RD Sharma Solutions Class 9 Maths Chapter 10 – Congruent Triangles (Updated for 2021-22)

RD Sharma Solutions Class 9 Maths Chapter 10 Congruent Triangles: Start practisng for your Class 9 Maths exam with the RD Sharma Solutions Class 9 Maths Chapter 9. All the solutions are designed by subject matter experts and that too as per the latest CBSE syllabus. 

Download RD Sharma Solutions Class 9 Maths – Chapter 10 Congruent Triangles PDF

RD Sharma Solutions Class 9 Maths Chapter 10

RD Sharma Solutions Class 9 Maths Chapter 10 Congruent Triangles: Exercise-Wise 

RD Sharma Class 9 Chapter 10 Exercise 10.1

RD Sharma Class 9 Chapter 10 Exercise 10.2

RD Sharma Class 9 Chapter 10 Exercise 10.3

RD Sharma Class 9 Chapter 10 Exercise 10.4

RD Sharma Class 9 Chapter 10 Exercise 10.5

RD Sharma Class 9 Chapter 10 Exercise 10.6

Detailed Exercise-wise Explanation with Important Topics in the Exercise

RD Sharma Class 9 Solutions Chapter 10 Ex 10A

In the RD Sharma Class 9 Solutions Chapter 10 Ex 10A the students will have to deal with the questions from the Congruent Triangles part. The experts have given extra bit of attention to Congruence of Line segments, Congruence of Angles, Congruence of Triangles, Congruence Relation, and Congruence Criteria-

-SAS (Side-Angle-Side)

-SSS (Side-Side-Side)

-ASA (Angle-Side-Angle)

– RHS (Right angle- Hypotenuse-Side)

as well as Congruence Criterion for fulfilling the needs of the students. Apart from these demands, you will also get to see things like in-depth explanations. Each part of RD Sharma Class 9 Chapter 10A has been covered to meet the requirements of the students. All the concepts have been explained in form of analysis.

RD Sharma Class 9 Chapter 10 Exercise 10B

In Chapter 10 Ex-10B RD Sharma Class 9 the students will be able to examine their performances by attempting objective type questions which are associated with Congruent Triangles. After going through the solutions of RD Sharma Class 9 Chapter 10B the students will be able to take risks fearlessly.

The preparation style of the students will undergo a lot of changes which will be beneficial for their performances in the final Maths Exam. Hence, you must buy a copy for yourself right now. The main aspects are based on the topic- Criterion of Congruence – ASA ( Angle-Side-Angle).

RD Sharma Solutions Class 9 Chapter 10C

In the RD Sharma Solutions Class 9 Maths Chapter 10 Exercise 10C the students will find step by step explanations which have been done by different experts.  The main areas which the experts have tried to focus on are problems based on the Criterion of Congruence – SAS (Side-Angle-Side).

Once you examine the contents you will be able to figure out the parts which need your immediate attention.

RD Sharma Class 9 Solutions Chapter 10 Ex10D

In the RD Sharma Class 9 Chapter 10 Exercise 10D Solutions you will be exposed to all kinds of tricky problems from Congruent Triangles. Try to answer all the questions from RD Sharma Class 9 Chapter 10D without seeing the answers from Chapter 10 solutions. Assess your performance and make a note of it. After some time you check all the answers from RD Sharma Solutions Class 9 Chapter 10D.

If you make a habit you will get the maximum benefit in your final Maths exam. While going through the concept of Congruence Criterion – SSS (Side-Side-Side).from RD Sharma Class 9 Solutions Chapter 10 Exercise 10D look for the ways the concepts have been presented. You will be in a commendable position if you look for the little things at the time of practising questions from Class 9 RD Sharma Chapter 10D.

RD Sharma Class 9 Chapter 10 Exercise 10E

In the Chapter 10E Solution of RD Sharma Class 9, the students will get all kind of explanations which will clear the basic concepts of Congruent Triangles chapter. The main area which has been dealt with here are connected with the Congruence Criterion – Right angle- Hypotenuse-Side.

The step-wise explanations have been provided by the experts to satisfy the needs of the students. The important concepts will become crystal clear and the students will be confident enough to solve their own problems. The students are required to solve problems on a daily basis for maximum benefit.

RD Sharma Class 9 Solutions Chapter 10 Ex 10F

In the exercise 10F of RD Sharma Solutions, students will get to know about the techniques which can be applied in case of Congruent Triangles. Parts like Inequality Relations in a Triangle have been addressed by the experts with the extra bit of care and all you need to do is sit with the relevant parts after understanding the concepts. You can also include the parts in your study materials while examining the essential aspects of Congruent Triangles.

In the last exercise, you will have to focus on the various kinds of questions from Congruent Triangles. Students often face problems while answering objective-based questions and the RD Sharma Class 9 solutions will bring you back on track. The in-depth explanations and smart approach will help the students to plan ahead and score good marks in this part.

Acess answers of RD Sharma Solutions Class 9 Chapter 10

RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.1

Question 1.
Write the complement of each of the following angles:
(i) 20°
(ii) 35°
(iii) 90°
(iv) 77°
(v) 30°
Solution:
We know that two angles are complement to each other if their sum is 90°. Therefore,
(i) Complement of 20° is (90° – 20°) = 70°
(ii) Complement of 35° is (90° – 35°) = 55°
(iii) Complement of 90° is (90° – 90°) = 0°
(iv) Complement of 77° is (90° – 77°) = 13°
(v) Complement of 30° is (90° – 30°) = 60°

Question 2.
Write the supplement of each of the following angles:
(i) 54°
(ii) 132°
(iii) 138°
Solution:
We know that two angles are supplement to each other if their sum if 180°. Therefore,
(i) Supplement of 54° is (180° – 54°) = 126°
(ii) Supplement of 132° is (180° – 132°) = 48°
(iii) Supplement of 138° is (180° – 138°) = 42°

Question 3.
If an angle is 28° less than its complement, find its measure.
Solution:
Let required angle = x, then
Its complement = x + 28°
∴  x + x + 28° = 90° ⇒  2x = 90° – 28° = 62°
∴ x = 62∘2  = 31°
∴ Required angle = 31°

Question 4.
If an angle is 30° more than one half of its complement, find the measure of the angle.
Solution:
Let the measure of the required angle = x
∴  Its complement =  90° – x
∴  x = 12 (90° – x) + 30°
2x = 90° – x + 60°
⇒ 2x + x = 90° + 60°
⇒  3x = 150°
⇒ x =  150∘3  = 50°
∴ Required angle = 50°

Question 5.
Two supplementary angles are in the ratio 4 : 5. Find the angles.
Solution:
Ratio in two supplementary angles = 4 : 5
Let first angle = 4x
Then second angle = 5x
∴  4x + 5x = 180
⇒  9x = 180°
∴ x  = 180∘9 = 20°
∴  First angle = 4x = 4 x 20° = 80°
and second angle = 5x
= 5 x 20° = 100°

Question 6.
Two supplementary angles differ by 48°. Find the angles.
Solution:
Let first angle = x                        ”
Then second angle = x + 48°
∴  x + x + 48° = 180°⇒  2x + 48° = 180°
⇒  2x = 180° – 48° = 132°
x= 132∘2 =66°
∴  First angle = 66°
and second angle = x + 48° = 66° + 48° = 114°
∴ Angles are 66°, 114°

Question 7.
An angle is equal to 8 times its complement. Determine its measure.
Solution:
Let the required angle = x
Then its complement angle = 90° – x
∴ x = 8(90° – x)
⇒ x = 720° – 8x ⇒  x + 8x = 720°
⇒ 9x = 720° ⇒ x =  720∘9 = 80°
∴  Required angle = 80°

