the required angles are 105° or 75°.

Question 5.

x and y are angles on a line ‘l’ forming linear pairs. If twice of x is 30° less than y, find x and y.

Solution:

(∵ x and y are angles on a line l forming linear pairs)

∴ x + y = 180° ………….. (1)

According.to the question,

2x = y – 30° ⇒ 2x – y = – 30°………….. (2)

Solving (1) and (2), we get

x = 50°, y = 130°

Question 6.

In the adjoining figure, find ‘y’ if x = 30°.

Solution:

2x + 2y= 180° (Linear Pair Axiom)

⇒ 2(30°) + 2y = 180° [∵ x = 30° (given)]

⇒ 60°+ 2y = 180° ⇒ 2y = 120°

∴ y = 60°

Question 7.

In figure, m∠ADC = 80°.

If ∠ADB : ∠CDB = 1 : 3, find the degree measure of ∠ADB and ∠CDB.

Solution:

m∠ADC = 80° ………. (1) (Given)

⇒ ∠ADB + ∠CDB = 80° ……………. (2) (Linear Pair Axiom)

⇒ ∠ADB : ∠CDB = 1 : 3 (Given)

Sum of the ratios = 1 + 3 = 4

∴ ∠ADB = \(\frac{1}{4}\) × 80° = 20°

and ∠CDB = \(\frac{3}{4}\) × 80° = 60°

Question 8.

In figure, if AB || CD, CD || EF and x : y = 3 : 2, find z.

Solution:

∵ AB || CD and CD

∴ AB || EF

(∵ Lines which are parallel to the same line are parallel to each other and a transversal PQ intersects them)

∴ x + y = 180°

(∵ The sum of the consecutive interior angles on the same side of a transversal is 180°)

x : y = 3 : 2 (Given)

∴ Sum of the ratios = 3 + 2 = 5

∴ x = \(\frac{3}{5}\) × 180° = 108°

y = \(\frac{2}{5}\) × 180° = 72°

Again, z = x (Alternate Interior Angles)

⇒ z = 108°

Question 9.

In the given figure, AB || DC and AD || BC. Prove that ∠DAB = ∠DCB.

Solution:

(∵ AB || DC and a transversal AD intersectg them.)

∴ ∠DAB + ∠ADC = 180° ……………. (1)

(∵ The sum of the consecutive interior angles on the same side of a transversal is 180° and AD || BC)

and a transversal DC intersects them

∴ ∠ADC + ∠DCB = 180° ……………. (2)

(∵ The sum of the consecutive interior angles on the same side of a transversal is 180°)

Subtracting equation (2) from equation (1), we get

∠DAB – ∠DCB = 0° ⇒ ∠DAB = ∠DCB

Question 10.

The angles of a triangle are (x + 10)°, (x + 40°) and (2x – 30)°. Find the value of x.

Solution:

(∵ The sum of the angles of a triangle is 180°)

∴ (x + 10)° + (x + 40)° + (2x – 30)° = 180°

⇒ (4x + 20)° = 180° ⇒ 4x = 160° ⇒ x = 40°

Question 11.

One of the three angles of a triangle is twice the smallest and another is three times the smallest. Find the angles.

Solution:

Let the smallest angle be x°.

Then, second angle = 2x°

third angle = 3x°

(∵ The sum of the angles of a triangle is 180°)

⇒ x° + 2x° + 3x° = 180°

⇒ 6x° = 180° ⇒ x° = \(\frac{180^{\circ}}{6}\) = 30°

∴ 2x° = 2(30°) = 60°

⇒ 3x° = 3(30°) = 90°

Hence the angles are 30°, 60° and 90°.

Question 12.

An exterior angle is drawn to a triangle. If this exterior angle is acute, then what type of triangle will be formed ?

Solution:

Let ABC be a triangle in which an exterior angle ACD is drawn.

Then, ∠ACD is acute. (Given)

⇒ ∠ACD < 90° ……….. (1)

Now, ∠ACD + ∠ACB = 180° …………… (2)

(Linear Pair Axiom)

From (1) and (2), ∠ACB > 90°

⇒ ∠ACB is obtuse.

