Class 7 Exam  >  Class 7 Notes  >  RD Sharma Solutions for Class 7 Mathematics  >  RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics PDF Download

Question 15:

One of the angles forming a linear pair is a right angle. What can you say about its other angle?

Answer 15:

One angle of a linear pair is the right angle, i.e., 90°.
∴ The other angle = 180°​ - 90° = 90​°

Question 16:

One of the angles forming a linear pair is an obtuse angle. What kind of angle is the other?

Answer 16:

If one of the angles of a linear pair is obtuse, then the other angle should be acute; only then can their sum be 180°.

Question 17:

One of the angles forming a linear pair is an acute angle. What kind of angle is the other?

Answer 17:

In a linear pair, if one angle is acute, then the other angle should be obtuse. Only then their sum can be 180°.

Question 18:

Can two acute angles form a linear pair?

Answer 18:

No, two acute angles cannot form a linear pair because their sum is always less than 180°.

Question 19:

If the supplement of an angle is 65°; then find its complement.

Answer 19:

Let be the required angle.
Then, we have: 
x + 65° = 180°
⇒⇒x = 180° - 65° = 115°

The complement of angle cannot be determined.

 

Question 20:

Find the value of x in each of the following figures.

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Answer 20:

(i)
Since ∠BOA+∠BOC=180°                  (Linear pair)
∴ ∠x=180°−∠BOA=180°−60°=120°

(ii)
Since ∠QOP+∠QOR=180°         (Linear pair)

∴2x+3x=180°

⇒5x=180°

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

(v)
2x°+x°+2x°+3x°=180°⇒8x=180

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

(vi)
3x°=105°

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Question 21:

In Fig. 22, it being given that ∠1 = 65°, find all other angles.

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Answer 21:

∠1=∠3∠1=∠3          (Vertically opposite angles)
∴∠3=65°∴∠3=65°
Since ∠1+∠2=180°       (Linear pair)
∴∠2=180°−65°=115°
∠2=∠4         (Vertically opposite angles)
∴∠4=∠2=115° and ∠3=65°

 

Question 22:

In Fig., OA and OB are opposite rays:

 

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

(i) If x = 25°, what is the value of y?
(ii) If y = 35°, what is the value of x?

Answer 22:

∠AOC + ∠BOC = 180°                   (Linear pair)
⇒(2y+5)+3x=180°
⇒3x+2y=175°

i) If x = 25°, then
3×25°+2y=175°

⇒75°+2y=175°
⇒2y=175°−75°=100°

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

(ii) If y = 35°, then
3x+2×35°=175°
⇒3x+70°=175°
⇒3x=175°−70°=105°

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

 

Question 23:

In Fig., write all pairs of adjacent angles and all the linear pairs.

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

 

Answer 23:

Adjacent angles:

∠DOA and ∠DOC
∠DOC and ∠BOC

∠AOD and ∠DOB
∠BOC and ∠AOC

Linear pairs of angles:

∠AOD and ∠DOB
∠BOC and ∠AOC

 

Question 24:

In Fig. 25, find ∠x. Further find ∠BOC, ∠COD and ∠AOD.

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Answer 24:

∠AOD+∠DOC+∠COB=180°(Linear pair)

(x+10)°+x°+(x+20)°=180°
3x+30°=180°
3x=180°−30°
3x=150°

x=150°/3=50°

∠BOC=x+20°=50°+20°=70°
∠COD=x=50°
∠AOD=x+10°=50°+10°=60°

 

Question 25:

How many pairs of adjacent angles are formed when two lines intersect in a point?

Answer 25:

If two lines intersect at a point, then four adjacent pairs are formed, and those pairs are linear as well.

Question 26:

How many pairs of adjacent angles, in all, can you name in Fig.?

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Answer 26:

There are 10 adjacent pairs in the given figure; they are:
∠EOD and ∠DOC
∠COD and ∠BOC
∠COB and ∠BOA

∠AOB and ∠BOD
∠BOC and ∠COE
∠COD and ∠COA
∠DOE and ∠DOB

∠EOD and ∠DOA
∠EOC and ∠AOC
∠AOB and ∠BOE
 

Question 27:

In Fig., determine the value of x.

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Answer 27:

∠AOB+∠BOC=180°           (Linear pair)
⇒3x+3x=180°

⇒6x=180°

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

 

Question 28:

In Fig., AOC is a line, find x.

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Answer 28:

∠AOB+∠BOC=180°                (Linear pair)
⇒70°+2x=180°
⇒2x=180°−70°=110°

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

 

Question 29:

In Fig., POS is a line, find x.

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Answer 29:

∠QOP+∠QOR+∠ROS=180°       (Angles on a straight line)

⇒60°+4x+40°=180°
⇒100°+4x=180°
⇒4x=180°−100°=80°

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

 

Question 30:

In Fig., lines l1 and l2 intersect at O, forming angles as shown in the figure. If x = 45°, find the values of yzand u.

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Answer 30:

∠z=∠x=45°       (Vertically opposite angles)
Now,∠x+∠y=180°      (Linear pair)
⇒∠y=180°−45°=135°
∠u=∠y=135°       (Vertically opposite angles)

 

Question 31:

In Fig., three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of xyz and u.

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

 

Answer 31:

∠BOD + ∠DOF + ∠FOA = 180°        (Linear pair)
∴ ∠FOA = ∠= 180°−90°−50°=40
∠FOA=∠x=40°    (Vertically opposite angles)
∠BOD=∠z=90°    (Vertically opposite angles)
∠EOC=∠y=50°    (Vertically opposite angles)

 

Question 32:

In Fig., find the values of xy and z.

RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics

Answer 32:

∠y=25°       (Vertically opposite angles)
Since ∠x+∠y=180°         (Linear pair)
∴∠x=180°−25°=155°
∠z=∠x=155°        (Vertically opposite angles)

The document RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math | RD Sharma Solutions for Class 7 Mathematics is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
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FAQs on RD Sharma Solutions (Part - 2) - Ex-14.1, Lines and Angles, Class 7, Math - RD Sharma Solutions for Class 7 Mathematics

1. What are the types of angles?
Ans. There are four types of angles: acute angle (less than 90 degrees), right angle (exactly 90 degrees), obtuse angle (between 90 and 180 degrees), and straight angle (exactly 180 degrees).
2. How do you identify parallel lines?
Ans. Parallel lines are lines that never intersect each other. They have the same slope and can be identified by using the slope-intercept form of a linear equation.
3. What is the sum of interior angles of a triangle?
Ans. The sum of the interior angles of a triangle is always 180 degrees. This property is known as the Triangle Sum Theorem.
4. What is the difference between alternate interior angles and corresponding angles?
Ans. Alternate interior angles are formed when a transversal intersects two parallel lines. They are located on opposite sides of the transversal and are congruent. Corresponding angles, on the other hand, are located on the same side of the transversal and in the same relative position. They are also congruent when the lines being intersected are parallel.
5. How do you find the measure of an exterior angle of a triangle?
Ans. The measure of an exterior angle of a triangle is equal to the sum of the measures of the two interior angles that are not adjacent to it.
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