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Haryana State Board HBSE 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes InText Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes InText Questions

Try These (Page 110)

Question 1.
Take post card. Use divider to measure its two adjacent sides.
Solution:
Open the divider. Place the end point of one of its arms at A and the second end point at B.
Without disturbing the opening, lift the divider and place it pn the ruler. Ensure that one end point is at the zero mark of the ruler. Now read the mark against the other end point. Similarly, measure its breadth BC.

Question 2.
Select any three objects having a flat top. Measure all the sides of the top using a divider and a ruler.
Solution:
(i) Open the divider. Place the end point of one of its arms at A and the second end point at B. Without disturbing the opening, lift the divider and place it on the ruler. Ensure that one end point is at the zero mark of the ruler. Now read the mark against the other end point. Similarly measure its breadth BC.
(ii) We can measure length and breadth of a playing card in the same way as above.
(iii) Similarly, we can measure the three sides of a triangular sheet ABC.

Try These (Page 114)

Question 1.
What is the angle name for half a revolution ? ‘
Solution:
The name of the angle for half a revolution is straight angle.

Question 2.
What is the angle name for one fourth revolution ?
Solution:
The name of the angle for one fourth revolution is right angle.

Question 3.
Draw 5 other situations of one fourth, half and three fourth revolution on a clock.
Solution:

Note : There is no special name for three fourth of a revolution.

Try These (Page 116-117)

Question 1.
The clock moves from 12 to 5. Is the revolution of the hour- hand more than 1 right angle ? [Fig.]
Solution:
Yes, the revolution of the hour-hand is more than 1 right angle.

Question 2.
What does the angle look like ? The clock moves 5 to 9. Is the angle moved by hour-hand more than 1 right angle ? [Fig.]
Solution:
The angle looks like obtuse angle.
Yes, the angle moved by hour hand is more than 1 right angle.

Question 3.
Draw the following and check the angle with your RA tester.
(a) going from 12 to 2, (b) from 6 to 7, (c) from 4 to 8, (d) from 2 to 5.
Solution:

Haryana State Board HBSE 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Ex 5.4 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Exercise 5.4

Question 1.
What is the measure of (i) a right angle, (ii) a straight angle ?
Solution:
(i) The measure of a right angle is 90°.
(ii) The measure of a straight angle is 180°.

Question 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 = 530 and m ∠B 35°, then m∠A > m ∠B.
Solution:
(a) True,
(b) False,
(c) True,
(d) True,
(e) True.

Question 3.
Write down the measures of:
(a) some acute angles.
(b) some obtuse angles.
(Give two examples of each).
Solution:
(a) 30° and 60° are acute angles.
(b) 110° and 120° are obtuse angles.

Question 4.
Measure the angles given below, using the Protractor and write down the measure.

Solution:
(a) 45°,
(b) 125°,
(c) 90°,
(d) ∠1 = 40°, ∠2 = 125°, ∠3 = 95°.

Question 5.
Which angle has a large measure ?

First estimate and then measure. Measure of Angle A =
Measure of Angle B =
Solution:
Measure of ∠A = 40° and measure of ∠B = 65°
∴ ∠B > ∠A.

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

Solution:
∠A = 45° and ∠B = 60°
∴ ∠B > ∠A.

Question 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, one of them should be ……………… or ………………
Solution:
(a) acute,
(b) obtuse,
(c) straight,
(d) acute,
(e) acute, obtuse.

Question 8.
Find the measure of the angle shown in each figure. (First estimate with your eye and then find the actual measure with a protractor).

Solution:
(a) 40°, (b) 65°, (c) 130°, (d) 135°.

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

Solution:
(a) 90°, (b) 30°, (d) 180°.

Question 10.
Investigate : In the given figures, protractor shows 30° Look at the same figure through a magnifying glass. Does the angle becomes larger ? Does the size of the angle change ?

Solution:
No, No.

Question 11.
Measure and classify each angle.

Solution:

Haryana State Board HBSE 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Ex 5.5 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Exercise 5.5

Question 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 U.
(d) The letter C.
Solution:
(a) The adjacent edges of a table and
(c) The line segments forming the letter ‘L’.