Hence, ΔABC is an obtuse angled triangle.

Question 13.

Ray OE bisects ∠AOB and OF is the ray opposite to OE. Show that ∠FOB = ∠FOA.

Solution:

∠FOB + ∠BOE = 180° ………………. (1)

(Linear Pair Axiom)

∠FOA + ∠AOE = 180° ……………… (2)

(Linear Pair Axiom)

From (1) and (2),

∠FOB + ∠BOE = ∠FOA + ∠AOE …………….. (3)

But ZBOE = ∠AOE ……………. (4)

(∵ Ray OE bisects ∠AOB)

∴ From (3) and (4),

∠FOB = ∠FOA

Question 14.

If two lines are perpendicular to the same line, prove that they are parallel to each other.

Solution:

Let each of the two lines m and n be perpendicular to the same line l. Then,

∠1 = 90°

∠2 = 90°

∴ ∠1 = ∠2

But these angles form a pair of equal corresponding angles, therefore, m || n.

Question 15.

In figure, if x = y and a = b, then prove that r|| n.

Solution:

r and m are two lines and a transversal p intersects, then such that

x = y

But these angles form a pair of equal corresponding angles

∴ r || m …………. (1)

Again, m and n are two lines and a transversal q intersects, then such that

a = b

But these angles form a pair of equal corresponding angles

∴ m || n …………. (2)

From (1) and (2),

r || n

(∵ Lines which are parallel to the same line are parallel to each other

∵ m || n and a transversal p intersects them)

Question 16.

The angles of a triangle are in the ratio 6 : 7 : 2. Find the angles of the triangle.

Solution:

Let the triangle be ABC.

Then according to the question,

∠A : ∠B : ∠C = 6 : 7 : 2

Sum of ratios = 6 + 7 + 2 = 15

We know that the sum of the angles of a triangle is 180°.

∴ ∠A + ∠B + ∠C = 180°

∴ ∠A = \(\frac{6}{15}\) × 180° = 72°

∠B = \(\frac{7}{15}\) × 180° = 84°

and ∠C = \(\frac{2}{15}\) × 180° = 24°

Question 17.

An exterior angle of a triangle is 110° and its two interior opposite angles are equal. Find each of these equal angles.

Solution:

We know that an exterior angle of a triangle is equal to the sum of its two interior opposite angles.

Let each interior opposite angle be x°. Then,

x° + x° = 110° ⇒ 2x° = 110° ⇒ x° = 55°

Hence, each of the two equal interior opposite angles of the triangle is 55°.

Question 18.

In the following figure, if AB || DE, ∠CDE = 53°, find ∠DCE.

Solution:

∵ AB || DE and AE is a transversal.

∴ ∠DEC = ∠EAB (Alternate interior angles)

⇒ ∠DEC = 35°

In Δ CDE,

∠DCE + ∠CDE + ∠DEC = 180° (∵ The sum of the angles of a triangle 180°)

⇒ ∠DCE + 53° + 35° = 180°

⇒ ∠DCE + 88° = 180°

⇒ ∠DCE = 180° – 88° = 92°

Question 19.

Students in a school are preparing flags as shown below for a rally to make people aware for saving water. In the diagram below, Δ ABC is shown with AC extended through point D.

If ∠BCD = 6x + 2, ∠BAC = 3x + 15 and ∠ABC = 2x – 1, what is the value of x ?

State the property used to solve the problem. What value are they exhibiting by doing so ?

Solution:

∠BCD = ∠BAC + ∠ABC

(∵ An exterior angle of a triangle is equal to the sum of its two interior opposite angles)

⇒ 6x + 2 = (3x + 15) + (2x – 1)

⇒ 6x + 2 = 5x + 14

⇒ 6x – 5x = 14 – 2

⇒ x = 12

The value they are exhibiting by doing so is creation of awareness in the public to save water. Water is a very important constituent for our life. It is always advisable to use it within limits and not to waste it at any cost. Otherwise our future generation will be in great trouble on account of its shortage and deficiency.

