Question

in exercises 1 - 4, find the value of x. then tell whether the side lengths form a pythagorean triple. examples 1 and 2 1. 2. 3. 4.

Explanation:

Step1: Use the Pythagorean theorem

For a right - triangle with legs \(a\) and \(b\) and hypotenuse \(c\), \(a^{2}+b^{2}=c^{2}\).

Problem 1:

• Here \(a = 7\), \(b = 11\), and \(c=x\) (hypotenuse).
• Substitute into the Pythagorean theorem: \(x^{2}=7^{2}+11^{2}\).
• Calculate \(7^{2}=49\) and \(11^{2}=121\).
• Then \(x^{2}=49 + 121=170\).
• So \(x=\sqrt{170}\approx13.04\).
• A Pythagorean triple consists of three positive integers. Since \(\sqrt{170}\) is not an integer, the side lengths \(7\), \(11\), \(\sqrt{170}\) do not form a Pythagorean triple.

Problem 2:

• Here \(a = 16\), \(b = 30\), and \(c=x\) (hypotenuse).
• Substitute into the Pythagorean theorem: \(x^{2}=16^{2}+30^{2}\).
• Calculate \(16^{2}=256\) and \(30^{2}=900\).
• Then \(x^{2}=256+900 = 1156\).
• So \(x=\sqrt{1156}=34\).
• Since \(16\), \(30\), \(34\) are all integers (\(16^{2}+30^{2}=256 + 900=1156=34^{2}\)), the side lengths form a Pythagorean triple.

Problem 3:

• Here \(a = 8\), \(c = 17\), and \(b=x\) (one of the legs).
• Using the Pythagorean theorem \(x^{2}=17^{2}-8^{2}\).
• Calculate \(17^{2}=289\) and \(8^{2}=64\).
• Then \(x^{2}=289 - 64=225\).
• So \(x=\sqrt{225}=15\).
• Since \(8\), \(15\), \(17\) are all integers (\(8^{2}+15^{2}=64 + 225=289=17^{2}\)), the side lengths form a Pythagorean triple.

Problem 4:

• Here \(a=x\), \(b = 9\), and \(c = 24\) (hypotenuse).
• Using the Pythagorean theorem \(x^{2}=24^{2}-9^{2}\).
• Calculate \(24^{2}=576\) and \(9^{2}=81\).
• Then \(x^{2}=576-81 = 495\).
• So \(x=\sqrt{495}\approx22.25\).
• Since \(\sqrt{495}\) is not an integer, the side lengths \(9\), \(24\), \(\sqrt{495}\) do not form a Pythagorean triple.

Answer:

1. \(x=\sqrt{7^{2}+11^{2}}=\sqrt{49 + 121}=\sqrt{170}\approx13.04\), not a Pythagorean triple.
2. \(x=\sqrt{16^{2}+30^{2}}=\sqrt{256+900}=\sqrt{1156}=34\), is a Pythagorean triple.
3. \(x=\sqrt{17^{2}-8^{2}}=\sqrt{289 - 64}=\sqrt{225}=15\), is a Pythagorean triple.
4. \(x=\sqrt{24^{2}-9^{2}}=\sqrt{576 - 81}=\sqrt{495}\approx22.25\), not a Pythagorean triple.