Class 6: Maths Chapter 13 solutions. Complete Class 6 Maths Chapter 13 Notes.
Contents
RS Aggarwal Solutions for Class 6 Maths Chapter 13–Angles and Their Measurement
RS Aggarwal 6th Maths Chapter 13, Class 6 Maths Chapter 13 solutions
Ex 13A Solutions
Question 1.
Solution:
Three examples are : Tongs, Scissors and Compasses.
Question 2.
Solution:
In the given angle ABC, the vertex is B and arms are AB−→− and BC−→− .
Question 3.
Solution:
(i) In the given figure, three angles are formed.
Names of the angles are :
∠ABC, ∠BAC and ∠ACB
(ii) In the given figure, four angles are formed.
∠ABC, ∠BCD, ∠CDA and ∠BAD
(iii) In the given figure, eight angles are formed.
Names of the angles are :
∠ABC, ∠BCD, ∠CDA, ∠BAD, ∠ABD, ∠DCB, ∠ADB and ∠BDC
Question 4.
Solution:
In given figure
(i) Points S and Q are in the interior of ∠AOB
(ii) Points P and R are in the exterior of ∠AOB.
(iii) Points A, O, B, N, T lie on ∠AOB.
Question 5.
Solution:
(i) False
(ii) True
(iii) False
(iv) True
(v) False
Question 6.
Solution:
In the given figure, another name for :
(i) ∠1 is ∠EPB
(ii) ∠2 is ∠PQC
(iii) ∠3 is FQD
Ex 13B Solutions
Question 1.
Solution:
(i) Obtuse angle
(ii) Right angle
(iii) straight angle
(iv) Reflex angle
(v) Acute angle
(vi) Complete angle
Question 2.
Solution:
We know that an acute angle is less than 90°
(ii) a right angle is equal to 90°
(iii) an obtuse angle is greater than 90° but less than 180°
(iv) an angle equal to 180° is a straight angle
(v) angle greater than 180° but less than 360° is called a reflex angle
(vi) angle equal to 360° is called a complete angle and angle equal to 0° is called a zero angle. Now the angles are :
(i) acute
(ii) obtuse
(iii) obtuse
(iv) right
(v) reflex
(vi) complete
(vii) obtuse
(viii) obtuse
(ix) acute
(x) acute
(xi) zero
(xii) acute Ans.
Question 3.
Solution:
(i) One right angle = 90°
(ii) Two right angles = (2 x 90)° = 180°
(iii) Three right angles = (3 x 90)° = 270°
(iv) Four right angles = (4 x 90)° = 360°
(v) 23 right angle = (23×90O) = 60°
(vi) 1½ right angle = (112×90O)
(32×90O) = 135°
Question 4.
Solution:
(i) When it is 3 o’ clock, the minute hand is at 12, and hour hand is at 3 as shown in the figure, clearly, the angle between the two hands 90°.
(ii) When it is 6 o’ clock, the minute hand is at 12 and the hour hand is at 6 as shown in the figure. Clearly, the angle between the two hands of the clock is a straight angle is i.e. 180°.
(iii) When it is 12 o’ clock, both the hands of the clock lie at 12 as shown in the figure. Clearly, the angle between the two hands = 0°.
(iv) When it is 9 o’ clock, the minute hand is at 12 and the hour hand is at 9 as shown in the figure. Clearly, the angle between the two hands = 90°.
Question 5.
Solution:
(i) Take the rular and draw any ray OA. Again using the rular, starting from O, draw a ray OB in such a way that the angle formed is less than 90°. Then, ∠AOB is the required acute angle.
(ii) Take the rular and draw any ray OA. Now, starting from O, draw another ray OB, with the help of the rular, such that the angle formed is greater than a right angle.
Then, ∠AOB is the required obtuse angle.
(iii) Take a rular and draw any ray OA. Now, starting from O, draw ray OB in the opposite direction of the ray OA. Then ∠AOB is the required straight angle.
Ex 13C Solutions
Question 1.
Solution:
(i) Place the protractor in such a way that its centre is exactly at the vertex O of the given angle AOB and the base line lies along the arm OA. Read off the mark through which the arm OB passes, starting from 0° on the side A.
We find that ∠AOB = 45°.
(ii) The given angle is ∠PQR. Place the protractor in such a way that its centre is exactly on the vertex Q of the given angle and the base line lies along the arm QR.Read off the mark through which the arm QP passes, starting from 0° on the side of R.
We find that ∠PQR = 67°
(in) The given angle is ∠DEF. Place the protractor in such a way that its centre is exactly on the vertex E of the given angle and the base line lies along the arm ED. Read off the mark through which the arm EF passes, starting from 0° on the side of D.
We find that ∠DEF = 130°
(iv) The given angle is ∠LMN. Place the protractor in such a way that its centre is exactly on the vertex M of the given angle and the base line lies along the arm ML. Read off the mark through which the arm MN passes, starting from 0° on the side of L.
We find that ∠LMN = 50°
(v) The given angle is ∠RST. Place the protractor in such a way that its centre is exactly on the vertex S of the given angle and the base line lies along the arm SR. Read off the mark through which the arm ST passes, starting from 0° on the side of R.
We find that the ∠RST = 130°.
(vi) The given angle is ∠GHI. Place the protractor in such a way that its centre is exactly on the vertex H of the given angle and the base line lies along the arm HI. Read off the mark through which the arm HG passes, starting from 0° on the side of I.
We find that ∠GHI = 70°
Question 2.
Solution:
(i) Draw a ray OA. Place the protractor in such a way that its centre lies exactly at O and the base line lies along OA. Starting from 0° on the side of A, look for the 25° mark on the protractor. Mark a point B at this 25° mark. Remove the protractor and draw the ray OB. Then ∠AOB is the required angle.