Question 20.

The degree measures of three angles of a triangle are x, y and z. If z = \(\frac{x+y}{2}\), then find the value of z.

Solution:

(∵ x, y and z are the degree measures of three angles of a triangle.)

∴ x + y + z = 180°

(∵ The sum of the angles of a triangle is 180°)

⇒ 2z + z = 180° (∵ z = \(\frac{x+y}{z}\))

⇒ x + y = 2z

⇒ 3z = 180°

⇒ z = \(\frac{180^{\circ}}{3}\) = 60°

Question 21.

In the given figure, find x and y.

Solution:

In ΔAOD,

∠AOD + ∠OAD + ∠ODA = 180° (∵ The sum of the angles of a triangle is 180°)

⇒ 90° + (x – 6°) + x = 180°

⇒ 2x + 84° = 180°

⇒ 2x = 180° – 84° = 96°

⇒ x = \(\frac{96^{\circ}}{2}\) = 48°

In ΔOBC,

∠BOC + ∠OCB + ∠OBC = 180° (∵ The sum of the angles of a triangle is 180°)

⇒ 90° + 4y + (3y + 6°) = 180°

⇒ 7y + 96° = 180°

⇒ 7y = 180°- 96° = 84

⇒ y = \(\frac{84^{\circ}}{7}\) = 12°

Question 22.

Find the value of ‘x’ from the below figure.

Solution:

Total angle at a point ’O’ is 360°.

From the figure,

2x° + x° + 2x° + 3x° = 360°

8x° = 360°

x = \(\frac{360}{8}\) = 45°

### 9th Class Maths Lines and Angles 4 Marks Important Questions

Question 1.

In figure, find the values of x and y and then show that AB || CD.

Solution:

∵ Ray AE stand on line GH.

∴ ∠AEG + ∠AEH = 180° (Linear Pair Axiom)

⇒ 50° + x = 180°

⇒ x = 180° – 50° = 130° …………….. (1)

⇒ y = 130° …………… (2)

(Vertically Opposite Angles)

From (1) and (2), we conclude that

x = y

But these are alternate interior angles and they are equal.

So, we can say that AB || CD.

Question 2.

Prove “If two lines intersect each other, then the vertically opposite angles are equal.”

Solution:

Given : Two lines AB and CD intersect each other at O.

To Prove : (i) ∠AOC = ∠BOD and (ii) ∠AOD = ∠BOC

Proof : (i) ∵ Ray OA stands on line CD

∴ ∠AOC + ∠AOD = 180° ………….. (1)

(Linear Pair Axiom)

∵ Ray OD stands on line AB.

∴ ∠AOD + ∠BOD = 180° ……………. (2)

(Linear Pair Axiom)

From (1) and (2),

∠AOC + ∠AOD = ∠AOD + ∠BOD

⇒ ∠AOC = ∠BOD

(ii) ∵ Ray OC stands on line AB.

∴ ∠AOC + ∠BOC = 180° ………….. (3)

(Linear Pair Axiom)

∴ OA stands on line CD

∴ ∠AOC + ∠AOD = 180° ……………. (4)

(Linear Pair Axiom)

From (3) and (4),

∠AOC + ∠AOD = ∠AOC + ∠BOC

⇒ ∠AOD = ∠BOC

Question 3.

In figure, if AOB is a line, OP bisects ∠BOC and OQ bisects ∠AOC, show that ∠POQ is a right angle.

Solution:

Given : AOB is a line. OP bisects ∠BOC and OQ bisects ∠AOC.

To Prove : ∠POQ = 90°

Proof : ∵ OP bisects ∠BOC

∴ ∠POC = \(\frac{1}{2}\) ∠BOC ……………. (1)

∵ OQ bisects ∠AOC

∴ ∠COQ = \(\frac{1}{2}\) ∠AOC …………….. (2)

Now, ∠POQ = ∠POC + ∠COQ

= \(\frac{1}{2}\) ∠BOC + \(\frac{1}{2}\) ∠AOC [From (1) and (2)]

= \(\frac{1}{2}\) (∠BOC + ∠AOC) = \(\frac{1}{2}\) (180°) (Linear Pair Axiom)

Question 4.

