NCERT Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes

Exercise 5.1: Measuring Line Segments

This exercise focuses on measuring line segments accurately using a ruler. Students learn to measure different lengths and understand the importance of precision in geometry.

Exercises 5.2, 5.3, and 5.4: Types of Angles

Students explore angles, such as acute, obtuse, right, straight, reflex, and complete angles. They learn to identify and draw these angles using a protractor.

Exercise 5.5: Perpendicular Lines

This exercise focuses on identifying and drawing perpendicular lines. Students learn the concept of perpendicularity, how to use a set square to draw perpendicular lines, and the properties of perpendicular lines in various shapes.

Exercise 5.6: Classification of Triangles

This exercise involves classifying triangles based on their sides (equilateral, isosceles, scalene) and angles (acute-angled, obtuse-angled, right-angled). Students learn to identify and differentiate between these types.

Exercise 5.7: Understanding Quadrilaterals

Students are introduced to different types of quadrilaterals, such as squares, rectangles, parallelograms, and rhombuses. They learn to identify these shapes based on their properties.

Exercise 5.8: Polygons

This exercise helps students recognize and categorize polygons based on the number of sides. They learn about regular and irregular polygons and understand their properties.

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

Exercise 5.1

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

Ans: One of the main disadvantage in comparing the line segments by mere observation is that there will be high chances of miscalculation.

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

Ans: We know that rulers are very thick. Hence, when we will measure the length, then we can misread the readings due to the thickness. Therefore, it is better to use a divider than a ruler.

3. Draw any line segment, say \[AB\]. Take any point \[C\] lying in between \[A\] and \[B\]. Measure the lengths of \[AB\], \[BC\] and \[AC\]. Is \[AB=AC+BC\]?

Ans: When we will draw any line segment $AB=7cm$ where $C$ is a point lying between $A$ and $B$ as shown in the figure below –

We have $AC=3cm$ and

$BC=4cm$.

Therefore, $AB=AC+BC$

$\Rightarrow AB=3+4$

$\Rightarrow AB=7cm$.

Yes, $AB=AC+BC$.

4. If \[A,\ B,\ \text{C}\] are three points on a line such that \[AB=5cm\] , \[BC=3cm\] and \[AC=8cm\] , which one of them lies between the other two?

Ans: As, the sum of $AB$ and $BC$ is equal to the length of the line segment \[AC\]. i.e., $AC=AB+BC$

$\Rightarrow 8=5+3$.

Therefore, the point $B$ lies in between the line segment \[AC\].

5. Verify whether \[D\] is the mid-point of \[\overline{AG}\].

Ans: From the figure we can observe that $AD=3cm$ and

$DG=3cm$. Therefore, $D$ is the mid-point.

6. If \[B\] is the mid-point of \[AC\] and \[C\] is the mid-point of \[BD\], where \[A\], \[B\], \[C\], \[D\] lie on a straight line, say why \[AB=CD\] ?

Ans:

As, given we have $B$ is the midpoint of $AC$.

$\Rightarrow AB=BC$

Similarly, $C$ is the midpoint of $BD$.

$\Rightarrow BC=CD$

From, the above conditions, we have $\Rightarrow AB=CD$. Hence, proved.

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.

Ans: No, from the given figures below we can conclude that the sum of two sides of a triangle can never be less than the third side of that triangle.

Exercise 5.2

1. What fraction of a clockwise revolution does the hour hand of a clock turn through, 

(i) when it goes from \[3\] to \[9\].

Ans: When the hour hand of a clock goes from $3$ to $9$, the number of divisions between them will be $6$, so the fraction will be $\frac{6}{12}=\frac{1}{2}$.

(ii) when it goes from \[4\] to \[7\].

Ans: When the hour hand of a clock goes from $4$ to $7$, the number of divisions between them will be $3$, so the fraction will be $\frac{3}{12}=\frac{1}{4}$.

(iii) when it goes from \[7\] to \[10\].

Ans: When the hour hand of a clock goes from $7$ to $10$, the number of divisions between them will be $3$, so the fraction will be $\frac{3}{12}=\frac{1}{4}$.

(iv) when it goes from \[12\] to \[9\].

Ans: When the hour hand of a clock goes from $12$ to $9$, the number of divisions between them will be $9$, so the fraction will be $\frac{9}{12}=\frac{3}{4}$.

(v) when it goes from \[1\] to \[10\].

