NCERT Solutions Class 6 Mathematics Chapter 5 Understanding Elementary Shapes have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 6 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 6 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 6 Mathematics are an important part of exams for Class 6 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 6 Mathematics and also download more latest study material for all subjects. Chapter 5 Understanding Elementary Shapes is an important topic in Class 6, please refer to answers provided below to help you score better in exams

## Chapter 5 Understanding Elementary Shapes Class 6 Mathematics NCERT Solutions

Class 6 Mathematics students should refer to the following NCERT questions with answers for Chapter 5 Understanding Elementary Shapes in Class 6. These NCERT Solutions with answers for Class 6 Mathematics will come in exams and help you to score good marks

### Chapter 5 Understanding Elementary Shapes NCERT Solutions Class 6 Mathematics

**Understanding Elementary Shapes**

**EXERCISE 5.1**

**Q.1) What is the disadvantage in comparing line segments by mere observation?**

**Sol.1)** When we compare two line segments by observation, the minute difference between the two of them cannot be observed and hence there are more chances of errors.

**Q.2) Why is it better to use a divider than a ruler, while measuring the length of a line.**

**Sol.2)** A ruler and a divider can be used to measure the length of a line segment. But, it is better to use a divider for an accurate measure. The thickness of the ruler and the positioning of the eye may cause difficulties in reading the correct measure.

**Q.3) Draw any line segment, say C lying in between A. Measure the lengths of AB, BC and AC.**

**Is π΄π΅ = π΄πΆ + πΆπ΅? [Note: If A,B,C are any three points on a line such that π΄πΆ +**

**πΆπ΅ = π΄π΅, then we that C lies between A and B.]**

**Sol.3)** It is given that point C is lying somewhere in between A & B. Therefore, all these points lying on the same line segment π΄π΅. Hence, for every situation in which point πΆ is lying in between π΄ & π΅ , it may be say that π΄π΅ = π΄πΆ + πΆπ΅

Consider a line segment π΄π΅ of length = 8 ππ.

Now, mark a point C at π΄πΆ = 5 ππ

Measure πΆπ΄.

We see that the length of πΆπ΄ = 3 ππ

i.e., π΄π΅ = π΄πΆ + πΆπ΅

Therefore, for any point C lying in between A and B on the line AB, the above is true.

**Q.4) If A, B, C are three points on a line such that AB = 5 cm, BC = 3 cm and one of them lies ****between the other two?**

**Sol.4)** It is given that π΄π΅ = 5 ππ, π΅πΆ = 3 ππ and π΄πΆ = 8 ππ.

From the given measures, we can see that

π΄π΅ + π΅πΆ = 8 ππ = π΄πΆ

β‘π΄ πΆ is the longest line segment, thus B is the point between A and C.

We can say this because π΄πΆ > π΄π΅ and π΄πΆ > π΅πΆ.

**Q.5) Verify, whether D is the midpoint of AB ^{β}**

**Sol.5)** From the figure above,

π΄π· = 4 β 1 = 3 ππ

π·πΊ = 7 β 4 = 3 ππ

Since, AD = DG = 3 cm, D is the mid-point of the line segment AG.

**Q.6) If B is the mid-point of AC and C is the mid-point of BD, where π΄, π΅, πΆ, π· lie on a straight ****line, say why π΄π΅ = πΆπ·?**

**Sol.6)** B is the mid-point of AC.

β΄ AB = BC β¦β¦β¦. (i)

And C is the mid-point of BD.

