GSEB Class 6 Maths Solutions Chapter 5 Understanding Elementary Shapes Exercise 5.1

Question 1. What is the disadvantage in comparing line segments by mere observation?
Answer: Just comparing line segments by only looking at them might not be exact. For example, two line segments might look the same length, but when measured, they could be slightly different. This means relying only on sight can cause errors.

In simple words: When you just look at line segments to compare them, it's easy to make mistakes because they might appear equal even if they are not.

Exam Tip: Visual estimation is often unreliable in geometry. Always use proper measuring tools for accuracy, especially for lines that appear deceptively similar in length.

Question 2. Why is it better use a divider than a ruler while measuring the length of a line segment?
Answer: Measuring the length of a line segment using a ruler can come with some errors because of two main reasons: (i) the thickness of the ruler itself, and (ii) angular viewing, which means looking at it from an angle. These kinds of mistakes can be avoided by measuring a line segment with a divider. Therefore, using a divider is better than a ruler for exact measurements.
In simple words: A divider is better than a ruler for measuring lines because rulers can cause errors due to their thickness or if you look at them from the wrong angle. Dividers help you avoid these mistakes.

Exam Tip: Always make sure your eye is directly above the mark you are reading on a ruler to prevent parallax error (angular viewing error).

Question 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 that AC + CB = AB, then we can be sure that C lies between A and B.
Answer: When we measure the lengths of the segments AB, BC, and AC, we find the following:
\( \overline{\mathrm{AC}} = 6.3 \text{ cm} \)
\( \overline{\mathrm{BC}} = 2.7 \text{ cm} \)
\( \overline{\mathrm{AB}} = 9.0 \text{ cm} \)
Then, \( \text{AC} + \text{BC} = 6.3 \text{ cm} + 2.7 \text{ cm} = 9.0 \text{ cm} \).
Since \( \text{AB} = 9.0 \text{ cm} \), we can see that \( \text{AC} + \text{BC} = \text{AB} \). This is thus confirmed.

In simple words: If a point C is exactly between A and B on a straight line, then adding the length of AC and the length of CB will always give you the total length of AB.

Exam Tip: This property is fundamental to understanding line segments. It essentially states that parts add up to form the whole on a straight line.

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?
Answer: We are given the lengths of the segments as:
\( \text{AB} = 5 \text{ cm} \)
\( \text{BC} = 3 \text{ cm} \)
\( \text{AC} = 8 \text{ cm} \)
We notice that \( \text{AB} + \text{BC} = 5 \text{ cm} + 3 \text{ cm} = 8 \text{ cm} \).
Since \( \text{AC} \) is also \( 8 \text{ cm} \), it means that \( \text{AB} + \text{BC} = \text{AC} \).
When the sum of two smaller segments equals the longest segment, the common point between the two smaller segments is located between the other two points. In this instance, point B lies between points A and C.
In simple words: When the length of two small parts of a line add up to the length of the whole line, the point they share in common is in the middle. So, B is between A and C.

Exam Tip: To find which point lies between the others, always look for the relationship where the sum of two segment lengths equals the length of the longest segment.

Question 5. Verify, whether D is the mid-point of \( \overline{\mathrm{AG}} \).
Answer: We are given points on a number line. Let's assume point A is at 0 and point G is at 6 (based on the provided image and calculation \( \text{AG} = 7 \text{ cm} - 1 \text{ cm} = 6 \text{ cm} \)). From the image, D is at position 3. The midpoint of a segment from 0 to 6 would be at \( \frac{0+6}{2} = 3 \).
We can also verify using the segment lengths as given in the problem context: Given \( \text{AG} = 6 \text{ cm} \) and \( \text{AD} = 3 \text{ cm} \).
If D is the midpoint of AG, then AD should be equal to DG, and \( \text{AD} + \text{DG} = \text{AG} \).
Since \( \text{AD} = 3 \text{ cm} \) and \( \text{AG} = 6 \text{ cm} \), then \( \text{DG} = \text{AG} - \text{AD} = 6 \text{ cm} - 3 \text{ cm} = 3 \text{ cm} \).
So, \( \text{AD} = 3 \text{ cm} \) and \( \text{DG} = 3 \text{ cm} \), which means \( \text{AD} = \text{DG} \).
Therefore, D is the mid-point of AG.

