Tuesday, December 3, 2024

Understanding Elementary Shapes

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NCERT Solutions for Class 6 Maths Chapter 5

The chapter Understanding Elementary Shapes delves into the foundation of geometric shapes, equipping you with the vocabulary and concepts to describe and analyze them. Here’s a comprehensive breakdown of the key areas you’ll encounter:

1. Building Blocks:

  • Points: The most fundamental element in geometry, considered to have no size or dimension. Imagine a tiny dot marking a specific location in space.
  • Lines: A one-dimensional shape with infinite length in opposite directions. Lines have no thickness and are defined solely by their direction. Think of a perfectly straight, infinitely long path extending in both directions.

2. Line Segments and Rays:

  • Line Segment: A portion of a line with two distinct endpoints. It has a definite length and represents a finite part of the infinitely long line. Imagine a segment cut out of a straight line, with two clear starting and ending points.
  • Ray: Similar to a line segment, but with one endpoint and extending infinitely in one direction. Imagine a line segment where one end has been stretched out infinitely long.

3. Curves:

  • Curves: A continuous path that can be traced without lifting your pen. Curves can be either straight (like a portion of a circle) or genuinely curved. This chapter likely focuses on simple, closed curves that form a complete loop. Imagine drawing a smooth line without stopping until you return to your starting point.

4. Polygons: The Stars of the Show:

  • Polygons: Closed shapes formed by connecting straight line segments in a specific order. They have a definite number of sides, angles, and vertices (corners where the sides meet). The number of sides is a crucial feature for classifying polygons. Here are some common examples:
    • Triangle (3 sides)
    • Quadrilateral (4 sides) – further classified into squares, rectangles, parallelograms, etc. based on additional properties
    • Pentagon (5 sides)
    • Hexagon (6 sides), and so on

5. Properties of Polygons:

  • Number of Sides: As mentioned earlier, this is a fundamental characteristic that defines the type of polygon.
  • Angles: The measure of the “turn” formed where two sides of a polygon meet at a vertex. Angles are typically measured in degrees.
  • Vertices: The points where two or more sides of a polygon come together. Each vertex has an associated angle formed by the sides meeting at that point.

6. Additional Concepts (may vary depending on the curriculum):

  • Open vs. Closed Curves: We already explored curves, but it’s important to distinguish between open and closed ones. Open curves have two distinct endpoints and do not enclose an area. Imagine a curved path that starts and stops at two different points, not forming a complete loop. In contrast, closed curves form a complete loop, enclosing a region within them. Think of a circle or a square – they have no endpoints and completely surround an area.
  • Diagonals: Line segments that connect non-consecutive vertices within a polygon (optional). Not all polygons have diagonals, but some (like quadrilaterals) can have them. Imagine drawing a line segment within a square to connect opposite corners – that’s a diagonal.
  • Symmetry: A property where a shape can be divided into two equal halves in a specific way. There are two main types of symmetry:
    • Reflective Symmetry: Imagine folding the shape in half along a line, and the two halves perfectly overlap. This line is called the axis of symmetry.
    • Rotational Symmetry: Imagine rotating the shape around a fixed point, and it looks exactly the same after a certain angle of rotation. This angle is called the rotational symmetry angle.

7. Exploring Relationships:

This chapter likely encourages you to visualize and analyze shapes, identify their properties, and understand how they relate to each other. You might learn about:

  • Classifying polygons based on their number of sides.
  • Identifying and measuring angles within polygons.
  • Recognizing symmetrical shapes and understanding the different types of symmetry.
  • Simple constructions using basic tools like rulers and compasses (optional).

NCERT Solutions for Class 6 Maths Chapter 5

Exercise 5.1

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

Ans : Our eyes aren’t great at judging small differences in line segment lengths. Meter observation only tells us the length, not position or direction. For accurate measurements and additional information, we rely on rulers.

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

Ans : 

Ruler Thickness: Ruler width can cause slight errors in reading the exact length. Dividers have sharp points for precise endpoint marking.
Awkward Placement: Dividers can be adjusted to fit any line segment location, unlike rulers limited by their edge.

