NCERT Solutions: Understanding Elementary Shapes

# NCERT Solutions for Class 6 Maths Chapter 5 - Understanding Elementary Shapes

 Table of contents Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 Exercise 5.6 Exercise 5.7 Exercise 5.8 Exercise 5.9 (Old NCERT)

## Exercise 5.1

Q1. What is the disadvantage in comparing line segments by mere observation?
Ans: We will be using the concept of line segments to solve this.
In the given figure, line segment PQ and RS seem to be equal but actually, they are not equal.
The disadvantage of comparing the lengths of two line segments by mere observation is that the lengths might not be accurate. Hence, a divider is used to compare the lengths of the line segments.

Q2. Why is it better to use a divider than a ruler, while measuring the length of a line segment?
Ans:  It is better to use a divider than a ruler while measuring the length of a line segment because a divider allows for more precise and accurate measurements. A divider has two sharp points that can be adjusted to mark the endpoints of the line segment. By opening or closing the divider, one can compare and transfer the length of the line segment to other distances or objects. This method ensures a more accurate measurement as compared to using a ruler, which may not provide the same level of precision.
A ruler may not be as suitable as a divider for measuring the length of a line segment due to the following reasons:

• Limited precision
• Difficulty in marking endpoints
• Inflexibility

Q3. Draw any line segment, say . Take any point C lying in between A and B. Measure the lengths of AB, BC and AC. Is AB = AC + CB?
[Note : If A,B,C are any three points on a line such that AC + CB = AB, then we can be sure that C lies between A and B]
Ans: We will be using the concept of the line segment to solve this.
Let us consider the line AB with point C being in the center of A and B and AB = 7 cm.
AC = 3 cm
CB = 4 cm
Therefore, AC + CB = 3 cm + 4 cm
AC + CB = 7 cm.
But, AB = 7 cm.
Hence, AB = AC + CB

Q4. If A, B, C are three points on a line such that AB = 5 cm, BC = 3 cm and AC = 8 cm, which one of them lies between the other two?
Ans: We will be using the concept of line segments to solve this.
We have, AB = 5 cm
BC = 3 cm
Therefore, AB + BC = 5 + 3
AB + BC = 8 cm
But, AC = 8 cm
Hence, B lies between A and C.

Q5. Verify, whether D is the mid point of .

Ans: We will be using the concept of midpoint to solve this.
From the given figure,
We have,
AG = 7 cm – 1 cm
AG = 6 cmAD = 4 cm – 1 cm
DG = 7 cm – 4 cm
DG = 3 cm
AD + DG = 3cm + 3cm
AD + DG = 6 cm
Therefore, AG = AD + DG
Where, AD = DG = 3cm
Hence, D is the midpoint of AG

Q6. If B is the mid-point of  and C is the midpoint of , where A, B, C, D lie on a straight line, say why AB = CD?
Ans: We will be using the concept of midpoint to solve this.
We have been given that, B is the midpoint of AC
⇒ AB = BC ----------------- (1)
C is the midpoint of BD
⇒ BC = CD --------------- (2)
Hence, we can say that AB = CD [From equation(1) and equation(2)]

