NCERT Solutions For Class 6 Maths Chapter 5 Understanding Elementary Shapes
In NCERT Solutions For Class 6 Maths Chapter 5 Understanding Elementary we learn about different shaps. Ncert Solutions are provided by our expert teachers. In NCERT Solutions Class 6 Maths Chapter 5 there are 9 exerrcise.
we will discuss corners, edges, planes, open curves and closed curves in our surroundings. We organise them into line segments, angles, triangles, polygons and circles.
NCERT Solutions For Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercises 5.1
Class 6th Maths Chapter 5 Solutions ex 5.1
Q.1. What is the disadvantage in comparing line segments by mere observation?
Ans:-
Comparing the lengths of two line segments simply by ‘observation’ may not be accurate. So we use a divider to compare the length of the given line segments.
Q.2. Why is it better to use a divider than a ruler, while measuring the length of a line segment?
Ans:-
While using a ruler, chances of error occur due to the thickness of the ruler and angular viewing. Hence, using divider accurate measurement is possible.
Q.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 + CB?
[Note: If A, B, C are any three points on a line such AC + CB = AB, then we can be sure that C lies between A and B]
Ans:-
A, B and C such that C lies between A and B and AB = 7 cm.
AC = 3 cm, CB = 4 cm.
∴ AC + CB = 3 cm + 4 cm = 7 cm.
But, AB = 7 cm.
So, AB = AC + CB.
Q.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?
Ans:-
AB = 5 cm
BC = 3 cm
AB + BC = 5 + 3 = 8 cm
But, AC = 8 cm
Hence, B lies between A and C.
5. Verify, whether D is the midpoint of AG¯¯¯¯¯¯¯¯.
Ans:-
From the given figure, we have
AG = 7 cm – 1 cm = 6 cm
AD = 4 cm – 1 cm = 3 cm
and DG = 7 cm – 4 cm = 3 cm
∴ AG = AD + DG.
Hence, D is the midpoint of AG¯¯¯¯¯¯¯¯.
Q.6. If B is the midpoint of AC¯¯¯¯¯¯¯¯ and C is the midpoint of BD¯¯¯¯¯¯¯¯, where A, B, C, D lie on a straight line, say why AB = CD?
Ans:-
given B is the midpoint of AC.
∴ AB = BC …(i)
C is the mid-point of BD.
BC = CD…(ii)
From Eq.(i) and (ii), We have
AB = CD
Q.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.
Ans:-
Case 1. In triangle ABC
Let AB = 2.5 cm
BC = 4.8 cm
and AC = 5.2 cm
AB + BC = 2.5 cm + 4.8 cm
= 7.3 cm
Since, 7.3 > 5.2
So, AB + BC > AC
Hence, sum of any two sides of a triangle is greater than the third side.
Case 2. In triangle PQR
Let PQ = 2 cm
QR = 2.5 cm
and PR = 3.5 cm
PQ + QR = 2 cm + 2.5 cm = 4.5 cm
Since, 4.5 > 3.5
So, PQ + QR > PR
Hence, sum of any two sides of a triangle is greater than the third side.
Case 3. In triangle XYZ
Let XY = 5 cm
YZ = 3 cm
and ZX = 6.8 cm
XY + YZ = 5 cm + 3 cm
= 8 cm
Since, 8 > 6.8
So, XY + YZ > ZX
Hence, the sum of any two sides of a triangle is greater than the third side.
Case 4. In triangle MNS
Let MN = 2.7 cm
NS = 4 cm
MS = 4.7 cm
and MN + NS = 2.7 cm + 4 cm = 6.7 cm
Since, 6.7 >4.7
So, MN + NS > MS
Hence, the sum of any two sides of a triangle is greater than the third side
Case 5. In triangle KLM
Let KL = 3.5 cm
LM = 3.5 cm
KM = 3.5 cm
and KL + LM = 3.5 cm + 3.5 cm = 7 cm
7 cm > 3.5 cm
Solution:
(i) For one-fourth revolution, we have
So, KL + LM > KM
Hence, the sum of any two sides of a triangle is greater than the third side.
Hence, we conclude that the sum of any two sides of a triangle is never less than the third side.