Question 8.
If the angles (2x – 10)° and (x – 5)° are complementary angles, find x.
Solution:
First complementary angle = (2x – 10°) and second = (x – 5)°
∴ 2x – 10° + x – 5° = 90°
⇒ 3x – 15° = 90° ⇒  3x = 90° + 15° = 105°
∴ x = 105∘3 = 35°
∴  First angle = 2x – 10° = 2 x 35° – 10°
= 70° – 10° = 60°
and second angle = x – 5 = 35° – 5 = 30°

Question 9.
If an angle differ from its complement by 10°, find the angle.
Solution:
Let required angle = x°
Then its complement angle = 90° – x°
∴ x – (90° – x) = 10
⇒  x – 90° + x = 10°⇒  2x = 10° + 90° = 100° 100°
⇒ x =  100∘2 = 50°
∴ Required angle = 50°

Question 10.
If the supplement of an angle is two-third of itself Determine the angle and its supplement.
Solution:
Let required angle = x
Then its supplement angle = 180° – x
∴  (180°-x)= 23x
540° – 3x = 2x ⇒ 2x + 3x = 540°
⇒ 5x = 540°⇒  x = 540∘5 = 108°
-. Supplement angle = 180° – 108° = 72°

Question 11.
An angle is 14° more than its complementary angle. What is its measure?
Solution:
Let required angle = x
Then its complementary angle = 90° – x
∴  x + 14° = 90° – x
x + x = 90° – 14° ⇒  2x = 76°
⇒ x =  76∘2 = 38°
∴  Required angle = 38°

Question 12.
The measure of an angle is twice the measure of its supplementary angle. Find its measure.
Solution:
Let the required angle = x
∴  Its supplementary angle = 180° – x
∴  x = 2(180°-x) = 360°-2x
⇒  x + 2x = 360°
⇒ 3x = 360°
⇒  x = 360∘3 = 120°
∴  Required angle = 120°

Question 13.
If the complement of an angle is equal to the supplement of the thrice of it. Find the measure of the angle.
Solution:
Let required angle = x
Then its complement angle = 90° – x
and supplement angle = 180° – x
∴  3(90° – x) = 180° – x
⇒ 270° – 3x = 180° – x
⇒270° – 180° = -x + 3x => 2x = 90°
⇒ x = 45°
∴  Required angle = 45°

Question 14.
If the supplement of an angle is three times its complement, find the angle.
Solution:
Let required angle = x
Then its complement = 90°-  x
and supplement = 180° – x
∴  180°-x = 3(90°-x)
⇒  180° – x = 270° – 3x
⇒  -x + 3x = 270° – 180°
⇒ 2x = 90° ⇒ x = 90∘2 =45°
∴ Required angle = 45°

Class 9 RD Sharma Solutions Chapter 10 Congruent Triangles Ex 10.2

Question 1.
In the figure, OA and OB are opposite rays:
(i) If x = 25°, what is the value of y?
(ii) If y = 35°, what is the value of x?

Solution:
(i) If x = 25°
∴ 3x = 3 x 25° = 75°
But ∠AOC + ∠BOC = 180° (Linear pair)
⇒ ∠AOC + 75° = 180°
⇒ ∠AOC = 180° – 75°
⇒ ∠AOC = 105°
∴ 2y + 5 = 105° ⇒ 2y= 105° – 5° = 100°
⇒ y = 100∘2  = 50°
∴ If x = 25° then y = 50°
(ii) If y = 35°, then ∠AOC = 2y + 5
∴ 2y + 5 = 2 x 35° + 5 = 70° + 5 = 75°
But ∠AOC + ∠BOC = 180° (Linear pair)
⇒ 75° + ∠BOC = 180°
⇒ ∠BOC = 180°-75°= 105°
∴ 3x = 105° ⇒ x = 105∘3 = 35°
∴ x = 35°

Question 2.
In the figure, write all pairs of adjacent angles and all the linear pairs.

Solution:
In the given figure,
(i) ∠AOD, ∠COD; ∠BOC, ∠COD; ∠AOD, ∠BOD and ∠AOC, ∠BOC are the pairs of adjacent angles.
(ii) ∠AOD, ∠BOD and ∠AOC, ∠BOC are the pairs of linear pairs.

Question 3.
In the figure, find x, further find ∠BOC, ∠COD and ∠AOD.

Solution:
In the figure,
AOB is a straight line
∴ ∠AOD + ∠DOB = 180° (Linear pair)
⇒ ∠AOD + ∠DOC + ∠COB = 180°
⇒ x+ 10° + x + x + 20 = 180°
⇒ 3x + 30° = 180°
⇒ 3x= 180° -30°= 150°
⇒ x = 150∘3 = 50°
∴ x = 50°
Now ∠BOC =x + 20° = 50° + 20° = 70°
∠COD = x = 50°
and ∠AOD = x + 10° = 50° + 10° = 60°

Question 4.
In the figure, rays OA, OB, OC, OD and OE have the common end point O. Show that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.

Solution:
Produce AO to F such that AOF is a straight line
Now ∠AOB + ∠BOF = 180° (Linear pair)

⇒ ∠AOB + ∠BOC + ∠COF = 180° …(i)
Similarly, ∠AOE + ∠EOF = 180°
⇒ ∠AOE + ∠EOD + ∠DOF = 180° …(ii)
Adding (i) and (ii)
∠AOB + ∠BOC + ∠COF + ∠DOF + ∠EOD + ∠AOE = 180° + 180°
⇒ ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°
Hence proved.

Question 5.
In the figure, ∠AOC and ∠BOC form a linear pair. If a – 2b = 30°, find a and b.

Solution:
∠AOC + ∠BOC = 180° (Linear pair)
⇒ a + 6 = 180° …(i)
and a – 2b = 30° …(ii)
Subtracting (ii) from (i), 3b = 150°
⇒ b = 150∘3 = 50°
and a + 50° = 180°
⇒ a= 180°-50°= 130°
∴ a = 130°, b = 50°

Question 6.
How many pairs of adjacent angles are formed when two lines intersect in a point?
Solution:
If two lines AB and CD intersect each other at a point O, then four pairs of linear pairs are formed i.e.
∠AOC, ∠BOC; ∠BOC, ∠BOD; ∠BOD, ∠AOD and ∠AOD, ∠AOC

Question 7.
How many pairs of adjacent angles, in all, can you name in the figure.

Solution:
In the given figure 10 pairs of adjacent angles are formed as given below:
∠AOB, ∠BOC; ∠AOB, ∠BOD; ∠AOC, ∠COD; ∠AOD, ∠BOE; ∠AOB, ∠BOE; ∠AOC, ∠COF; ∠BOC, ∠COD; ∠BOC, ∠COE; ∠COD, ∠DOE and ∠BOD, ∠DOE

Question 8.
In the figure, determine the value of x.

Solution:
In the figure
∠AOC + ∠COB + ∠BOD +∠AOD = 360 (angles at a point )

⇒ 3x + 3x + x + 150° = 360°
⇒ 7x – 360° – 150° = 210°
⇒ x = 210∘7 = 30°
∴ x = 30°

Question 9.
In the figure, AOC is a line, find x.

Solution:
In the figure,
∠AOB + ∠BOC = 180° (Linear pair)
⇒ 70° + 2x = 180°
⇒ 2x = 180° – 70°
⇒ 2x = 110°⇒x = 110∘2 = 55°
∴ x = 55°

Question 10.
In the figure, POS is a line, find x.

Solution:
In the figure, POS is a line
∴ ∠POQ + ∠QOS = 180° (Linear pair)
⇒ ∠POQ + ∠QOR + ∠ROS = 180°
⇒ 60° + 4x + 40° = 180°
⇒ 4x + 100° – 180°
⇒ 4x = 180° – 100° = 80°
⇒ x = 80∘4 =20°
∴ x = 20°

Question 11.
In the figure, ACB is a line such that ∠DCA = 5x and ∠DCB = 4x. Find the value of x.