Q7. Draw five triangles and measure their sides. Check in each case, if the sum of the lengths of any two sides is always less than the third side.
Ans: We will be using the concept of triangles to solve this.
Case I. In ∆ABC
Consider:
AB = 2.5 cm
BC = 4.8 cm
AC = 5.2 cm
AB + BC = 2.5 cm + 4.8 cm
AB + BC = 7.3 cm
Since, 7.3 > 5.2 , AB + BC > AC
Therefore, the sum of any two sides of a triangle is greater than the third side.
Case II. In ∆PQR,
Consider:
PQ = 2 cm
QR = 2.5 cm
PR = 3.5 cm
PQ + QR = 2 cm + 2.5 cm
PQ + QR = 4.5 cm
Since, 4.5 > 3.5, PQ + QR > PR
Therefore, the sum of any two sides of a triangle is greater than the third side.
Case III. In ∆XYZ,
Consider
XY = 5 cm
YZ = 3 cm
ZX = 6.8 cm
XY + YZ = 5 cm + 3 cm
XY + YZ = 8 cm
Since, 8 > 6.8, XY + YZ > ZX
Therefore, the sum of any two sides of a triangle is greater than the third side.
Case IV. In ∆MNS,
Consider
MN = 2.7 cm
NS = 4 cm
MS = 4.7 cm
MN + NS = 2.7 cm + 4 cm
MN + NS = 6.7 cm
Since, 6.7 >4.7, MN + NS > MS
Hence, the sum of any two sides of a triangle is greater than the third side.
Case V. In ∆KLM,
Consider
KL = 3.5 cm
LM = 3.5 cm
KM = 3.5 cm
L + LM = 3.5 cm + 3.5 cm
L + LM = 7 cm
7 cm > 3.5 cm, KL + LM > KM
Therefore, the sum of any two sides of a triangle is greater than the third side. Hence, we conclude that the sum of any two sides of a triangle is never less than the third side.

## Exercise 5.2

Q1. What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from
(a) 3 to 9
(b) 4 to 7
(c) 7 to 10
(d) 12 to 9
(e) 1 to 10
(f) 6 to 3

Hint: To find the portion of the clock swept by the hour hand between any two numbers(hours) say 'a' and 'b' where b > a will be found by subtracting 'a' from 'b' i.e, b - a.
Also, to find the fraction of a clockwise revolution between 'a' and 'b' we now divide the portion of the clock swept by the hour hand between them by 12 (since, the clock is divided into 12 equal divisions) Hence, the fraction will be (b - a) / 12 Using this concept, let's solve the given problems.

(a) 3 to 9
Ans: Lets first subtract 3 from 9 and then divide the answer by 12
9 – 3 = 6 ÷ 12 = 1/2 of a clockwise revolution

(b) 4 to 7
Ans: Lets first subtract 4 from 7 and then divide the answer by 12
7 – 4 = 3 ÷ 12 = 1/4 of a clockwise revolution

(c) 7 to 10
Ans: Lets first subtract 7 from 10 and then divide the answer by 12
10 – 7 = 3 ÷ 12 = 1/4 of a clockwise revolution

(d) 12 to 9
Ans: i.e., 0 to 9 [Since, 12 is at 0th position of the clock]
Lets first subtract 0 from 9 and then divide the answer by 12
9 – 0 = 9 ÷ 12 = 3/4 of a clockwise revolution

(e) 1 to 10
Ans: Lets first subtract 1 from 10 and then divide the answer by 12
10 – 1 = 9 ÷ 12 = 3/4 of a clockwise revolution

(f) 6 to 3
Ans: i.e., 6 to 12 and then 12 to 3
6 to 12 = 12 – 6 = 6 and 12 to 3 = 0 to 3 = 3 – 0 = 3
6 + 3 = 9 ÷ 12 = 3/4 of a clockwise revolution

Q2. Where will the hand of a clock stop if it
(a) starts at 12 and makes 1/2 of a revolution, clockwise?
Ans: When the clock hand starts from 12 and makes ½ of a revolution, the clock hand stops at 6.
1 revolution = 12 hours
1/2 revolution = 6 hours
Hence, the clock will land at 6.

(b) starts at 2 and makes 1/2 of a revolution, clockwise?
Ans: When the clock hand starts from 2 and makes ½ of a revolution, the clock hand stops at 8
1 revolution = 12 hours
1/2 revolution = 6 hours
Hence, the clock will land at 2 + 6 = 8.

(c) starts at 5 and makes 1/4 of a revolution, clockwise?
Ans: When the clock hand starts from 5 and makes 1/4 of a revolution, the clock hand stops at 8
1 revolution = 12 hours
1/4 revolution = 3 hours
Hence, the clock will land at (5 + 3) = 8.

