# NCERT Solutions for Class 6 Maths Chapter 5 - Understanding Elementary Shapes

Check NCERT solutions for class 6 Maths chapter5 available in downloadable format. Get latest and free solutions for all questions givenin this chapter. NCERT Solutions for Class 6 Maths Chapter 5 - Understanding Elementary Shapes

NCERT solutions for class 6 Maths chapter 5 will prove to be very helpful to understand the concepts and excel in the Mathematics subject. Here you will find the latest updated solutions to prepare for the current academic session 2019-2020.

Exercise 5.1

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

Solution.

Comparing the lengths of two line segments by mere observation cannot tell us accurately that which line is shorter or which is longer.

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

Solution.

While using a ruler, due to the incorrect position of eye, there are chances of error in the reading whereas, with ruler this error can be minimised and we can get an accurate result. [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]

Solution. Here, AB = 5 cm and a point C lies between A and B such that

AC = 2 cm, CB = 3cm.

&there4; AC + CB = 2 cm + 3 cm = 5 cm.

But, AB = 5 cm.

So, AB = AC + CB.

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?

Solution.

Given, AB = 5 cm

BC = 3 cm

AC = 8 cm

Now AB + BC = 5 + 3 = 8 cm which is equal to AC.

Hence, B lies between A and C. Solution.

From the given figure, we have

AG = 7 – 1 = 6 units

AD = 4– 1 = 3 units

And DG = 7– 4 = 3 units

As D is the point lying on AG such that AD = DG

Hence, D is the mid-point of AG. Since, B is the mid-point of AC

Therefore, AB = BC               …(i)

Again C is the mid-point of BD

Therefore, BC = CD               …(ii)

From equations (i) and (ii), we have

AB = CD

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

Solution.

Students may draw five triangles of different measures. One has been drawn below for reference: To check: The sum of the length of any two sides of ∆ABC is less than its third side.

Case 1.Here, In△ABC

AB = 2.5 cm

AC = 5.5 cm

BC = 6 cm

AB + AC = 2.5 cm + 5.5 cm

= 8 cm

Since, 8 cm>6 cm

So, AB + BC > AC

Hence, sum of any two sides of a triangle is always greater than the third side.

Students may calculate the same for their other triangles and verify the statement.

Exercise 5.2

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

Solution.

(a) 3 to 9 (b) 4 to 7 (c) 7 to 10 (d) 12 to 9 (e) 1 to 10 (f) 6 to 3 2.Where will the hand of a clock stop if it Solution. 3.Which direction will you face if you start facing (Should we specify clockwise or anticlockwise for this last question? Why not?)

Solution.

(a) When we start facing east and make 1/2 of a revolution (180o) clockwise, we will face west direction.  (c)When we start facing west and make 3/4 of a revolution (270 o = 180 o + 90o) anticlockwise, we will first reach the east after 180 o turn and then finally north after taking another 90o turn. (d) When we start facing south and make one full revolution in either clockwise or anticlockwise direction, we will reach back to the starting position that is south. 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?

Solution.

Remember: A complete revolution is of 360o and two adjacent directions are inclined at 90O (1/4 of revolution). Two opposite directions form an angle 180o which is 1/2 of a revolution.

(a) If we start from east and turn clockwise to reach north we have turned through 1/2 of a revolution first to reach west and then a 1/4 revolution to reach north.  (b) If we start from south and turn clockwise to face east, we will first reach north by turning through 1/2 of a revolution and then move by 1/4 of a revolution to reach east. (c) If we start from west and turn clockwise to face east, we will turn through 1/2 of a revolution. 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

Solution.

(a) 3 to 6

Here, the hour hand turns through 1 right angle. (b) 2 to 8

Here, the hour hand turns through 2 right angles. (c) 5 to 11

Here, the hour hand turns through 2 right angles. (d) 10 to 1

Here, the hour hand turns through 1 right angle. (e) 12 to 9

Here, the hour hand turns through 3 right angles. (f) 12 to 6

Here, the hour hand turns through 2 right angles. 6.How many right angles do you make if you start facing

(a) south and turn clockwise to west?

