**Download NCERT Solutions for Class 6 Maths Chapter 5 – Understanding Elementary Shapes**

**NCERT Solutions for Class 6 Maths Chapter 5 – Understanding Elementary Shapes comprises of the 9 Exercises**

**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**

**NCERT Solutions for Class 6 Maths Chapter 5 Exercise 5.1**

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

**Sol. 1.** There are a lot of chances of error, due to improper viewing, while comparing line segments by mere observation.

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

**Sol. 2.** A divider gives accurate measurement of the length of a line segment as compared to that obtained by a ruler.

**Ques. 3. 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.]**

**Sol. 3. **

Yes, AB = AC + CB. This is because C lies between A and B.

**Ques. 4. 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?**

**Sol. 4. **AC = AB + BC

Since, 8 = 5 + 3

Hence, point B lies between A and C.

**Ques. 5. Verify, whether D is the mid-point of ****.**

**Sol. 5. **

= 7-1 = 6 units

= 4-1 = 3 units

= 7-4 = 3 units

= +

Hence, D is the mid-point of .

**Ques. 6. If B is the mid-point of and C is the mid-point of , where A,B,C,D lie on a straight line, say why AB = CD?**

**Sol. 6. **B is the mid-point of AC, means AB = BC……(1)

C is the mid-point of BD, means BC = CD……(2)

From equation (1) & (2),

AB = BC = CD

Hence, AB = CD.

**Ques. 7. 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.**

**Sol. 7.** The following are the five triangles along with their measurements.

In all the above five cases, we can see that the sum of any two sides of a triangle is always greater than the third side.

For example, in the first triangle, 4 + 5 = 9 > 3 or 3 + 4 = 7 > 5 or 5 + 3 = 8 > 4.

Similarly, in all the other triangles, the same rule follows.

**NCERT Solutions for Class 6 Maths Chapter 5 Exercise 5.2**

**Ques. 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. **

**(a)** One revolution of clock, i.e., 12 hours of a clock takes 360 ˚.

So, in one hour, the hour hand of the clock rotate by = 30˚.

From 3 to 9, i.e., in 6 hours, the hour hand of the clock turns through

30×6=180˚.

Hence, the fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from 3 to 9 = = .

**(b)** One revolution of clock, i.e., 12 hours of a clock take 360˚.

So, in one hour, the hour hand of the clock rotate by = 30˚.

From 4 to 7, i.e., in 3 hours, the hour hand of the clock turns through

30×3 = 90˚.

Hence, the fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from 4 to 7 = = .

**(c)** One revolution of clock, i.e., 12 hours of a clock take 360˚.

So, in one hour, the hour hand of the clock rotate by = 30˚.

From 7 to 10, i.e., in 3 hours, the hour hand of the clock turns through

30×3 = 90˚.

Hence, the fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from 7 to 10 = = .

**(d)** One revolution of clock, i.e., 12 hours of a clock take 360˚.

So, in one hour, the hour hand of the clock rotate by = 30˚.

From 12 to 9, i.e., in 9 hours, the hour hand of the clock turns through

30×9 = 270˚.

Hence, the fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from 12 to 9 = = .

**(e)** One revolution of clock, i.e., 12 hours of a clock take 360˚.

So, in one hour, the hour hand of the clock rotate by = 30˚.

From 1 to 10, i.e., in 9 hours, the hour hand of the clock turns through

30×9 = 270˚.

Hence, the fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from 1 to 10 = = .

**(f)** One revolution of clock, i.e., 12 hours of a clock take 360˚.

So, in one hour, the hour hand of the clock rotate by = 30˚.

From 6 to 3, i.e., in 9 hours, the hour hand of the clock turns through

30×9 = 270˚.

Hence, the fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from 6 to 3 = = .

**Ques. 2. Where will the hand of a clock stop if it****(a) starts at 12 and makes of a revolution, clockwise?****(b) starts at 2 and makes of a revolution, clockwise?****(c) starts at 5 and makes of a revolution, clockwise?****(d) starts at 5 and makes of a revolution, clockwise?**

**Sol. 2. (a)** of a revolution implies of 360˚, i.e., 180˚.

Starting at 12 and making of a revolution, clockwise means that, it has turned 180˚ clockwise.

