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 20192020.
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.
∴ 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 midpoint of AG.
Since, B is the midpoint of AC
Therefore, AB = BC …(i)
Again C is the midpoint 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.
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 (180^{o}) clockwise, we will face west direction.
(c)When we start facing west and make 3/4 of a revolution (270^{ o} = 180^{ o} + 90^{o}) anticlockwise, we will first reach the east after 180^{ o} turn and then finally north after taking another 90^{o} 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 360^{o} and two adjacent directions are inclined at 90^{O }(1/4 of revolution). Two opposite directions form an angle 180^{o }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.
1.Match the following:
(i) Straight angle (ii) Right angle (iii) Acute angle (iv) Obtuse angle (v) Reflex angle 
(a) Less than onefourth of a revolution (b) More than half a revolution (c) Half of a revolution (d) Onefourth 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 (= 180^{o}) 
(c) Half of a revolution (= 180^{o}) 
(ii) Right angle (= 90^{o}) 
(d) Onefourth of a revolution (= 90^{o}) 
(iii) Acute angle (less than 90^{o}) 
(a) Less than onefourth of a revolution (less than 90^{o}) 
(iv) Obtuse angle (greater than 90^{o} and less than180^{o}) 
(e) Between 1/4 and 1/2 of a revolution (Between 90^{o} and 180^{o}) 
(v) Reflex angle (greater than 180^{o} and less than360^{o}) 
(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 90^{o})
(b) Obtuse angle(greater than 90^{o} and less than180^{o})
(c) Right angle (equal to 90^{o})
(d) Reflex angle (greater than 180^{o} and less than360^{o})
(e) Straight angle (equal to 180^{o})
(f) Acute angle (less than 90^{o})
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 = 180^{o})
(d) acute (as each angle is less than 90^{o})
(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 
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 setsquares 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 setsquares in the box are given below:
Measure angles of small setsquare: 45°, 45° and 90°.
Measure angles of larger setsquare: 30°, 60° and 90°.
Yes, the angle of measure 90° is common between both the squaresets.
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
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, 1^{st} side + 2^{nd} side = 1+1 = 2 = 3^{rd} 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.
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 90^{o}.
 Opposite sides are equal in length.
In a square:
 All interior angles are equal to 90^{o}.
 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 90^{o}
A parallelogram with all its angles equal to 90^{o}becomes 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 90^{o}
In a rhombus:
 All sides are equal
 Opposite sides are parallel
Thus, a rhombus with all its angles equal to 90^{o} 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 (90^{o}).
Hence, square is a regular quadrilateral.
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.
(a) Quadrilateral
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
△ADE – Scalene 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
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
 Icecream 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
(a) Your instrument box?
(b) A brick?
(b) A matchbox?
(d) A roadroller?
(e) A sweet laddu?
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 roadroller is cylinder.
(e) Shape of a sweet laddu is sphere.
Check chapterwise 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 3 Playing with 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.