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November 21, 2024Sum of the Lengths of Two Sides of a Triangle: A polygon is a simple closed figure made up of only line segments. A triangle is the smallest polygon formed by joining three-line segments. These three-line segments are called the sides of the triangle. Any three-line segments cannot form a triangle. A triangle can be drawn only if the sum of any two sides of the triangle is greater than the third side. This is stated as the triangle inequality.
This article will discuss the comparison of the sum of the lengths of two sides of a triangle with the third side to check whether the triangle can be constructed or not.
Look at the figures given below. Which of these are triangles?
Made up of three-line segments, a triangle is a simple closed figure. So, figures (i) and (iii) cannot be a triangle.
Figure (i) is not a closed one, and figure (iii) has four sides. The below-given figure is a triangle.
In \(\Delta PQR,\) the
Three sides are \(PQ,QR,RP\)
Three angles are \(\angle PQR,\angle QRP,\angle RPQ\)
Three vertices are \(P,Q,R\)
The side opposite to vertex \(P\) is \(QR.\)
Can you name the sides which are opposite to vertices \(Q\) and \(R\)? The side opposite to vertex \(Q\) is \(PR.\) The side opposite to vertex \(R\) is \(PQ.\)
Likewise, the side opposite to \(\angle QPR\) is \(QR.\) Can you name the side which is opposite to \(\angle PQR\)?
Triangles are classified according to their side and angle properties. Triangles are classified into three kinds based on their sides:
Based on the angles, triangles are again classified into three types:
Sum of the Lengths of Two Sides of a Triangle: “Inequality” is a word that means “not equal.” Consider the following scenario. Consider a \(\Delta ABC\) with sides \(AB,BC\) and \(CA\) as shown in the below figure.
The sides of a triangle meet the following important property:
Property: The sum of any two sides of a triangle is greater than the third side. That is, in a \(\Delta ABC,\) we have:
\(AB + BC < CA,BC + CA > AB\) and \(CA + AB > BC\)
This important property of a triangle is termed triangle inequality.
To validate the above property, let us try the following activity.
Draw any three triangles as shown in figures \(\left({\text{i}} \right),\left({{\text{ii}}} \right)\) and \(\left({{\text{iii}}}\right).\) Name each triangle as \(\Delta ABC\) whose sides are \(AB,BC\) and \(CA\)
Triangle | \(BC\) | \(CA\) | \(AB\) | \(CA + AB\) | \(CA + AB – BC\) | \(AB + BC\) | \(AB + BC – CA\) | \(BC + CA\) | \(BC + CA – AB\) |
(i) | \(3\) | \(4\) | \(6\) | \(10\) | \(7\) | \(8\) | \(5\) | \(7\) | \(1\) |
(ii) | \(4\) | \(6\) | \(9\) | \(15\) | \(11\) | \(13\) | \(7\) | \(10\) | \(1\) |
(iii) | \(7\) | \(8\) | \(10\) | \(18\) | \(11\) | \(15\) | \(7\) | \(15\) | \(10\) |
From the above table, we will observe that:
The value of \(CA + AB – BC\) is positive
The value of \(AB + BC – CA\) is positive
The value of \(BC + CA – AB\) is positive.
Now,
\(CA + AB – BC\) is positive \( \Rightarrow CA + AB – BC > 0 \Rightarrow CA + AB > BC\)
\( AB + BC – CA\) is positive \( \Rightarrow AB + BC – CA > 0 \Rightarrow AB + BC > CA\)
\(BC + CA – AB\) is positive \( \Rightarrow BC + CA – AB > 0 \Rightarrow BC + CA > AB\)
As a result, the following geometrical statement has been confirmed:
Any two sides of a triangle add up to more than the third side.
Remark 1: Based on the preceding explanation, three-line segments with lengths equal to three specified integers form the sides of a triangle if the total of the lengths of each pair of two of these is larger than the length of the third.
Remark 2: The greatest angle in a triangle is the angle opposite the largest side.
Theorem: The sum of any two sides of a triangle is greater than the third side of a triangle.