In 1 hour, the hand of the clock takes 30˚.

So, 180˚ can be turned through in 6 hours in clockwise direction.

Hence, the hand of the clock stops at 6 hours clockwise from 12, i.e., at 6.

**(b)** of a revolution implies of 360˚, i.e., 180˚.

Starting at 2 and making of a revolution, clockwise means that, it has turned 180˚ clockwise.

In 1 hour, the hand of the clock takes 30˚.

So, 180˚ can be turned through in 6 hours in clockwise direction.

Hence, the hand of the clock stops at 6 hours clockwise from 2, i.e., at 8.

**(c)** of a revolution implies of 360˚, i.e., 90˚.

Starting at 5 and making of a revolution, clockwise means that, it has turned 90˚ clockwise.

In 1 hour, the hand of the clock takes 30˚.

So, 90˚ can be turned through in 3 hours in clockwise direction.

Hence, the hand of the clock stops at 3 hours clockwise from 5, i.e., at 8.

**(d)** of a revolution implies of 360˚, i.e., 270˚.

Starting at 5 and making of a revolution, clockwise means that, it has turned 270˚ clockwise.

In 1 hour, the hand of the clock takes 30˚.

So, 270˚ can be turned through in 9 hours in clockwise direction.

Hence, the hand of the clock stops at 9 hours clockwise from 5, i.e., at 2.

**Ques. 3. Which direction will you face if you start facing**

**(a) east and make of a revolution clockwise**

**(b) east and make of a revolution clockwise?**

**(c) west and make of a revolution anti-clockwise?**

**(d) south and make one full revolution?**

**(Should we specify clockwise or anticlockwise for this last question? Why not?)**

**Sol. 3.** **(a)** Facing east turning of a revolution clockwise, i.e., 180˚ clockwise from east, we will face the west direction.

**(b)** Facing east turning of a revolution clockwise, i.e., 360+180=180˚ clockwise from east, we will face the west direction. (In first 1 revolution we reach the original point. In next revolution we turn 180˚ further.)

**(c)** Facing west turning of a revolution anti-clockwise, i.e., 270˚ anti-clockwise from west, we will face the north direction.

**(d)** Facing south making one full revolution clockwise, or anticlockwise, i.e., 360˚, we will reach the original point of start, i.e., south.

**Ques. 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. (a)**If we stand facing east and turn clockwise to face north, we turn of a revolution.

**(b)**If we stand facing south and turn clockwise to face east, we turn of a revolution.

**(c)**If we stand facing west and turn clockwise to face east, we turn of a revolution.

**Ques. 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.** **(a)** From 3 to 6, i.e., in 3 hours, the hour hand of the clock turns 30˚ ×3=90˚, i.e., 1 right angle.**(b)** From 2 to 8, i.e., in 6 hours, the hour hand of the clock turns 30˚ ×6=180˚, i.e., 2 right angles.**(c)** From 5 to 11, i.e., in 6 hours, the hour hand of the clock turns 30˚ ×6=180˚, i.e., 2 right angles.**(d)** From 10 to 1, i.e., in 3 hours, the hour hand of the clock turns 30˚ ×3=90˚, i.e., 1 right angle.**(e)** From 12 to 9, i.e., in 9 hours, the hour hand of the clock turns 30˚ ×9=270˚, i.e., 3 right angles.**(f)** From 12 to 6, i.e., in 6 hours, the hour hand of the clock turns 30˚ ×6=180˚, i.e., 2 right angles.

**Ques. 6. How many right angles do you make if you start facing****(a) south and turn clockwise to west?****(b) north and turn anti-clockwise to east?****(c) west and turn to west?****(d) south and turn to north?**

**Sol. 6. (a)** Facing south and turning clockwise to west, it takes 90˚, i.e., 1 right angle.**(b)** Facing north and turning anti-clockwise to east, it takes 270˚, i.e., 3 right angles.**(c)** Facing west and turning to west, it takes 360˚, i.e., 4 right angles.**(d)** Facing south and turning to north, it takes 180˚, i.e., 2 right angles.