(b) north and turn anticlockwise to east?

(c) west and turn to west?

(d) south and turn to north?

Solution.

(a) If I start facing south and turn clockwise to west, I will make 1 right angle. (b) If I start facing north and turn anticlockwise to east, I will make 3 right angles. (c) If I start facing west and turn to west, I will make 4 right angles in both clockwise and anticlockwise directions. (d) If I start facing south and turn to north, I will make 2 right angles in both clockwise and anticlockwise directions. 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?

Solution.

Hint: Students may use the trick that every 15 minutes round of the hour hand is equal to 1 right angle turn.

(a) The hour hand will stop at 9. (b) The hour hand will stop at 2. (c) The hour hand will stop at 7. (d) The hour hand will stop at 7. Exercise 5.3

1.Match the following:

 (i) Straight angle (ii) Right angle (iii) Acute angle (iv) Obtuse angle (v) Reflex angle (a) Less than one-fourth of a revolution (b) More than half a revolution (c) Half of a revolution (d) One-fourth of a revolution (e) Between1/4 and 1/2 of a revolution (f) One complete revolution.

Solution.

Correct match of the two given columns is given below:

 (i) Straight angle (= 180o) (c) Half of a revolution (= 180o) (ii) Right angle (= 90o) (d) One-fourth of a revolution (= 90o) (iii) Acute angle (less than 90o) (a) Less than one-fourth of a revolution (less than 90o) (iv) Obtuse angle (greater than 90o and less than180o) (e) Between  1/4 and 1/2 of a revolution (Between 90o and 180o) (v) Reflex angle (greater than 180o and less than360o) (b) More than half a revolution

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

(a) Acute angle (less than 90o)

(b) Obtuse angle(greater than 90o and less than180o)

(c) Right angle (equal to 90o)

(d) Reflex angle (greater than 180o and less than360o)

(e) Straight angle (equal to 180o)

(f) Acute angle (less than 90o)

Exercise 5.4

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

Solution.

(i) Measure of a right angle = 90°

(ii) Measure of a straight angle =180°

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 ∠B = 35°, then m∠A>m∠B.

Solution.

(a) True

(b) False

(c) True

(d) True

(e) True

3.Write down the measures of

(a) some acute angles

(b) some obtuse angles

(give at least two examples of each).

Solution.

(a) Acute angles: 35°, 80°

(b) Obtuse angles: 100°, 170°

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

Let’s name the given angles as follows: (a) Measure of angle P = 45°

(b) Measure of ∠Q= 120°

(c) Measure of ∠R= 90°

(d)Measure of ∠S = 60°,

Measure of ∠T = 90° and

Measure of ∠U = 125°

5.Which angle has a large measure? First estimate and then measure. Measure of Angle A =

Measure of Angle B =

Solution.

Angle B is wider than angle A so it must have a larger measure than angle A.

Measure of angle A = 40°

Measure of angle B = 60°

Thus, ∠B >∠A.

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

Given angle are: ∠B is wider than ∠A hence it has a larger measure than ∠A.

Measure of angle A = 45°

Measure of angle B = 60°

Thus, ∠B >∠A.

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 ……… .

Solution.

(a) acute

(b) obtuse

(c) straight (sum of the two right angles = 180o)

(d) acute (as each angle is less than 90o)

(e) 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). Solution.

Given angles are: • Measure of ∠a = 40°
• Measure of ∠b = 130°
• Measure of ∠c = 65°
• Measure of ∠d = 135°

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

(i) Angle between hands of the clock in first figure = 90°

(ii) Angle between hands of the clock in second figure = 30°

(iii) Angle between hands of the clock in third figure = 180°.

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?

Solution.

The measure of the angle remains the same even when observed through magnifying glass.

11.Measure and classify each angle: Angle Measure Type ∠AOB ∠AOC ∠BOC ∠DOC ∠DOA ∠DOB

Solution.

Given angles are classified as follows:

 Angle Measure Type ∠AOB 40° Acute angle ∠AOC 125° Obtuse angle ∠BOC 85° Acute angle ∠DOC 95° Obtuse angle ∠DOA 140° Obtuse angle ∠DOB 180° Straight angle

Exercise 5.5

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 a letter ‘L’.

