NCERT Solutions Class 6 Mathematics Chapter 5 Understanding Elementary Shapes

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 𝐴𝐵 = 𝐴𝐶 + 𝐶𝐵? [Note: If A,B,C are any three points on a line such that 𝐴𝐶 +
𝐶𝐵 = 𝐴𝐵, 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 𝐴𝐵. Hence, for every situation in which point 𝐶 is lying in between 𝐴 & 𝐵 , it may be say that 𝐴𝐵 = 𝐴𝐶 + 𝐶𝐵
Consider a line segment 𝐴𝐵 of length = 8 𝑐𝑚.
Now, mark a point C at 𝐴𝐶 = 5 𝑐𝑚
Measure 𝐶𝐴.
We see that the length of 𝐶𝐴 = 3 𝑐𝑚
i.e., 𝐴𝐵 = 𝐴𝐶 + 𝐶𝐵
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 𝐴𝐵 = 5 𝑐𝑚, 𝐵𝐶 = 3 𝑐𝑚 and 𝐴𝐶 = 8 𝑐𝑚.
From the given measures, we can see that
𝐴𝐵 + 𝐵𝐶 = 8 𝑐𝑚 = 𝐴𝐶
⃡𝐴 𝐶 is the longest line segment, thus B is the point between A and C.
We can say this because 𝐴𝐶 > 𝐴𝐵 and 𝐴𝐶 > 𝐵𝐶.

Q.5) Verify, whether D is the midpoint of AB^↔

Sol.5) From the figure above,
𝐴𝐷 = 4 – 1 = 3 𝑐𝑚
𝐷𝐺 = 7 – 4 = 3 𝑐𝑚
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 𝐴, 𝐵, 𝐶, 𝐷 lie on a straight line, say why 𝐴𝐵 = 𝐶𝐷?
Sol.6) B is the mid-point of AC.
∴ AB = BC ………. (i)
And C is the mid-point of BD.
∴ BC = CD ………. (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 𝑢𝑛𝑖𝑡
Third side of triangle = 5.2 𝑐𝑚
In figure (b),
Sum of two sides = 15 + 8 = 23 𝑢𝑛𝑖𝑡𝑠
Third side of triangle = 17 𝑐𝑚
In figure (c),
Sum of two sides = 10 + 9 = 19 𝑢𝑛𝑖𝑡
Third side of triangle = 15 𝑐𝑚
In figure (d),
Sum of two sides = 5 + 5 = 10 𝑢𝑛𝑖𝑡𝑠
Third side of triangle = 8 𝑐𝑚

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, 360° which is one revolution.
When it moves from 12 to 6, it moves two right angles, that is, 180°. This is 180/360 = ½ of one revolution.

When the hour hand of the clock goes from 6 to 3, it rotates by 3 right angles, that is = 270°. Fraction of revolution = 270°/360° = 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 360°. 
1/2 of a revolution is 360°'/2 = 180°. 
1/2 of a revolution is 0°. 3/4 of a revolution is 270°
a) The hand of the clock starts at 12 and makes 1/2 of a revolution. That is, it starts at 12 and revolves 180° 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 180° and stops at 8.
c) When the hand starts at 5 and makes 1/4 of a revolution, that is, 90°, it stops at 9.
d) When the hand starts at 5 and makes 3/4 of a revolution, that is 270°, 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 ¾ 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 180°. 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, 180°. Hence, you would be facing west.
c) When you start facing west and make ¾ of a revolution anti-clockwise, you revolve by 270°. This is by 180°and then by 90°. When you revolve by 180°anti-clockwise you face east and then you revolve by 90° anti-clockwise, you face north.
d) When you start facing south and make a full revolution, that is 360°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. 90°.
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 90° + 90° + 90° = 270° from East to North in the clock-wise direction. This is 270°/360° = 3/4 th of the complete revolution.
b) You start from south and turn clockwise to face east.

So, you have turned 90° + 90° = 180° from West to East in the clock-wise direction.
This is 180°/360° = 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, 360°.
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. 90°.

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 360°, 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 ¼ and ½ of a revolution
f) One complete revolution
Sol.1) Straight angle is 180° which is ½ of a complete revolution. So,     i) ↔ c)
Right angle is 90° which is ¼ of a complete revolution. So,     ii) ↔ d)
Acute angles are angles less than 90°. This is less than one-fourth of a revolution. So,     iii) ↔ a)
Obtuse angles are angles greater than 90° but less than 180°. This is between ¼ and ½ of a complete revolution. So,     iv) ↔ e)
Reflex angles are angles greater than 180° but less than 360°, that is, more than ½ of a complete revolution. So,    v) ↔ 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 90° and hence that of a straight angle is 180°. 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 90°
b) A straight angle measures 180°.

Q.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 = 90°.
(e) If 𝑚∠𝐴 = 50° and 𝑚∠𝐵 = 35°, then 𝑚∠𝐴 > 𝑚∠𝐵.
Sol.2) a) True
b) False, an obtuse angle is greater than 90°
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 90°. Some acute angles are 35°., 43°.
b) Obtuse angles are angles greater than 90°. and less than 180°.. Some obtuse angles are 135°., 143°.

Q.4) Measure the angles given below, using the protractor and write down the measure:
Sol.4) (a) 45°
(b) 120°
(c) 90°
(d) 60°, 130°, 90°

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

Sol.5) ∠𝐵 has larger measure.
∠𝐴 = 40° and ∠𝐵 = 65°

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

Sol.6) The angles measure 75° and 70°
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) 45°
(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 ‘L’.
(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 PQ¯ be the perpendicular to the line segment XY¯. Let PQ¯ and XY¯ intersect in the point A. What is the measure of ∠𝑃𝐴𝑌.
Sol.2) 

PQ is perpendicular to XY and it intersects at point A.
So, ∠𝑃𝐴𝑌 = 90°.

Q.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 measures 90°, 60°, 30°
Other set-square measures 90°, 45°, 45°
90° is a common measure.

Q.4) Study the diagram. The line 𝑙 is perpendicular to line 𝑚
(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) 𝐴𝐶 > 𝐹𝐺
(ii) 𝐶𝐷 = 𝐺𝐻
(iii) 𝐵𝐶 < 𝐸𝐻.
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 𝑐𝑚, 8 𝑐𝑚 𝑎𝑛𝑑 9 𝑐𝑚.
(b) Δ𝐴𝐵𝐶 with 𝐴𝐵 = 8.7𝑐𝑚, 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, 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 – Equilateral triangle (e)
ii) 2 sides of equal length – Isosceles (g)
iii) All sides of different length – Scalene (a)
iv) 3 acute angles – Acute angled (f)
v) 1 right angle – Right angled (d)
vi) 1 obtuse angle – Obtuse angled (c)
vii) 1 right angle with 2 sides of equal length – 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 90° & 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