Solution:
ACB is a line, ∠DCA = 5x and ∠DCB = 4x
∠ACD + ∠DCB = 180° (Linear pair)
⇒ 5x + 4x = 180° ⇒ 9x = 180°
⇒ x = 180∘9 = 20°
∴ x = 20°
∴ ∠ACD = 5x = 5 x 20° = 100° and ∠DCB = 4x = 4 x 20° = 80°

Question 12.
Given ∠POR = 3x and ∠QOR = 2x + 10°, find the value ofx for which POQ will be a line.

Solution:
∠POR = 3x and ∠QOR = 2x + 10°
If POQ is a line, then
∠POR + ∠QOR = 180° (Linear pair)
⇒ 3x + 2x + 10° = 180°
⇒ 5x = 180° – 10° = 170°
⇒ x = 170∘5 = 34°
∴ x = 34°

Question 13.
What value ofy would make AOB, a line in the figure, if ∠AOC = 4y and ∠BOC = (6y + 30).

Solution:
In the figure,
AOB is a line if
∠AOC + ∠BOC = 180°
⇒ 6y + 30° + 4y= 180°
⇒ 10y= 180°-30°= 150°
150°
⇒ y = 150∘10 = 15°
∴ y = 15°

Question 14.
In the figure, OP, OQ, OR and OS are four rays. Prove that: ∠POQ + ∠QOR + ∠SOR + ∠POS = 360° [NCERT]

Solution:
In the figure, OP, OQ, OR and OS are the rays from O
To prove : ∠POQ + ∠QOR + ∠SOR + ∠POS = 360°
Construction : Produce PO to E

Proof: ∠POQ + ∠QOE = 180° (Linear pair)
Similarly, ∠EOS + ∠POS = 180°
Adding we get,
∠POQ + ∠QOR + ∠ROE + ∠EOS + ∠POS = 180° + 180° ,
⇒ ∠POQ + ∠QOR + ∠ROS + ∠POS = 360° Hence ∠POQ + ∠QOR + ∠SOR + ∠POS = 360°

Question 15.
In the figure, ray OS stand on a line POQ. Ray OR and ray OT are angle bisectors of ∠POS and ∠SOQ respectively. If ∠POS = x, find ∠ROT. [NCERT]

Solution:
Ray OR stands on a line POQ forming ∠POS and ∠QOS

OR and OT the angle bisects of ∠POS and ∠QOS respectively. ∠POS = x
But ∠POS + ∠QOS = 180° (Linear pair)
⇒ x + ∠QOS = 180°
⇒ ∠QOS = 180° – x
∵ OR and OT are the bisectors of angle

Question 16.
In the figure, lines PQ and RS intersect each other at point O. If ∠POR : ∠ROQ = 5 : 7. Find all the angles. [NCERT]

Solution:
Lines PQ and PR, intersect each other at O
∵ Vertically opposite angles are equal
∴ ∠POR = ∠QOS and ∠ROQ = ∠POS
∠POR : ∠ROQ = 5:7
Let ∠POR = 5x and ∠ROQ = 7x
But ∠POR + ∠ROQ = 180° (Linear pair)
∴ 5x + 7x = 180° ⇒ 12x = 180°
⇒ x = 180∘12 = 15°
∴ ∠POR = 5x = 5 x 15° = 75°
and ∠ROQ = 7x = 7 x 15° = 105°
But ∠QOS = POR = 75° (Vertically opposite angles)
and ∠POS = ∠ROQ = 105°

Question 17.
In the figure, a is greater than b by one third of a right-angle. Find the values of a and b.

Solution:
In the figure,
∠AOC + ∠BOC = 180° (Linear pair)
⇒ a + b =180° …(i)
But a = b + 13 x 90° = b + 30°
⇒ a – b = 30° …(ii)
Adding (i) and (ii)
210°
2a = 210° ⇒ a = 210∘2 = 105°
and 105° + b = 180°
⇒ b = 180° – 105°
∴ b = 75°
Hence a = 105°, b = 75°

Question 18.
In the figure, ∠AOF and ∠FOG form a linear pair.
∠EOB = ∠FOC = 90° and ∠DOC = ∠FOG = ∠AOB = 30°

(i) Find the measures of ∠FOE, ∠COB and ∠DOE.
(ii) Name all the right angles.
(iii) Name three pairs of adjacent complementary angles.
(iv) Name three pairs of adjacent supplementary angles.
(v) Name three pairs of adjacent angles.
Solution:
In the figure,
∠AOF and ∠FOG form a linear pair
∠EOB = ∠FOC = 90°
∠DOC = ∠FOG = ∠AOB = 30°

(i) ∠BOE = 90°, ∠AOB = 30°
But ∠BOE + ∠AOB + ∠EOG = 180°
⇒ 30° + 90° + ∠EOG = 180°
∴∠EOG = 180° – 30° – 90° = 60°
But ∠FOG = 30°
∴ ∠FOE = 60° – 30° = 30°
∠COD = 30°, ∠COF = 90°
∴ ∠DOF = 90° – 30° = 60°
∠DOE = ∠DOF – ∠EOF
= 60° – 30° = 30°
∠BOC = BOE – ∠COE
= 90° – 30° – 30° = 90° – 60° = 30°
(ii) Right angles are,
∠AOD = 30° + 30° + 30° = 90°
∠BOE = 30° + 30° + 30° = 90°
∠COF = 30° + 30° + 30° = 90°
and ∠DOG = 30° + 30° + 30° = 90°
(iii) Pairs of adjacent complementaiy angles are ∠AOB, ∠BOD; ∠AOC, ∠COD; ∠BOC, ∠COE
(iv) Pairs of adjacent supplementary angles are ∠AOB, ∠BOG; ∠AOC, ∠COG and ∠AOD, ∠DOG
(v) Pairs of adjacent angles are ∠BOC, ∠COD; ∠COD, ∠DOE and ∠DOE, ∠EOF.

Question 19.
In the figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 12(∠QOS – ∠POS). [NCERT]

Solution:
Ray OR ⊥ ROQ. OS is another ray lying between OP and OR.

To prove : ∠ROS = 12(∠QOS – ∠POS)
Proof : ∵ RO ⊥ POQ
∴ ∠POR = 90°
⇒ ∠POS + ∠ROS = 90° …(i)
⇒ ∠ROS = 90° – ∠POS
But ∠POS + ∠QOS = 180° (Linear pair)
= 2(∠POS + ∠ROS) [From (i)]
∠POS + ∠QOS = 2∠ROS + 2∠POS
⇒ 2∠ROS = ∠POS + ∠QOS – 2∠POS
= ∠QOS – ∠POS
∴ ROS = 12 (∠QOS – ∠POS)

Class 9 Maths Chapter 10 Congruent Triangles RD Sharma Solutions Ex 10.3

Question 1.
In the figure, lines l₁ and l₂ intersect at O, forming angles as shown in the figure. If x = 45, find the values of y, z and n.

Solution:
Two lines l₁ and l₂ intersect each other at O ∠x = 45°
∵ ∠z = ∠x (Vertically opposite angles)
= 45°
But x + y = 180° (Linear pair)
⇒45° + y= 180°
⇒ y= 180°-45°= 135°
But u = y (Vertically opposite angles)
∴ u = 135°
Hence y = 135°, z = 45° and u = 135°

Question 2.
In the figure, three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u.

Solution:
Three lines AB, CD and EF intersect at O
∠BOD = 90°, ∠DOF = 50°
∵ AB is a line
∴ ∠BOD + ∠DOF + FOA = 180°
⇒ 90° + 50° + u = 180°
⇒ 140° + w = 180°
∴ u= 180°- 140° = 40°
But x = u (Vertically opposite angles)
∴ x = 40°
Similarly, y = 50° and z = 90°
Hence x = 40°, y = 50°, z = 90° and u = 40°

Question 3.
In the figure, find the values of x, y and z.