β΄ BC = CD β¦β¦β¦. (ii)

From eq. (i) and (ii),

AB = CD

**Q.7) Draw five triangles and measure their sides. Check in each case, of the sum of the ****lengths of any two sides is always less than the third side.**

**Sol.7)** Yes, sum of two sides of a triangle is always greater than the third side.

In figure (a),

Sum of two sides = 5.2 + 5.2 = 10.4 π’πππ‘

Third side of triangle = 5.2 ππ

In figure (b),

Sum of two sides = 15 + 8 = 23 π’πππ‘π

Third side of triangle = 17 ππ

In figure (c),

Sum of two sides = 10 + 9 = 19 π’πππ‘

Third side of triangle = 15 ππ

In figure (d),

Sum of two sides = 5 + 5 = 10 π’πππ‘π

Third side of triangle = 8 ππ

**Exercise 5.2**

**Q.1) 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**

**Sol.1)** When the hand of a clock moves from one position to another, it turns through an angle.

When it moves from 12 to 12, it moves four right angles, that is, 360Β° which is one revolution.

When it moves from 12 to 6, it moves two right angles, that is, 180Β°. This is 180/360 = Β½ of one revolution.

When the hour hand of the clock goes from 6 to 3, it rotates by 3 right angles, that is = 270Β°. Fraction of revolution = 270Β°/360Β° = 3/4

**Q.2) Where will the hand of a clock stop if it?**

**(a) starts at 12 and makes 1/2 of a revolution, clockwise?**

**(b) starts at 2 and makes 1/2 of a revolution, clockwise?**

**(c) starts at 5 and makes 1/4 of a revolution, clockwise?**

**(d) starts at 5 and makes 3/4 of a revolution, clockwise?**

**Sol.2)** A complete revolution is measured as 360Β°.

1/2 of a revolution is 360Β°'/2 = 180Β°.

1/2 of a revolution is 0Β°. 3/4 of a revolution is 270Β°

a) The hand of the clock starts at 12 and makes 1/2 of a revolution. That is, it starts at 12 and revolves 180Β° and stops at 6.

b) The hand starts at 2 and makes 1/2 of a revolution. That is, it starts at 2 and revolves 180Β° and stops at 8.

c) When the hand starts at 5 and makes 1/4 of a revolution, that is, 90Β°, it stops at 9.

d) When the hand starts at 5 and makes 3/4 of a revolution, that is 270Β°, it stops at 2.

**Q.3) Which direction will you face if you start facing
(a) East and make 1/2 of a revolution clockwise?
(b) East and make 1(1/2) of a revolution clockwise?
(c) West and make ΒΎ of a revolution anti-clockwise?
(d) South and make one full revolution?
(Should we specify clockwise or anti-clockwise for this last question? Why not?)
Sol.4**

a) When you start facing east and make 1/2 of a revolution, you revolve by 180Β°. Hence, you would be facing west.

b) When you start facing east and make 1 revolution clockwise, you face east again. Then you make another 1/2 of a revolution, that is, 180Β°. Hence, you would be facing west.

c) When you start facing west and make ΒΎ of a revolution anti-clockwise, you revolve by 270Β°. This is by 180Β°and then by 90Β°. When you revolve by 180Β°anti-clockwise you face east and then you revolve by 90Β° anti-clockwise, you face north.

d) When you start facing south and make a full revolution, that is 360Β°you will again face the south. The direction, whether clockwise or anticlockwise, does not matter in this case because a full revolution will bring back to the initial position.

**Q.4) What part of a revolution have you turned through if you stand facing
(a) east and turn clockwise to face north?
(b) south and turn clockwise to face east?
(c) west and turn clockwise to face east?
Sol.4)** The following image shows the directions, north (N), south (S), east (E) and west (W).

When you stand facing north and turn clockwise to face east, you have turned by a right angle, i.e. 90Β°.

So, you have made 1/4th of a complete revolution

a) You stand facing east and turn clockwise to face north.

So, you have turned 90Β° + 90Β° + 90Β° = 270Β° from East to North in the clock-wise direction. This is 270Β°/360Β° = 3/4 th of the complete revolution.

b) You start from south and turn clockwise to face east.

So, you have turned 90Β° + 90Β° = 180Β° from West to East in the clock-wise direction.