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 : Yes, if you draw any line segment AB and take any point C lying between A and B, then measuring the lengths will confirm that 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?

Ans : In this scenario, B lies between A and C.

Here’s why:

We are given that AB = 5 cm, BC = 3 cm, and AC = 8 cm.

If C were between A and B, then the total length of AC would be the sum of AB and BC (AC = AB + BC). However, this is not the case here (8 cm ≠ 5 cm + 3 cm).

On the other hand, if B is between A and C, then the total length of AC would be the sum of AB and BC (AC = AB + BC). In this case, it holds true (8 cm = 5 cm + 3 cm).

Therefore, based on the property that the whole length of a line segment is the sum of the lengths of its two segments when a point divides it, we can confirm that B lies between A and C.

5. Verify, whether D is the mid point of AG

understanding elementary shapes class 6

Ans : 

  • The total length of segment AG is 7 centimeters minus 1 centimeter, which equals 6 centimeters. (AG = 7 cm – 1 cm = 6 cm)
  • The length of segment AD is 4 centimeters minus 1 centimeter, resulting in 3 centimeters. (AD = 4 cm – 1 cm = 3 cm)
  • Similarly, the length of segment DG is calculated as 7 centimeters minus 4 centimeters, which is also 3 centimeters. (DG = 7 cm – 4 cm = 3 cm)

Interestingly, we observe that the total length of AG (6 centimeters) is the sum of the lengths AD and DG (3 centimeters + 3 centimeters). (AG = AD + DG)

This observation suggests that point D might be the midpoint of segment AG.

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?

Ans : 

B is the mid point of  AC

∴ AB = BC …(i)

C is the mid-point of BD

BC = CD ….(ii)

From Eq.(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.

Ans : 

Case : I

PQ = 2 cm

QR = 2.5 cm

PR = 3.5 cm

We can calculate the sum of two sides:

  • PQ + QR = 2 cm + 2.5 cm = 4.5 cm

Now, let’s compare this sum to the length of the third side:

  • PR = 3.5 cm

Based on this example, the Triangle Inequality applies. The sum of two sides (PQ and QR) is greater than the length of the third side (PR) for this specific triangle.

Case : II

PQ = 2 cm

QR = 2.5 cm

PR = 3.5 cm

We can calculate the sum of two sides:

  • PQ + QR = 2 cm + 2.5 cm = 4.5 cm

Now, let’s compare this sum to the length of the third side:

  • PR = 3.5 cm

Based on this example, the Triangle Inequality applies. The sum of two sides (PQ and QR) is greater than the length of the third side (PR) for this specific triangle.

Case III : 

XY = 5 cm

YZ = 3 cm

ZX = 6.8 cm

Calculate the sum of two sides: We’ll add the lengths of XY and YZ.

  • XY + YZ = 5 cm + 3 cm = 8 cm

Compare the sum to the third side: Now, compare the sum (8 cm) to the length of the remaining side ZX (6.8 cm).

We see that 8 cm (the sum of XY and YZ) is indeed greater than 6.8 cm (the length of ZX). 

Case IV :

MN = 2.7 cm

NS = 4 cm

MS = 4.7 cm

However, based on the given measurements:

MN + NS = 2.7 cm + 4 cm = 6.7 cm

MS = 4.7 cm

We see that 6.7 cm (MN + NS) is not greater than 4.7 cm (MS). This contradicts the Triangle Inequality.