Ans: When the hour hand of a clock goes from $1$ to $10$, the number of divisions between them will be $9$, so the fraction will be $\frac{9}{12}=\frac{3}{4}$.

(vi) when it goes from \[6\] to \[3\].

Ans: When the hour hand of a clock goes from $6$ to $3$, the number of divisions between them will be $9$, so the fraction will be $\frac{9}{12}=\frac{3}{4}$.

2. Where will the hand of a clock stop if it:

(a) starts at \[12\] and make \[\frac{1}{2}\] of a revolution, clockwise?

Ans: When the hand of a clock starts at $12$ and make $\frac{1}{2}$ of a revolution, clockwise then it will stop at $\frac{12}{2}=6$.

(b) starts at \[2\] and makes \[\frac{1}{2}\] of a revolution, clockwise?

Ans: When the hand of a clock starts at $2$ and make $\frac{1}{2}$ of a revolution, clockwise then it will stop at $2+\frac{12}{2}=8$.

(c) starts at \[5\] and makes \[\frac{1}{4}\] of a revolution, clockwise?

Ans: When the hand of a clock starts at $5$ and make $\frac{1}{4}$ of a revolution, clockwise then it will stop at $5+\frac{12}{4}=8$.

(d) starts at \[5\] and makes \[\frac{3}{4}\] of a revolution, clockwise?

Ans: When the hand of a clock starts at $5$ and make $\frac{3}{4}$ of a revolution, clockwise then it will stop at $5+12\times \frac{3}{4}=14$ i.e., at $2$ as one revolution is $=12$.

3. Which direction will you face if you start facing:

(a) East and make \[\frac{1}{2}\] of a revolution clockwise?

Ans:

When we will make $\frac{1}{2}$ of a revolution, clockwise from east, then we will face the west direction.

(b) East and make \[1\frac{1}{2}\] of a revolution clockwise?

Ans: When we will make $1\frac{1}{2}$ of a revolution, clockwise from east, then we will face the west direction.

(c) West and makes \[\frac{3}{4}\] of a revolution clockwise?

Ans:

When we will make $\frac{3}{4}$ of a revolution, clockwise from west, then we will face the north direction.

(d) South and make one full revolution?

Ans:

When we will make a one full revolution from south, then we will face the south direction. There will be no need for mentioning clockwise or anti-clockwise because anyhow one full revolution will bring us back to the original direction.

4. What part of a revolution have you turned through if you stand facing –

(a) East and turn clockwise to face north?

Ans:

When we will turn clockwise from east to face north, then the revolution will be $\frac{3}{4}$.

(b) South and turn clockwise to face east?

Ans:

When we will turn clockwise from south to face east, then the revolution will be $\frac{3}{4}$.

(c) West and turn clockwise to face east?

Ans:

When we will turn clockwise from west to face east, then the revolution will be $\frac{1}{2}$.

5. Find the number of right angles turned through by the hour hand of a clock 

(a) when it goes from \[3\] to \[6\].

Ans: When the hour hand of a clock goes from $3$ to $6$ then the number of right angles turned will be one.

(b) when it goes from \[2\] to \[8\].

Ans: When the hour hand of a clock goes from $2$ to $8$ then the number of right angles turned will be two.

(c) when it goes from \[5\] to \[11\].

Ans: When the hour hand of a clock goes from $5$ to $11$ then the number of right angles turned will be two.

(d) when it goes from \[10\] to \[1\].

Ans: When the hour hand of a clock goes from $10$ to $1$ then the number of right angles turned will be one.

(e) when it goes from \[12\] to \[9\].

Ans: When the hour hand of a clock goes from $12$ to $9$ then the number of right angles turned will be three.

(f) when it goes from \[12\] to \[6\].

Ans: When the hour hand of a clock goes from $12$ to $6$ then the number of right angles turned will be two.

6. How many right angles do you make if you start facing –

(a) South and turn clockwise to west?

Ans: When we start facing south and turn clockwise to west then the number of right angles turned will be one.

(b) North and turn anti-clockwise to east?

Ans: When we start facing north and turn anti-clockwise to east then the number of right angles turned will be three.

(c) West and turn to west?

Ans: When we start facing west and turn to west then the number of right angles turned will be four.

(d) South and turn to north?

Ans: When we start facing south and turn to north then the number of right angles turned will be two.

7. Where will the hour hand of a clock stop if it starts: 

(a) from \[6\] and turns through \[1\] right angle?