NCERT Solutions For Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercises 5.2
Ch 5 Class 6 Maths Solutions ex 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
Ans:-
9 – 3 = 6 ÷ 12 = 1/2 of a revolution
(b) 4 to 7
Ans:-
7 – 4 = 3 ÷ 12 = 1/4 of a revolution
(c) 7 to 10
Ans:-
10 – 7 = 3 ÷ 12 = 1/4 of a revolution
(d) 12 to 9
Ans:-
9 – 0 = 9 ÷ 12 = 3/4 of a revolution
(e) 1 to 10
Ans:-
10 – 1 = 9 ÷ 12 = 3/4 of a revolution
(f) 6 to 3
Ans:-
6 to 3 i.e., 6 to 12 and then 12 to 3
6 to 12 = 12 – 6 = 6 and 12 to 3 = 0 to 3 = 3 – 0 = 3
6 + 3 = 9 ÷ 12 = 3/4 of a revolution
Q.2. Where will the hand of a clock stop if it
(a) starts at 12 and makes 1 / 2 of a revolution, clockwise?
Ans:-
Starting from 12 and making 1/2 of a revolution, the clock hand stops at 6.
(b) starts at 2 and makes 1 / 2 of a revolution, clockwise?
Ans:-
Starting from 2 and making 1/2 of a revolution, the clock hand stops at 8.
(c) starts at 5 and makes 1 / 4 of a revolution, clockwise?
Ans:-
Starting from 5 and making 1/2 of a revolution, the clock hand stops at 8.
(d) starts at 5 and makes 3 / 4 of a revolution, clockwise?
Ans:-
Starting from 5 and making 1/2 of a revolution, the clock hand 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 ½ of a revolution clockwise?
(c) west and make 3 / 4 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?)
Taking one full revolution we will reach back to the original (starting) position. Therefore, it makes no difference whether we turn clockwise or anticlockwise.
Q.4. What part of a revolution have you turned through if you stand facing
(a) east and turn clockwise to face north?
Ans:-
If we start from east and reach north (turning clockwise) 3/4 of a revolution is required
(b) south and turn clockwise to face east
Ans:-
If we start from south turning clockwise to face east, 3/4 of a revolution is required.
(c) west and turn clockwise to face east?
Ans:-
If we start from west turning clockwise to face east, 1/2 of a revolution is required.
5. Find the number of right angles turned through by the hour hand of a clock when it goes from
(a) 3 to 6
Ans:-
Starting from 3 to 6, the hour hand turns through 1 right angle.
(b) 2 to 8
Ans:-
Starting from 2 to 8, the hour hand turns through 2 right angles.
(c) 5 to 11
Ans:-
Starting from 5 to 11, the hour hand turns through 2 right angles
(d) 10 to 1
Ans:-
Starting from 10 to 1, the hour hand turns through 1 right angle
(e) 12 to 9
Ans:-
Starting from 12 to 9, the hour hand turns through 3 right angles.
(f) 12 to 6
Ans:-
Starting from 12 to 6, the hour hand turns through 2 right angles.
Q.6. How many right angles do you make if you start facing
(a) south and turn clockwise to the west?
(b) north and turn anti-clockwise to the east?
(c) west and turn to west?
(d) south and turn to the north?
Ans:-
Q.7. Where will the hour hand of a clock stop if it starts
(a) from 6 and turns through 1 right angle?
Ans:-
Starting from 6 and turning through 1 right angle, the hour hand stops at 9
(b) from 8 and turns through 2 right angles?
Ans:-
Starting from 8 and turning through 2 right angles, the hour hand stops at 2.
(c) from 10 and turns through 3 right angles?
Ans:-
Starting from 10 and turning through 3 right angles, the hour hand stops at 7.
(d) from 7 and turns through 2 straight angles?
Ans:-
Starting from 7 and turning through 2 right angles, the hour hand stops at 7
NCERT Solutions For Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercises 5.3
Class 6th Maths Chapter 5 Solution ex 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 1/4 and 1/2 of a revolution.
– (f) One complete revolution
Ans:-
(i) Straight angle ↔ (c) Half of a revolution.
(ii) Right angle ↔ (d) One-fourth of a revolution.
(iii) Acute angle ↔ (a) Less than one-fourth of a revolution.
(iv) Obtuse angle ↔ (e) Between 1/4 and 1/2 of a revolution.