In figure, l || m.

Show that ∠1 + ∠2 – ∠3 = 180°.

Solution:

Given l || m

To Prove : ∠1 + ∠2 – ∠3 = 180°

Construction : Through C, draw CF || l || m

Proof : l || CF (By construction and a transversal BC intersects them)

∴ ∠1 + ∠FCB = 180° …………… (1)

(∵ The sum of the consecutive interior angles on the same side of a transversal is 180° )

⇒ ∠1 + (∠2 – ∠FCD) = 180° ……………. (2)

∵ FC || DE (By construction and a transversal CD intersects them)

∴ ∠FCD = ∠3 ………………. (3)

(Alternate Interior Angles)

From (2) and (3),

∠1 + ∠2 – ∠3 = 180°

Question 5.

In given figure, AB || CD and DE || FG. Determine ∠PDE, ∠AFD and ∠DFG.

Solution:

∵ AB || CD and a transversal DF intersects them.

∴ ∠PDC = ∠AFD (Corresponding angles)

⇒ 55° = ∠AFD ⇒ ∠AFD = 55° ……………. (1)

∵ AFB is a straight line.

⇒ ∠AFD + ∠DFG + ∠GFB = 180° (A straight angle measures 180°)

⇒ 55° + ∠DFG + 70° = 180° [From (1)]

⇒ ∠DFG + 125° = 180°

⇒ ∠DFG = 180° – 125

⇒ ∠DFG = 55° ……………… (2)

∵ DE || FG and a transversal PF intersects them.

∴ ∠PDE = ∠PFG (Corresponding angles)

⇒ ∠PDE = 55° ………… (3) [From (2)]

Question 6.

Find x°, y° and z° in the given figure, if AB || CD.

Solution:

∵ AB || CD

and a transversal BC intersects them.

∴ ∠ABC = ∠BCD (Alternate interior angles)

⇒ x° = 70° ………….. (1)

In ΔABC,

∠ABC + ∠BAC + ∠ACB = 180° (∵ The sum of the angles of a triangle is 180°)

⇒ x° + 53° + z° = 180°

⇒ 70° + 53° + z° = 180° [from (1)]

⇒ 123° + z° = 180°

⇒ z° = 180° – 123° = 57° …………….. (2)

∵ ACE is a straight line.

∴ ∠ACB + ∠BCD + ∠DCE = 180° (∵ A straight angle measures 180°)

⇒ 57° + 70° + y° = 180° [from (2)]

⇒ 127° + y° = 180°

⇒ y° = 180° – 127°

= 53° ……………. (3)

Question 7.

Prove that the sum of the angles of a triangle is 180°. Also, find the angles of a triangle if they are in the ratio 5 : 6 : 7.

(OR)

State and prove the angle sum property of a triangle.

(OR)

Prove that the sum of the angles of a triangle is 180°.

Solution:

First part :

Given : A triangle PQR and ∠1, ∠2 and ∠3 are the angles of Δ PQR.

To Prove : ∠1 + ∠2 + ∠3 = 180°

Construction : Draw a line XPY parallel to QR through the opposite vertex P.

Proof : ∵ XPY is a line.

∴ ∠4 + ∠1 + ∠5 = 180° ………… (1)

(A straight angle measures 180°)

But XPY || QR

and PQ, PR are transversals.

So, ∠4 = ∠2 and ∠5 = ∠3 [Pairs of alternate interior angles]

Substituting for ∠4 and ∠5 in (1), we get,

∠2 + ∠1 + ∠3 = 180°

∠1 + ∠2 + ∠3 = 180°

Second Part :

∠1 : ∠2 : ∠3 = 5 : 6 : 7 (Given)

Sum of ratios = 5 + 6 + 7 = 18

∠1 + ∠2 + ∠3 = 180°

∴ ∠1 = \(\frac{5}{18}\) × 180° = 50°

∠2 = \(\frac{6}{18}\) × 180° = 60°

∠3 = \(\frac{7}{18}\) × 180° = 70°

Question 8.