(d) starts at 5 and makes 3/4 of a revolution, clockwise?
Ans: When the clock hand starts from 5 and makes 3/4 of a revolution, the clock hand stops at 2.
1 revolution = 12 hours
3/4 revolution = 9 hours
Hence, the clock will land at (5 + 9) = 14

Which is 2 hours added to 12 so the answer is 2 o'clock.

Q3. Which direction will you face if you start facing
(a) East and make ½ of a revolution clockwise?
(b ) East and make 1 ½ of a revolution clockwise?
(c) West and make ¾ of a revolution anti-clockwise?
(d) South and make one full revolution?
(Should we specify clockwise or anticlockwise for this last question? Why not?)

(a) East and make ½ of a revolution clockwise?
Ans: If we start facing East and make ½ of a revolution clockwise, we will be facing the West direction.
1 revolution = 360 degrees
1/2 revolution = 180 degrees
Thus, we start from East, move 180 degrees clockwise and stop at West.

(b) East and make 1 ½ of a revolution clockwise?
Ans: If we start facing East and make 1 ½ revolution clockwise, we will be facing the West direction.
1 revolution = 360 degrees
1/2 revolution = 180 degrees
Hence, 1 ½ revolution = 360 + 180 = 1 revolution + another 1/2 revolution
Thus, we start from East, move 360 degrees clockwise to complete 1 revolution, and continue moving for another 180 degrees to cover 1/2 revolution and stop at West.

(c) West and make ¾ of a revolution anti-clockwise?
Ans: If we start facing the West and make ¾ revolution anticlockwise, we will be facing the North direction.
1 revolution = 360 degrees
3/4 revolution = 270 degrees = 1/2 revolution + 1/4 revolution (180 degrees + 90 degrees)
Thus, we start from West, move 180 degrees anti-clockwise to complete 1/2 revolution till East, and continue moving for another 90 degrees to cover 1/4 revolution anti-clockwise and stop at North.

(d) South and make one full revolution?
Ans: It does not matter if we turn clockwise or anticlockwise, taking one full revolution will make it a full revolution and reach back to the starting position.
Thus, after 1 complete revolution that is 360 degrees, we stop at the same point from where we had started i.e, South.

Q4. What part of a revolution have you turned through if you stand facing
(a) east and turn clockwise to face north?
Ans: If we start from east and turn clockwise to face north,3/4 of a revolution is required, as seen in the image below.

(b) south and turn clockwise to face east?
Ans: If we start from the south and turn clockwise to face east, 3/4 of a revolution is required.

(c) west and turn clockwise to face east?
Ans: If we start from the west and turn clockwise to face east, 1/2 of a revolution is required.

Q5. Find the number of right angles turned through by the hour hand of a clock when it goes from
(a) 3 to 6
Ans: 3 to 6 - The hour hand turns through 1 right angle starting from 3 to 6.

(b) 2 to 8
Ans: 2 to 8 - The hour hand turns through 2 right angles starting from 2 to 8.

(c) 5 to 11
Ans: 5 to 11- The hour hand turns through 2 right angles starting from 5 to 11.

(d) 10 to 1
Ans: 10 to 1 - The hour hand turns through 1 right angle starting from 10 to 1.

(e) 12 to 9
Ans: 2 to 9 - The hour hand turns through 3 right angles starting from 12 to 9.

(f) 12 to 6
Ans: 12 to 6 - The hour hand turns through 2 right angles starting from 12 to 6.

Q6. How many right angles do you make if you start facing
(a) south and turn clockwise to west?
Ans: The number of right angles made while facing the south and turning clockwise to the west is 1 right angle.

(b) north and turn anticlockwise to east?
Ans: The number of right angles made while facing the north and turning anticlockwise to the east is 3 right angles.

(c) west and turn to west?
Ans: The number of right angles made while facing the west and turning to the west is 4 right angles.

(d) south and turn to north?
Ans: The number of right angles made while facing the south and turning to the north is 2 right angles.