Question 2.
Let \(\overline{\mathrm{PQ}}\) be the perpendicular to the line segment \(\overline{\mathrm{XY}}\). Let \(\overline{\mathrm{PQ}}\) and \(\overline{\mathrm{XY}}\) intersect in the point A.
(a) How is the point A placed in \(\overline{\mathrm{XY}}\) ?
(b) What is the mea¬sure of ∠PAY ?

Solution:
(a) Point A lies between X and Y.
(b) Measure of ∠PAY = 90°.

Question 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:

(i) 90°, 60° and 30°,
(ii) 90°, 45° and 45°, common angle = 90°.

Question 4.
Study the diagram. The line T is perpendicular to line im\
(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, (Hi) BC < EH.

Solution:
(a) Yes, CE = EG. .
(b) Yes, PE bisects CG.
(c) PE is the perpendicular bisector of DF and BH.
(d) (i) T, (ii) T, (Hi) T.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Ex 5.6 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Exercise 5.6

Question 1.
Name the types of following triangles :
(a) Triangle with lengths of sides 7 cm, 8 cm and. 9 cm.
(b) ΔABC with AB = 8.7 cm, AC = 7 cm arid 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°.
Solution:
(a) Scalene triangle.
(b) Scalene triangle.
(c) Equilateral triangle.
(d) Right-angled triangle.
(e) Isosceles right-angled triangle.
(f) Acute-angled triangle.

Question 2.
Match :

Solution:
(i) → (e),
(ii) →(g),
(iii) → (a),
(iv) → (f),
(v) → (d),
(vi) → (c),
(vii) → (b).

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


Solution:
(a) Isosceles, Acute-angled.
(b) Scalene, Right-angled.
(c) Isosceles, Obtuse-angled.
(d) Isosceles, Right-angled.
(e) Equilateral, Acute-angled.
(f) Scalene, Obtuse-angled.

Question 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.
Solution:
(a) Yes, we can make an equilateral triangle with 3 matchsticks. [Fig. (i)]
(b) No, we cannot make a triangle with 4 matchsticks.
(c) Yes, we can make an isosceles triangle with 5 matchsticks. [Fig. (ii)]
(d) Yes, we can make an equilateral triangle with 6 matchsticks. [Fig. (iii)]

Haryana State Board HBSE 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Ex 5.7 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Exercise 5.7

Question 1.
Say True or False :
(a) Eachangleofarectangleisarightangle.
(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,
(f) False.

Question 2.
Give reasons for the following :
(а) 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.
Solution:
(a) All the properties of a rectangle are there in a square.
(b) All the properties of a || gm are there in a rectangle.
(c) All the properties of a rhombus are there in a square.
(d) All are four-sided closed plane figures.
(e) All the properties of a || gm are there in a square.

Question 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:
A square is a regular quadrilateral.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Ex 5.8 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Exercise 5.8

Question 1.
Examine whether the following are polygons. If any one among them is not, say why.

Solution:
(a) Not a polygon, since it is not closed.
(a) Yes, it is a polygon.
(b) Not a polygon, since it is not closed by line segments.
(c) Not a polygon, since it is not closed by line segments.

Question 2.
Name each polygon:

Solution:
(a) Quadrilateral,
(b) Triangle,
(c) Pentagon,
(d) Hexagon,
(e) Octagon,
(f) Decagon.

Question 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:
Equilateral triangle BDF.

Question 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.
Solution:
Quadrilateral CDGH is the required rectangle.

Question 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:
\(\overline{\mathrm{AC}}, \overline{\mathrm{AD}}, \overline{\mathrm{BD}}, \overline{\mathrm{BE}}\) and \(\overline{\mathrm{CE}}\) are the required five diagonals of pentagon ABCDE.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Ex 5.9 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Exercise 5.9

Question 1.
Match the following :


Solution:
(a) (ii),
(b) (iv),
(c) (v),
(d) (iii),
(e) (i).