This is 180Β°/360Β° = 1/2 of the complete revolution.

**Q.5) Find the number of right angles turned through by the hour hand of a clock when it goes from
(a) 3 to 6 (b) 2 to 8 (c) 5 to 11
(d) 10 to 1 (e) 12 to 9 (f) 12 to 6
Sol.5)** When the hour hand of a clock moves from 12 to 3, from 3 to 6, from 6 to 9 and from 9 to 12, it makes a right angle. So, when the hour hand of a clock turns four right angles, it is said to make a complete revolution, 360Β°.

a) 3 to 6

The hour hand turned 1 right angle.

b) 2 to 8

**Q.6) How many right angles do you make if you start facing?**

**(a) south and turn clockwise to west?**

**(b) north and turn anti**

**(c) west and turn to west?**

**(d) south and turn to north?**

**Sol.6)** The following image shows the directions, north (N), south (S), east (E) and west (W).

When you stand facing north and turn clockwise to face east, you have turned by a right angle, i.e. 90Β°.

c) You start facing west and turn to west you have made one complete revolution, which is four right angles.

d) When you start facing south and turn to north you have made two right angles.

**Q.7) 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?**

**Sol.7)** a) When the hour hand starts from 6 and turns through 1 right angle, it will stop at 9.

b) When the hour hand starts from 8 and turns through 2 right angles, it will stop at 2. [8 to 11 is one right angle, 11 to 2 is the second right angle]

c) When the hour hand starts from 10 and turns through 3 right angles, it stops at 7. [10 to 1 is one right angle, 1 to 4 is the second and 4 to 7 is the third right angle.]

d) When the hour hand starts from 7 and turns through 2 straight angles, that is 360Β°, it stops at 7.

**Exercise 5.3**

**Q.1) Match the following:**

**i) Straight angle a) Less than one-fourth of a revolution
ii) Right angle b) More than half a revolution**

**iii) Acute angle c) Half of a revolution**

**iv) Obtuse angle d) One-fourth of a revolution**

**v) Reflex angle e) Between ΒΌ and Β½ of a revolution**

**f) One complete revolution**

**Sol.1)**Straight angle is 180Β° which is Β½ of a complete revolution. So,

**i) β c)**

Right angle is 90Β° which is ΒΌ of a complete revolution. So,

**ii) β d)**

Acute angles are angles less than 90Β°. This is less than one-fourth of a revolution. So,

**iii) β a)**

Obtuse angles are angles greater than 90Β° but less than 180Β°. This is between ΒΌ and Β½ of a complete revolution. So,

**iv) β e)**

Reflex angles are angles greater than 180Β° but less than 360Β°, that is, more than Β½ of a complete revolution. So,

**v) β b)**

**Q.2) Classify each one of the following angles as right, straight, acute, obtuse or reflex:**

**Sol.2)** The measure of a right angle is 90Β° and hence that of a straight angle is 180Β°. An angle is acute if its measure is smaller than that of a right angle and is obtuse if its measure is greater than that of a right angle and less than a straight angle. A reflex angle is larger than a straight angle.

a) Acute angle

b) Obtuse angle

c) Right angle

d) Reflex angle

e) Straight angle

f) Acute angle

**Exercise 5.4**

**Q.1) What is the measure of :**

**(i) a right angle? (ii) a straight angle?**

**Sol.1)** a) A right angle measures 90Β°

b) A straight angle measures 180Β°.

**Q.2) Say True or False:**

**(a) The measure of an acute angle < 90Β°.**

**(b) The measure of an obtuse angle < 90Β°.**

**(c) The measure of a reflex angle > 180Β°.**

**(d) The measure of one complete revolution = 90Β°.**

**(e) If πβ π΄ = 50Β° and πβ π΅ = 35Β°, then πβ π΄ > πβ π΅.**

**Sol.2)** a) True

b) False, an obtuse angle is greater than 90Β°

c) True

d) True

e) True

**Q.3) Write down the measures of**

**(a) some acute angles. (b) some obtuse angles.**

**(give at least two examples of each).**

**Sol.3)** a) Acute angles are angles less than 90Β°. Some acute angles are 35Β°., 43Β°.

b) Obtuse angles are angles greater than 90Β°. and less than 180Β°.. Some obtuse angles are 135Β°., 143Β°.