**Ques. 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)** Starting from 6 and turning through 1 right angle, the hour hand of the clock turns 90˚ clockwise from 6, i.e., it stops at 9.**(b)** Starting from 8 and turning through 2 right angles, the hour hand of the clock turns 180˚ clockwise from 8, i.e., = 6 hours ahead. Hence, it stops at 2.**(c)** Starting from 10 and turning through 3 right angles, the hour hand of the clock turns 270˚ clockwise from 10, i.e., = 9 hours ahead. Hence, it stops at 7.**(d)** Starting from 7 and turning through 2 straight angles, the hour hand of the clock turns 2×180=360˚ clockwise (or anticlockwise) from 7, i.e., it stops at 7.

**NCERT Solutions for Class 6 Maths Chapter 5 Exercise 5.3**

**Ques. 1. Match the following:**

**Sol. 1. (i)** **⇒** (c) A straight angle is of 180 degree, i.e., half of a revolution.**(ii)** **⇒** (d) A right angle is of 90 degree which is one-fourth of a revolution.**(iii) ⇒** (a) Acute angle is an angle more than 0 degree and less than 90 degree. Hence it makes less than ¼ of a revolution.**(iv) ⇒** (e) Obtuse angle is an angle more than 90 degree and less than 180 degree. Hence, it makes between ¼ and ½ of a revolution.**(v) ⇒** (b) More than half of the revolution is a reflex angle.

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

**Sol. 2. (a)** Acute angle (Angle less than 90 degree)**(b)** Obtuse angle (Angle more than 90 degree)**(c)** Right angle (Angle equal to 90 degree)**(d)** Reflex angle (Angle more than 180 degree)**(e)** Straight angle (Angle equal to 180 degree)**(f)** Acute angle (Angle less than 90 degree)

**NCERT Solutions for Class 6 Maths Chapter 5 Exercise 5.4**

**Ques. 1. What is the measure of (i) a right angle? (ii) a straight angle?**

**Sol. 1. (i)** A right angle measures 90˚.**(ii)** A straight angle measures 180˚.

**Ques. 2. 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 one complete revolution = 360°.****(e) If m ∠A = 53° and m ∠B = 35°, then m ∠A > m ∠B.**

**Sol. 2. (a)** True**(b)** False. Obtuse angle measures more than 90˚ and less than 180˚.**(c)** True**(d)** True**(e)** True. (As, 53>35)

**Ques. 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 Angle = 64°, 85°**(b)** Obtuse Angle = 97°, 164°

**Ques. 4. Measure the angles given below using the Protractor and write down the measure.**

**Sol. 4.** Upon measuring using protractor, we find out that the measures are:

a) 40° (acute angle)

b) 130° (obtuse angle)

c) 90° (right angle)

d) 60° (acute angle)

___________________________________________________________________________

**Ques. 5. Which angle has a large measure? First estimate and then measure.**

**Measure of Angle A =****Measure of Angle B =**

**Sol. 5.** Upon estimating, we can see that Angle B has a larger measure than Angle A.

Upon measuring, we find out, ∠A=40°, ∠B=60°.

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

**Sol. 6.** Upon estimating, we can see that the second angle has a larger measure.

Upon measuring, we find out, ∠1=45°, ∠2=55°.

**Ques. 7. Fill in the blanks with acute, obtuse, right or straight:****(a) An angle whose measure is less than that of a right angle is______.****(b) An angle whose measure is greater than that of 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)** An angle whose measure is less than that of a right angle is __acute__.**(b)** An angle whose measure is greater than that of a right angle is __obtuse__.**(c)** An angle whose measure is the sum of the measures of two right angles is __straight__.**(d)** When the sum of the measures of two angles is that of a right angle, then each one of them is __acute__.**(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 __obtuse__.