(v) Reflex angle ↔ (f) One complete revolution, right, acute, obtuse or reflex
Q.2. Classify each one of the following angles as right, straight, acute, obtuse, or reflex:
Ans:-
(a) Acute angle
(b) Obtuse angle
(c) Right angle
(d) Reflex angle
(e) Straight angle
(f) Acute angle
NCERT Solutions For Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercises 5.4
Ch 5 Class 6 Maths Solutions ex 5.4
Q.1. What is the measure of
(i) a right angle?
Ans:-
Measure of a right angle = 90°
(ii) a straight angle
Ans:-
Measure of a straight angle = 180°
Q.2. Say True or False:
(a) The measure of an acute angle < 900
(b) The measure of an obtuse angle < 900
(c) The measure of a reflex angle > 1800
(d) The measure of one complete revolution = 3600
(e) If m ∠A = 530 and m ∠B = 350, then m ∠A > m ∠B.
Ans:-
(a) True
(b) False
(c) True
(d) True
(e) True
Q.3. Write down the measures of
(a) some acute angles
Ans:-
20°, 55°, and 70° are acute angles.
(b) some obtuse angles
Ans:-
110°, 150°, and 175° are obtuse angles.
Q.4. Measures the angles given below using the protractor and write down the measure.
(i) 45°
(ii) 125°
(iii) 90°
(iv) ∠1 = 60°, ∠2 = 90°, ∠3 = 125°
Q.5. Which angle has a large measure? First, estimate and then measure.
Measure of Angle A =
Measure of Angle B =
Ans:-
Measure of Angle A = 40°
Measure of Angle B = 60°.
Q.6. From these two angles which has larger measure? Estimate and then confirm by measuring them.
Ans:-
Measure of angle (a) = 45°
and the measure of angle (b) = 60°
Q.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:- acute
(b) An angle whose measure is greater than that of a right angle is ____
Ans:- obtuse
(c) An angle whose measure is the sum of the measures of two right angles is _______
Ans:- straight
(d) When the sum of the measures of two angles is that of a right angle, then each one of them is _____
Ans:- 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 ______
Ans:- obtuse
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).
(a) Measure of the angle = 40°
(b) Measure of the angle = 130°
(c) Measure of the angle = 65°
(d) Measure of the angle = 135°.
Q.9. Find the angle measure between the hands of the clock in each figure:
Ans:-
(i) The angle between the hour hand and minute hand of a clock at 9.00 a.m = 90°
(ii) The angle between the hour hand and minute hand of a clock at 1.00 p.m = 30°
(iii) The angle between the hour hand and minute hand of a clock at 6.00 p.m = 180°.
10. Investigate
In the given figure, the angle measure 30°. Look at the same figure through a magnifying glass. Does the angle become larger? Does the size of the angle change?
Ans:-
It is an activity. So try it yourself.
11. Measure and classify each angle:
Angle | Measure | Type |
<AOB | ||
<AOC | ||
<BOC | ||
<DOC | ||
<DOA | ||
<DOB |
Ans:-
Angle | Measure | Type |
<AOB | 400 | Acute |
<AOC | 1250 | Obtuse |
<BOC | 850 | Acute |
<DOC | 950 | Obtuse |
<DOA | 1400 | Obtuse |
<DOB | 1800 | Straight |
NCERT Solutions For Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercises 5.5
NCERT solutions for class 6 maths chapter 5 ex 5.5
Q.1. Which of the following are models for perpendicular lines:
(a) The adjacent edges of a table top.
Ans:-
Yes, the adjacent edges of a table top are the models of perpendicular lines.
(b) The lines of a railway track.
Ans:-
No, The lines of a railway track are parallel to each other. So they are not a model for perpendicular lines.
(c) The line segments forming the letter ‘L’.
Ans:-
Yes, the two-line segments of‘L’ are the model of perpendicular lines.
(d) The letter V.
Ans:-
No, the two line segments of ‘V’ are not a model for perpendicular lines.
Q.2 Let PQ¯¯ be the perpendicular to the line segment XY¯¯ Let PQ¯¯ and XY¯¯ intersect at in the point A. What is the measure of ∠PAY?
Ans:-
Since PQ¯¯ ⊥ XY
∴ ∠PAY = 90°
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:-
The measure of angles in one set square are 300, 600 and 900
The other set square has a measure of angles 450, 450 and 900
Yes, the angle of measure 900 is common in between them
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.