Question 2.
What shape is :
(a) Your instruments box ?
(b) A brick ?
(c) A matchbox ?
(d) A road-roller ?
(e) A sweet laddu ?
Solution:
(a) Cuboid,
(b) Cuboid,
(c) Cuboid,
(d) Cylinder,
(e) Sphere.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Ex 5.3 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Exercise 5.3

Question 1.
Match the following :
(i) Straight angle
(ii) Right angle
(iii) Acute angle
(iv) Obtuse angle
(v) Reflex angle
Solution:
(i) (c),
(ii) -(d),
(iii)- (a)
(iv) – (e),
(v) – (b)

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

Solution:
(a) Acute angle
(b) Obtuse angle
(c) Right angles
(d) Reflex angle
(e) Straight angle
(f) Acute angles.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Ex 5.2 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Exercise 5.2

Question 1.
What fraction of a revolution clockwise 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 :
\(\frac{1}{2}\) of a revolution.

(b) 4 to 7 :
\(\frac{1}{4}\) of a revolution.

(c) 7 to 10 :
\(\frac{1}{4}\) of a revolution.

(d) 12 to 9 :
\(\frac{3}{4}\) of a revolution.

(e) 1 to 10 :
\(\frac{3}{4}\) of a revolution.

(f) 6 to 3 :
\(\frac{3}{4}\) of a revolution.

Question 2.
Where will the hand of a clock stop if it
(a) Starts at 12 and makes 1/2 of 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 ?
(id) Starts at 5 and makes 3/4 of a revolution, clockwise ?
Sol. (a) The hand of the clock will stop at 6 in this case. [See Fig. (a)]
(b) The hand of the clock will stop at 8 in this case. [See Fig. (b)]
(c) The hand of the clock will stop at 8 in this case. [See Fig. (c)]
(d) The hand of the clock will stop at 2 in this case. [See Fig. (d)]

Question 3.
Which direction will you face if you start facing
(a) east and make \(\frac{1}{2}\) of a revolution clockwise ?
(b) east and make 1\(\frac{1}{2}\) revolution clockwise ?
(c) west and make \(\frac{3}{4}\) revolution anti-clockwise ?
(d) south and make one full revolution?
(Should we specify clockwise or anti-clockwise for this last question ? Why not ?)
Solution:
(a) In this case, you will face west direction. [See Fig. (a)]
(b) In this case,, also you will face west direction. [See Fig. (h)]
(c) In this case, you will face north direction. [See Fig. (c)]
(d) In this case, you will face south direction. [See Fig. (d)]

(Not necessary, since it is always a complete angle.)

Question 4.
What part of 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:
(a) In this case, you have turned 3/4 of a revolution. [See Fig. (a)]
(b) In this case, you have turned 3/4 of a revolution. [See Fig. (b)]
(c) In this case, you have turned 1/2 of a revolution. [See Fig. (c)]

Question 5.
Find the number of right angles 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 : In this case, the hour hand turns through 1 right angle. [See Fig.(a)]
(b) 2 to 8 : In this case, the hour hand turns 2 right angles. [See Fig. (h)]
(c) 5 to 11 : In this case the hour hand turns through 2 right angles. [See Fig.(c)]
(d) 10 to 1 : In this case the hour hand turns through 1 right angle.[See Fig.(d)]
(e) 12 to 9 : In this case the hour hand turns through 3 right angles.[See Fig.(e)]
(f) 12 to 6 : In this case the hour hand turns through 2 right angles.[See Fig. (f)]

Question 6.
How many right angles do you make if you start facing :
(a) south and turn clockwise to west.
(b) north and turn anti-clockwise to east.
(c) west and turn to west.
(d) south and turn to north.
Solution:
(a) 1 right angle in this case. [See Fig. (a)]
(b) 3 right angles in this case. [See Fig. (b)]
(c)

4 right angles in both cases, clockwise as well as anti-clockwise direction. [See Fig. 5.20(c)]

(d)

2 right angles in both the cases, clockwise as well as anti-clockwise. [See Fig.(d)]

Question 7.
Where will the hour hand of a clock stop if it starts
(а) 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:
(a) In this case, the hour-hand will stop at 9. [See Fig. (a)]
(b) In this case, the hour hand will stop at 2. [See Fig. (5)]
(c) In this case, the hour hand will stop at 7. [See Fig. (c)]
(d) In this case, the hour hand will stop at 7. [See Fig. (d)]

Haryana State Board HBSE 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Ex 5.1 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Exercise 5.1

Question 1.
What is the disadvantage in comparing line segments by mere observation ?
Solution:
In comparing line segments by mere observation, we cannot always be sure about the usual judgement. For example, look at A the following segments :
The difference in lengths between these two may not be obvious. Actually in this figure \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{PQ}}\) have the same lengths. This is not quite obvious.

