Question

two sides of an obtuse triangle measure 12 inches and 14 inches. the longest side measures 14 inches. what is the greatest possible whole - number length of the unknown side? 2 inches, 3 inches, 7 inches, 9 inches

Explanation:

Step1: Recall triangle inequality and obtuse triangle condition

For a triangle with sides \(a\), \(b\), \(c\) (where \(c\) is the longest side), the triangle inequality is \(a + b>c\), \(a + c>b\), \(b + c>a\). For an obtuse triangle, if \(c\) is the longest side, then \(a^{2}+b^{2}<c^{2}\) (when \(c\) is opposite the obtuse angle). Here, the longest side is 14, so let the unknown side be \(x\). We need to consider two cases: either \(x\) is the longest side (but 14 is given as the longest, so \(x\leq14\)) or 14 is the longest side (so \(x < 14\)). Also, from triangle inequality, \(|12 - 14|<x<12 + 14\), so \(2<x<26\). But since 14 is the longest, \(x<14\). Now, for obtuse triangle with longest side 14, \(12^{2}+x^{2}<14^{2}\) (if 14 is opposite the obtuse angle) or \(x^{2}+14^{2}<12^{2}\) (but \(x^{2}+14^{2}\) is always greater than \(12^{2}\) since \(x>0\), so only the first case: \(144 + x^{2}<196\), so \(x^{2}<52\), so \(x<\sqrt{52}\approx7.21\). Also, from triangle inequality, \(x>2\). So \(x\) must be an integer greater than 2 and less than 7.21. The greatest whole - number in this range is 7? Wait, no, wait. Wait, maybe I made a mistake. Wait, if the longest side is 14, then the other two sides are 12 and \(x\). So the obtuse angle could be opposite 14 (so \(12^{2}+x^{2}<14^{2}\)) or opposite \(x\) (so \(12^{2}+14^{2}<x^{2}\)), but if opposite \(x\), then \(x\) would be longer than 14, but the problem says the longest side is 14, so \(x\leq14\), so the obtuse angle must be opposite 14. So \(12^{2}+x^{2}<14^{2}\), \(144 + x^{2}<196\), \(x^{2}<52\), \(x < \sqrt{52}\approx7.21\). Also, from triangle inequality, \(x>14 - 12 = 2\). So \(x\) is an integer between 3 (since \(x>2\)) and 7 (since \(x<7.21\)). Wait, but the options are 2,3,7,9. Wait, 9: let's check \(12^{2}+9^{2}=144 + 81 = 225\), \(14^{2}=196\). \(225>196\), so if \(x = 9\), then \(12^{2}+9^{2}>14^{2}\), so the triangle would be acute or obtuse? Wait, if \(c\) is the longest side, then if \(a^{2}+b^{2}>c^{2}\), it's acute; if \(a^{2}+b^{2}=c^{2}\), right; if \(a^{2}+b^{2}<c^{2}\), obtuse. Wait, maybe I mixed up the cases. Wait, maybe the unknown side is the longest side? But the problem says "the longest side measures 14 inches", so the unknown side \(x\leq14\). So if \(x\) is the longest side, then \(x = 14\), but the problem says the longest side is 14, so maybe \(x<14\). Wait, let's re - evaluate the triangle inequality: the sum of any two sides must be greater than the third side. So \(12 + x>14\) (so \(x>2\)), \(12 + 14>x\) (so \(x<26\)), \(x + 14>12\) (always true since \(x>0\)). Now, for obtuse triangle:

Case 1: 14 is the longest side (opposite obtuse angle). Then \(12^{2}+x^{2}<14^{2}\) => \(x^{2}<196 - 144=52\) => \(x<\sqrt{52}\approx7.21\).