**Q.4) Measure the angles given below, using the protractor and write down the measure:**

**Sol.4)** (a) 45Β°

(b) 120Β°

(c) 90Β°

(d) 60Β°, 130Β°, 90Β°

**Q.5) Which angle has a large measure?**

**First estimate and then measure.**

**Sol.5)** β π΅ has larger measure.

β π΄ = 40Β° and β π΅ = 65Β°

**Q.6) From these two angles which has larger measure? Estimate and then confirm by ****measuring them:**

**Sol.6)** The angles measure 75Β° and 70Β°

Hence, the 1st angle is greater

**Q.7) Fill in the blanks with acute, obtuse, ****straight :**

**(a) An angle whose measure is less right angle is______.**

**(b) An angle whose measure is greater a right angle is ______.**

**(c) An angle whose measure is the sum of the measures of two right angles is _____.**

**(d) When the sum of the measures of two angles is that of a right angle, then each one of ****them is ______.**

**(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 _______.**

**Sol.7)** (a) acute angle

(b) obtuse angle

(c) straight angle

(d) 45Β°

(e) Obtuse angle

**Q.8) Find the measure of the angle shown in each figure. (First estimate with your eyes and ****then find the actual measure with a protractor**

**Exercise 5.5**

**Q.1) 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 βLβ.**

**(d) The letter V.**

**Sol.1)** a) Adjacent edges of table top make perpendicular angle.

b) Lines of railway track do not make perpendicular angle. They are parallel to each other.

c) Yes, right angles are formed in 'L'.

d) Right angle is not formed when we write 'V'.

**Q.2) Let PQΒ― be the perpendicular to the line segment XYΒ―. Let PQΒ― and XYΒ― intersect in the ****point A. What is the measure of β ππ΄π.**

**Sol.2) **

PQ is perpendicular to XY and it intersects at point A.

So, β ππ΄π = 90Β°.

**Q.3) There are two βset-squaresβ 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?**

**Sol.3)** One set-square measures 90Β°, 60Β°, 30Β°

Other set-square measures 90Β°, 45Β°, 45Β°

90Β° is a common measure.

**Q.4) Study the diagram. The line π is perpendicular to line π**

**(a) Is CE = EG?**

**(b) Does PE bisect CG? segment XY.**

**(c) Identify any two line segments for which PE is the perpendicular bisector.**

**(d) Are these true?**

**(i) π΄πΆ > πΉπΊ**

**(ii) πΆπ· = πΊπ»**

**(iii) π΅πΆ < πΈπ».**

**Sol.4)**

a) CE = 2 units and EG = 2 units. So, CE = EG.

b) Yes, as CE = EG, we say PE bisects CG.

c) DF and BH are line segments for which PE is the perpendicular bisector.

d)

i) True, as AC is 2 units but FG is 1 unit.

ii) True, CD = GH = 1 unit.

iii) True, BC is 1 unit and EH is 3 units.`

**Exercise 5.6**

**Q.1) Name the types of following triangles:**

**(a) Triangle with lengths of sides 7 ππ, 8 ππ πππ 9 ππ.**

**(b) Ξπ΄π΅πΆ with π΄π΅ = 8.7ππ, π΄πΆ = 7 ππ and π΅πΆ = 6 ππ.**

**(c) Ξπππ
such that ππ = ππ
= ππ
= 5 ππ.**

**(d) Ξπ·πΈπΉ with πβ π· = 90Β°**

**(e) Ξπππ with πβ π = 90Β° and ππ = ππ.**

**(f) ΞπΏππ with πβ πΏ = 30Β°, πβ π = 70Β° and πβ π = 80Β°.**

**Sol.1)** a) Scalene triangle, as the triangle has three unequal sides.

b) Scalene triangle, as the triangle has three unequal sides.

c) Equilateral triangle, as the triangle has three equal sides.

d) Right-angled triangle as one angle is a right angle.

e) Right-angled isosceles triangle as one angle is right angle and two sides are equal.

f) Acute-angled triangle as each angle is less than a right angle.

