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22. the sum of the lengths of any two sides of a triangle must be great…

Question

  1. the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

triangle image with sides 6, x, 11
a. for the triangle shown, anna said that any value of x greater than 5 is possible. explain anna’s error.
b. write a compound inequality that represents all possible values of x.

Explanation:

Response
Part a

Step1: Recall triangle inequality

The triangle inequality states that the sum of any two sides must be greater than the third side. So we have three inequalities here: \(6 + x>11\), \(6 + 11>x\), and \(x + 11>6\) (the last one is always true for positive \(x\)).

Step2: Analyze Anna's error

From \(6 + x>11\), we get \(x > 5\). But from \(6+11>x\), we get \(x < 17\). Anna only considered the lower bound (\(x>5\)) but forgot the upper bound (\(x < 17\)). So \(x\) can't be any value greater than 5, it also has to be less than 17.

Step1: Apply first inequality

For \(6 + x>11\), subtract 6 from both sides: \(x>11 - 6\), so \(x > 5\).

Step2: Apply second inequality

For \(6+11>x\), simplify: \(17>x\) or \(x < 17\).

Step3: Combine inequalities

Since \(x\) must satisfy both \(x > 5\) and \(x < 17\), the compound inequality is \(5 < x < 17\).

Answer:

Anna forgot to consider the upper bound from the triangle inequality \(6 + 11>x\) (i.e., \(x<17\)). So \(x\) must be greater than 5 AND less than 17, not just greater than 5.

Part b