**Ques. 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).**

**Sol. 8.** Upon estimating and then measuring we find out that the measurements are:

(i) 30°

(ii) 120°

(iii) 60°

(iv) 150°

**Ques. 9. Find the angle measure between the hands of the clock in each figure:**

**Sol. 9. ** One complete revolution = 360° (in 12 hours)

Angle subtended in one hour = 360/12 = 30°

(i) Angle between 9 and 12 = (12-9) × 30° = 3×30° = 90°

(ii) Angle between 12 and 1 = 30°

(iii) Angle between 12 and 6 = 6 × 30° = 180°

**Ques. 10. Investigate In the given figure, the angle measures 30°. Look at the same figure through a magnifying glass. Does the angle become larger? Does the size of the angle change?**

**Sol. 10. **

No, the angle remains the same.

**Ques. 11. Measure and classify each angle:**

**Sol. 11.**

**NCERT Solutions for Class 6 Maths Chapter 5 Exercise 5.5**

**Ques. 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 ‘L’.****(d) The letter V.**

**Sol. 1. (a)** The adjacent edges of a table top form perpendicular lines.**(b)** The lines of a railway track are parallel to each other. So, they don’t form perpendicular lines.**(c)** The line segments forming the letter ‘L’ is a model for perpendicular lines.**(d)** The letter V forms an acute angle, and hence, does not represent perpendicular line model.

Hence, only (a) and (c) are models for perpendicular lines.

**Ques. 2. Let be the perpendicular to the line segment . Let and intersect in the point A. What is the measure of ∠PAY?**

**Sol. 2.** The point of intersection of and forms 90 degrees, since, is perpendicular to .

Hence, ∠PAY = 90°.

**Ques. 3. 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?**

**Sol. 3. **One set square has the measure of 30°, 60° and 90°.

The other one measures 45°, 45° and 90°.

The measure of one angle is common, i.e., 90°.

**Ques. 4. 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.**

**Sol. 4.** **(a)** Yes. CE = EG = 2 units**(b)** Yes. PE bisects CG at E, such that CE = EG.**(c)** and are the line segments for which PE is the perpendicular bisector.**(d)** (i) True. (AC = 2units, FG = 1unit. Hence, AC>FG )

(ii) True. (CD = GH = 1unit)

(iii) True. (BC = 1unit, EH = 3 units. Hence, BC<EH)

**NCERT Solutions for Class 6 Maths Chapter 5 Exercise 5.6**

**Ques. 1. 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 = 90°****(e) ΔXYZ with m ∠Y = 90° and XY = YZ.****(f) ΔLMN with m ∠L = 30°, m ∠M = 70° and m ∠N = 80°.**

**Sol. 1. (a)** Scalene triangle (all sides are different in length)**(b)** Scalene triangle (all sides are different in length)**(c)** Equilateral triangle (all side lengths are same)**(d)** Right angled triangle (one angle is 90°)**(e)** Isosceles right angled triangle (one angle is 90° and two side lengths are equal)**(f)** Acute angled triangle (All angles are less than 90°)

**Ques. 2. Match the following:**

**Sol. 2. ** (i)**⇒** (e) [An equilateral triangle has same length of all the three sides.]

(ii)**⇒** (g) [An isosceles triangle has two of its side lengths same.]

(iii)**⇒** (a) [Scalene triangle has all side lengths different.]

(iv)**⇒** (f) [Acute angled triangles have all three angles acute, i.e., less than 90°.]

(v)**⇒** (d) [Right angled triangle has one of its angle = 90°, i.e., it has one right angle.]

(vi)**⇒** (c) [Obtuse angled triangle has one of its angle more than 90°, i.e., one obtuse angle.]

(vii)**⇒** (b) [An Isosceles right angled triangle has two of its sides equal, with the angle between them = 90°.]

**Ques. 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 and acute angled triangle [Since, two of its side lengths are same, and all the angles are acute.]**(b)** Scalene and Right Angled triangle [Since, all side lengths are different, and one angle is 90°.]**(c)** Isosceles and Obtuse angled triangle [Since, two of its side lengths are same, and one of the angles is obtuse.]**(d)** Isosceles and Right Angled triangle [Since, two of its side lengths are same, and one angle is 90°.]**(e)** Equilateral and acute angled triangle [Since, all the side lengths are same, and all the angles are acute.]**(f)** Scalene and Obtuse angled triangle [Since, all side lengths are different, and one of the angles is obtuse.]