(d) The letter V.

Solution.

(a) Yes, the adjacent edges of a table top are a model of perpendicular lines.

(b) No, the lines of a railway tracks are not a model of perpendicular lines as they are parallel to each other.

(c) Yes, the line segments forming a letter ‘L’ are a model of perpendicular lines.

(d) No, the two line segments forming a letter ‘V’ are not a model of perpendicular lines. Since  PQ⊥ XY

Therefore, ∠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?

Solution.

Figures of two set-squares in the box are given below: Measure angles of small set-square: 45°, 45° and 90°.

Measure angles of larger set-square: 30°, 60° and 90°.

Yes, the angle of measure 90° is common between both the square-sets.

4.Study the diagram. The line l is perpendicular to line m. (a) Is CE = EG?

(b) Does PE bisects 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

Solution.

(a) Yes,CE = EG = 2 units

(b) Yes, PE bisects CG as CE = EG (d) (i) True

Since AC = 2 units and FG = 1 unit

Therefore, AC > FG

(ii) True; CD = GH = 1 unit

(iii) True

Since BC = 1 unit and EH = 3 units

Therefore, BC < EH

Exercise 5.6

1. Name the types of following triangles:

(а) Triangle with lengths of sides 7 cm, 8 cm and 9 cm.

(b) ∆ABC with AB = 8.7 cm, AC = 7 cm and BC = 6 cm.

(c) ∆PQR such that PQ = QR = PR = 5 cm.

(d) ∆DEF with m∠D = 90°

(e) ∆XYZ with m∠Y = 90° and XY = YZ.

(f) ∆LMN with m∠L = 30° m∠M = 70° and m∠N = 80°.

Solution.

(a) Since, the measures of all three sides of triangle are different

Hence, it is a scalene triangle.

(b) Since, the measures of all three sides of ∆ABC are different

Hence, it is a scalene triangle.

(c) Since all sides of ∆PQR are equal

Hence, it is an equilateral triangle.

(d) Since measure of one angle of ∆DEF is 90°

Hence it is a right angled triangle.

(e) Since measure of one angle of ∆XYZ is 90° and its two sides are equal

Hence it is anisosceles right angledtriangle.

(f) Since all three angles of ∆LMN are less than 90°

Hence it is an acute angled triangle.

2. Match the following:

 Measure 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

Solution.

 Measure of triangle Type of triangle (i) 3 sides of equal length (e) Equilateral (ii) 2 sides of equal length (g) Isosceles (iii) All sides are of different length (a) Scalene (iv) 3 acute angles (f) Acute angled (v) 1 right angle (d) Right angled (vi) 1 obtuse angle (c) Obtuse angled (vii) 1 right angle with two sides of equal length (b) Isosceles right angled

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

(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.

4. Try to construct triangles using matchsticks. 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, give of reasons for it.

Solution.

(a) Yes, we can make a triangle with 3 matchsticks as shown below: Since, 1 matchstick forms each side. So, all the sides of this triangle are equal.

Hence, it is an equilateral triangle.

(b) No, we cannot make a triangle with 4 matchsticks.

Explanation: If we make a triangle with 4 matchsticks and each matchstick measures 1 unit then:

Larger side of triangle measures 2 units

Smaller sides of triangle measures 1 unit each.

Then, 1st side + 2nd side = 1+1 = 2 = 3rd side

But we know that sum of any two sides of a triangle is always greater than its third side.

Hence proved.

(c) Yes, we can make a triangle with 5 matchsticks as shown below: Since, two sides of this triangle are formed by 2 equal sized matchsticks

Hence, it is an isosceles triangle.

(d) Yes, we can make a triangle with 6 matchsticks as shown below: Since three match sticks form three sides of this triangle. So, all the sides of this triangle are equal.

Hence, it is an equilateral triangle.

Exercise 5.7

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.

Solution.

(a) True

(b) True

(c) True

(d) True

(e) False; only opposite sides of a parallelogram are equal in length.

(f) False; only one pair of opposite sides of a trapezium are parallel.