Solution:
Two lines l₁ and l₂ intersect each other at O
∴ Vertically opposite angles are equal,
∴ y = 25° and x = z
Now 25° + x = 180° (Linear pair)
⇒ x= 180°-25°= 155°
∴ z = x = 155°
Hence x = 155°, y = 25°, z = 155°

Question 4.
In the figure, find the value of x.

Solution:
∵ EF and CD intersect each other at O
∴ Vertically opposite angles are equal,
∴ ∠1 = 2x
AB is a line
3x + ∠1 + 5x = 180° (Angles on the same side of a line)
⇒ 3x + 2x + 5x = 180°
⇒ 10x = 180° ⇒ x = 180∘10  = 18°
Hence x = 18°

Question 5.
If one of the four angles formed by two intersecting lines is a right angle, then show that each of the four angles is a right angle.
Solution:
Given : Two lines AB and CD intersect each other at O. ∠AOC = 90°

To prove: ∠AOD = ∠BOC = ∠BOD = 90°
Proof : ∵ AB and CD intersect each other at O
∴ ∠AOC = ∠BOD and ∠BOC = ∠AOD (Vertically opposite angles)
∴ But ∠AOC = 90°
∴ ∠BOD = 90°
∴ ∠AOC + ∠BOC = 180° (Linear pair)
⇒ 90° + ∠BOC = 180°
∴ ∠BOC = 180° -90° = 90°
∴ ∠AOD = ∠BOC = 90°
∴ ∠AOD = ∠BOC = ∠BOD = 90°

Question 6.
In the figure, rays AB and CD intersect at O.
(i) Determine y when x = 60°
(ii) Determine x when y = 40°

Solution:
In the figure,
AB is a line
∴ 2x + y = 180° (Linear pair)
(i) If x = 60°, then
2 x 60° + y = 180°
⇒ 120° +y= 180°
∴ y= 180°- 120° = 60°
(ii) If y = 40°, then
2x + 40° = 180°
⇒ 2x = 180° – 40° = 140°
⇒ x= 140∘2  =70°
∴ x = 70°

Question 7.
In the figure, lines AB, CD and EF intersect at O. Find the measures of ∠AOC, ∠COF, ∠DOE and ∠BOF.

Solution:
Three lines AB, CD and EF intersect each other at O
∠AOE = 40° and ∠BOD = 35°
(i) ∠AOC = ∠BOD (Vertically opposite angles)
= 35°
AB is a line
∴ ∠AOE + ∠DOE + ∠BOD = 180°
⇒ 40° + ∠DOE + 35° = 180°
⇒ 75° + ∠DOE = 180°
⇒ ∠DOE = 180°-75° = 105°
But ∠COF = ∠DOE (Vertically opposite angles)
∴ ∠COF = 105°
Similarly, ∠BOF = ∠AOE (Vertically opposite angles)
⇒ ∠BOF = 40°
Hence ∠AOC = 35°, ∠COF = 105°, ∠DOE = 105° and ∠BOF = 40°

Question 8.
AB, CD and EF are three concurrent lines passing through the point O such that OF bisects ∠BOD. If ∠BOF = 35°, find ∠BOC and ∠AOD.
Solution:
AB, CD and EF intersect at O. Such that OF is the bisector of
∠BOD ∠BOF = 35°

∵ OF bisects ∠BOD,
∴ ∠DOF = ∠BOF = 35° (Vertically opposite angles)
∴ ∠BOD = 35° + 35° = 70°
But ∠BOC + ∠BOD = 180° (Linear pair)
⇒ ∠BOC + 70° = 180°
⇒ ∠BOC = 180°-70°= 110°
But ∠AOD = ∠BOC (Vertically opposite angles)
= 110°
Hence ∠BOC = 110° and ∠AOD =110°

Question 9.
In the figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Solution:
In the figure, AB and CD intersect each other at O
∠AOC + ∠BOE = 70°
∠BOD = 40°

AB is a line
∴ ∠AOC + ∠BOE + ∠COE = 180° (Angles on one side of a line)
⇒ 70° + ∠COE = 180°
⇒ ∠COE = 180°-70°= 110°
and ∠AOC = ∠BOD (Vertically opposite angles)
⇒ ∠AOC = 40°
∴ ∠BOE = 70° – 40° = 30°
and reflex ∠COE = 360° – ∠COE
= 360°- 110° = 250°

Question 10.
Which of the following statements are true (T) and which are false (F)?
(i) Angles forming a linear pair are supplementary.
(ii) If two adjacent angles are equal, then each angle measures 90°.
(iii) Angles forming a linear pair can both be acute angles.
(iv) If angles forming a linear pair are equal, then each of these angles is of measure 90°.
Solution:
(i) True.
(ii) False. It can be possible if they are a linear pair.
(iii) False. In a linear pair, if one is acute, then the other will be obtuse.
(iv) True.

Question 11.
Fill in the blanks so as to make the following statements true:
(i) If one angle of a linear pair is acute, then its other angle will be …….. .
(ii) A ray stands on a line, then the sum of the two adjacent angles so formed is ……… .
(iii) If the sum of two adjacent angles is 180°, then the …… arms of the two angles are opposite rays.
Solution:
(i) If one angle of a linear pair is acute, then its other angle will be obtuse.
(ii) A ray stands on a line, then the sum of the two adjacent angles so formed is 180°.
(iii) If the sum of two adjacent angles is 180°, then the uncommon arms of the two angles are opposite rays.

Question 12.
Prove that the bisectors of a pair of vertically opposite angles are in the same straight line.
Solution:
Given : Lines AB and CD intersect each other at O.
OE and OF are the bisectors of ∠AOC and ∠BOD respectively

To prove : OE and OF are in the same line
Proof : ∵ ∠AOC = ∠BOD (Vertically opposite angles)
∵ OE and OF are the bisectors of ∠AOC and ∠BOD
∴ ∠1 = ∠2 and ∠3 = ∠4
⇒ ∠1 = ∠2 = 12 ∠AOC and
∠3 = ∠4 = 12 ∠BOD
∴ ∠1 = ∠2 = ∠3 = ∠4
∵ AOB is a line
∴ ∠BOD + ∠AOD = 180° (Linear pair)
⇒ ∠3 + ∠4 + ∠AOD = 180°
⇒ ∠3 + ∠1 + ∠AOD = 180° (∵ ∠1 = ∠4)
∴ EOF is a straight line

Question 13.
If two straight lines intersect each other, prove that the ray opposite to the bisector of one of the angles thus formed bisects the vertically opposite angle.
Solution:
Given : AB and CD intersect each other at O. OE is the bisector of ∠AOD and EO is produced to F.

To prove : OF is the bisector of ∠BOC
Proof : ∵ AB and CD intersect each other at O
∴ ∠AOD = ∠BOC (Vertically opposite angles)
∵OE is the bisector of ∠AOD
∴ ∠1 = ∠2
∵ AB and EF intersect each other at O
∴∠1 = ∠4 (Vertically opposite angles) Similarly, CD and EF intersect each other at O
∴ ∠2 = ∠3
But ∠1 = ∠2
∴ ∠3 = ∠4
OF is the bisector of ∠BOC

RD Sharma Class 9 Maths Book Questions Chapter 10 Congruent Triangles Ex 10.4

Question 1.
In the figure, AB || CD and ∠1 and ∠2 are in the ratio 3 : 2. Determine all angles from 1 to 8.

