GSEB Class 7 Maths Solutions Chapter 6 The Triangles and Its Properties Exercise 6.4

Question 1. Is it possible to have a triangle with the following sides?
(i) 2 cm, 3 cm, 5 cm
(ii) 3 cm, 6 cm, 7 cm
(iii) 6 cm, 3 cm, 2 cm
Answer:
(i) For 2 cm, 3 cm, 5 cm:
We add the first two sides: \( 2 \text{ cm} + 3 \text{ cm} = 5 \text{ cm} \). The third side is \( 5 \text{ cm} \).
Here, the total of the lengths of two sides equals the length of the third side, which is not allowed for a triangle.
Therefore, a triangle cannot be formed with these sides.

(ii) For 3 cm, 6 cm, 7 cm:
First, \( 3 \text{ cm} + 6 \text{ cm} = 9 \text{ cm} \). Since \( 9 \text{ cm} > 7 \text{ cm} \).
Next, \( 3 \text{ cm} + 7 \text{ cm} = 10 \text{ cm} \). Since \( 10 \text{ cm} > 6 \text{ cm} \).
Then, \( 6 \text{ cm} + 7 \text{ cm} = 13 \text{ cm} \). Since \( 13 \text{ cm} > 3 \text{ cm} \).
Because the total of any two sides is always greater than the third side, a triangle can be formed with these sides.

(iii) For 6 cm, 3 cm, 2 cm:
First, \( 6 \text{ cm} + 3 \text{ cm} = 9 \text{ cm} \). Since \( 9 \text{ cm} > 2 \text{ cm} \).
Next, \( 3 \text{ cm} + 2 \text{ cm} = 5 \text{ cm} \). Since \( 5 \text{ cm} < 6 \text{ cm} \).
Then, \( 2 \text{ cm} + 6 \text{ cm} = 8 \text{ cm} \). Since \( 8 \text{ cm} > 3 \text{ cm} \).
Since the sum of two sides (3 cm + 2 cm = 5 cm) is not greater than the third side (6 cm), a triangle cannot be formed with these sides.
In simple words: To make a triangle, if you add up the length of any two sides, the answer must always be bigger than the length of the third side. If this rule isn't followed for even one pair of sides, then you cannot make a triangle.

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. Check all three possible pairs.

Question 2. Take any point O in the interior of a triangle PQR. Is
(i) OP + OQ> PQ?
(ii) OQ + OR > QR?
(iii) OR + OP > RP?
Answer:
(i) Yes, \( \text{OP} + \text{OQ} > \text{PQ} \). This is true because the total of the lengths of any two sides of a triangle is greater than the length of the third side. Here, \( \triangle \text{POQ} \) is a triangle.

(ii) Yes, \( \text{OQ} + \text{OR} > \text{QR} \). This is also true for the same reason. Here, \( \triangle \text{QOR} \) is a triangle.

(iii) Yes, \( \text{OR} + \text{OP} > \text{RP} \). This is also true for the same reason. Here, \( \triangle \text{ROP} \) is a triangle.
In simple words: If you pick any point inside a triangle and draw lines from it to the corners, those lines create smaller triangles. In any triangle, two sides added together are always longer than the third side. This rule applies to the smaller triangles formed inside too.

Exam Tip: This question tests your understanding of the triangle inequality theorem, applied to smaller triangles formed within a larger one. Any three non-collinear points form a triangle, and their sides must satisfy the inequality.

Question 3. AM is a median of a triangle ABC. Is AB + BC + CA > 2AM? (Consider the sides of triangles \( \triangle \text{ABM} \) and \( \triangle \text{AMC} \).)
Answer:
Since the sum of the lengths of any two sides of a triangle is greater than the length of the third side:

In \( \triangle \text{ABM} \), we have:
\( \text{AB} + \text{BM} > \text{AM} \) ... (1)

Similarly, in \( \triangle \text{ACM} \), we have:
\( \text{CA} + \text{CM} > \text{AM} \) ... (2)

Adding equations (1) and (2), we get:
\( (\text{AB} + \text{BM}) + (\text{CA} + \text{CM}) > \text{AM} + \text{AM} \)

\( \implies [\text{AB} + (\text{BM} + \text{CM}) + \text{CA}] > 2\text{AM} \)

Since M is the midpoint of BC (because AM is a median), \( \text{BM} + \text{CM} = \text{BC} \).
\( \implies [\text{AB} + (\text{BC}) + \text{CA}] > 2\text{AM} \)

Thus, it is true that \( (\text{AB} + \text{BC} + \text{CA}) > 2\text{AM} \).
In simple words: Yes, the sum of all three sides of the original triangle (AB + BC + CA) is always greater than two times the length of its median (2AM). We show this by using the triangle rule on the two smaller triangles that the median creates.

Exam Tip: When proving inequalities involving medians, break the main triangle into two smaller triangles using the median, then apply the triangle inequality theorem to each smaller triangle and sum the results. Remember that a median bisects the side it connects to.

Question 4. ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD?
Answer:
The sum of the lengths of any two sides of a triangle is always greater than the length of the third side.