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes

Exercise 5.2

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

Ans : 

(a) 3 to 9: The hour hand moves 6 hours (from 3 to 9). Fraction = (Hours moved) / (Total hours per revolution) = 6 hours / 12 hours = 1/2 revolution

(b) 4 to 7: The hour hand moves 3 hours (from 4 to 7). Fraction = 3 hours / 12 hours = 1/4 revolution

(c) 7 to 10: The hour hand moves 3 hours (from 7 to 10). Fraction = 3 hours / 12 hours = 1/4 revolution

(d) 12 to 9: The hour hand moves 9 hours (as it goes past the 12 and reaches 9). Fraction = 9 hours / 12 hours = 3/4 revolution

(e) 1 to 10: The hour hand moves 9 hours (from 1 to 10). Fraction = 9 hours / 12 hours = 3/4 revolution

(f) 6 to 3: The hour hand moves 9 hours (as it goes past the 12 and reaches 3). Fraction = 9 hours / 12 hours = 3/4 revolution

In conclusion:

  • (a) 3 to 9: 1/2 revolution
  • (b) 4 to 7: 1/4 revolution
  • (c) 7 to 10: 1/4 revolution
  • (d) 12 to 9: 3/4 revolution
  • (e) 1 to 10: 3/4 revolution
  • (f) 6 to 3: 3/4 revolution

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 1/2 of a revolution, clockwise?

(d) starts at 5 and makes 1/2 of a revolution, clockwise?

Ans : 

(a) Starts at 12 and makes 1/2 revolution, clockwise:

  • One full revolution corresponds to 12 hours movement on the clock face.
  • Half a revolution (1/2) is equal to 12 hours / 2 = 6 hours.

Therefore, the hour hand will stop at 6.

(b) Starts at 2 and makes 1/2 revolution, clockwise:

Following the same logic as above:

  • Half a revolution is equal to 6 hours.

Since the hand starts at 2 and moves 6 hours clockwise, it will end at 8.

(c) Starts at 5 and makes 1/2 revolution, clockwise:

  • Half a revolution is equal to 6 hours.

The hand starts at 5 and moves 6 hours clockwise. However, there are only 7 hours (including 5) on the clock face before reaching 12 again. So, the hand will move past the 12 and continue for another hour.

Therefore, it will stop at 1.

(d) Starts at 5 and makes 1/2 revolution, clockwise (duplicate):

The answer for (d) is the same as (c). Starting at 5 and making 1/2 revolution clockwise will end the hand at 1.

3. Which direction will you face if you start facing

(a) east and make 1/2 of a revolution clockwise? z

(b ) east and make 1 1/2 of a revolution clockwise? z

(c) west and make 3/4 of a revolution anticlockwise?

(d) south and make one full revolution? (Should we specify clockwise or anticlockwise for this last question? Why not?)

Ans : 

(a) East and make 1/2 of a revolution clockwise:

  • One full revolution is 360 degrees.
  • Half a revolution (1/2) is equal to 360 degrees / 2 = 180 degrees.

Since clockwise represents a right turn, 180 degrees from facing east will put you facing west.

(b) East and make 1 1/2 of a revolution clockwise:

  • One and a half revolutions (1 ½) is equal to 1 full revolution (1) + half a revolution (½).
  • 1 full revolution = 360 degrees.
  • ½ revolution = 180 degrees (as established in part (a)).

Therefore, 1 ½ revolutions is equal to 360 degrees + 180 degrees = 540 degrees.

Following the logic from part (a), a 180-degree turn (half revolution) clockwise from facing east will put you facing west.

(c) West and make 3/4 of a revolution anticlockwise:

  • One full revolution is 360 degrees.
  • Three-quarters of a revolution (3/4) is equal to 360 degrees / 4 = 270 degrees.

Therefore, you’ll be facing north.

(d) In this case, making a full revolution (360 degrees) from facing south will return you to facing south. Since a complete rotation brings you back to the original position, the direction of rotation (clockwise or anticlockwise) doesn’t matter for a full circle.

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?

Ans : 

(a) East and turn clockwise to face north:

  • Standing east and facing north requires a turn of 90 degrees clockwise.
  • A complete revolution is 360 degrees.

Therefore, you’ve turned through (90 degrees) / (360 degrees per revolution) = one-quarter (1/4) of a revolution.

(b) South and turn clockwise to face east:

Similar to (a), facing east from south requires a 90-degree turn clockwise. So, you’ve turned through one-quarter (1/4) of a revolution.