Solution:
AB || CD and l is transversal ∠1 : ∠2 = 3 : 2
Let ∠1 = 3x
Then ∠2 = 2x
But ∠1 + ∠2 = 180° (Linear pair)
∴ 3x + 2x = 180° ⇒ 5x = 180°
⇒ x = 180∘5  = 36°
∴ ∠1 = 3x = 3 x 36° = 108°
∠2 = 2x = 2 x 36° = 72°
Now ∠1 = ∠3 and ∠2 = ∠4 (Vertically opposite angles)
∴ ∠3 = 108° and ∠4 = 72°
∠1 = ∠5 and ∠2 = ∠6 (Corresponding angles)
∴ ∠5 = 108°, ∠6 = 72°
Similarly, ∠4 = ∠8 and
∠3 = ∠7
∴ ∠8 = 72° and ∠7 = 108°
Hence, ∠1 = 108°, ∠2= 72°
∠3 = 108°, ∠4 = 72°
∠5 = 108°, ∠6 = 72°
∠7 = 108°, ∠8 = 12°

Question 2.
In the figure, l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find ∠l, ∠2 and ∠3.

Solution:
l || m || n and p is then transversal which intersects then at X, Y and Z respectively ∠4 = 120°
∠2 = ∠4 (Alternate angles)
∴ ∠2 = 120°
But ∠3 + ∠4 = 180° (Linear pair)
⇒ ∠3 + 120° = 180°
⇒ ∠3 = 180° – 120°
∴ ∠3 = 60°
But ∠l = ∠3 (Corresponding angles)
∴ ∠l = 60°
Hence ∠l = 60°, ∠2 = 120°, ∠3 = 60°

Question 3.
In the figure, if AB || CD and CD || EF, find ∠ACE.

Solution:
Given : In the figure, AB || CD and CD || EF
∠BAC = 70°, ∠CEF = 130°
∵ EF || CD
∴ ∠ECD + ∠CEF = 180° (Co-interior angles)
⇒ ∠ECD + 130° = 180°
∴ ∠ECD = 180° – 130° = 50°
∵ BA || CD
∴ ∠BAC = ∠ACD (Alternate angles)
∴ ∠ACD = 70° (∵ ∠BAC = 70°)
∵ ∠ACE = ∠ACD – ∠ECD = 70° – 50° = 20°

Question 4.
In the figure, state which lines are parallel and why.

Solution:
In the figure,
∵ ∠ACD = ∠CDE = 100°
But they are alternate angles
∴ AC || DE

Question 5.
In the figure, if l || m,n|| p and ∠1 = 85°, find ∠2.

Solution:
In the figure, l || m, n|| p and ∠1 = 85°
∵ n || p
∴ ∠1 = ∠3 (Corresponding anlges)
But ∠1 = 85°
∴ ∠3 = 85°
∵ m || 1
∠3 + ∠2 = 180° (Sum of co-interior angles)
⇒ 85° + ∠2 = 180°
⇒ ∠2 = 180° – 85° = 95°

Question 6.
If two straight lines are perpendicular to the same line, prove that they are parallel to each other.
Solution:

Question 7.
Two unequal angles of a parallelogram are in the ratio 2:3. Find all its angles in degrees.
Solution:
In ||gm ABCD,

∠A and ∠B are unequal
and ∠A : ∠B = 2 : 3
Let ∠A = 2x, then
∠B = 3x
But ∠A + ∠B = 180° (Co-interior angles)
∴ 2x + 3x = 180°
⇒ 5x = 180°
⇒ x = 180∘5  = 36°
∴ ∠A = 2x = 2 x 36° = 72°
∠B = 3x = 3 x 36° = 108°
But ∠A = ∠C and ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠C = 72° and ∠D = 108°
Hence ∠A = 72°, ∠B = 108°, ∠C = 72°, ∠D = 108°

Question 8.
In each of the two lines is perpendicular to the same line, what kind of lines are they to each other?
Solution:
AB ⊥ line l and CD ⊥ line l

∴ ∠B = 90° and ∠D = 90°
∴ ∠B = ∠D
But there are corresponding angles
∴ AB || CD

Question 9.
In the figure, ∠1 = 60° and ∠2 = (23)3 a right angle. Prove that l || m.

Solution:
In the figure, a transversal n intersects two lines l and m
∠1 = 60° and
∠2 = 23 rd of a right angle 2
= 23 x 90° = 60°
∴ ∠1 = ∠2
But there are corresponding angles
∴ l || m

Question 10.
In the figure, if l || m || n and ∠1 = 60°, find ∠2.

Solution:
In the figure,

l || m || n and a transversal p, intersects them at P, Q and R respectively
∠1 = 60°
∴ ∠1 = ∠3 (Corresponding angles)
∴ ∠3 = 60°
But ∠3 + ∠4 = 180° (Linear pair)
60° + ∠4 = 180° ⇒ ∠4 = 180° – 60°
∴ ∠4 = 120°
But ∠2 = ∠4 (Alternate angles)
∴ ∠2 = 120°

Question 11.
Prove that the straight lines perpendicular to the same straight line are parallel to one another.
Solution:
Given : l is a line, AB ⊥ l and CD ⊥ l

Question 12.
The opposite sides of a quadrilateral are parallel. If one angle of the quadrilateral is 60°, find the other angles.
Solution:
In quadrilateral ABCD, AB || DC and AD || BC and ∠A = 60°

∵ AD || BC and AB || DC
∴ ABCD is a parallelogram
∴ ∠A + ∠B = 180° (Co-interior angles)
60° + ∠B = 180°
⇒ ∠B = 180°-60°= 120°
But ∠A = ∠C and ∠B = ∠D (Opposite angles of a ||gm)
∴ ∠C = 60° and ∠D = 120°
Hence ∠B = 120°, ∠C = 60° and ∠D = 120°

Question 13.
Two lines AB and CD intersect at O. If ∠AOC + ∠COB + ∠BOD = 270°, find the measure of ∠AOC, ∠COB, ∠BOD and ∠DOA.
Solution:
Two lines AB and CD intersect at O
and ∠AOC + ∠COB + ∠BOD = 270°
But ∠AOC + ∠COB + ∠BOD + ∠DOA = 360° (Angles at a point)

∴ 270° + ∠DOA = 360°
⇒ ∠DOA = 360° – 270° = 90°
But ∠DOA = ∠BOC (Vertically opposite angles)
∴ ∠BOC = 90°
But ∠DOA + ∠BOD = 180° (Linear pair)
⇒ 90° + ∠BOD = 180°
∴ ∠BOD= 180°-90° = 90° ,
But ∠BOD = ∠AOC (Vertically opposite angles)
∴ ∠AOC = 90°
Hence ∠AOC = 90°,
∠COB = 90°,
∠BOD = 90° and ∠DOA = 90°

Question 14.
In the figure, p is a transversal to lines m and n, ∠2 = 120° and ∠5 = 60°. Prove that m || n.

Solution:
Given : p is a transversal to the lines m and n
Forming ∠l, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8
∠2 = 120°, and ∠5 = 60°
To prove : m || n
Proof : ∠2 + ∠3 = 180° (Linear pair)
⇒ 120°+ ∠3 = 180°
⇒ ∠3 = 180°- 120° = 60°
But ∠5 = 60°
∴ ∠3 = ∠5
But there are alternate angles
∴ m || n

Question 15.
In the figure, transversal l, intersects two lines m and n, ∠4 = 110° and ∠7 = 65°. Is m || n?

Solution:
A transversal l, intersects two lines m and n, forming ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8
∠4 = 110° and ∠7 = 65°
To prove : Whether m || n or not
Proof : ∠4 = 110° and ∠7 = 65°
∠7 = ∠5 (Vertically opposite angles)
∴ ∠5 = 65°
Now ∠4 + ∠5 = 110° + 65° = 175°
∵ Sum of co-interior angles ∠4 and ∠5 is not 180°.
∴ m is not parallel to n

Question 16.
Which pair of lines in the figure are parallel? Give reasons.