In simple words: D is the midpoint of AG if the distance from A to D is the same as the distance from D to G. Since AD is 3 cm and DG is 3 cm, D is indeed the middle point.

Exam Tip: A midpoint divides a line segment into two equal parts. To verify, calculate the lengths of the two smaller segments and check if they are identical.

Question 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?
Answer: Given that B is the mid-point of AC, it means the length of AB is equal to the length of BC.
\( \text{AB} = \text{BC} \)....(i)
Similarly, we are given that C is the mid-point of BD, which means the length of BC is equal to the length of CD.
\( \text{BC} = \text{CD} \)....(ii)
From equations (i) and (ii), since both AB and CD are equal to BC, we can conclude that AB must be equal to CD.
\( \text{AB} = \text{CD} \).

In simple words: If B is the middle of AC, then AB and BC are the same length. If C is the middle of BD, then BC and CD are the same length. Because BC is equal to both AB and CD, it means AB and CD must also be equal to each other.

Exam Tip: When points are midpoints, they create equal segments. Use these equivalences to prove relationships between other segments by substitution.

Question 7. Draw five triangles and measure their sides. Check in each case, if the sum of the lengths of any two sides is always greater than the third side.
Answer: We will draw triangles with different sets of lengths and check the triangle inequality theorem.

(i) In \( \triangle\text{ABC} \), we have:
\( \text{AB} = 3.7 \text{ cm} \)
\( \text{BC} = 3.7 \text{ cm} \)
\( \text{AC} = 3.7 \text{ cm} \)
Check 1: \( \text{AB} + \text{BC} = 3.7 \text{ cm} + 3.7 \text{ cm} = 7.4 \text{ cm} \). We see that \( 7.4 \text{ cm} > 3.7 \text{ cm} \), so \( (\text{AB} + \text{BC}) > \text{AC} \).
Check 2: \( \text{BC} + \text{AC} = 3.7 \text{ cm} + 3.7 \text{ cm} = 7.4 \text{ cm} \). We see that \( 7.4 \text{ cm} > 3.7 \text{ cm} \), so \( (\text{BC} + \text{AC}) > \text{AB} \).
Check 3: \( \text{AB} + \text{AC} = 3.7 \text{ cm} + 3.7 \text{ cm} = 7.4 \text{ cm} \). We see that \( 7.4 \text{ cm} > 3.7 \text{ cm} \), so \( (\text{AB} + \text{AC}) > \text{BC} \).

(ii) In \( \triangle\text{PQR} \), we have:
\( \text{PQ} = 4 \text{ cm} \)
\( \text{PR} = 2.7 \text{ cm} \)
\( \text{RQ} = 4.7 \text{ cm} \)
Check 1: \( \text{PQ} + \text{RQ} = 4 \text{ cm} + 4.7 \text{ cm} = 8.7 \text{ cm} \). We see that \( 8.7 \text{ cm} > 2.7 \text{ cm} \), so \( (\text{PQ} + \text{RQ}) > \text{PR} \).
Check 2: \( \text{PQ} + \text{PR} = 4 \text{ cm} + 2.7 \text{ cm} = 6.7 \text{ cm} \). We see that \( 6.7 \text{ cm} > 4.7 \text{ cm} \), so \( (\text{PQ} + \text{PR}) > \text{RQ} \).
Check 3: \( \text{RQ} + \text{PR} = 4.7 \text{ cm} + 2.7 \text{ cm} = 7.4 \text{ cm} \). We see that \( 7.4 \text{ cm} > 4 \text{ cm} \), so \( (\text{RQ} + \text{PR}) > \text{PQ} \).