In the given figure side BC of ΔABC is produced in both the directions. Prove that the sum of the two exterior angles so formed is greater than 180°.

Solution:

Given : Side BC of ΔABC is produced in both the directions.

Tot Prove : ∠4 + ∠5 > 180°

Proof : ED is a line

∴ ∠4 + ∠2 = 180° …………….. (1)

(Linear Pair Axiom)

and ∠5 + ∠3 = 180° …………… (2)

(Linear Pair Axiom)

Adding (1) and (2), we get

∠4 + ∠5 + ∠2 + ∠3 = 360° …………… (3)

Now, ∠1 + ∠2 + ∠3 = 180° …………… (4)

(∵ The sum of the angles of a triangle is 180°)

⇒ ∠2 + ∠3 = 180° – ∠1 ……………. (5)

From (3) and (5),

∠4 + ∠5 + (180° – ∠1) = 360°

⇒ ∠4 + ∠5 = 180° + ∠1

⇒ ∠4 + ∠5 > 180°

Question 9.

In figure, PS is the bisector of ∠QPR and PT ⊥ QR. Show that

∠TPS = \(\frac{1}{2}\)(∠Q – ∠R).

Solution:

Given : In figure, PS is the bisector of ∠QPR and PT ⊥ QR.

To Prove : ∠TPS = \(\frac{1}{2}\)(∠Q – ∠R).

Proof : PS is the bisector of ∠QPR

∴ ∠QPS = ∠RPS

⇒ ∠1 + ∠TPS = ∠2 ………… (1)

In Δ PQT,

∠1 + ∠Q + ∠PTQ = 180°

(∵ The sum of the angles of a triangle is 180°)

⇒ ∠1 + ∠Q + 90° = 180° (∵ PT ⊥ QR)

⇒ ∠1 + ∠Q = 90° ⇒ ∠Q = 90° – ∠1 ………….. (2)

In Δ PRT,

∠TPR + ∠PTR + ∠R = 180° (∵ The sum of the angles of a triangle is 180°)

⇒ (∠TPS + ∠2) + 90° + ∠R = 180° (∵ PT ⊥ QR)

⇒ TPS + ∠2 + ∠R = 90° ……………. (3)

From (2) and (3),

∠Q = (∠TPS + ∠2 + ∠R) – 1

⇒ ∠Q – ∠R = ∠TPS + (∠2 – ∠1)

⇒ ∠Q – ∠R = ∠TPS + ∠TPS [∵ From (1)]

⇒ ∠Q – ∠R = 2∠TPS

⇒ ∠TPS = \(\frac{1}{2}\)(∠Q – ∠R)

Question 10.

In the given figure p || q and ‘t’ is a transversal.

Based on the above information, answer the following questions:

i) If ∠1 = 100°, then ∠6 = _________

A) 100° (corresponding angles)

B) 100° (exterior alternate angles)

C) 80° (co-interior angles)

D) 80°(corresponding angles)

Answer:

A) 100° (corresponding angles)

ii) If ∠2 = 80°, then ∠7 = ___________

A) 80° (corresponding angles)

B) 80° (exterior alternate angles)

C) 80° (interior alternate angles)

D) 100° (exterior alternate angles)

Answer:

B) 80° (exterior alternate angles)

iii) If ∠3 = 128°, then ∠5 = _____________

A) 56° (corresponding angles)

B) 128° (co-interior angles)

C) 52° (co-interior angles)

D) 64° (interior alternate angles)

Answer:

C) 52° (co-interior angles)

iv) If ∠1 : ∠2 = 9 : 4, then ∠7 : ∠8 =

A) 3 : 2

B) 9 : 4

C) 4 : 9

D) 2 : 3

Answer:

C) 4 : 9

v) If ∠3 + ∠6 = 200°, then ∠7 =

A) 50°

B) 100°

C) 120°

D) 80°

Answer:

D) 80°

### 9th Class Maths Lines and Angles 8 Marks Important Questions

Question 1.