Ans: One right angle = One-fourth of complete rotation of a clock $\frac{1}{4}\times 12=3$

When the hour hand of a clock starts from $6$ and turns through $1$ right angle, then it will stop at $6+3=9$.

(b) from \[8\] and turns through \[2\] right angles?

Ans: One right angle = One-fourth of complete rotation of a clock $\frac{1}{4}\times 12=3$

When the hour hand of a clock starts from $8$ and turns through $2$ right angles, then it will stop at $8+3+3=14$ i.e., at $2$.

(c) from \[10\] and turns through \[3\] right angles?

Ans: One right angle = One-fourth of complete rotation of a clock $\frac{1}{4}\times 12=3$

When the hour hand of a clock starts from $10$ and turns through $3$ right angles, then it will stop at $10+3+3+3=19$ i.e., at 7$.

(d) from \[7\] and turns through \[2\] straight angles?

Ans: One Straight angle = 2 right angles

When the hour hand of a clock starts from $7$ and turns through $2$ straight angles, then it will stop at $7+6+6=19$ i.e., at 7.

Exercise 5.3

1. Match the following:

Ans:

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

(a) 

Ans: It is an acute angle.

(b) 

Ans: It is an obtuse angle.

(c)

Ans: It is a right angle.

(d)

Ans: It is a reflex angle.

(e)

Ans: It is a straight angle.

(f) 

Ans: It is an acute angle.

Exercise 5.4

1. What is the measure of –

(i) a right angle?

Ans: The measure of a right angle is $90{}^\circ $.

(ii) a straight angle?

Ans: The measure of a straight angle is $180{}^\circ $.

2. Say True or False:

(a) The measure of an acute angle \[<90{}^\circ \].

Ans: True.

As, we measure any acute angle, then it will always be less than a right angle.

(b) The measure of an obtuse angle \[<90{}^\circ \].

Ans: False.

When we measure an obtuse angle, then it will always be more than a right angle.

(c) The measure of a reflex angle \[>180{}^\circ \].

Ans: True.

A reflex angle is always more than the measure of a straight line.

(d) The measure of a complete revolution \[=360{}^\circ \].

Ans: True.

When we complete one revolution, it is measured as turn of two straight lines.

(e) If \[m\angle A=53{}^\circ \] and 

\[m\angle B=35{}^\circ \], then 

\[m\angle A>m\angle B\].

Ans: True.

3. Write down the measure of:

(a) some acute angles

Ans: As, an acute angle is measured as less than a right angle. Therefore, some of the acute angles will be $45{}^\circ $, $30{}^\circ $, and etc.

(b) some obtuse angles

Ans: As, an obtuse angle is measured as more than a right angle. Therefore, some of the obtuse angles will be $105{}^\circ $, $140{}^\circ $, and etc.

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

(a)

Ans: After using the protractor, we can conclude that the given figure is an acute angle measured as $40{}^\circ $.

(b)

Ans: After using the protractor, we can conclude that the given figure is an obtuse angle measured as $130{}^\circ $.

(c) 

Ans: After using the protractor, we can conclude that the given figure is a right angle measured as $90{}^\circ $.

(d)

Ans: After using the protractor, we can conclude that the given figure is an acute angle measured as $60{}^\circ $.

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

Ans: After estimating we can conclude that the measure of angle $B$ is larger than the measure of angle $A$.

Measure of angle \[A=\]

The measure of angle $A$ is $40{}^\circ $.

Measure of angle \[B=\]

The measure of angle $B$ is $65{}^\circ $.

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

Ans: From the given figure we can estimate that the second angle has a larger measure than the first angle. After using the protractor, the measure of first angle is $45{}^\circ $ and the measure of second angle is $60{}^\circ $.

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 __________.

Ans: An angle whose measure is less than that of a right angle is acute angle.

(b) An angle whose measure is greater than that of a right angle is __________. 

Ans: An angle whose measure is greater than that of a right angle is obtuse angle.

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

Ans: An angle whose measure is the sum of the measures of two right angles is 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: When the sum of the measures of two angles is that of a right angle, then each one of them is 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: 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 angle.

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).

Ans: After estimating the measure of each angle shown in figure, we have –

(i) $30{}^\circ $

(ii) $120{}^\circ $

(iii) $60{}^\circ $

(iv) $150{}^\circ $

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

Ans: From the given figure, as the revolution is of $\frac{1}{4}$ from $9$ to $12$ hence, the measure of the angle in first figure will be of $90{}^\circ $, second figure has angle of measure $30{}^\circ $ as each division is equal and the third figure has angle of measure $180{}^\circ $ as the revolution is of $\frac{1}{2}$ which will form a straight line.