**Q.2) Match the following**

**Measures of triangle Type of triangle**

**i) 3 sides of equal length (a) Scalene triangle**

**ii) 2 sides of equal length (b) Isosceles right angled**

**iii) All sides of different length (c) Obtuse angled**

**iv) 3 acute angles (d) Right angled**

**v) 1 right angle (e) Equilateral**

**vi) 1 obtuse angle (f) Acute angled**

**vii) 1 right angle with 2 sides of equal length (g) Isosceles**

**Sol.**2) i) 3 sides of equal length β Equilateral triangle (e)

ii) 2 sides of equal length β Isosceles (g)

iii) All sides of different length β Scalene (a)

iv) 3 acute angles β Acute angled (f)

v) 1 right angle β Right angled (d)

vi) 1 obtuse angle β Obtuse angled (c)

vii) 1 right angle with 2 sides of equal length β Isosceles right angled (b)

**Q.3) Name each of the following triangles in two different ways: (you may judge the nature of ****the angle by observation).**

**Sol.3) **

a) Isosceles triangle, as two sides are of equal length.

Acute angled triangle, as angle measures are less than right angle.

b) Scalene triangle, as three sides are of unequal length.

Right angled triangle, as one angle measures 90 degrees.

c) Isosceles triangle, as two sides are of equal length.

Obtuse angled triangle, as one angle is obtuse.

d) Isosceles triangle, as two sides are of equal length.

Right angled triangle, as one angle measures 90 degrees.

e) Equilateral triangle, as three sides are of equal length.

Acute angled triangle, as angle measures are less than right angle.

f) Scalene triangle, as three sides are of unequal length.

Obtuse angled triangle, as one angle is obtuse

**Q.4) 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)
Name the type of triangle in each case. If you cannot make a triangle, think of reasons for it.**

**Sol.4)** (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.

**Exercise 5.7**

**Q.1) 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**

**Sol.1)** (i) In a rectangle, each angle is a right angle.** TRUE**

(ii) In a rectangle, the opposite sides are equal in length. **TRUE**

(iii) In a square, the diagonals are perpendicular to each another. **TRUE**

(iv) In a rhombus, all the sides are equal in length. **TRUE**

(v) In a parallelogram, all the sides are equal in length. **FALSE**

(vi) In a trapezium, the opposite sides are parallel. **FALSE**

**Q.2) 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 quadrilaterals.
(e) Square is also a parallelogram.**

**Sol.2**) a) When a rectangle is drawn with all equal sides, it becomes a square. Hence it can be thought of as a special rectangle.

b) When a parallelogram is drawn with all the angles as a right angle, it becomes a rectangle. Hence, a rectangle is a special parallelogram.

c) A rhombus and a square have equal sides. When a rhombus is drawn with all the angles as a right angle, it becomes a square.

d) Squares, rectangle, parallelograms are all closed shapes made of 4 line segments and are all quadrilaterals.

e) A parallelogram has two opposite sides that are parallel and equal. Similarly, a square has all sides that are equal and the opposite sides are parallel. Hence, square can be thought of as a special parallelogram.

**Q.3) 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?**

**Sol.3)** In a square, all the interior angles are of 90Β° & all the sides are of the same length.

Therefore, a square is a regular quadrilateral

**Exercise 5.8**

**Q.1) Examine whether the following are polygons. If anyone among them is not, say why?**

**Sol.1)** 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.

**Q.2) Name each polygon.**

**Male 2 more examples of each these.**

**Q.3) 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.
Sol.3)**

ABCDEFGH is a regular octagon and CDGH is a rectangle.

