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Answer :
To determine the range in which the length of the third side of a triangle must lie, we use the triangle inequality theorem. This theorem states that for any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side.
Given sides are:
- [tex]\(a = 29\)[/tex] units
- [tex]\(b = 40\)[/tex] units
Let's denote the third side by [tex]\(x\)[/tex]. According to the triangle inequality theorem, we have the following inequalities:
1. [tex]\(a + b > x\)[/tex]
2. [tex]\(a + x > b\)[/tex]
3. [tex]\(b + x > a\)[/tex]
We need to solve these inequalities to find the range for [tex]\(x\)[/tex]:
1. [tex]\( a + b > x \)[/tex]
[tex]\[
29 + 40 > x \implies 69 > x \implies x < 69
\][/tex]
2. [tex]\( a + x > b \)[/tex]
[tex]\[
29 + x > 40 \implies x > 40 - 29 \implies x > 11
\][/tex]
3. [tex]\( b + x > a \)[/tex]
[tex]\[
40 + x > 29 \implies x > 29 - 40 \implies x > -11
\][/tex]
However, since [tex]\(x\)[/tex] must be positive and the inequality [tex]\(x > 11\)[/tex] already covers all positive values greater than 11, the [tex]\(x > -11\)[/tex] condition is redundant.
Thus, combining these valid inequalities, we find:
[tex]\[
11 < x < 69
\][/tex]
So, the third side [tex]\(x\)[/tex] must lie between 11 and 69 units. Therefore, the correct answer is:
C. [tex]\(11 < x < 69\)[/tex]
Given sides are:
- [tex]\(a = 29\)[/tex] units
- [tex]\(b = 40\)[/tex] units
Let's denote the third side by [tex]\(x\)[/tex]. According to the triangle inequality theorem, we have the following inequalities:
1. [tex]\(a + b > x\)[/tex]
2. [tex]\(a + x > b\)[/tex]
3. [tex]\(b + x > a\)[/tex]
We need to solve these inequalities to find the range for [tex]\(x\)[/tex]:
1. [tex]\( a + b > x \)[/tex]
[tex]\[
29 + 40 > x \implies 69 > x \implies x < 69
\][/tex]
2. [tex]\( a + x > b \)[/tex]
[tex]\[
29 + x > 40 \implies x > 40 - 29 \implies x > 11
\][/tex]
3. [tex]\( b + x > a \)[/tex]
[tex]\[
40 + x > 29 \implies x > 29 - 40 \implies x > -11
\][/tex]
However, since [tex]\(x\)[/tex] must be positive and the inequality [tex]\(x > 11\)[/tex] already covers all positive values greater than 11, the [tex]\(x > -11\)[/tex] condition is redundant.
Thus, combining these valid inequalities, we find:
[tex]\[
11 < x < 69
\][/tex]
So, the third side [tex]\(x\)[/tex] must lie between 11 and 69 units. Therefore, the correct answer is:
C. [tex]\(11 < x < 69\)[/tex]
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