10. Investigate:

In the given figure, the angle measure \[30{}^\circ \]. Look at the same figure through a magnifying glass. Does the angle become larger? Does the size of the angle change?

Ans: No, the size of the angle does not change and hence, remains same even after using the magnifying glass.

11. Measure and classify each angle:

Ans: The measure of each angle can be measured by the help of a protractor. Hence, the angles will be –

Exercise 5.5

1. Which of the following are models for perpendicular lines –

(a) The adjacent edges of a tabletop.

Ans: Yes, adjacent edges of a table top are perpendicular.

(b) The lines of a railway track.

Ans: No, the lines of a railway track are not perpendicular.

(c) The line segments forming the letter ‘\[L\]’.

Ans: Yes, the line segments forming the letter $L$ are perpendicular.

(d) The letter \[V\].

Ans: No, the letter $V$ is not perpendicular.

2. Let \[PQ\] be the perpendicular to the line segment \[XY\]. Let \[PQ\] and \[XY\] intersect in the point \[A\]. What is the measure of \[\angle PAY\].

Ans: As, $PQ$ and $XY$ are perpendicular line segments to each other and intersect at point $A$ can be shown in the figure below –

Therefore, the measure of $\angle PAY=90{}^\circ $.

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?

Ans: From the two “set-squares” in our box, we have one set-square as $30{}^\circ $, $60{}^\circ $, $90{}^\circ $ and another set-square as $45{}^\circ $, $45{}^\circ $, $90{}^\circ $. Therefore, the angle that is common is $90{}^\circ $.

4. Study the diagram. The line \[l\] is perpendicular to line \[m\].

(a) Is \[CE=EG\]?

Ans: As, $CE=2\ \text{units}$ and

$EG=2\ \text{units}$. Therefore, 

$CE=EG$.

(b) Does \[PE\] bisect \[CG\]?

Ans: From the given figure we can conclude that yes, $PE$ bisects $CG$.

(c) Identify any two-line segments for which \[PE\] is the perpendicular bisector.

Ans: Any two-line segments for which $PE$ is the perpendicular bisector are $DF$ and $CG$.

(d) Are these true? 

(i) \[AC>FG\]

Ans: Yes, it’s true. As, $AC=2\ \text{units}$ and

$FG=1\ \text{unit}$. Hence, $AC>FG$.

(ii) \[CD=GH\]

Ans: Yes, it’s true. As, both $CD$ and $GH$ are of $\text{1}\ \text{unit}$.

(iii) \[BC<EH\]

Ans: Yes, it’s true. As, $BC=1\ unit$ and

$EH=3\ \text{units}$. Hence, $BC<EH$.

Exercise 5.6

1. Name the types of following triangles: 

(a) Triangle with lengths of sides \[7cm\], \[8cm\] and \[9cm\].

Ans: As, all the three sides of the triangle have different lengths,

therefore, it is a scalene triangle.

(b) \[\Delta ABC\] with \[AB=8.7cm\],  \[AC=7cm\] and \[BC=6cm\].

Ans: As, all the three sides of the triangle have different lengths, therefore, it is a scalene triangle.

(c) \[\Delta PQR\] such that \[PQ=QR=PR=5cm\].

Ans: As, all the three sides of the triangle have same lengths, therefore, it is an equilateral triangle.

(d) \[\Delta DEF\] with \[m\angle D=90{}^\circ \].

Ans: Since, one of the angles of the triangle is $90{}^\circ $, therefore, it is a right-angled triangle.

(e) \[\Delta XYZ\] with \[m\angle Y=90{}^\circ \] and  \[XY=YZ\].

Ans: Since, one of the angles of the triangle is $90{}^\circ $, and two sides are of same length therefore, it is an isosceles right-angled triangle.

(f) \[\Delta LMN\] with \[m\angle L=30{}^\circ \], \[m\angle M=70{}^\circ \] and  \[m\angle N=80{}^\circ \].

Ans: Since, all the angles of the triangle are less than right angle, therefore, it is an acute triangle.

2. Match the following: Measure of Triangle Types of Triangle 

Ans:

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

Ans: The triangles in two different ways will be as –

(a) Acute angled triangle and Isosceles triangle.

(b) Right-angled triangle and Scalene triangle.

(c) Obtuse-angled triangle and Isosceles triangle.

(d) Right-angled triangle and Isosceles triangle.