(iii) In \( \triangle\text{BCD} \), we have:
\( \text{BC} = 3 \text{ cm} \)
\( \text{CD} = 4 \text{ cm} \)
\( \text{BD} = 5 \text{ cm} \)
Check 1: \( \text{BC} + \text{CD} = 3 \text{ cm} + 4 \text{ cm} = 7 \text{ cm} \). We see that \( 7 \text{ cm} > 5 \text{ cm} \), so \( (\text{BC} + \text{CD}) > \text{BD} \).
Check 2: \( \text{BD} + \text{CD} = 5 \text{ cm} + 4 \text{ cm} = 9 \text{ cm} \). We see that \( 9 \text{ cm} > 3 \text{ cm} \), so \( (\text{BD} + \text{CD}) > \text{BC} \).
Check 3: \( \text{BC} + \text{BD} = 3 \text{ cm} + 5 \text{ cm} = 8 \text{ cm} \). We see that \( 8 \text{ cm} > 4 \text{ cm} \), so \( (\text{BC} + \text{BD}) > \text{CD} \).

(iv) In \( \triangle\text{XYZ} \), we have:
\( \text{XY} = 4.5 \text{ cm} \)
\( \text{YZ} = 3.8 \text{ cm} \)
\( \text{XZ} = 2 \text{ cm} \)
Check 1: \( \text{XY} + \text{YZ} = 4.5 \text{ cm} + 3.8 \text{ cm} = 8.3 \text{ cm} \). We see that \( 8.3 \text{ cm} > 2 \text{ cm} \), so \( (\text{XY} + \text{YZ}) > \text{XZ} \).
Check 2: \( \text{YZ} + \text{XZ} = 3.8 \text{ cm} + 2 \text{ cm} = 5.8 \text{ cm} \). We see that \( 5.8 \text{ cm} > 4.5 \text{ cm} \), so \( (\text{YZ} + \text{XZ}) > \text{XY} \).
Check 3: \( \text{XZ} + \text{XY} = 2 \text{ cm} + 4.5 \text{ cm} = 6.5 \text{ cm} \). We see that \( 6.5 \text{ cm} > 3.8 \text{ cm} \), so \( (\text{XZ} + \text{XY}) > \text{YZ} \).

(v) In \( \triangle\text{KLM} \), we have:
\( \text{KL} = 4.5 \text{ cm} \)
\( \text{KM} = 3.4 \text{ cm} \)
\( \text{LM} = 6 \text{ cm} \)
Check 1: \( \text{KL} + \text{KM} = 4.5 \text{ cm} + 3.4 \text{ cm} = 7.9 \text{ cm} \). We see that \( 7.9 \text{ cm} > 6 \text{ cm} \), so \( (\text{KL} + \text{KM}) > \text{LM} \).
Check 2: \( \text{KM} + \text{LM} = 3.4 \text{ cm} + 6 \text{ cm} = 9.4 \text{ cm} \). We see that \( 9.4 \text{ cm} > 4.5 \text{ cm} \), so \( (\text{KM} + \text{LM}) > \text{KL} \).
Check 3: \( \text{KL} + \text{LM} = 4.5 \text{ cm} + 6 \text{ cm} = 10.5 \text{ cm} \). We see that \( 10.5 \text{ cm} > 3.4 \text{ cm} \), so \( (\text{KL} + \text{LM}) > \text{KM} \).

In all the above cases, we have seen that the sum of the lengths of any two sides is always greater than the third side. Thus, the sum of the lengths of any two sides of a triangle is never less than its third side.
In simple words: For any triangle, if you pick any two sides and add their lengths together, that total length will always be longer than the length of the third side. This rule always holds true for triangles.

Exam Tip: Remember the Triangle Inequality Theorem: The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. This is a crucial condition for any three segments to form a triangle.