(c) West and turn clockwise to face east:

  • To face east from west while turning clockwise, you need to cover a larger angle. Imagine a compass. You’d move from west through north and then towards east.
  • In this case, the total turn is 270 degrees (west to north is 90 degrees, and north to east is another 180 degrees).

Therefore, you’ve turned through (270 degrees) / (360 degrees per revolution) = three-quarters (3/4) 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

Ans :  

(a) 3 to 6:

The hour hand moves 3 hours (from 3 to 6). Angle covered = (Movement in hours) x (Degrees per hour) = (3 hours) x (360 degrees / 12 hours) = 90 degrees Number of right angles = Angle covered / Degrees per right angle = 90 degrees / 90 degrees/right angle = 1 right angle

(b) 2 to 8:

The hour hand moves 6 hours (from 2 to 8). Angle covered = (6 hours) x (360 degrees / 12 hours) = 180 degrees Number of right angles = Angle covered / Degrees per right angle = 180 degrees / 90 degrees/right angle = 2 right angles

(c) 5 to 11:

The hour hand moves 6 hours (from 5 to 11). Number of right angles = Angle covered / Degrees per right angle = (180 degrees) / (90 degrees/right angle) = 2 right angles

(d) 10 to 1:

The hour hand moves 3 hours (from 10 to 1). Number of right angles = Angle covered / Degrees per right angle = (90 degrees) / (90 degrees/right angle) = 1 right angle

(e) 12 to 9:

The hour hand moves 9 hours (as it goes past the 12 and reaches 9). Number of right angles = Angle covered / Degrees per right angle = (270 degrees) / (90 degrees/right angle) = 3 right angles

(f) 12 to 6:

The hour hand moves 6 hours (as it goes past the 12 and reaches 6). Number of right angles = Angle covered / Degrees per right angle = (180 degrees) / (90 degrees/right angle) = 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?

Ans : 

(a) South and turn clockwise to west:

  • Turning from south to west while going clockwise requires a 90-degree turn.
  • A right angle is also equal to 90 degrees.

Therefore, you make 1 right angle.

(b) North and turn anticlockwise to east:

  • Turning from north to east while going anticlockwise requires a 90-degree turn again.

Therefore, you make 1 right angle.

(c) West and turn to west?

  • In this scenario, you aren’t changing direction at all. Since a right angle represents a 90-degree turn, turning zero degrees means you make 0 right angles.

(d) South and turn to north:

  • Turning from south to north requires a 180-degree turn.

A key point: While a 180-degree turn makes you face the opposite direction, it doesn’t involve any right angles. A right angle is a specific quarter turn of 90 degrees.

Therefore, turning south to north makes 0 right angles.

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?

Ans : 

(a) Move the hand 90 degrees clockwise from 6. Since there are hour markers every 30 degrees, this movement lands exactly at 9.

(b) The hand moves 180 degrees clockwise from 8. Counting by hour markers (every 30 degrees), this movement takes us past 9, 10, and 11, landing at 2.

(c) The hand will move past all the hour markers and end up back at its starting position, which is 10.

(d) The hand will end up back at its starting position, which is 7.

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes

Exercise 5.3

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 ↔ (f) One complete 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 ↔ (b) More than half a revolution

2. Classify each one of the following angles

class 6 maths ncert solutions Chapter 5

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

Exercise 5.4

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

Ans : 

(i) A right angle measures 90 degrees.

(ii) A straight angle measures 180 degrees.

2. Write 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.

Ans : 

(a) True: The measure of an acute angle is less than 90 degrees. 

(b) False: The measure of an obtuse angle is between 90 degrees and 180 degrees (not less than 90 degrees).

 (c) True: The measure of a reflex angle is greater than 180 degrees and less than 360 degrees. 

(d) True: One complete revolution around a circle is equal to 360 degrees. 

(e) True: If m∠A = 53° and ∠B = 35°, then 53 degrees is indeed greater than 35 degrees.