(ii) Draw a ray OA. Place the protractor in such a way that its centre lies exactly at O and the base line lies along OA. Starting from 0° on the side of A, look for the 72° mark on the protractor. Mark a point B at this 72° mark. Remove the protractor and draw the ray OB. Then,
∠AOB is the required angle of measure 72°.
(iii) Draw a ray OA. Place the protractor in such a way that its centre lies exactly at O and the base line lies along OA. Starting from 0° on the side of A, look for the 90° mark on the protractor. Mark a point B at this 90° mark. Remove the protractor and draw the ray OB. Then ∠AOB is the required angle whose measure is 90°.
(iv) Draw a ray OA. Place the protractor in such a way that its centre exactly lies at O and the base line lies along OA. Starting from 0° on the side of A, look for the 117° mark on the protractor. Mark a point B at this 117° mark. Remove the protractor and draw the ray OB. Then ∠AOB is the required angle whose measure is 117°.
(v) Draw a ray OP. Place the protractor in such a way that its centre lies exactly at O and the base line lies along OP. Starting from 0° on the side of P, look for the 165° mark on the protractor. Mark a point Q at this 165° mark. Remove the protractor and draw the ray OQ. Then, ∠POQ is the required angle whose measure is 165°.
(vi) Draw a ray OP. Place the protractor in such a way that its centre lies exactly at O and the base line lies along OP. Starting from 0° on the side of P, look for the 23° mark on the protractor. Mark a point Q at this 23° mark. Remove the protractor and draw the ray OQ. Then ∠POQ is the required angle whose measure is 23°.
(vii) Draw a ray OA. Place the protractor in such a way that its centre lies exactly at O and the base line lies along OA. Starting from 0° on the side of A,Took for the 180° mark on the protractor. Mark a point B on this 180° mark. Remove the protractor and draw the ray OB. Then ∠AOB is the required angle whose measure is 180°.
(viii) Draw a ray Rs. Place the protractor in such away that its centre lies exactly at R and the base line lies along RS. Starting from 0° on the side of S, look for the 48° mark on the protractor. Mark a point T at this 48° mark. Remove the protractor and draw the ray RT. Then, ∠SRT is the required angle whose measure is 48°.
Question 3.
Solution:
On measuring the given angle ABC with the help of a protractor, it is 50°
Now, place the protractor on EF in such a way that its centre lies on E exactly and base with the line EF.
Now read off the mark through with the arm ED passes at 50°.
Join DE,
Then ∠DEF is equal to 50° i.e. equal to ∠ABC.
Question 4.
Solution:
Steps of construction :
(i) Draw a line segment AB = 6 cm.
(ii) Take a point C on AB such that AC = 4 cm.
(iii) Place protractor with its centre at C and base along CB.
(iv) Mark a point D against 90°.
(v) Remove the protractor and join DC. Then DC ⊥ AB. Ans.
Ex 13D Solutions
Objective questions
Mark against the correct answer in each of following.
Question 1.
Solution:
(c) vertex of an angle lie on it.
Question 2.
Solution:
(c) an angle.
Question 3.
Solution:
(c) A straight angle has 180°
Question 4.
Solution:
An angle measuring 90° is called a right angle. (b)
Question 5.
Solution:
An angle measuring 91° is an obtuse angle as it is more than 90° and less than 180°.(b)
Question 6.
Solution:
An angle measuring 270° is a reflex angles as it is greater than 180° and less than 360°. (d)
Question 7.
Solution:
(c) A straight angle is equal to 180°
Question 8.
Solution:
(c) A reflex angles is greater than 180° but less than 360°
Question 9.
Solution:
(d) A complete angle is equal to 360°.
Question 10.
Solution:
(b) A reflex angle is greater than 180° but less than 360°.
Question 11.
Solution:
Two right angles = (2 x 90)°
= 180° (b)
Question 12.
Solution:
32 of a right angle = 32 x 90° = 135° as 1 right angle = 90° (b)
Question 13.
Solution:
36 spokes has 360°
Angle between two adjacent spokes
= 360O36O = 10° (c)
RS Aggarwal Solutions for Class 6 Maths Chapter 13: Download PDF
RS Aggarwal Solutions for Class 6 Maths Chapter 13–Angles and Their Measurement
Download PDF: RS Aggarwal Solutions for Class 6 Maths Chapter 13–Angles and Their Measurement PDF
Chapterwise RS Aggarwal Solutions for Class 6 Maths :
- Chapter 1–Number System
- Chapter 2–Factors and Multiples
- Chapter 3–Whole Numbers
- Chapter 4–Integers
- Chapter 5–Fractions
- Chapter 6–Simplification
- Chapter 7–Decimals
- Chapter 8–Algebraic Expressions
- Chapter 9–Linear Equations in One Variable
- Chapter 10–Ratio, Proportion and Unitary Method
- Chapter 11–Line Segment, Ray and Line
- Chapter 12–Parallel Lines
- Chapter 13–Angles and Their Measurement
- Chapter 14–Constructions (Using Ruler and a Pairs of Compasses)
- Chapter 15–Polygons
- Chapter 16–Triangles
- Chapter 17–Quadrilaterals
- Chapter 18–Circles
- Chapter 19–Three-Dimensional Shapes
- Chapter 20–Two-Dimensional Reflection Symmetry (Linear Symmetry)
- Chapter 21–Concept of Perimeter and Area
- Chapter 22–Data Handling
- Chapter 23–Pictograph
- Chapter 24–Bar Graph
About RS Aggarwal Class 6 Book
Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal’s book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.
He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.
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