Given: A \(\Delta ABC.\)
To prove: \(AB + AC > BC\)
Construction: Extend the side \(AB\) to point \(D\) such that \(AD = AC.\) Join \(CD\)
Proof: In \(\Delta ACD\)
\(AD = AC\) (By construction)
Therefore, \(\angle 1 = \angle 2\) (because the opposite angles of equal sides of a triangle are equal)
Now, \(\angle BCD = \angle BCA + \angle 1\)
\(\angle BCA = \angle 2\left({\angle 1 = \angle 2} \right)\)
Now, in \(\Delta BCD\)
\(\angle BCD > \angle 2\)
\( \Rightarrow BD > BC\) (Because the side opposite to a greater angle is longer)
\( \Rightarrow AB + AD > BC\,\left({BD = AB + AD} \right)\)
\( \Rightarrow AB + AC > BC\) (By construction \(AD = AC\))
Similarly, we can also prove that \(AB + BC > AC\) and \(AC + BC > AB.\)
Take the same triangles as in the above example and tabulate the results as follows:
leela4Triangles | \(BC\) | \(CA\) | \(AB\) | \(CA – AB\) | \(CA – AB – BC\) | \(AB – BC\) | \(AB – BC – CA\) | \(BC – CA\) | \(BC – CA – AB\) |
(i) | \(3\) | \(4\) | \(6\) | \( – 2\) | \( – 5\) | \(3\) | \( – 1\) | \( – 1\) | \( – 7\) |
(ii) | \(4\) | \(6\) | \(9\) | \( – 3\) | \( – 7\) | \(5\) | \( – 1\) | \( – 2\) | \( – 11\) |
(iii) | \(7\) | \(8\) | \(10\) | \( – 2\) | \( – 9\) | \(3\) | \( – 5\) | \( – 1\) | \( – 11\) |
From the above table, we will observe that:
Each value of \(CA – AB – BC\) is negative
Each value of \(AB – BC – CA\) is negative
Each value of \(BC – CA – AB\) is negative.
Now,
\(CA – AB – BC\) is negative \( \Rightarrow CA – AB – BC < 0 \Rightarrow CA – AB < BC\)
\(AB – BC – CA\) is negative \( \Rightarrow AB – BC – CA < 0 \Rightarrow AB – BC < CA\)
\(BC – CA – AB\) is negative \( \Rightarrow BC – CA – AB < 0 \Rightarrow BC – CA < AB\)
Thus, we have confirmed the following geometrical truth:
The difference between any two sides of a triangle is smaller than the third side.
Q.1. Marie has three sticks of lengths \(4\,{\text{units}},\,8\,{\text{units}},\) and \(2\,{\text{units}}{\text{.}}\) Using the sum of the length of the two sides property, check to see whether Marie can make a triangle using these sticks.
Ans: Let’s see if the third side is bigger than the sum of the two sides.
1. \(4 + 8 > 2 \Rightarrow 12 > 2 \Rightarrow {\text{TRUE}}\)
2. \(2 + 8 > 4 \Rightarrow 10 > 4 \Rightarrow {\text{TRUE}}\)
3. \(4 + 2 > 8 \Rightarrow 6 > 8 \Rightarrow {\text{FALSE}}\)
As a result, the triangle inequality theorem is violated by the lengths of the sticks. As a result, Marie is unable to create a triangle with sticks of the specified lengths.
Q.2. John wants to tie a ribbon around his triangle flag. The flag has \(6\,{\text{units}}\) on one side and \(2\,{\text{units}}\) on the other. Determine the amount of ribbon necessary for the third side using the triangle inequality theorem.
Ans: The length of the third side must be smaller than the sum of the other two sides, according to the triangle inequality theorem.
So, the third side is less than \(6 + 2 = 8\,{\text{units}}\)
Furthermore, the difference between the other two sides must be smaller than the third side.
So, the third side is greater than \(6 – 2 = 4\,{\text{units}}{\text{.}}\)
Thus, the length of the ribbon can be \(5,6,\) or \(7\,{\text{units}}{\text{.}}\)
Q.3. If \(4\,{\text{cm}},\,8\,{\text{cm}}\) and \(2\,{\text{cm}}\) are the measures of three lines segments. Is it possible to draw a triangle using it?