Case 2: \(x\) is the longest side (but problem says longest is 14, so \(x\leq14\), but if \(x>14\), it contradicts. So only case 1. So \(x\) must be greater than 2 (from \(12 + x>14\)) and less than 7.21. The whole numbers in this range are 3,4,5,6,7. The greatest whole - number is 7? But wait, the options include 9. Wait, maybe I made a mistake in the obtuse angle case. Wait, maybe the obtuse angle is opposite \(x\), but then \(x\) would be longer than 14, but the problem says the longest side is 14, so that's not possible. Wait, let's check the option 9. If \(x = 9\), sides are 12,9,14. Check triangle inequality: \(12 + 9>14\) (21>14), \(12 + 14>9\) (26>9), \(9 + 14>12\) (23>12). Now check for obtuse: \(9^{2}+12^{2}=81 + 144 = 225\), \(14^{2}=196\). Since \(9^{2}+12^{2}>14^{2}\), the angle opposite 14 is acute, and the angle opposite \(x = 9\)? No, wait, the longest side is 14, so the angle opposite 14 is the largest angle. Wait, if \(a^{2}+b^{2}>c^{2}\), the angle opposite \(c\) is acute. So if \(12^{2}+9^{2}>14^{2}\), then the angle opposite 14 is acute, so the triangle is acute? But we need an obtuse triangle. Wait, maybe I had the formula reversed. The correct formula for an obtuse triangle: if \(c\) is the longest side, then the triangle is obtuse if \(a^{2}+b^{2}<c^{2}\) (angle opposite \(c\) is obtuse) or if \(a^{2}+c^{2}<b^{2}\) (if \(b\) is longer than \(c\)) or \(b^{2}+c^{2}<a^{2}\) (if \(a\) is longer than \(c\)). So in our case, longest side is 14, so \(c = 14\), \(a = 12\), \(b=x\). So obtuse when \(a^{2}+b^{2}<c^{2}\) (angle opposite \(c\) is obtuse) or if \(x\) is longer than 14 (but \(x\leq14\)), so only \(a^{2}+b^{2}<c^{2}\). So \(12^{2}+x^{2}<14^{2}\) => \(x^{2}<196 - 144 = 52\) => \(x<\sqrt{52}\approx7.21\). Also, from triangle inequality, \(x>14 - 12=2\). So \(x\) must be an integer between 3 and 7 (inclusive, since \(x\) is whole number). So the greatest possible whole - number is 7? But wait, the options are 2,3,7,9. Wait, maybe the problem has a mistake, or I misread. Wait, maybe the longest side is not 14? Wait, the problem says "Two sides of an obtuse triangle measure 12 inches and 14 inches. The longest side measures 14 inches." So 14 is the longest, so \(x\leq14\). Wait, let's check \(x = 9\): sides 12,9,14. \(12^{2}+9^{2}=144 + 81 = 225\), \(14^{2}=196\). Since \(225>196\), the angle opposite 14 is acute. Now, if we consider \(x\) as the longest side, but the problem says 14 is the longest, so \(x<14\). Wait, maybe the question is wrong, or I made a mistake. Wait, let's check the triangle inequality again. For \(x = 7\): \(12^{2}+7^{2}=144 + 49 = 193\), \(14^{2}=196\). \(193<196\), so \(12^{2}+7^{2}<14^{2}\), so the angle opposite 14 is obtuse. So triangle with sides 12,7,14 is obtuse. For \(x = 9\): \(12^{2}+9^{2}=225>196 = 14^{2}\), so angle opposite 14 is acute, and since 14 is the longest side, all other angles are acute? No, wait, if \(a^{2}+b^{2}>c^{2}\), the angle opposite \(c\) is acute, but there could be another angle that is obtuse? Wait, no, in a triangle, only one angle can be obtuse (since sum of angles is 180). So if the longest side is \(c\), the largest angle is opposite \(c\). So if \(a^{2}+b^{2}<c^{2}\), angle opposite \(c\) is obtuse. If \(a^{2}+b^{2}>c^{2}\), angle opposite \(c\) is acute, so the triangle is acute (since the largest angle is acute). So for triangle with sides 12,9,14, the largest angle is opposite 14, which is acute, so the triangle is acute. For triangle with sides 12,7,14, largest angle opposite 14 is obtuse (since \(12^{2}+7^{2}<14^{2}\)). For \(x = 3\): \(12^{2}+3^{2}=144 + 9 = 153<196\), so angle opposite 14 is obtuse, but \(x = 3\) is less than 7. For \(x = 7\), it's obtuse, \(x = 9\) is acute. So the greatest whole - number length of \(x\) (unknown side) such that the triangle is obtuse and 14 is the longest side is 7? But wait, the options include 9. Maybe the problem meant that the longest side is not necessarily 14? Wait, let's re - read the problem: "Two sides of an obtuse triangle measure 12 inches and 14 inches. The longest side measures 14 inches. What is the greatest possible whole - number length of the unknown side?" So 14 is the longest, so unknown side \(x<14\). From obtuse condition, \(12^{2}+x^{2}<14^{2}\) => \(x<\sqrt{52}\approx7.21\). So \(x\) can be 3,4,5,6,7. The greatest whole number is 7. So the answer should be 7 inches.

Step2: Evaluate the options

- Option 2: \(x = 2\), but \(2\) does not satisfy \(x>2\) (from triangle inequality \(|12 - 14|<x\) => \(2<x\)), so invalid.
- Option 3: Valid, but not the greatest.
- Option 7: Valid, and the greatest among valid options (since \(x<7.21\)).
- Option 9: \(x = 9\) gives an acute triangle (as shown above), so invalid.

Answer:

7 inches (the option corresponding to 7 inches)