Ans:-
(a) Yes, since, CE = 2 units and EG = 2 units respectively
(b) Yes. Since, CE = EG as both are of 2 units. Hence PE bisect CG
(c) and are the line segments for which PE is the perpendicular bisector
(d) (i) True. Since AC = 2 units and FG = 1 unit
∴ AC > FG
(ii) True because both are of 1 unit
(iii) True. Since, BC = 1 unit and EH = 3 units
∴ BC < EH
NCERT Solutions For Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercises 5.6
class 6 maths chapter 5 solutions ex 5.6
1. Name the types of following triangles:
(a) Triangle with lengths of sides 7 cm, 8 cm and 9 cm.
Ans:-
All sides of the given triangle are different
Hence, it is a Scalene triangle.
(b) ∆ABC with AB = 8.7 cm, AC = 7 cm and BC = 6 cm.
Ans:-
All sides of the given triangle are different
Hence, it is a Scalene triangle.
(c) ∆PQR such that PQ = QR = PR = 5 cm.
Ans:-
All sides are equal.
Hence, it is an equilateral triangle.
(d) ∆DEF with ∠D = 90°
Ans:-
Given that: In ∆DEF, m∠D = 90°
Hence it is a right-angled triangle
(e) ∆XYZ with ∠Y = 90° and XY = YZ.
Ans:-
Given that: In ∆XYZ, m∠Y = 90° and XY = YZ
Hence it is a right-angled triangle
(f) ∆LMN with ∠L = 30°, ∠M = 70° and ∠N = 80°
Ans:-
Given that: ∆LMN, m∠L = 30°, m ∠M = 70° and m∠N = 80°.
Hence it is an acute-angled triangle.
Q.2. Match the following:
Measures of Triangle Type of Triangle
(i) 3 sides of equal length (a) Scalene
(ii) 2 sides of equal length (b) Isosceles right-angled
(iii) All sides are 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 two sides of equal length (g) Isosceles
Ans:-
(i) ↔ (e)
(ii) ↔ (g)
(iii) ↔ (a)
(iv) ↔ (f)
(v) ↔ (d)
(vi) ↔ (c)
(vii) ↔ (b)
Q.3. Name each of the following triangles in two different ways: (you may judge the nature of the angle by observation)
(a) (i) Acute angled triangle
(ii) Isosceles triangle
(b) (i) Right-angled triangle
(ii) Scalene triangle
(c) (i) Obtuse angled triangle
(ii) Isosceles triangle
(d) (i) Right-angled triangle
(ii) Isosceles triangle
(e) (i) Acute angled triangle
(ii) Equilateral triangle
(f) (i) Obtuse angled triangle
(ii) Scalene triangle.
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
Ans:-
(a) Yes, we can make an equilateral triangle with 3 matchsticks
(b) No, we cannot make a triangle with 4 matchsticks
(c) Yes, we can make an equilateral triangle with 5 matchsticks
(d) Yes, we can make an equilateral triangle with 6 matchsticks
NCERT Solutions For Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercises 5.7
Ch 5 class 6 maths solutions ex 5.7
Q.1. Say True or False:
(a) Each angle of a rectangle is a right angle.
Ans:-True,
(b) The opposite sides of a rectangle are equal in length.
Ans:-True,
(c) The diagonals of a square are perpendicular to one another.
Ans:-True,
(d) All the sides of a rhombus are of equal length.
Ans:- True,
(e) All the sides of a parallelogram are of equal length.
Ans:-False,
(f) The opposite sides of a trapezium are parallel.
Ans:-False,
Q.2. Give reasons for the following:
(a) A square can be thought of as a special rectangle.
Ans:-
A square has all the properties of a rectangle.
So, it is a special rectangle.
(b) A rectangle can be thought of as a special parallelogram.
Ans:-
A rectangle has the same properties as that of a parallelogram.
So, it is a special parallelogram
(c) A square can be thought of as a special rhombus.
Ans:-
A square has the same properties as that of a rhombus.
So, it is a special rhombus.
(d) Squares, rectangles, parallelograms are all quadrilaterals.
Ans:-
Square, rectangles, and parallelogram are all quadrilateral as they are all enclosed by four sides
(e) Square is also a parallelogram.