**Ques. 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)** Yes. 3 matchsticks will give us equilateral triangle with all sides equal.**(b)** No. We can’t make a triangle with 4 matchsticks because sum of two sides must be greater than the third side in a triangle. (1+1=2 is not greater than 2)**(c)** Yes. 5 matchsticks will give us isosceles triangle with two sides equal.**(d)** Yes. 6 matchsticks will give us equilateral triangle with all sides equal.

**NCERT Solutions for Class 6 Maths Chapter 5 Exercise 5.7**

**Ques. 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. (a)** True**(b)** True**(c)** True**(d)** True**(e)** False. [Opposite sides lengths are equal in a parallelogram. All sides are not necessarily equal.]**(f)** False. [Only one pair of opposite sides of a trapezium are parallel]

**Ques. 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)** A square can be thought of as a special rectangle with adjacent sides equal.**(b)** A rectangle can be thought of as a special parallelogram with all the angles equal to 90°.**(c)** A square can be thought of as a special rhombus with all the angles equal to 90°.**(d)** Squares, rectangles, parallelograms are all quadrilaterals as they all are polygons having 4 sides with the sum of interior angles equal to 360°.**(e)** Square is also a parallelogram because its opposite sides are equal and parallel.

**Ques. 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.** A regular quadrilateral has all sides equal and all angles equal giving the sum equal to 360°. Such a quadrilateral will have each angle equal to 90°. Hence, it is a square.

**NCERT Solutions for Class 6 Maths Chapter 5 Exercise 5.8**

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

**Sol. 1.** **(a)** Not a polygon (It is not a closed figure. End points are open.)**(b)** It is a polygon (closed curve bounded by 6 line segments).**(c)** Not a polygon (It doesn’t have line segments, although it is closed.)**(d)** Not a polygon (It doesn’t have line segments in all sides, although it is closed.)

**Ques. 2. Name each polygon. Make two more examples of each of these.**

**Sol. 2. (a)** Quadrilateral

**(b)** Triangle

**(c)** Pentagon

**(d)** Octagon

**Ques. 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.**

In a regular hexagon, all 6 sides are equal. Connecting vertices A, E and F, the triangle so formed is AEF, with sides AF = EF. Hence, the triangle so formed is isosceles triangle.

**Ques. 4. 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.**

**Sol. 4**

Joining the vertices C, D, G and H of the regular octagon ABCDEFGH, we get a rectangle, CDGH.

**Ques. 5. 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.**

**Sol. 5.**

Diagonals of pentagon ABCDE is given as AC, AD, BD, BE and CE.

**NCERT Solutions for Class 6 Maths Chapter 5 Exercise 5.9**

**Ques. 1. Match the following:**

**Give two new examples of each shape.**

**Sol. 1.** (a)**⇒** (ii)

(b)**⇒** (iv)

(c)**⇒** (v)

(d)**⇒** (iii)

(e)**⇒** (i)

**Ques. 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)** Instrument box **⇒** Cuboid

(b) A brick **⇒** Cuboid

(c) A match box **⇒** Cuboid

(d) A road-roller **⇒** Cylinder

(e) A sweet laddu **⇒** Sphere

**This Chapter contains the Exercises relating to the following topics , which are discussed in Chapter 5 – Understanding Elementary Shapes Class 6 NCERT book : –**

- Exercise 5.1 Introduction
- Exercise 5.2 Measuring Line Segments
- Exercise 5.3 Angles – ‘Right’ and ‘Straight’
- Exercise 5.4 Angles – ‘Acute’, ‘Obtuse’ and ‘Reflex’
- Exercise 5.5 Measuring Angles
- Exercise 5.6 Perpendicular Lines
- Exercise 5.7 Classification of Triangles
- Exercise 5.8 Quadrilaterals
- Exercise 5.9 Polygons
- Exercise 5.10 Three Dimensional Shapes

**Download NCERT Solutions for Class 6 Maths Chapter 5 – Understanding Elementary Shapes**

**Click here for Maths Class 6 NCERT Solutions**

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