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 ∠x, ∠z 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 2.

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

Solution:

∠QRT = ∠RQS + ∠QSR

(∵ An exterior angle of a triangle is equal to the sum of its two interior opposite angles) 1

⇒ 65° = 28° + ∠QSR

⇒ ∠QSR = 65° – 28° = 37°

∵ PQ ⊥ SP

∴ ∠QPS = 90°

∵ PQ || SR

∴ ∠QPS + ∠PSR = 180° (∵ The sum of consecutiye interior angles on the same side of the transversal is 180°)

⇒ 90° + ∠PSR = 180°

⇒ ∠PSR = 180° – 90° = 90°

⇒ ∠PSR + ∠QSR = 80° ⇒ y + 37° = 90°

⇒ y = 90° – 37° = 53°

In Δ PQS, ∠PQS + ∠QSP + ∠QPS = 180°

(∵ The sum of all the angles of a tri ⇒ x + y + 90°= 180° angle is 180°)

⇒ x + 53° + 90° = 180° ⇒ x + 143° = 180°z

⇒ x = 180° – 143° = 37°

Question 3.

In figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at piont T, then prove that

∠QTR = \(\frac{1}{2}\) ∠QPR.

OR

2 ∠QTR = ∠QPR

Solution:

∵ ∠TRS is an exterior angle of ΔTQR.

∴ ∠TRS = ∠TQR + ∠QTR ……………… (1)

[ ∵ An exterior angle of a triangle is equal to the sum of its two interior opposite angles.]

∴ ∠PRS is an exterior anlge of ΔPQR .

∴ ∠PRS = ∠PQR + ∠QPR ……………. (2)

(∵ An exterior angle of a triangle is equal to the sum of its two interior opposite angles)

⇒ 2∠TRS = 2∠TQR + ∠QPR (∵ QT is the bisector of ∠PQR and RT is the bisector of ∠PRS)

⇒ 2(∠TRS – ∠TQR) = ∠QPR ……………… (3)

From (1),

∠TRS – ∠TQR) = ∠QTR …………… (4)

From (3) and (4), we obtain

2∠QTR = ∠QPR ⇒ ∠QTR = \(\frac{1}{2}\) ∠QPR

Question 4.

Prove that the sum of the angles of a hexagon is 720°.

Solution:

Given : A hexagon ABCDEF

To Prove : ∠A + ∠B + ∠C – ∠D + ∠E + ∠F = 720°

Construction : Join AD, BE. and CF

Proof : In ΔOAB,

∠1 + ∠8 + ∠9 = 180° ……………… (1)

∵ The sum of the angles of a triangle is 180°

In ΔOBC,

∠2 + ∠10 + ∠11 = 180° ……………… (2)

(∵ The sum of the angles of a triangle is 180°)

In ΔOCD, ∠3 + ∠12 + ∠13 = 180° ……………….. (3)

(∵ The sum of the angles of a triangle is 180°)

In ΔODE, ∠4 + ∠14 + ∠15 = 180° ………….. (4)

(∵ The sum of the angles of a triangle is 180°)

In ΔOEF, ∠5 + ∠16 + ∠17 = 180° ……………… (5)

(∵ The sum of the angles of a triangle is 180°)

In A OFA, ∠6 + ∠18 + ∠7 = 180° ………………. (6)

( ∵ The sum of the angles of a triangle is 180°)

Adding (1), (2), (3), (4), (5) and (6), we get

(∠1 + ∠2 + ∠3 + ∠4 + v∠5 + ∠6) + (∠7 + ∠8) + (∠9 + ∠10) + (∠11 + ∠12) + (∠13 + ∠14) + (∠15 + ∠16) = 1080°

⇒ 360° + ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 1080° (∵ The sum of the angles round a point is equal 360°)

⇒ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 1080° – 360° = 720°

**AP 9th Class Maths Important Questions Chapter 6 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°