(e) Equilateral triangle and acute angled triangle.

(f) Obtuse-angled triangle and scalene triangle.

4. Can you make a triangle with

(a) \[3\] matchsticks – Triangle possible

(b) \[4\] matchsticks – Triangle not-possible

(c)  \[5\] matchsticks – Triangle Possible 

This is an acute angle triangle, and it is possible to make triangle with five matchsticks.

(d) \[6\] matchsticks – Triangle Possible

Exercise 5.7

1. Say true or false: 

(a) Each angle of a rectangle is a right angle.

Ans:

True. Each side of a rectangle is a right angle.

(b) The opposite sides of a rectangle are equal in length.

Ans:

True. Opposite sides of a rectangle are always equal and parallel to each other.

(c) The diagonals of a square are perpendicular to one another.

Ans:

True. Square has diagonals that are bisecting each other at $90{}^\circ $.

(d) All the sides of a rhombus are of equal length.

Ans:

True. Rhombus has all sides of equal length.

(e) All the sides of a parallelogram are of equal length.

Ans:

False. Not all sides of a parallelogram are of equal length.

(f) The opposite sides of a trapezium are parallel.

Ans: False. Opposite sides of a trapezium are never equal.

2. Give reasons for the following: 

(a) A square can be thought of as a special rectangle.

Ans: As, opposite sides of a rectangle are equal, and all angles are $90{}^\circ $. Therefore, we can say square as a special rectangle with all sides of equal length and all angles as $90{}^\circ $.

2. A rectangle can be thought of as a special parallelogram.

Ans: As, both rectangle and parallelogram have opposite sides equal and parallel to each other. Therefore, we can say that rectangle can be a special parallelogram.

(c) A square can be thought of as a special rhombus.

Ans: Because both square and rhombus has all sides of same length and also their diagonals bisect each other at $90{}^\circ $. Therefore, a square can be a special rhombus.

(d) Squares, rectangles, parallelograms are all quadrilateral.

Ans: A quadrilateral is a shape that has at least $4$ sides. Therefore, squares, rectangles, parallelograms are all types of quadrilaterals.

(e) Square is also a parallelogram.

Ans: Since, in parallelogram opposite sides are always equal and parallel to each other. Therefore, a square can also be a special parallelogram.

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?

Ans: A square has its all sides in equal length and all angles are also equal. Therefore, a square is a regular quadrilateral.

Exercise 5.8

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

Ans: As, the figure in $(a)$ is not close, hence, it cannot be a polygon. Whereas figure $(b)$ is a closed shape, hence, it is a polygon. Similarly, in figure $(c)$ and $(d)$ as the shape is not enclosed by line segments, therefore, they are also not polygons.

2. Name each polygon:

Ans: The polygon in figure $(a)$ is a quadrilateral as it has four sides, in figure $(b)$ it is a triangle as it has three sides, in figure $(c)$ it is a pentagon as it has five sides and in figure $(d)$ it is a octagon as it has eight sides.

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.

Ans: A rough sketch of a regular hexagon with connecting any three of its vertices and drawing a triangle will be as –

So, we have joined three vertices $A$, $E$, and $F$. Therefore, the triangle formed is an isosceles triangle as two sides are equal.

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.

Ans: A rough sketch of a regular octagon when a rectangle is drawn using any of its four vertices will be as –

Here, $CDGH$ is a rectangle.

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.

Ans: A rough sketch of a pentagon with its diagonals will be as –

Here, $ABCDE$ is a pentagon which has diagonals as $AD$, $AC$, $BE$, and $BD$.

Overview of Deleted Syllabus for CBSE Class 6 Maths Understanding Elementary Shapes

Class 6 Maths Chapter 5: Exercises Breakdown

Conclusion

Chapter 5 of Class 6 Maths, "Understanding Elementary Shapes," provides students with a solid foundation in geometry. This chapter covers a wide range of topics, including measuring line segments, types of angles, classification of triangles and quadrilaterals, polygons, three-dimensional shapes, perpendicular lines, and symmetry. Each topic is designed to help students develop a comprehensive understanding of geometric concepts, which are essential for their further studies in mathematics.

In previous years' exams, Class 6th Maths Chapter 5 has been a significant part of the curriculum, with around 10-15 questions typically asked. 

Other Study Material for CBSE Class 6 Maths Chapter 5

Chapter-Specific NCERT Solutions for Class 6 Maths

Given below are the chapter-wise NCERT Solutions for Class 6 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.