**Q.5) the polygon. Draw a rough sketch of a pentagon and draw its diagonals.
Sol.5)** ABCDE is the required pentagon and its diagonals are AD, AC, BE and BD.

**Exercise 5.9**

**Q.1) Match the following:**

**Sol.1**) a) Cone ii)

b) Sphere iv)

c) Cylinder v)

d) Cuboid iii)

e) Pyramid i)

**Q.2) What shape is**

**(a) Your instrument box?**

**(b) A brick?**

**(c) A match box?**

**(d) A road-roller?**

**(e) A sweet laddu?**

**Sol.2)** a) Cuboid

b) Cuboid

c) Cuboid

d) Cylinder

e) Sphere

**Question. A triangle which has two sides equal is called an 'Isosceles triangle'. Four points are shown below:
If all the points are joined to all others, which triangle will be Isosceles?**

(a) βPQR

(b) βPQS

(c) βQRS

(d) βSPR

**Answer : D**

**Question. How many squares (of any size) are there in the figure that have a dot inside?**

(a) 10

(b) 4

(c) 2

(d) 1

**Answer : B**

**Question. A cuboidal block is made up of smaller blocks as shown. If the shaded blocks are removed, we can say that:**

(a) 5% of the solid is removed

(b) 20% of the solid is removed

(c) 25% of the solid is removed

(d) 75% of the solid is removed.

**Answer : B**

**Question. What is the length of the dark line next to the ruler below?**

(a) About 3. 5 units

(b) About 4 units

(c) About 4. 5 units

(d) About 5 units

**Answer : A**

**Question. At which of the following times is the angle between the hands of a clock exactly 120 degrees?**

(a) 05:00

(b) 03:35

(c) 08:00

(d) 09:05

**Answer : C**

**Question. A triangle ABC is cut out into three parts and then rejoined as shown. From this, we can say that**

(a) The angles of this triangle are equal.

(b) The angles of this triangle meet at a point

(c) The angles of this triangle together make a straight angle.

(d) One of the angles of this triangle is obtuse.

**Answer : C**

**Question. In the figure shown, the size of angle AOB is 30 degrees. Ray OA is rotated by an angle of 60 degrees in the clockwise direction. What's the minimum rotation that ray OB has to make (in either direction) so that the angle between them becomes 15 degrees?**

(a) 0Β°

(b) 45Β° anticlockwise

(c) 15Β° anticlockwise

(d) 30Β° clockwise

**Answer : C**

**Question. How many corners will a solid cube have after a triangular cut is made at one corner as shown below?**

(a) 8

(b) 9

(c) 10

(d) 11

**Answer : C**

**Question. Which of these angles measures 60 ^{o}?**

**Answer : B**

**Question. An equilateral triangle is one having all sides equal. Which set of sticks shown here can form an equilateral triangle?**

**Answer : A**

**Question. How many right angles are there in this figure?**

(a) 1

(b) 2

(c) 4

(d) 5

**Answer : C**

**Question. A line-segment is named using its two extreme points. The line-segment above is named, how many line-segments are there WHICH CAN BE NAMED?**

(a) 4

(b) 5

(c) 6

(d) 11

**Answer : B**

**Question. Shown below is a grid with two shapes P and Q marked on it.
How many triangles exactly like P will be needed to cover shape Q completely with NO OVERLAP?**

