Question

c. choose three lengths that could be used to form a triangle but would not form a right triangle. select all that apply. a. 9 in., 12 in., 16 in. b. 9 in., 12 in., 15 in. c. 9 in., 40 in., 41 in. d. 24 in., 45 in., 53 in. e. 24 in., 45 in., 51 in.

Explanation:

Step1: Recall triangle - inequality theorem

For three side - lengths \(a\), \(b\), and \(c\) of a triangle, \(a + b>c\), \(a + c>b\), and \(b + c>a\). Also, for a right - triangle, the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) (where \(c\) is the longest side) holds.

Step2: Check option A

For \(a = 9\), \(b = 12\), \(c = 16\).
First, check the triangle - inequality: \(9+12 = 21>16\), \(9 + 16=25>12\), \(12 + 16 = 28>9\).
Then, check the Pythagorean theorem: \(9^{2}+12^{2}=81 + 144=225\), and \(16^{2}=256\). Since \(9^{2}+12^{2}
eq16^{2}\), it is a non - right triangle.

Step3: Check option B

For \(a = 9\), \(b = 12\), \(c = 15\).
Triangle - inequality: \(9+12 = 21>15\), \(9 + 15=24>12\), \(12 + 15 = 27>9\).
Pythagorean theorem: \(9^{2}+12^{2}=81 + 144 = 225=15^{2}\), so it is a right triangle.

Step4: Check option C

For \(a = 9\), \(b = 40\), \(c = 41\).
Triangle - inequality: \(9+40 = 49>41\), \(9 + 41=50>40\), \(40 + 41 = 81>9\).
Pythagorean theorem: \(9^{2}+40^{2}=81+1600 = 1681=41^{2}\), so it is a right triangle.

Step5: Check option D

For \(a = 24\), \(b = 45\), \(c = 53\).
Triangle - inequality: \(24+45 = 69>53\), \(24 + 53=77>45\), \(45 + 53 = 98>24\).
Pythagorean theorem: \(24^{2}+45^{2}=576+2025 = 2601\), and \(53^{2}=2809\). Since \(24^{2}+45^{2}
eq53^{2}\), it is a non - right triangle.

Step6: Check option E

For \(a = 24\), \(b = 45\), \(c = 51\).
Triangle - inequality: \(24+45 = 69>51\), \(24 + 51=75>45\), \(45 + 51 = 96>24\).
Pythagorean theorem: \(24^{2}+45^{2}=576 + 2025=2601=51^{2}\), so it is a right triangle.

Answer:

A. 9 in., 12 in., 16 in.
D. 24 in., 45 in., 53 in.