In \( \triangle \text{ABC} \), we have:
\( (\text{AB} + \text{BC}) > \text{AC} \) ... (1)

Similarly, in \( \triangle \text{ADC} \), we have:
\( (\text{CD} + \text{DA}) > \text{AC} \) ... (2)

Adding equations (1) and (2), we get:
\( (\text{AB} + \text{BC}) + (\text{CD} + \text{DA}) > \text{AC} + \text{AC} \)
\( \implies (\text{AB} + \text{BC} + \text{CD} + \text{DA}) > 2\text{AC} \) ... (3)

Again, in \( \triangle \text{ABD} \), we have:
\( (\text{AB} + \text{DA}) > \text{BD} \) ... (4)

In \( \triangle \text{BCD} \), we have:
\( (\text{BC} + \text{CD}) > \text{BD} \) ... (5)

Adding equations (4) and (5), we get:
\( (\text{AB} + \text{DA}) + (\text{BC} + \text{CD}) > \text{BD} + \text{BD} \)
\( \implies (\text{AB} + \text{BC} + \text{CD} + \text{DA}) > 2\text{BD} \) ... (6)

Now, adding inequalities (3) and (6), we get:
\( (\text{AB} + \text{BC} + \text{CD} + \text{DA}) + (\text{AB} + \text{BC} + \text{CD} + \text{DA}) > 2\text{AC} + 2\text{BD} \)
\( \implies 2(\text{AB} + \text{BC} + \text{CD} + \text{DA}) > 2(\text{AC} + \text{BD}) \)

Dividing by 2 on both sides:
\( \implies (\text{AB} + \text{BC} + \text{CD} + \text{DA}) > (\text{AC} + \text{BD}) \)
So, yes, the statement is true.
In simple words: Yes, for any four-sided shape (quadrilateral), if you add up the lengths of all its outer sides, that total will always be greater than if you add up the lengths of its two inner diagonal lines. You can prove this by breaking the quadrilateral into several triangles and applying the triangle inequality rule to each one.

Exam Tip: When dealing with quadrilaterals, often the strategy is to divide them into triangles using diagonals and apply the triangle inequality theorem to each of these smaller triangles to establish relationships between the sides and diagonals.

Question 5. ABCD is quadrilateral. Is AB + BC + CD + DA < 2(AC + BD)?
Answer:
Since the sum of the lengths of any two sides of a triangle is greater than the length of the third side.

Consider the point of intersection of the diagonals AC and BD as O.

In \( \triangle \text{OAB} \), we have:
\( (\text{OA} + \text{OB}) > \text{AB} \) ... (1)

Similarly, in \( \triangle \text{OBC} \), we have:
\( (\text{OB} + \text{OC}) > \text{BC} \) ... (2)

In \( \triangle \text{OCD} \), we have:
\( (\text{OC} + \text{OD}) > \text{CD} \) ... (3)

In \( \triangle \text{OAD} \), we have:
\( (\text{OA} + \text{OD}) > \text{AD} \) ... (4)

Adding equations (1), (2), (3) and (4), we get:
\( (\text{OA} + \text{OB}) + (\text{OB} + \text{OC}) + (\text{OC} + \text{OD}) + (\text{OA} + \text{OD}) > (\text{AB} + \text{BC} + \text{CD} + \text{DA}) \)
\( \implies 2[\text{OA} + \text{OB} + \text{OC} + \text{OD}] > (\text{AB} + \text{BC} + \text{CD} + \text{DA}) \)

We can rearrange the left side:
\( \implies 2[(\text{OA} + \text{OC}) + (\text{OB} + \text{OD})] > (\text{AB} + \text{BC} + \text{CD} + \text{DA}) \)

Since \( \text{OA} + \text{OC} = \text{AC} \) and \( \text{OB} + \text{OD} = \text{BD} \), we substitute these into the inequality:
\( \implies 2[\text{AC} + \text{BD}] > (\text{AB} + \text{BC} + \text{CD} + \text{DA}) \)

Rewriting the inequality, we get:
\( (\text{AB} + \text{BC} + \text{CD} + \text{DA}) < 2(\text{AC} + \text{BD}) \)
So, yes, the given statement is true.
In simple words: Yes, the total length of all four outer sides of a quadrilateral is always less than two times the sum of its two diagonal lengths. We can show this by using the triangle rule on the four small triangles formed by the diagonals and their intersection point.

Exam Tip: For quadrilateral problems involving diagonals, consider the point where the diagonals intersect. This point, along with the vertices, forms four small triangles. Applying the triangle inequality theorem to each of these small triangles is a common technique to prove relationships about the quadrilateral's sides and diagonals.

Question 6. The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?
Answer:
For a triangle to exist, the sum of the lengths of any two sides must be greater than the length of the third side.

First, let's consider the upper limit for the third side. The sum of the two given sides must be greater than the third side:
\( 12 \text{ cm} + 15 \text{ cm} > \text{Third side} \)
\( \implies 27 \text{ cm} > \text{Third side} \)
This means the Third side must be less than \( 27 \text{ cm} \).

Next, let's find the lower limit. The difference between the lengths of any two sides must be less than the length of the third side.
\( |15 \text{ cm} - 12 \text{ cm}| < \text{Third side} \)
\( \implies 3 \text{ cm} < \text{Third side} \)
This means the Third side must be greater than \( 3 \text{ cm} \).

Combining both conditions, we find that the third side should be any length between \( 3 \text{ cm} \) and \( 27 \text{ cm} \).
In simple words: For a triangle to be made, its third side has to be longer than the difference between the other two sides, but also shorter than the sum of those two sides. So, the third side's length must fall somewhere in between these two values.

Exam Tip: Remember the two parts of the triangle inequality theorem for problems like this: (1) The sum of any two sides must be greater than the third side, and (2) The difference of any two sides must be less than the third side. These two conditions define the range for the possible length of the unknown side.