Solution:
Given : In the figure, ∠A = 115°, ∠B = 65°, ∠C = 115° and ∠D = 65°
∵ ∠A + ∠B = 115°+ 65°= 180°
But these are co-interior angles,
∴ AD || BC
Similarly, ∠A + ∠D = 115° + 65° = 180°
∴ AB || DC

Question 17.
If l, m, n are three lines such that l ||m and n ⊥ l, prove that n ⊥ m.
Solution:
Given : l, m, n are three lines such that l || m and n ⊥ l

To prove : n ⊥ m
Proof : ∵ l || m and n is the transversal.
∴ ∠l = ∠2 (Corresponding angles)
But ∠1 = 90° (∵ n⊥l)
∴ ∠2 = 90°
∴ n ⊥ m

Question 18.
Which of the following statements are true (T) and which are false (F)? Give reasons.
(i) If two lines are intersected by a transversal, then corresponding angles are equal.
(ii) If two parallel lines are intersected by a transversal, then alternate interior angles are equal.
(iii) Two lines perpendicular to the same line are perpendicular to each other.
(iv) Two lines parallel to the same line are parallel to each other.
(v) If two parallel lines are intersected by a transversal, then the interior angles on the same side of the transversal are equal.
Solution:
(i) False. Because if lines are parallel, then it is possible.
(ii) True.
(iii) False. Not perpendicular but parallel to each other.
(iv) True.
(v) False. Sum of interior angles on the same side is 180° not are equal.

Question 19.
Fill in the blanks in each of the following to make the statement true:
(i) If two parallel lines are intersected by a transversal then each pair of corresponding angles are ……..
(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are …….
(iii) Two lines perpendicular to the same line are ……… to each other.
(iv) Two lines parallel to the same line are ……… to each other.
(v) If a transversal intersects a pair of lines in such away that a pair of alternate angles are equal, then the lines are …….
(vi) If a transversal intersects a pair of lines in such away that the sum of interior angles on the same side of transversal is 180°, then the lines are …….
Solution:
(i) If two parallel lines are intersected by a transversal, then each pair of corresponding angles are equal.
(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are supplementary.
(iii) Two lines perpendicular to the same line are parallel to each other.
(iv) Two lines parallel to the same line are parallel to each other.
(v) If a transversal intersects a pair of lines in such away that a pair of alternate angles are equal, then the lines are parallel.
(vi) If a transversal intersects a pair of lines in such away that the sum of interior angles on the same side of transversal is 180°, then the lines are parallel.

Question 20.
In the figure, AB || CD || EF and GH || KL. Find ∠HKL.

Solution:
In the figure, AB || CD || EF and KL || HG Produce LK and GH

∵ AB || CD and HK is transversal
∴ ∠1 = 25° (Alternate angles)
∠3 = 60° (Corresponding angles)
and ∠3 = ∠4 (Corresponding angles)
= 60°
But ∠4 + ∠5 = 180° (Linear pair)
⇒ 60° + ∠5 = 180°
⇒ ∠5 = 180° – 60° = 120°
∴ ∠HKL = ∠1 + ∠5 = 25° + 120° = 145°

Question 21.
In the figure, show that AB || EF.

Solution:
Given : In the figure, AB || EF
∠BAC = 57°, ∠ACE = 22°
∠ECD = 35° and ∠CEF =145°
To prove : AB || EF,
Proof : ∠ECD + ∠CEF = 35° + 145°
= 180°
But these are co-interior angles
∴ EF || CD
But AB || CD
∴ AB || EF

Question 22.
In the figure, PQ || AB and PR || BC. If ∠QPR = 102°. Determine ∠ABC. Give reasons.

Solution:
In the figure, PQ || AB and PR || BC
∠QPR = 102°
Produce BA to meet PR at D

∵ PQ || AB or DB
∴ ∠QPR = ∠ADR (Corresponding angles)
∴∠ADR = 102° or ∠BDR = 102°
∵ PR || BC
∴ ∠BDR + ∠DBC = 180°
(Sum of co-interior angles) ⇒ 102° + ∠DBC = 180°
⇒ ∠DBC = 180° – 102° = 78°
⇒ ∠ABC = 78°

Question 23.
Prove that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary.
Solution:
Given : In two angles ∠ABC and ∠DEF AB ⊥ DE and BC ⊥ EF
To prove: ∠ABC + ∠DEF = 180° or ∠ABC = ∠DEF
Construction : Produce the sides DE and EF of ∠DEF, to meet the sides of ∠ABC at H and G.

Proof: In figure (i) BGEH is a quadrilateral
∠BHE = 90° and ∠BGE = 90°
But sum of angles of a quadrilateral is 360°
∴ ∠HBG + ∠HEG = 360° – (90° + 90°)
= 360° – 180°= 180°
∴ ∠ABC and ∠DEF are supplementary
In figure (if) in quadrilateral BGEH,
∠BHE = 90° and ∠HEG = 90°
∴ ∠HBG + ∠HEG = 360° – (90° + 90°)
= 360°- 180° = 180° …(i)
But ∠HEF + ∠HEG = 180° …(ii) (Linear pair)
From (i) and (ii)
∴ ∠HEF = ∠HBG
⇒ ∠DEF = ∠ABC
Hence ∠ABC and ∠DEF are equal or supplementary

Question 24.
In the figure, lines AB and CD are parallel and P is any point as shown in the figure. Show that ∠ABP + ∠CDP = ∠DPB.

Solution:
Given : In the figure, AB || CD
P is a point between AB and CD PD
and PB are joined
To prove : ∠APB + ∠CDP = ∠DPB
Construction : Through P, draw PQ || AB or CD

Proof: ∵ AB || PQ
∴ ∠ABP = BPQ …(i) (Alternate angles)
Similarly,
CD || PQ
∴ ∠CDP = ∠DPQ …(ii)
(Alternate angles)
Adding (i) and (ii)
∠ABP + ∠CDP = ∠BPQ + ∠DPQ
Hence ∠ABP + ∠CDP = ∠DPB

Question 25.
In the figure, AB || CD and P is any point shown in the figure. Prove that:
∠ABP + ∠BPD + ∠CDP = 360°

Solution:
Given : AB || CD and P is any point as shown in the figure
To prove : ∠ABP + ∠BPD + ∠CDP = 360°
Construction : Through P, draw PQ || AB and CD

Proof : ∵ AB || PQ
∴ ∠ABP+ ∠BPQ= 180° ……(i) (Sum of co-interior angles)
Similarly, CD || PQ
∴ ∠QPD + ∠CDP = 180° …(ii)
Adding (i) and (ii)
∠ABP + ∠BPQ + ∠QPD + ∠CDP
= 180°+ 180° = 360°
⇒ ∠ABP + ∠BPD + ∠CDP = 360°

Question 26.
In the figure, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF. Prove that ∠ABC = ∠DEF.

Solution:
Given : In ∠ABC and ∠DEF. Their arms are parallel such that BA || ED and BC || EF
To prove : ∠ABC = ∠DEF
Construction : Produce BC to meet DE at G
Proof: AB || DE
∴ ∠ABC = ∠DGH…(i) (Corresponding angles)

BC or BH || EF
∴ ∠DGH = ∠DEF (ii) (Corresponding angles)
From (i) and (ii)
∠ABC = ∠DEF

Question 27.
In the figure, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF. Prove that ∠ABC + ∠DEF = 180°.

Solution:
Given: In ∠ABC = ∠DEF
BA || ED and BC || EF
To prove: ∠ABC = ∠DEF = 180°
Construction : Produce BC to H intersecting ED at G

Proof: ∵ AB || ED
∴ ∠ABC = ∠EGH …(i) (Corresponding angles)
∵ BC or BH || EF
∠EGH || ∠DEF = 180° (Sum of co-interior angles)
⇒ ∠ABC + ∠DEF = 180° [From (i)]
Hence proved.

RD Sharma Class 9 Solution Chapter 10 Congruent Triangles VSAQS

Question 1.
Define complementary angles.
Solution:
Two angles whose sum is 90°, are called complementary angles.

Question 2.
Define supplementary angles.
Solution:
Two angles whose sum is 180°, are called supplementary angles.