(a) 3

(b) 4

(c) 6

(d) 8

**Answer : C**

**Question. What is the length of the dark line next to the ruler below?**

(a) 2

(b) 1.5

(c) 1

(d) 0.5

**Answer : A**

NCERT Solutions Class 6 Mathematics Chapter 1 Knowing our Numbers |

NCERT Solutions Class 6 Mathematics Chapter 2 Whole Numbers |

NCERT Solutions Class 6 Mathematics Chapter 3 Playing with Numbers |

NCERT Solutions Class 6 Mathematics Chapter 4 Basic Geometrical Ideas |

NCERT Solutions Class 6 Mathematics Chapter 5 Understanding Elementary Shapes |

NCERT Solutions Class 6 Mathematics Chapter 6 Integers |

NCERT Solutions Class 6 Mathematics Chapter 7 Fractions |

NCERT Solutions Class 6 Mathematics Chapter 8 Decimals |

NCERT Solutions Class 6 Mathematics Chapter 9 Data Handling |

NCERT Solutions Class 6 Mathematics Chapter 10 Mensuration |

NCERT Solutions Class 6 Mathematics Chapter 11 Algebra |

NCERT Solutions Class 6 Mathematics Chapter 12 Ratio and Proportion |

NCERT Solutions Class 6 Mathematics Chapter 13 Symmetry |

NCERT Solutions Class 6 Mathematics Chapter 14 Practical Geometry |

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### NCERT Solutions Class 6 Mathematics Chapter 5 Understanding Elementary Shapes

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### Chapter 5 Understanding Elementary Shapes Class 6 Mathematics NCERT Solutions

The Class 6 Mathematics NCERT Solutions Chapter 5 Understanding Elementary Shapes are designed in a way that will help to improve the overall understanding of students. The answers to each question in Chapter 5 Understanding Elementary Shapes of Mathematics Class 6 has been designed based on the latest syllabus released for the current year. We have also provided detailed explanations for all difficult topics in Chapter 5 Understanding Elementary Shapes Class 6 chapter of Mathematics so that it can be easier for students to understand all answers.

**NCERT Solutions Chapter 5 Understanding Elementary Shapes Class 6 Mathematics**

Class 6 Mathematics NCERT Solutions Chapter 5 Understanding Elementary Shapes is a really good source using which the students can get more marks in exams. The same questions will be coming in your Class 6 Mathematics exam. Learn the Chapter 5 Understanding Elementary Shapes questions and answers daily to get a higher score. Chapter 5 Understanding Elementary Shapes of your Mathematics textbook has a lot of questions at the end of chapter to test the students understanding of the concepts taught in the chapter. Students have to solve the questions and refer to the step-by-step solutions provided by Mathematics teachers on studiestoday to get better problem-solving skills.

**Chapter 5 Understanding Elementary Shapes Class 6 NCERT Solution Mathematics**

These solutions of Chapter 5 Understanding Elementary Shapes NCERT Questions given in your textbook for Class 6 Mathematics have been designed to help students understand the difficult topics of Mathematics in an easy manner. These will also help to build a strong foundation in the Mathematics. There is a combination of theoretical and practical questions relating to all chapters in Mathematics to check the overall learning of the students of Class 6.

**Class 6 NCERT Solution Mathematics Chapter 5 Understanding Elementary Shapes**

NCERT Solutions Class 6 Mathematics Chapter 5 Understanding Elementary Shapes detailed answers are given with the objective of helping students compare their answers with the example. NCERT solutions for Class 6 Mathematics provide a strong foundation for every chapter. They ensure a smooth and easy knowledge of Revision notes for Class 6 Mathematics. As suggested by the HRD ministry, they will perform a major role in JEE. Students can easily download these solutions and use them to prepare for upcoming exams and also go through the Question Papers for Class 6 Mathematics to clarify all doubts

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All questions given in the end of the chapter Chapter 5 Understanding Elementary Shapes have been answered by our teachers

NCERT solutions for Mathematics Class 6 can help you to build a strong foundation, also access free study material for Class 6 provided on our website.

Carefully read the solutions for Class 6 Mathematics, understand the concept and the steps involved in each solution and then practice more by using other questions and solutions provided by us