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An A-frame restaurant is shaped as a triangle with two side lengths of 20 m and 30 m. Complete the inequality below to describe the range of possible length [tex]\( x \)[/tex] of the third side of the restaurant.

[tex]\[ \square \ \textless \ x \ \textless \ \square \][/tex]


Sagot :

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]

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