Question 3.
Define adjacent angles.
Solution:
Two angles which have common vertex and one arm common are called adjacent angles.

Question 4.
The complement of an acute angles is…….
Solution:
The complement of an acute angles is an acute angle.

Question 5.
The supplement of an acute angles is………
Solution:
The supplement of an acute angles is a obtuse angle.

Question 6.
The supplement of a right angle is…….
Solution:
The supplement of a right angle is a right angle.

Question 7.
Write the complement of an angle of measure x°.
Solution:
The complement of x° is (90° – x)°

Question 8.
Write the supplement of an angle of measure 2y°.
Solution:
The supplement of 2y° is (180° – 2y)°

Question 9.
If a wheel has six spokes equally spaced, then find the measure of the angle between two adjacent spokes.
Solution:
Total measure of angle around a point = 360°
Number of spokes = 6
∴ Angle between the two adjacent spokes = 360∘6 = 60°

Question 10.
An angle is equal to its supplement. Determine its measure.
Solution:
Let required angle = x°
Then its supplement angle = 180° – x
x = 180° – x
⇒ x + x = 180°
⇒  2x = 180° ⇒  x = 180∘2 = 90°
∴ Required angle = 90°

Question 11.
An angle is equal to five times its complement. Determine its measure.
Solution:
Let required measure of angle = x°
∴  Its complement angle = 90° – x
∴  x = 5(90° – x)
⇒  x = 450° – 5x
⇒  x + 5x = 450°
⇒  6x = 450°
⇒ x = 450∘6 = 75°
∴ Required angle = 75°

Question 12.
How many pairs of adjacent angles are formed when two lines intersect in a point?
Solution:
If two lines AB and CD intersect at a point O, then pairs of two adjacent angles are, ∠AOC and ∠COB, ∠COB and ∠BOD, ∠BOD and DOA, ∠DOA and ∠ZAOC
i.e, 4 pairs

RD Sharma Solutions Class 9 Chapter 10 Congruent Triangles MCQS

Mark the correct alternative in each of the following:
Question 1.
One angle is equal to three times its supplement. The measure of the angle is
(a) 130°
(b) 135°
(c) 90°
(d) 120°
Solution:
Let required angle = x
Then its supplement = (180° – x)
x = 3(180° – x) = 540° – 3x
⇒ x + 3x = 540°
⇒ 4x = 540°
⇒ x = 540∘4  = 135°
∴ Required angle = 135° (b)

Question 2.
Two straight lines AB and CD intersect one another at the point O. If ∠AOC + ∠COB + ∠BOD = 274°, then ∠AOD =
(a) 86°
(b) 90°
(c) 94°
(d) 137°
Solution:
Sum of angles at a point O = 360°
Sum of three angles ∠AOC + ∠COB + ∠BOD = 274°

∴ Fourth angle ∠AOD = 360° – 274°
= 86° (a)

Question 3.
Two straight lines AB and CD cut each other at O. If ∠BOD = 63°, then ∠BOC =
(a) 63°
(b) 117°
(c) 17°
(d) 153°
Solution:
CD is a line
∴ ∠BOD + ∠BOC = 180° (Linear pair)

⇒ 63° + ∠BOC = 180°
⇒ ∠BOC = 180° – 63°
∴ ∠BOC =117° (b)

Question 4.
Consider the following statements:
When two straight lines intersect:
(i) adjacent angles are complementary
(ii) adjacent angles are supplementary
(iii) opposite angles are equal
(iv) opposite angles are supplementary Of these statements
(a) (i) and (iii) are correct
(b) (ii) and (iii) are correct
(c) (i) and (iv) are correct
(d) (ii) and (iv) are correct
Solution:
Only (ii) and (iii) arc true. (b)

Question 5.
Given ∠POR = 3x and ∠QOR = 2x + 10°. If POQ is a striaght line, then the value of x is
(a) 30°
(b) 34°
(c) 36°
(d) none of these
Solution:
∵ POQ is a straight line
∴ ∠POR + ∠QOR = 180° (Linear pair)
⇒ 3x + 2x + 10° = 180°
⇒ 5x = 180 – 10° = 170°
∴ x = 170∘5  = 34° (b)

Question 6.
In the figure, AOB is a straight line. If ∠AOC + ∠BOD = 85°, then ∠COD =
(a) 85°
(b) 90°
(c) 95°
(d) 100°

Solution:
AOB is a straight line,
OC and OD are rays on it
and ∠AOC + ∠BOD = 85°
But ∠AOC + ∠BOD + ∠COD = 180°
⇒ 85° + ∠COD = 180°
∠COD = 180° – 85° = 95° (c)

Question 7.
In the figure, the value of y is
(a) 20°
(b) 30°
(c) 45°
(d) 60°

Solution:
In the figure,

y = x (Vertically opposite angles)
∠1 = 3x
∠2 = 3x
∴ 2(x + 3x + 2x) = 360° (Angles at a point)
2x + 6x + 4x = 360°
12x = 360° ⇒ x = 360∘12  = 30°
∴ y = x = 30° (b)

Question 8.
In the figure, the value of x is
(a) 12
(b) 15
(c) 20
(d) 30

Solution:
∠1 = 3x+ 10 (Vertically opposite angles)
But x + ∠1 + ∠2 = 180°

⇒ x + 3x + 10° + 90° = 180°
⇒ 4x = 180° – 10° – 90° = 80°
x = 80∘4 = 20   (c)

Question 9.
In the figure, which of the following statements must be true?
(i) a + b = d + c
(ii) a + c + e = 180°
(iii) b + f= c + e
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (ii) and (iii) only

Solution:
In the figure,
(i) a + b = d + c
a° = d°
b° = e°
c°= f°
(ii) a + b + e = 180°
a + e + c = 180°
⇒ a + c + e = 180°
(iii) b + f= e + c
∴ (ii) and (iii) are true statements (d)

Question 10.
If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2:3, then the measure of the larger angle is
(a) 54°
(b) 120°
(c) 108°
(d) 136°
Solution:
In figure, l || m and p is transversal

= 35 x 180° = 108° (c)

Question 11.
In the figure, if AB || CD, then the value of x is
(a) 20°
(b) 30°
(c) 45°
(d) 60°

Solution:
In the figure, AB || CD,
and / is transversal
∠1 = x (Vertically opposite angles)
and 120° + x + ∠1 = 180° (Co-interior angles)

Question 12.
Two lines AB and CD intersect at O. If ∠AOC + ∠COB + ∠BOD = 270°, then ∠AOC =
(a) 70°
(b) 80°
(c) 90°
(d) 180°
Solution:
Two lines AB and CD intersect at O

∠AOC + ∠COB + ∠BOD = 270° …(i)
But ∠AOC + ∠COB + ∠BOD + ∠DOA = 360° …(ii)
Subtracting (i) from (ii),
∠DOA = 360° – 270° = 90°
But ∠DOA + ∠AOC = 180°
∴ ∠AOC = 180° – 90° = 90° (c)

Question 13.
In the figure, PQ || RS, ∠AEF = 95°, ∠BHS = 110° and ∠ABC = x°. Then the value of x is
(a) 15°
(b) 25°
(c) 70°
(d) 35°

Solution:
In the figure,

PQ || RS, ∠AEF = 95°
∠BHS = 110°, ∠ABC = x
∵ PQ || RS,
∴ ∠AEF = ∠1 = 95° (Corresponding anlges)
But ∠1 + ∠2 = 180° (Linear pair)
⇒ ∠2 = 180° – ∠1 = 180° – 95° = 85°
In ∆AGH,
Ext. ∠BHS = ∠2 +x
⇒ 110° = 85° + x
⇒ x= 110°-85° = 25° (b)

Question 14.
In the figure, if l1 || l2, what is the value of x?
(a) 90°
(b) 85°
(c) 75°
(d) 70°

