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Type the correct answer in each box. Use numerals instead of words.

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 lengths [tex]$x$[/tex] of the third side of the restaurant.

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

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To determine the values to fill in the boxes, you can use the triangle inequality theorem, which states:

[tex]\[ a + b \ \textgreater \ c \][/tex]
[tex]\[ a + c \ \textgreater \ b \][/tex]
[tex]\[ b + c \ \textgreater \ a \][/tex]

For side lengths 20 m and 30 m, we have:

[tex]\[ 20 + 30 \ \textgreater \ x \implies x \ \textless \ 50 \][/tex]
[tex]\[ 20 + x \ \textgreater \ 30 \implies x \ \textgreater \ 10 \][/tex]
[tex]\[ 30 + x \ \textgreater \ 20 \implies x \ \textgreater \ -10 \][/tex] (This inequality is always true since [tex]x[/tex] must be positive)

So the correct inequality describing the range of possible lengths [tex]x[/tex] of the third side is:

[tex]\[ 10 \ \textless \ x \ \textless \ 50 \][/tex]

Therefore, the boxes should be filled as follows:

[tex]\[ 10 \ \textless \ x \ \textless \ 50 \][/tex]



Answer :

GinnyAnswer
To determine the range of possible lengths [tex]\( x \)[/tex] of the third side of a triangle with side lengths of 20 meters and 30 meters, we use the triangle inequality theorem. The theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

This gives us three inequalities:
1. The sum of the first two sides must be greater than the third side:
[tex]\[ 20 + 30 > x \][/tex]
2. The sum of the first side and the unknown side must be greater than the second side:
[tex]\[ 20 + x > 30 \][/tex]
3. The sum of the second side and the unknown side must be greater than the first side:
[tex]\[ 30 + x > 20 \][/tex]

Now, let's simplify each of these inequalities:

1. Simplify the first inequality:
[tex]\[ 50 > x \][/tex]
This can be rewritten as:
[tex]\[ x < 50 \][/tex]

2. Simplify the second inequality:
[tex]\[ 20 + x > 30 \][/tex]
[tex]\[ x > 10 \][/tex]

3. Simplify the third inequality:
[tex]\[ 30 + x > 20 \][/tex]
[tex]\[ x > -10 \][/tex]
Since side lengths cannot be negative, this inequality doesn't restrict our solution further.

Combining the valid inequalities from above, we get:
[tex]\[ 10 < x < 50 \][/tex]

Thus, the range of possible lengths [tex]\( x \)[/tex] of the third side of the triangle is:
[tex]\[ 10 < x < 50 \][/tex]

Therefore, the completed inequality is:

[tex]\[ 10 < x < 50 \][/tex]

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