Ans: Let us check if the sum of the two sides is greater than the third side.
1. \(4 + 8 > 2 \Rightarrow 12 > 2 \Rightarrow {\text{True}}\)
2. \(8 + 2 > 4 \Rightarrow 10 > 4 \Rightarrow{\text{True}}\)
3. \(4 + 2 > 8 \Rightarrow 6 > 8 \Rightarrow {\text{False}}\)
As a result, the triangle’s sides do not fulfil the inequality theorem. As a result, we can’t make a triangle out of these three-line segments.
Q.4. Can a triangle be formed with side lengths of \(6\,{\text{cm}},7\,{\text{cm}}\) and \(5\,{\text{cm}}\)?
Ans: If \(6\,{\text{cm}},7\,{\text{cm}}\) and \(5\,{\text{cm}}\) are the sides of the triangle, then they should satisfy
The inequality \(b + c > a.\)
Hence,
1. \(6 + 7 > 5 \Rightarrow 13 > 5 \Rightarrow {\text{True}}\)
2. \(7 + 5 > 6 \Rightarrow 12 > 6 \Rightarrow {\text{True}}\)
3. \(6 + 5 > 7 \Rightarrow 11 > 7 \Rightarrow {\text{True}}\)
All three conditions are satisfied. Therefore, a triangle can be formed with side lengths as \(6\,{\text{cm}},7\,{\text{cm}}\) and \(5\,{\text{cm}}\)
Q.5. If the two sides of a triangle are 12 and 17, find all the possible lengths of the third side.
Ans: We can use the formula to get the possible values for the triangle’s third side: Difference of the length of two sides \( < \) Unknown side \( < \) Sum of the length of two sides
\(17 – 12 < x < 17 + 12\)
\(5 < x < 29\)
There could be any value for the third side between \(5\) and \(29,\) like \(6,7,7,7……………..,27,28,29.\)
This article has discussed the basic definition of triangles. Triangles are the smallest polygons made of three-line segments. The triangles are classified based on their sides and angles. Also, we discussed the relationship between the sides of the triangle.
We found that a triangle can be created when the sum of the triangle’s two sides is more than the third side or the difference between the two sides is less than the triangle’s third side. Any three measures of lengths can be the sides of the triangle, if and only if the measures meet these two inequalities.
Q.1. What is the sum of the lengths of the sides of any figure called?
Ans: The sum of the lengths of the sides of any figure is called the perimeter.
Q.2. Does the sum of the lengths of two sides of a triangle property apply to all triangles?
Ans: Yes, the property of the sum of the lengths of two triangle sides applies to all triangles. Any one of the triangle’s sides must be shorter than the sum of the other two. It is not a triangle if one of the sides is more than or equal to the sum of the other two sides.
Q.3. Give an example of the difference between the lengths of two sides of a triangle?
Ans: Triangle with side lengths \(a = 2,b = 3\) and \(c = 4\) units exist, as lengths of all sides satisfy the theorem.
1. \(b – c – a = 3 – 4 – 2 = \, – 3 \Rightarrow b – c < a \Rightarrow {\text{TRUE}}\)
2. \(c – a – b = 4 – 2 – 3 = \,- 1 \Rightarrow c – a < b \Rightarrow {\text{TRUE}}\)
3. \(a – b – c = 2 – 3 – 4 = \,- 5 \Rightarrow a – b < c \Rightarrow {\text{TRUE}}\)
Q.4. What are the different side lengths that a triangle may have?
Ans: According to the triangle inequality theorem, the lengths of any two sides of a triangle must add up to greater than the length of the third side. This indicates that a triangle with side lengths of \(2,7,\) and \(12\) cannot be drawn since \(2 + 7\) is less than \(12.\left({3,4,5} \right),\left({5,6,7} \right),\) etc., can be the sides.
Q.5. What does the triangle inequality theorem mean?
Ans: According to the triangle inequality theorem, the sum of any two triangle sides is larger than the third side, and if the sum of any two triangle sides is not higher than the third side, the triangle does not exist.
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