Solution:
In the figure,

∠1 = 58° (Vertically opposite angles)
Similarly, ∠2 = 37°
∵ l₁ || l₂, EF is transversal
∠GEF + EFD = 180° (Co-interior angles)
⇒ ∠2 + ∠l +x = 180°
⇒ 37° + 58° + x = 180°
⇒ 95° + x= 180°
x = 180°-95° = 85° (b)

Question 15.
In the figure, if l₁ || l₂, what is x + y in terms of w and z?
(a) 180-w + z
(b) 180° + w- z
(c) 180 -w- z
(d) 180 + w + z

Solution:
In the figure, l₁ || l₂

p and q are transversals
∴ w + x = 180° ⇒ x = 180° – w (Co-interior angle)
z = y (Alternate angles)
∴ x + y = 180° – w + z (a)

Question 16.
In the figure, if l₁ || l₂, what is the value of y?
(a) 100
(b) 120
(c) 135
(d) 150

Solution:
In the figure, l₁ || l₂ and l₃ is the transversal

Question 17.
In the figure, if l1 || l2 and l3 || l4 what is y in terms of x?
(a) 90 + x
(b) 90 + 2x
(c) 90 – x2
(d) 90 – 2x

Solution:
In the figure,

l₁ || l₂ and l₃ || l₄ and m is the angle bisector
∴ ∠2 = ∠3 = y
∵ l₁ || l₂
∠1 = x (Corresponding angles)
∵ l₃ || l₄
∴ ∠1 + (∠2 + ∠3) = 180° (Co-interior angles)
⇒ x + 2y= 180°
⇒ 2y= 180°-x
⇒ y = 540∘−x4
= 90° – x2 (c)

Question 18.
In the figure, if 11| m, what is the value of x?
(a) 60
(b) 50
(c) 45
d) 30

Solution:
In the figure, l || m and n is the transversal

⇒ y = 25°
But 2y + 25° = x+ 15°
(Vertically opposite angles) ⇒ x = 2y + 25° – 15° = 2y+ 10°
= 2 x 25°+10° = 50°+10° = 60° (a)

Question 19.
In the figure, if AB || HF and DE || FG, then the measure of ∠FDE is
(a) 108°
(b) 80°
(c) 100°
(d) 90°

Solution:
In the figure,
AB || HF, DE || FG

∴ HF || AB
∠1 =28° (Corresponding angles)
But ∠1 + ∠FDE + 72° – 180° (Angles of a straight line)
⇒ 28° + ∠FDE + 72° = 180°
⇒ ∠FDE + 100° = 180°
⇒ ∠FDE = 180° – 100 = 80° (b)

Question 20.
In the figure, if lines l and m are parallel, then x =
(a) 20°
(b) 45°
(c) 65°
(d) 85°

Solution:
In the figure, l || m

∴ ∠1 =65° (Corresponding angles)
In ∆BCD,
Ext. ∠1 = x + 20°
⇒ 65° = x + 20°
⇒ x = 65° – 20°
⇒ x = 45° (b)

Question 21.
In the figure, if AB || CD, then x =
(a) 100°
(b) 105°
(c) 110°
(d) 115°

Solution:
In the figure, AB || CD
Through P, draw PQ || AB or CD

∠A + ∠1 = 180° (Co-interior angles)
⇒ 132° + ∠1 = 180°
⇒ ∠1 = 180°- 132° = 48°
∴ ∠2 = 148° – ∠1 = 148° – 48° = 100°
∵ DQ || CP
∴ ∠2 = x (Corresponding angles)
∴ x = 100° (a)

Question 22.
In tlie figure, if lines l and in are parallel lines, then x =
(a) 70°
(b) 100°
(c) 40°
(d) 30°

Solution:
In the figure, l || m

∠l =70° (Corresponding angles)
In ∆DEF,
Ext. ∠l = x + 30°
⇒ 70° = x + 30°
⇒ x = 70° – 30° = 40° (c)

Question 23.
In the figure, if l || m, then x =
(a) 105°
(b) 65°
(c) 40°
(d) 25°

Solution:
In the figure,
l || m and n is the transversal

∠1 = 65° (Alternate angles)
In ∆GHF,
Ext. x = ∠1 + 40° = 65° + 40°
⇒ x = 105°
∴ x = 105° (a)

Question 24.
In the figure, if lines l and m are parallel, then the value of x is
(a) 35°
(b) 55°
(c) 65°
(d) 75°

Solution:
In the figure, l || m
and PQ is the transversal

∠1 = 90°
In ∆EFG,
Ext. ∠G = ∠E + ∠F
⇒ 125° = x + ∠1 = x + 90°
⇒ x = 125° – 90° = 35° (a)

Question 25.
Two complementary angles are such that two times the measure of one is equal to three times the measure of the other. The measure of the smaller angle is
(a) 45°
(b) 30°
(c) 36°
(d) none of these
Solution:
Let first angle = x
Then its complementary angle = 90° – x
∴ 2x = 3(90° – x)
⇒ 2x = 270° – 3x
⇒ 2x + 3x = 270°
⇒ 5x = 270°
⇒ x = 270∘5  = 54°
∴ second angle = 90° – 54° = 36°
∴ smaller angle = 36° (c)

Question 26.

Solution:

Question 27.
In the figure, AB || CD || EF and GH || KL.
The measure of ∠HKL is
(a) 85°
(b) 135°
(c) 145°
(d) 215°

Solution:
In the figure, AB || CD || EF and GH || KL and GH is product to meet AB in L.

∵ AB || CD
∴ ∠1 = 25° (Alternate angle)
and GH || KL
∴ ∠4 = 60° (Corresponding angles)
∠5 = ∠4 = 60° (Vertically opposite angle)
∠5 + ∠2 = 180° (Co-interior anlges)
∴ ⇒ 60° + ∠2 = 180°
∠2 = 180° – 60° = 120°
Now ∠HKL = ∠1 + ∠2 = 25° + 120°
= 145° (c)

Question 28.
AB and CD are two parallel lines. PQ cuts AB and CD at E and F respectively. EL is the bisector of ∠FEB. If ∠LEB = 35°, then ∠CFQ will be
(a) 55°
(b) 70°
(c) 110°
(d) 130°
Solution:
AB || CD and PQ is the transversal EL is the bisector of ∠FEB and ∠LEB = 35°

∴ ∠FEB = 2 x 35° = 70°
∵ AB || CD
∴ ∠FEB + ∠EFD = 180°
(Co-interior angles)
70° + ∠EFD = 180°
∴ ∠EFD = 180°-70°= 110°
But ∠CFQ = ∠EFD
(Vertically opposite angles)
∴ ∠CFQ =110° (c)

Question 29.
In the figure, if line segment AB is parallel to the line segment CD, what is the value of y?
(a) 12
(b) 15
(c) 18
(d) 20

Solution:
In the figure, AB || CD
BD is transversal
∴ ∠ABD + ∠BDC = 180° (Co-interior angles)
⇒y + 2y+y + 5y = 180°
⇒ 9y = 180° ⇒ y = 180∘9  = 20° (d)

Question 30.
In the figure, if CP || DQ, then the measure of x is
(a) 130°
(b) 105°
(c) 175°
(d) 125°

Solution:
In the figure, CP || DQ
BA is transversal
Produce PC to meet BA at D

∵ QB || PD
∴ ∠D = 105° (Corresponding angles)
In ∆ADC,
Ext. ∠ACP = ∠CDA + ∠DAC
⇒ x = ∠1 + 25°
= 105° + 25° = 130° (a)

Important concepts from RD Sharma Solutions Class 9 Maths Chapter 10 Congruent Triangles

• Congruence of Line segments
• Congruence of Angles
• Congruence of Triangles
• Congruence Relation
• Some inequality Relations in a Triangle

This is the complete blog on RD Sharma Solution Class 9 Maths Chapter 9. If you have any doubts regarding the CBSE Class 9 Maths exam, ask in the comments.

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