Answer :
To determine the range of possible lengths [tex]\( x \)[/tex] for the third side of a triangle with side lengths 20 meters and 30 meters, we need to apply the triangle inequality theorem. The theorem states that the sum of any two sides of a triangle must be greater than the length of the third side.
1. First, we consider the sum of the given sides:
[tex]\[ 20 + 30 > x \implies x < 50 \][/tex]
This gives us the upper bound for [tex]\( x \)[/tex].
2. Next, we need to ensure that the side [tex]\( x \)[/tex] and each one of the given sides must sum to more than the remaining given side:
[tex]\[ 20 + x > 30 \implies x > 30 - 20 \implies x > 10 \][/tex]
[tex]\[ 30 + x > 20 \implies x > 20 - 30 \implies x > -10 \][/tex]
Since a side length cannot be negative, only the inequality [tex]\( x > 10 \)[/tex] is relevant for our purposes.
Therefore, combining these results, we can describe the range of possible lengths [tex]\( x \)[/tex] for the third side of the triangle as:
[tex]\[ 10 < x < 50 \][/tex]
To summarize, the correct answer for the inequality is:
[tex]\[ 10 < x < 50 \][/tex]
1. First, we consider the sum of the given sides:
[tex]\[ 20 + 30 > x \implies x < 50 \][/tex]
This gives us the upper bound for [tex]\( x \)[/tex].
2. Next, we need to ensure that the side [tex]\( x \)[/tex] and each one of the given sides must sum to more than the remaining given side:
[tex]\[ 20 + x > 30 \implies x > 30 - 20 \implies x > 10 \][/tex]
[tex]\[ 30 + x > 20 \implies x > 20 - 30 \implies x > -10 \][/tex]
Since a side length cannot be negative, only the inequality [tex]\( x > 10 \)[/tex] is relevant for our purposes.
Therefore, combining these results, we can describe the range of possible lengths [tex]\( x \)[/tex] for the third side of the triangle as:
[tex]\[ 10 < x < 50 \][/tex]
To summarize, the correct answer for the inequality is:
[tex]\[ 10 < x < 50 \][/tex]
Answer:
10 < x < 50
Step-by-step explanation:
The two sides of a triangle are given as 20 and 30.
The smallest the third side can be is when we subtract the two given sides.
30-20 = 10
The largest is when we add the sides.
30+20 = 50
10 < x < 50