3. Write down the measures of

(a) some acute angles

(b) some obtuse angles

Ans : 

(a) Acute Angles:

  • 30 degrees
  • 45 degrees
  • 62 degrees
  • 78 degrees
  • Any angle measure between 0 degrees (not inclusive) and 90 degrees (exclusive) is considered acute.

(b) Obtuse Angles:

  • 105 degrees
  • 130 degrees
  • 157 degrees
  • 170 degrees
  • Any angle measure between 90 degrees (exclusive) and 180 degrees (inclusive) is considered obtuse.

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

class 6 maths solution Chapter 5

Ans : 

(a) 45°

(b) 125°

(c) 90°

(d) ∠1 = 60°, ∠2 = 90°, ∠3 = 125°

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

Measure of Angle A =

Measure of Angle B =

ncert solution class 6 maths Chapter 5

Ans : 

Measure of Angle A = 40°
Measure of Angle B = 60°.

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

Ans : 

(a) = 45°

(b) = 60°

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 ……… .

Ans : 

(a) An angle whose measure is less than that of a right angle is acute.

 (b) An angle whose measure is greater than that of a right angle is obtuse

(c) An angle whose measure is the sum of the measures of two right angles is straight.

 (d) When the sum of the measures of two angles is that of a right angle, then each one of them is right

(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 obtuse

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

ncert class 6 maths solutions Chapter 5
ncert solutions for class 6 maths

Ans : 

(a) Measure of the angle = 40°
(b) Measure of the angle = 130°
(c) Measure of the angle = 65°
(d) Measure of the angle = 135°.

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

class 6 maths Understanding Elementary Shapes

Ans : 

Clock 1 (9:00 AM): 90 degrees

Clock 2 (1:00 PM): 30 degrees

Clock 3 (6:00 PM): 180 degrees

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

ncert class 6 maths Understanding Elementary Shapes

Ans : 

No, looking at the figure through a magnifying glass will not change the size of the angle itself. The angle will still measure 30 degrees. Here’s why:

  • A magnifying glass enlarges the image we see, but it doesn’t distort the actual size or shape of the object itself.
  • The angle is a measure of the rotation between two lines, independent of the physical size of those lines.

11. Measure and classify each angle:

ncert solution for class 6 maths Understanding Elementary Shapes
AngleMeasureType
<AOB
<AOC
< BOC
<DOC
<DOA
<DOB

Ans : 

AngleMeasureType
<AOB40°Acute angle
<AOC125°Obtuse angle
< BOC85°Acute angle
<DOC95°Obtuse angle
<DOA140°Obtuse angle
<DOB180°Straight angle

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes

Exercise 5.5 

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.

Ans : 

Adjacent edges of a table top (a): These edges might be parallel (never meeting) or slightly bent, not necessarily perpendicular.

Railway tracks (b): Although they appear parallel from a distance, railway tracks are actually laid with a slight outward tilt for stability. So, they are not perpendicular.

Letter V (d): The lines in a V do intersect, but not at a right angle. They typically create a smaller acute angle at the bottom.

Letter L (c): The line segments in a capital L form a right angle at the corner, making it a perfect model for perpendicular lines.

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 : The measure of ∠PAY is 90 degrees.

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 : 

30-60-90 triangle: This set square has angles measuring 30 degrees, 60 degrees, and 90 degrees.

45-45-90 triangle: This set square has angles measuring 45 degrees (twice) and 90 degrees.

Both types of set squares share a common angle measure of 90 degrees.

4. Study the diagram. The line l is perpendicular to line m.

(a) Is CE = EG?

(b) Does PE bisects CG?

class 6 maths solution Understanding Elementary Shapes

(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 : 

  1. Yes
  2. Yes
  3. BG and DF
  4. (i) True (ii) True (iii) True

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes

Exercise 5.6 

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°.

Ans : 

(a) This triangle could be scalene.

(b) This triangle is most likely acute.

(c) This triangle is equilateral.

(d) This triangle is a right triangle.

(e) This triangle is an isosceles right triangle

(f) This triangle is most likely acute.