Ans:-
All 4 sides are of same length. Therefore a square is 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?
Ans:-
Square is only the regular quadrilateral with equal sides and equal angles.
Therefore, the square is a regular quadrilateral.
NCERT Solutions For Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercises 5.8
class 6th maths chapter 5 solutions ex 5.8
Q.1. Examine whether the following are polygons. If anyone among them is not, say why?
(a) It is not a closed figure. Hence, it is not a polygon.
(b) It is a polygon made of six sides
(c) No it is not a polygon because it is not made of line segments.
(d) It is not a polygon as it is not made of line segments.
Q.2. Name each polygon.
(a) A quadrilateral:- A shape has four sides is know as Quadrilateral.
More Examples:-
(b) A Triangle :- A shape has three sides is know as Triangle.
More Examples:-
(c) A Pentagon :- A shape has five sides is know as Pentagon.
More Examples:-
(d) A Octagon :- A shape has 8 sides is know as Octagon.
More Examples:-
Q.3. Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of triangle you have drawn.
Ans:-
abcdef is a rough sketch of a regular hexagon. If we join any three vertices like a, b, and c, we get a scalene triangle abc.
Q.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:-
PQRSTUVW is a rough sketch of a regular octagon. PQTU is the rectangle formed by joining the four vertices of the given octagon.
Q.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:-
From the figure we may find ac, ad, bd, be and ce are the diagonals
NCERT Solutions For Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercises 5.9
NCERT Class 6 Maths Chapter 5 Solutions ex 5.9
- Match the following :
Ans:-
(a) 4 ↔ (ii)
Examples:- An ice-cream cone And Birthday cap
(b) ↔ (iv)
Examples:- Tennis ball And Cricket ball
(c) ↔ (v)
Examples:- A road roller And A lawn roller
(d) ↔ (iii)
Examples:- Math book And A brick
(e) ↔ (i)
Examples:- A diamond And Egypt-Pyramids
Q.2. What shape is
(a) Your instrument box?
Ans:-
The shape of instrument box is cuboid.
(b) A brick?
Ans:-
The shape of a brick is cuboid.
(c) A match box?
Ans:-
The shape of a matchbox is cuboid.
(d) A road-roller?
Ans:-
The shape of a road-roller is a cylinder.
(e) A sweet laddu?
Ans:-
The shape of a sweet laddu is a sphere.
Class 6 Science NCERT Notes | Class 6 Complete Study Material |
NCERT Solutions for Class 6 Science | Class 6 Maths Chapter 6 Solutions |
NCERT Solutions For Class 6 Maths Chapter 5 Understanding Elementary Shapes are based on NCERT Books. You can download ncert book for class 6 maths chapter 5.
What we learn ?
- The distance between the end points of a line segment is its length.
- A graduated ruler and the divider are useful to compare lengths of line
segments. - When a hand of a clock moves from one position to another position we have
an example for an angle. One full turn of the hand is 1 revolution. A right angle is ¼ revolution and a straight angle is ½ a revolution . We use a protractor to measure the size of an angle in degrees. 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.Two intersecting lines are perpendicular if the angle between them is 90°. - The perpendicular bisector of a line segment is a perpendicular to the line
- segment that divides it into two equal parts.
- Triangles can be classified as follows based on their angles:
Nature of angles in the triangle | Name |
Each angle is acute One angle is a right angle One angle is obtuse | Acute angled triangle Right angled triangle Obtuse angled triangle |
7. Triangles can be classified as follows based on the lengths of their sides:
Nature of sides in the triangle | Name |
All the three sides are of unequal length Any two of the sides are of equal length All the three sides are of equal length | Scalene triangle Isosceles triangle Equilateral triangle |
8. Polygons are named based on their sides.
Number of sides | Name of the Polygon |
3 4 5 6 8 | Triangle Quadrilateral Pentagon Hexagon Octagon |
9. Quadrilaterals are further classified with reference to their properties.
Properties | Name of the Quadrilateral |
One pair of parallel sides Two pairs of parallel sides Parallelogram with 4 right angles Parallelogram with 4 sides of equal length A rhombus with 4 right angles | Trapezium Parallelogram Rectangle Rhombus Square |
- We see around us many three dimensional shapes. Cubes, cuboids, spheres,
cylinders, cones, prisms and pyramids are some of them.