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) Square, rectangles, parallelograms are all quadrilaterals.

(e) Square is also a parallelogram.

Solution.

(a) (a) In a rectangle:

• All interior angles are equal to 90o.
• Opposite sides are equal in length.

In a square:

• All interior angles are equal to 90o.
• All sides are equal in length.

Thus, a rectangle with all sides equal becomes a square. So, square is a special rectangle.

(b) In a parallelogram:

• Opposite sides are equal
• Opposite sides are parallel

In a rectangle:

• Opposite sides are equal
• Opposite sides are parallel
• All angles are equal to 90o

A parallelogram with all its angles equal to 90obecomes a rectangle.

Hence, a rectangle can be thought of as a special parallelogram.

(c) In a square:

• All sides are equal
• Opposite sides are parallel
• All angles are equal to 90o

In a rhombus:

• All sides are equal
• Opposite sides are parallel

Thus, a rhombus with all its angles equal to 90o becomes a square. So, a square can be thought of as a special rhombus

(d) A quadrilateral is a polygon which has four sides.

Since, squares, rectangles and parallelogram all are made of 4 sides so they are all quadrilaterals.

(e) In a square opposite sides are equal and parallel to each other.

Also, in a parallelogram, opposite sides are equal and parallel to each other.

Hence, a square is also a 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?

Solution.

Since, square is the only quadrilateral with all its sides of same length and angles of same measure (90o).

Hence, square is a regular quadrilateral.

Exercise 5.8

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

A polygon is a closed plane figure enclosed with more than two line segments.

(a) This figure is not closed so, it is not a polygon.

(b) This figure is a polygon with six sides.

(c) This figure is not enclosed with line segments so it is not a polygon.

(d) This figure is enclosed by only line segments but one arc and two line segments so it is not a polygon.

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

Solution.

Two more examples of a quadrilateral are: (b) Triangle

Two more examples of atriangle are: (c) Pentagon

Two more examples of a pentagon are: (d) Octagon

Two more examples of an octagon are: 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.

Solution.

A regular hexagon is drawn as follows with three of its vertices connected to form a triangle: Different types of triangles can be formed by connecting any three vertices of a hexagon. These are:

△ABF – Isosceles triangle

△ACE – Isosceles triangle

△BCF – Right angled triangle

(Students may draw any one triangle)

4.Draw a rough sketch of a regular octagon. (Using squared paper if you wish). Draw a rectangle by joining exactly four of the vertices of the octagon.

Solution.

A regular octagon is drawn as follows: ADEH is the rectangle formed by joining exactly four vertices of the given octagon.

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.

Solution.

A rough sketch of a pentagon can be drawn as follows: By joining its any two vertices, we get the following diagonals:

AC, AD, BE, BD and CE

Exercise 5.9

1.Match the following: Give two examples of each shape.

Solution.

Correct match and examples of each shape are given below:

(a) ↔ (ii)

Examples of cone:

• Birthday cap
• Ice-cream cone

(b) ↔ (iv)

Examples of sphere:

• Football
• Cricket ball

(c) ↔ (v)

Examples of cylinder:

• A piece of chalk
• Pencil

(d) ↔ (iii)

Examples of cuboid:

• Brick
• Match box

(e) ↔ (i)

Examples of pyramid:

• The great pyramids of Eygpt
• Tent

2.What shape is

(b) A brick?

(b) A matchbox?

Solution.

(a) Shape of instrument box is cuboid.

(b) Shape of a brick is cuboid.

(c) Shape of a matchbox is cuboid.

(d) Shape of a road-roller is cylinder.

(e) Shape of a sweet laddu is sphere.

## Check chapter-wise NCERT Solutions for Class 6 Maths from the links given below:

NCERT Solutions for Class 6 Maths Chapter 1 - Knowing Our Numbers

NCERT Solutions for Class 6 Maths Chapter 2 - Whole Numbers

NCERT Solutions for Class 6 Maths Chapter 4 - Basic Geometrical Ideas

NCERT solutions for other chapters will be provided here very soon. Check here for the detailed and appropriate solutions.