Q7. Where will the hour hand of a clock stop if it starts
(a) from 6 and turns through 1 right angle?
(b) from 8 and turns through 2 right angles?
(c) from 10 and turns through 3 right angles?
(d) from 7 and turns through 2 straight angles?

(a) from 6 and turns through 1 right angle?
Ans: We know that in 1 complete revolution in either clockwise or anticlockwise direction, hour hand of a clock will rotate by 360o or 4 right angles
(a) If hour hand of a clock starts from 6 and turns through 1 right angle, it will stop at 9

(b) If hour hand of a clock starts from 8 and turns through 2 right angles, it will stop at 2

(c) If hour hand of a clock starts from 10 and turns through 3 right angles, it will stop at 7

(d) If hour hand of a clock starts from 7 and turns through 2 straight angles, it will stop at 7

## Exercise 5.3

Q1. Match the following:

Ans: We will use the concept of angles to solve this.

Q2. Classify each one of the following angles as right, straight, acute, obtuse or reflex:
Ans: We will use the concept of angles to solve this.
(a) Acute angle - It is less than 90°
(b) Obtuse angle - It is greater than 90° but less than 180°
(c) Right angle - It is equal to 90°
(d) Reflex angle - It is greater than 180° but less than 360°
(e) Straight angle - It is equal to 180°
(f) Acute angle - It is less than 90°

## Exercise 5.4

Q1. What is the measure of
(i) a right angle?

Ans: The measure of a right angle is 90o
(ii) a straight angle?

Ans: The measure of a straight angle is 180o

Q2. Say True or False:
(a) The measure of an acute angle < 90°.
(b) The measure of an obtuse angle < 90°.
(c) The measure of a reflex angle > 180°.
(d) The measure of on complete revolution = 360°.
(e) If m∠A = 53o and m∠B = 35o then m∠A > m∠B.
Ans:
(a) True; the measure of an acute angle is less than 90o
(b) False; the measure of an obtuse angle is more than 90o but less than 180o
(c) True; the measure of a reflex angle is more than 180o
(d) True; the measure of one complete revolution is 360o
(e) True; A is greater than B

Q3. Write down the measure of:
(a) some acute angles
(b) some obtuse angles
(give at least two examples of each)
Ans: (a) The measures of an acute angle are 50°, 65°
(b) The measures of obtuse angle are 110°, 175°

Q4. Measure the angles given below using the Protractor and write down the measure:

Ans: (a) The measure of an angle is 45°
(b) The measure of an angle is 120°
(c) The measure of an angle is 90°
(d) The measures of angles are 60°, 90° and 130°

Q5. Which angle has a large measure? First estimate and then measure:

Measure of angle A = ?
Measure of angle B = ?
Ans: ∠B has larger measure.
∠A = 40o and ∠B = 68o

Q6. From these two angles which has larger measure? Estimate and then confirm by measuring them:
Ans: Second angle has larger measure.
The measures of these angles are 45o and 55o.

Q7. Fill in the blanks with acute, obtuse, right or straight:
(a) An angle whose measure is less than that of a right angle is ________________.
Ans: acute angle
(b) An angle whose measure is greater than that of a right angle is ________________.
Ans: obtuse angle
(c) An angle whose measure is the sum of the measures of two right angles is ________________.
Ans: straight angle
(d) When the sum of the measures of two angles is that of a right angle, then each one of them is ________________.
Ans: acute angle
(e) When the sum of the measures of two angles is that of a straight angle and if one of them is acute then the other should be ________________.
Ans: obtuse angle

Q8. Find the measure of the angle shown in each figure. (First estimate with your eyes and then find the actual measure with a protractor.)

Ans: The measures of the angles shown in the above figures are 40o, 130o, 65o and 135o

Q9. Find the angle measure between the hands of the clock in each figure:

Ans: (i) 90 (Right angle)
(ii) 30 (Acute angle)
(iii) 180 (Straight angle)

Q10. Investigate:
In the given figure, the angle measure 30o. Look at the same figure through a magnifying glass. Does the angle becomes larger? Does the size of the angle change?