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

Ans : 

(i)  3 sides of equal length    ↔ (e) Equilateral

(ii) 2 sides of equal length   ↔ (g) 2 sides of equal length  

(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)

ncert solution class 6 maths Understanding Elementary Shapes

Ans : 

(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?

cbse 6th class maths textbook solutions Understanding Elementary Shapes

(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.

Ans : 

(a) 3 matchsticks:

  • Yes, you can make a triangle!
  • Arrange the three matchsticks to form an equilateral triangle.
  • An equilateral triangle has all three sides equal in length.

(b) 4 matchsticks:

  • No, you cannot make a triangle.
  • The Triangle Inequality Theorem states that the sum of any two sides of a triangle must be greater than the third side.
  • In this case, with only four matchsticks, it’s impossible to create three segments that satisfy this rule.

(c) 5 matchsticks:

  • Yes, you can make a triangle!
  • Arrange three matchsticks to form a base line. Then, use the remaining two matchsticks to create equal legs leaning against the base, forming an isosceles triangle.
  • An isosceles triangle has two sides with equal length.

(d) 6 matchsticks:

  • Yes, you can make a triangle!
  • Arrange the six matchsticks to form a large equilateral triangle.
  • As mentioned earlier, an equilateral triangle has all three sides equal in length.

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes

Exercise 5.8

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

NCERT Solutions for Class 6 Maths Chapter 5 all exercise

Ans : 

Shape (a) – Square: This is a polygon. It’s a quadrilateral (four sides) with all sides equal in length and all angles right angles (90 degrees). It’s a specific type of polygon called a square.

Shape (b) – Hexagon: This is a polygon. It’s a hexagon (six sides) with all sides equal in length and all angles presumably measuring 120 degrees (though it’s difficult to say for sure from the image).

Shape (c) – Circle: This is not a polygon because it’s a curved shape, not made of straight lines.

Shape (d) – Cone: This is not a polygon because it’s a 3D shape, not a flat two-dimensional shape.

2. Name the polygon.

NCERT Solutions for Class 6 Maths Chapter 5 exercise 5.8 free study

Make two more examples of each of these.

Ans : 

(a) A quadrilateral Examples:

(b) A Triangle
Examples:

(c) A Pantagon
Examples:

(d) A Octagon
Examples:

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.

Ans : 

ABCDEF is a rough sketch of a regular hexagon. If we join any three vertices like D, A and B, we get a scalene triangle DAB.
NCERT Solutions for Class 6 Maths Chapter 5 ex. 5.8 free to download
But if we join the alternate vertices, we get an equilateral triangle EAC.

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.

Ans  : 

ABCDEFGH is a rough sketch of regular octagon. GHCD is the rectangle formed by joining the four vertices of the given octagon.

NCERT Solutions for Class 6 Maths Chapter 5 exercise 5.8 in english medium

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 : 

A B C D E is the rough sketch of a pentagon.

By joining its any two vertices, we get, the following diagonals.

AD, AC , BE, BDand CE

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes

FAQ’s

What topics are covered in NCERT Class 6 Maths Chapter 5?

NCERT Class 6 Maths Chapter 5 Understanding Elementary Shapes covers various topics related to understanding elementary shapes, including basic geometrical concepts, properties of shapes, and their applications.

How can NCERT solutions help in understanding elementary shapes?

NCERT solutions provide step-by-step explanations, illustrations, and practice questions that help students grasp fundamental geometric concepts effectively.

Are there any real-life applications of understanding elementary shapes?

Yes, understanding elementary shapes is crucial in various real-life applications such as construction, architecture, engineering, and design.

How can I use NCERT solutions to improve my understanding of geometry?

By regularly practicing NCERT solutions and solving the exercises provided, you can strengthen your understanding of elementary shapes and improve your geometry skills.

Can NCERT solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes help me prepare for exams?

Yes, NCERT solutions offer comprehensive coverage of the chapter’s topics and provide ample practice questions to help you prepare effectively for exams.

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