QUESTION IMAGE
Question
question
which of the following sets of numbers could not represent the three sides of a triangle?
answer
{11,17,26} {10,14,25}
{14,26,38} {11,13,23}
Step1: Recall Triangle Inequality Theorem
The triangle inequality theorem states that for three side lengths \(a\), \(b\), and \(c\) (where \(c\) is the longest side), \(a + b>c\) must hold. We will check each set of numbers.
Step2: Check \(\{11, 17, 26\}\)
Longest side is \(26\). Check \(11 + 17\): \(11+17 = 28\). Since \(28>26\), this can be a triangle.
Step3: Check \(\{10, 14, 25\}\)
Longest side is \(25\). Check \(10 + 14\): \(10 + 14=24\). Since \(24<25\), this does not satisfy the triangle inequality. But we need to check all options to be sure.
Step4: Check \(\{14, 26, 38\}\)
Longest side is \(38\). Check \(14 + 26\): \(14+26 = 40\). Since \(40>38\), this can be a triangle.
Step5: Check \(\{11, 13, 23\}\)
Longest side is \(23\). Check \(11 + 13\): \(11+13 = 24\). Since \(24>23\), this can be a triangle. Wait, but we saw \(\{10,14,25\}\) fails. Wait, maybe I miscalculated. Wait \(10 + 14 = 24\), which is less than \(25\). Let's re - check other options. Wait, maybe I made a mistake. Wait, let's re - check \(\{10,14,25\}\): \(10+14 = 24<25\), so it can't be a triangle. Let's check again:
For a triangle with sides \(a\), \(b\), \(c\) ( \(c\geq a,b\) ), \(a + b>c\), \(a + c>b\), \(b + c>a\). The most restrictive is \(a + b>c\) when \(c\) is the largest.
- For \(\{11,17,26\}\): \(11 + 17=28>26\), \(11+26 = 37>17\), \(17 + 26=43>11\)
- For \(\{10,14,25\}\): \(10+14 = 24<25\), so this fails the triangle inequality.
- For \(\{14,26,38\}\): \(14 + 26=40>38\), \(14+38 = 52>26\), \(26 + 38=64>14\)
- For \(\{11,13,23\}\): \(11+13 = 24>23\), \(11+23 = 34>13\), \(13+23 = 36>11\)
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\(\{10, 14, 25\}\)