Ans: No, the measure of angle will be same.

Q11. Measure and classify each angle:

Ans:

Exercise 5.5

Q1. Which of the following are models for perpendicular lines:
(a) The adjacent edges of a table top.
(b) The lines of a railway track.
(c) The line segments forming the letter ‘L’.
(d) The letter V.

Ans:

(a) Perpendicular
(b) Not perpendicular
(c) Perpendicular
(d) Not perpendicular

Q2. Let   be the perpendicular to the line segment . Let  and intersect in the point A. What is the measure of ∠PAY?
Ans: We will be using the concept of perpendicular lines to solve this.
Since PQ ⊥ XY Therefore, ∠PAY = 90°

Q3. There are two “set-squares” in your box. What are the measures of the angles that are formed at their corners? Do they have any angle measure that is common?

Ans: One set-square has 60°,90°, 30° and other set-square has 45° ,90° ,45°. They have 90° as common angle.

Q4. Study the diagram. The line l is perpendicular to line m.

(a) Is CE = EG?
(b) Does PE bisect CG?
(c) Identify any two line segments for which PE is the perpendicular bisector.
(d) Are these true?
(i) AC > FG
(ii) CD = GH
(iii) BC < EH
Ans: (a) Yes, both measure 2 units.
(b) Yes, because CE = EG
(d) (i) True, (ii) True, (iii) True

Exercise 5.6

Q1. Name the types of following triangles:
(a) Triangle with lengths of sides 7 cm, 8 cm and 9 cm.
(b) ΔABC with AB = 8.7 cm, AC = 7 cm and BC = 6 cm.
(c) ΔPQR such that PQ = QR = PR = 5 cm.
(d) ΔDEF with m∠D = 90o
(e) ΔXYZ with m∠Y = 90o and XY = YZ
(f) ΔLMN with m∠L = 30o , m∠M = 70o and m∠N = 80o
Ans: (a) Scalene triangle
(b) Scalene triangle
(c) Equilateral triangle
(d) Right-angled triangle
(e) Isosceles right-angled triangle
(f) Acute-angled triangle

Q2. Match the following:

Ans: (i) → (e)
(ii) → (g)
(iii) → (a)
(iv) → (f)
(v) → (d)
(vi) → (c)
(vii) → (b)

Q3. Name each of the following triangles in two different ways: (You may judge the nature of angle by observation)

Ans: (i) Acute-angled and isosceles triangle
(ii) Right-angled and scalene triangle
(iii) Obtuse-angled and isosceles triangle
(iv) Right-angled and isosceles triangle
(v) Equilateral and acute-angled triangle
(vi) Obtuse-angled and scalene triangle

Q4. Try to construct triangles using match sticks. Some are shown here.

Can you make a triangle with:
(a) 3 matchsticks?
(b) 4 matchsticks?
(c) 5 matchsticks?
(d) 6 matchsticks?
(Remember you have to use all the available matchsticks in each case)
If you cannot make a triangle, think of reasons for it.
Ans: (a) 3 matchsticks This is an acute angle triangle and it is possible with 3 matchsticks to make a triangle because sum of two sides is greater than third side.
(b) 4 matchsticks This is a square, hence with four matchsticks we cannot make triangle.

(c) 5 matchsticks This is an acute angle triangle and it is possible to make triangle with five matchsticks, in this case sum of two sides is greater than third side.

(d) 6 matchsticks This is an acute angle triangle and it is possible to make a triangle with the help of 6 matchsticks because sum of two sides is greater than third side.