Question 2.
Why is it better to use a divider with a ruler, while measuring the length of a linesegment ? ‘
Solution:
It is better to use a divider and a ruler, while measuring the length of a line segment, because by this method we can find the exact length of a line segment.

Question 3.
Draw a line segment, say \(\overline{\mathrm{AB}}\). Take any point C lying in between A and B. Measure the lengths of AB, BC and AC. Is AC = AB + BC ? [Note: If A, B, C are any three points on a line such that AC + CB = AB, then we can be sure that C lies between A and B.]
Solution:
On measuring, we find that
AB = 5 cm
BC = 2 cm
and AC = 3 cm
AB + BC = 5 cm + 2 cm = 7 cm
∴ AC ≠ AB + BC

But AC + CB = 3 cm + 2 cm
= 5 cm = AB.
Thus, AC + CB = AB.

Question 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:
AB = 5 cm
BC = 3cm
and AC = 8 cm

AB + BC = 5 cm + 3 cm
= 8 cm = AC
AB + BC = AC
∴ B lies between A and C.

Question 5.
Verify whether D is the mid-point of AG.
Solution:
AD = 4 – 1 = 3 units
DG = 7 – 4 = 3 units
∵ AD = DG

⇒ D is the mid-point of AG.

Question 6.
If B is the mid-point of \(\overline{\mathrm{AC}}\) and C is the mid-point of \(\overline{\mathrm{BD}}\) , where A, B, C, D lie on a straight line, say why AB = CD ?
Solution:
B is the mid-point of \(\overline{\mathrm{AC}}\)
⇒ AB = BC ………… (i)
and C is the mid-point of \(\overline{\mathrm{BD}}\)
⇒ BC = CD …(ii)

Comparing (i) and (ii), we get AB = CD.

Question 7.
Draw five triangles and measure their sides. Check in each case if the sum of two sides is ever less than the third side.
Solution:
(i) AB = BC = AC = 3 cm

AB + BC = 3 cm + 3 cm = 6 cm
∴ AB + BC > AC
BC + AC = 3 cm + 3 cm = 6 cm
∴ BC + AC > AB
AC + AB = 3 cm + 3 cm = 6 cm
∴ AC + AB > BC.

(ii) AB = AC = 3.5 cm
and BC = 3 cm

AB + BC = 3.5 cm + 3 cm
= 6.5 cm
∴ AB + BC > AC
BC + AC = 3 cm + 3.5 cm
= 6.5 cm
∴ BC + AC > AB
AC + AB = 3.5 cm + 3.5 cm
= 7 cm .
∴ AC + AB > BC.

(iii) Scalene Triangle :
AB = 2.5 cm, BC = 4 cm
AC = 3.5 cm

AB + BC = 2.5 cm + 4 cm = 6.5 cm
∴ AB + BC > AC
BC + AC = 4 cm + 3.5 cm = 7.5 cm
∴ BC + AC > AB
AC + AB = 3.5 cm + 2.5 cm = 6 cm
∴ AC + AB > BC.

(iv) AB = 4 cm, BC = 3 cm, AC = 5 cm
AB + BC = 4 cm + 3 cm = 7 cm
AB + BC > AC
BC + AC = 3 cm + 5 cm = 8 cm
BC + AC > AB
AC + AB = 5 cm + 4 cm = 9 cm
AC + AB > BC.

(v) AB = 2.5 cm
BC = 3 cm
AC = 5 cm
AB + BC = 2.5 cm + 3 cm = 5.5 cm AB + BC > AC
BC + AC = 3 cm + 5 cm = 8 cm BC + AC > AB
AC + AB = 5 cm + 2.5 cm = 7.5 cm

AC + AB > BC.