Question

pythagorean theorem converse hw1.2b #10. which of the following could be the perimeters of the three squares below? a 12 ft², 16 ft², and 20 ft² b 20 ft², 16 ft² and 24 ft² c both a and b d neither a nor b

Explanation:

Step1: Recall Pythagorean theorem for squares.

If the side - lengths of the three squares are \(a\), \(b\), and \(c\) (where \(c\) is the side - length of the largest square), then \(a^{2}+b^{2}=c^{2}\) for the right - triangle formed by the sides of the squares. The perimeter of a square \(P = 4s\), and if \(P_1 = 4a\), \(P_2=4b\), \(P_3 = 4c\), then \((\frac{P_1}{4})^{2}+(\frac{P_2}{4})^{2}=(\frac{P_3}{4})^{2}\), or \(P_1^{2}+P_2^{2}=P_3^{2}\).

Step2: Check option A.

For \(P_1 = 12\), \(P_2 = 16\), \(P_3=20\). Calculate \(P_1^{2}+P_2^{2}\) and \(P_3^{2}\). \(P_1^{2}=12^{2}=144\), \(P_2^{2}=16^{2}=256\), \(P_1^{2}+P_2^{2}=144 + 256=400\), and \(P_3^{2}=20^{2}=400\). So \(P_1^{2}+P_2^{2}=P_3^{2}\).

Step3: Check option B.

For \(P_1 = 16\), \(P_2 = 20\), \(P_3 = 24\). Calculate \(P_1^{2}+P_2^{2}\) and \(P_3^{2}\). \(P_1^{2}=16^{2}=256\), \(P_2^{2}=20^{2}=400\), \(P_1^{2}+P_2^{2}=256+400 = 656\), and \(P_3^{2}=24^{2}=576\). Since \(P_1^{2}+P_2^{2}
eq P_3^{2}\).

Answer:

A. \(12\ ft\), \(16\ ft\), and \(20\ ft\)