## Exercise 5.7

Q1. Say true or false:
(a) Each angle of a rectangle is a right angle.
(b) The opposite sides of a rectangle are equal in length.
(c) The diagonals of a square are perpendicular to one another.
(d) All the sides of a rhombus are of equal length.
(e) All the sides of a parallelogram are of equal length.
(f) The opposite sides of a trapezium are parallel.
Ans: (a) True
(b) True
(c) True
(d) True
(e) False
(f) False

Q2. Give reasons for the following:
(a) A square can be thought of as a special rectangle.
(b) A rectangle can be thought of as a special parallelogram.
(c) A square can be thought of as a special rhombus.
(d) Squares, rectangles, parallelograms are all quadrilateral.
(e) Square is also a parallelogram.
Ans: (a) Because its all angles are right angle and opposite sides are equal.
(b) Because its opposite sides are equal and parallel.
(c) Because its four sides are equal and diagonals are perpendicular to each other.
(d) Because all of them have four sides.
(e) Because its opposite sides are equal and parallel.

Q3. A figure is said to be regular if its sides are equal in length and angles are equal in measure. Can you identify the regular quadrilateral?
Ans: A square is a regular quadrilateral because all the interior angles are 90o, and all sides are of the same length.

## Exercise 5.8

Q1. Examine whether the following are polygons. If anyone among these is not, say why?
Ans: (a) As it is not a closed figure, therefore, it is not a polygon.
(b) It is a polygon because it is closed by line segments.
(c) It is not a polygon because it is not made by line segments.
(d) It is not a polygon because it not made only by line segments, it has curved surface also.

Q2. Name each polygon:
Make two more examples of each of these.
(b) Triangle
(c) Pentagon
(d) Octagon

Q3. Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of the triangle you have drawn.
Ans: ABCDEF is a regular hexagon and triangle thus formed by joining AEF is an isosceles triangle.

Q4. Draw a rough sketch of a regular octagon. (Use squared paper if you wish). Draw a rectangle by joining exactly four of the vertices of the octagon.
Ans: ABCDEFGH is a regular octagon and CDGH is a rectangle.

Q5. A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon. Draw a rough sketch of a pentagon and draw its diagonals.
Ans: ABCDE is the required pentagon and its diagonals are AD, AC, BE and BD.

## Exercise 5.9 (Old NCERT)

Ques 1: Match the following:

 (a) Cone (b) Sphere (c) Cylinder (d) Cuboid (e) Pyramid

Give two example of each shape.
Ans:

 (a) Cone (b) Sphere (c) Cylinder (d) Cuboid (e) Pyramid

Ques 2: What shape is:
(b) A brick?
(c) A match box?
Ans: (a) Cuboid
(b) Cuboid
(c) Cuboid
(d) Cylinder
(e) Sphere

The document NCERT Solutions for Class 6 Maths Chapter 5 - Understanding Elementary Shapes is a part of the Class 6 Course Mathematics (Maths) Class 6.
All you need of Class 6 at this link: Class 6

## Mathematics (Maths) Class 6

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## FAQs on NCERT Solutions for Class 6 Maths Chapter 5 - Understanding Elementary Shapes

 1. What are elementary shapes?
Ans. Elementary shapes are simple geometric shapes like circles, squares, rectangles, triangles, and polygons that are the basic building blocks of more complex shapes and figures.
 2. How can elementary shapes be classified?
Ans. Elementary shapes can be classified based on their properties such as the number of sides, angles, and vertices they have. Common classifications include triangles, quadrilaterals, and polygons.
 3. Why is it important to understand elementary shapes?
Ans. Understanding elementary shapes is important as it forms the foundation for learning more advanced geometry concepts. It helps in visualizing and analyzing shapes in various contexts like art, architecture, and engineering.
 4. How can elementary shapes be used in real-life applications?
Ans. Elementary shapes are used in real-life applications such as designing buildings, creating maps, and calculating areas and volumes of objects. They provide a framework for understanding and representing the physical world.
 5. Can elementary shapes be combined to form more complex shapes?
Ans. Yes, elementary shapes can be combined and transformed to create more complex shapes and figures. By understanding the properties of individual shapes, one can manipulate them to generate new shapes and patterns.

## Mathematics (Maths) Class 6

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